The total electric potential energy is [tex]\frac{kq_{1} q_{3} }{r_{13} } + \frac{kq_{2} q_{3} }{r_{23} } + \frac{kq_{1} q_{2} }{r_{12} }[/tex].
Electric Potential Energy of a System of Charges :
The system's electric potential energy is equal to the amount of work necessary to create a system of charges by guiding them toward their designated locations from infinity against the electrostatic force without accelerating them. The symbol for it is U.U=W=qV. Electrostatic fields are conservative, therefore the work is independent of the path.
Assume three charges q₁ , q₂ and q₃ bring from infinity to point P.
To bring q₁ no work is done,
[tex]V_{p} = \frac{kq_{1} }{r_{1} }[/tex]
where, V = electric potential energy.
q = point charge.
r = distance between any point around the charge to the point charge.
k = Coulomb constant; k = 9.0 × 109 N.
Now bring q₂,
[tex]V_{2} = \frac{kq_{2} }{r_{2} }[/tex]
Work done by q₁ ;
[tex]W_{1} = V_{p} q_{2} = \frac{kq_{1}q_{2} }{r_{12} }[/tex]
Now bring q₃,
[tex]V_{3} = \frac{kq_{3} }{r_{3} }[/tex]
Work done on q₃ by q₁ and q₂
[tex]W= q_{3} [ V_{1} + V_{2} ][/tex]
[tex]=\frac{kq_{1} q_{3} }{r_{13} }[/tex][tex]+ \frac{kq_{2} q_{3} }{r_{23} } + \frac{kq_{1}q_{3} }{r_{12} }[/tex]
This work done is stored in the form of potential energy.
∴U=W= potential energy of three systems.
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The total electric potential energy is [tex]\frac{kq1q3}{r13} + \frac{kq2q3}{r23} + \frac{kq1q2}{r12}[/tex]
Electric Potential Energy of a System of Charges
The labor required to establish a system of charges by guiding them toward their intended positions from infinity against the electrostatic force without accelerating them is equal to the electric potential energy of the system. Its identifier is U.U=W=qV. Due to the conservatism of electrostatic fields, the work is independent of the path.
Consider having three charges. Q1, Q2, and Q3 bring point P from infinity.
No work has been done to bring q1,
[tex]V_{1} = \frac{kq1}{r1}[/tex]
where, V = electric potential energy.
q = point charge.
r = distance between any point around the charge to the point charge.
k = Coulomb constant; k = 9.0 × 109 N.
Now bring q₂,
[tex]V_{2} = \frac{kq2}{r2}[/tex]
Work done by q₁ ;
W1 = [tex]V_{p} q2[/tex] = [tex]\frac{kq1q2}{r12}[/tex]
Now bring q₃,
[tex]V_{3} = \frac{kq3}{r3}[/tex]
Work done on q₃ by q₁ and q₂
W= q3{[tex]V_{1} + V_{2}[/tex]}
W = [tex]\frac{kq1q3}{r13} + \frac{kq2q3}{r23} + \frac{kq1q2}{r12}[/tex]
This work done is stored in the form of potential energy.
∴U=W= potential energy of three systems.
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Find the speed of a satellite in a circular orbit around the Earth with a radius 3.57 times the mean radius of the Earth. (Radius of Earth = 6.37×103 km, mass of Earth = 5.98×1024 kg, G = 6.67×10-11 Nm2/kg2.)
The speed of a satellite in a circular orbit around the Earth is 4,188 m/s.
Speed of the satelliteThe speed of the satellite is calculated as follows;
v = √GM/r
where;
M is mass of Earthr is radius of satellitev = √[(6.67 x 10⁻¹¹ x 5.98 x 10²⁴) / (3.57 x 6.37 x 10⁶)]
v = 4,188 m/s
Thus, the speed of a satellite in a circular orbit around the Earth is 4,188 m/s.
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Part B
A roller coaster ride starts with the roller coaster car being pulled to the top of the first hill with pulley system. The car is
released from the top with an initial velocity close to zero, then accelerates downward. From that first hill, the roller coaster just
coasts; there is no driving force, other than gravity, to keep It going. Assuming no friction, what can you say about the height of
the other hills in the roller coaster ride?
The highest point of a roller coaster is almost always the first hill. In the majority of roller coasters, the hills get smaller as the train travels down the track.
To find the answer, we have to know more about the mechanical energy of a system.
How to find the answer?Since it influences the mechanical energy of the system, the first hill must be the highest.One of the fundamental tenets of physics is that, in the absence of friction, mechanical energy must be conserved. Mechanical energy is the product of kinetic energy and potential energy.When the vehicles ascend the first hill on the roller coaster, mechanical energy is provided to the system because the speed is zero at this point.Mechanical energy = U = mgh
Where m represents the car mass, g represents gravity, and h represents height
If the system is to continue moving, the other hills on the mountain must be lower than the first hill. When the vehicles are released, this energy is converted into kinetic and potential energy when it lowers and ascends, but the sum of these two cannot be larger than the starting energy.Finally, by applying the principle of energy conservation, we may determine that, the initial hill must be the highest.
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A 149-g baseball is dropped from a tree 15.0 m above the ground.
With what speed would it hit the ground if air resistance could be ignored?
Express your answer to three significant figures and include the appropriate units.
If it actually hits the ground with a speed of 9.00 m/s , what is the magnitude of the average force of air resistance exerted on it?
Express your answer to three significant figures and include the appropriate units.
1. The speed with which the ball hits the ground is 17.1 m/s
2. The magnitude of the average force of air resistance exerted on it is 0.77 N
1. How to determine the velocity with which the ball hits the groundInitial velocity (u) = 0 m/s Acceleration due to gravity (g) = 9.8 m/s²Height (h) = 15 m Final velocity (v) =?v² = u² + 2gh
v² = 2gh
Take the square root of both side
v = √(2 × 9.8 × 15)
v = 17.1 m/s
2. How to determine the forceWe'll begin by calculating the time to reach the ground. This is illustrated below:
Acceleration due to gravity (g) = 9.8 m/s²Height (h) = 15 m Time (t) =?h = ½gt²
15 = ½ × 9.8 × t²
15 = 4.9 × t²
Divide both side by 4.9
t² = 15 / 4.9
Take the square root of both side
t = √(15 / 4.9)
t = 1.75 s
Now we can determine the force. This can be obtained as illustrated below:
Mass (m) = 149 g = 149 / 1000 = 0.149 KgInitial velocity (u) = 0 m/sFinal velocity (v) = 9 m/sTime (t) = 5 ms = 1.75 sForce (F) = ?F = m(v –u) / t
F = 0.149(9 – 0) / 1.75
F = 0.77 N
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Two hockey pucks are moving towards each other. (Assume no friction.) The first one is 0.13 kg and moving at a speed of 1.11 m/s, while the second puck is 0.16 kg and moving at 1.21 m/s, and they collide. (Assume elastic collision.) After collision, the second puck ends up with a speed of 1.16m/s at an angle of 42 degrees below its original path, while the first puck ends up with an unknown speed at an angle above its original path. Find the final speed and angle of the first puck.
The final speed and angle of the first puck are 1.17 m/s and 54.5° respectively.
What happened in an Elastic Collision ?In an elastic collision, both momentum and energy are conserved. But only momentum is conserved in inelastic collision.
Given that two hockey pucks are moving towards each other. The first one is 0.13 kg and moving at a speed of 1.11 m/s, while the second puck is 0.16 kg and moving at 1.21 m/s, and they collide. (Assume elastic collision.) After collision, the second puck ends up with a speed of 1.16m/s at an angle of 42 degrees below its original path, while the first puck ends up with an unknown speed at an angle above its original path.
The given parameters are;
M1 = 0.13 kgM2 = 0.16 kgU1 = 1.11 KgU2 = 1.21 KgV1 = ?V2 = 1.16 kgФ1 = ?Ф2 = 42°The mathematical representation of the above question will be in two components.
Horizontal component
M1U1 - M2U2 = M1V1cosФ - M2V2cosФ
Substitute all the parameters
0.13 x 1.11 - 0.16 x 1.21 = 0.13 x V1 cosФ - 0.16 x 1.16cos42
0.1443 - 0.1936 = 0.13V1cosФ - 0.1379
0.13V1cosФ = 0.0886
V1cosФ = 0.0886/0.13
V1cosФ = 0.6815 ........ (1)
Vertical component
0 = M1V1sinФ - M2V2sinФ
M1V1sinФ = M2V2sinФ
Substitute all the parameters
0.13 x V1 sinФ = 0.16 x 1.16sin42
V1 sinФ = 0.1242/0.13
V1 sinФ = 0.9553 ......... (2)
Divide equation 2 by 1
V1 sinФ / V1 cosФ = 0.9553/ 0.6815
Tan Ф = 1.40
Ф = [tex]Tan^{-1}[/tex](1.4)
Ф = 54.5°
Substitute Ф into equation 2
V1 sin54.5 = 0.9553
V1 = 0.9553 / 0.8141
V1 = 1.17 m/s
Therefore, the final speed and angle of the first puck are 1.17 m/s and 54.5° respectively.
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The thermal emission of the human body has maximum intensity at a wavelength of approximately 9.5 μm.What photon energy corresponds to this wavelength?
Answer:
Explanation:
2.1 x 10^2 - 20J
The escape velocity of a bullet from the surface of planet Y is 1695.0 m/s. Calculate the escape velocity from the surface of the planet X if the mass of planet X is 1.59 times that of Y, and its radius is 0.903 times the radius of Y.
The escape velocity from the surface of the planet X is 2,249.2 m/s.
Escape velocity of planet X[tex]v = \sqrt{\frac{2GM}{r} } \\\\v^2 = \frac{2GM}{r}\\\\v^2r = 2GM\\\\G = \frac{v^2r}{2M}[/tex]
where;
M is mass of the planetr is radius of the planetG is universal gravitation constant[tex]\frac{v_x^2 \ r_x}{2M_x} = \frac{v_y^2 \ r_y}{2M_y} \\\\\frac{v_x^2 \ r_x}{M_x} = \frac{v_y^2 \ r_y}{M_y} \\\\v_x^2 = \frac{v_y^2 \ r_yM_x}{M_yr_x}\\\\v_x^2 = \frac{(1695)^2 (r_y)(1.59M_y)}{M_y(0.903r_y)} \\\\v_x^2 = 5,058,814.78\\\\v_x = \sqrt{5,058,814.78} \ \ = 2,249.2 \ m/s[/tex]
Thus, the escape velocity from the surface of the planet X is 2,249.2 m/s.
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Question 2
A photon of green light has a wavelength of 520 nm. Find the green photon's frequency in Hz?
Hints: C=fa ; this will give you the frequency in Hz; 1 nm = 1x10-⁹ nm
5.77 ×[tex]10^1^4[/tex] Hz is the green photon's frequency .
The distance between similar points (adjacent crests) in adjacent cycles of a waveform signal that is propagated in space is known as the wavelength. A wave's wavelength is often measured in meters (m), centimeters (cm), or millimeters (mm) (mm). The relationship between frequency and wavelength is inverse.
Given:Wavelength of green light = 520 nm
f = c / λ
where, f = Frequency
c = Speed of light = 3 × [tex]10^8[/tex] m/s
λ = Wavelength of light
∴ f = c / λ
f = [tex]\frac{3*10^8}{520 * 10^-^9}[/tex]
= 5.77 ×[tex]10^1^4[/tex] Hz
Therefore, 5.77 ×[tex]10^1^4[/tex] Hz is the green photon's frequency .
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a)Calculate the effective value of g, the acceleration of gravity, at 7900 m above the Earth's surface.
b)Calculate the effective value of g, the acceleration of gravity, at 7900 km above the Earth's surface.
The solution for the acceleration of gravity is given as
[tex]g_{1}=9.789 \mathrm{~m} / \mathrm{s}^{2}$[/tex][tex]g_2=1.955 \mathrm{~m} / \mathrm{s}^{2}$[/tex]This is further explained below.
What is the effective value of g, the acceleration of gravity, at 7900 km above the Earth's surface.?Generally,
Mass of earth [tex]$M=5.97 \times 10^{24} \mathrm{~kg}$[/tex]
Radius of earth [tex]$R=6371 \mathrm{~km}$[/tex]
Gravitational const. [tex]$G=6.67 \times 10^{-11} \mathrm{Nm}^{2} \mathrm{~kg}^{-2}$[/tex]
height [tex]$h_{1}=7900 \mathrm{~m}=7.9 \mathrm{~km}$$$[/tex]
[tex]R+h_{1}=6371+7.9 &\\\\R+h_{1}=6378.9 \mathrm{~km} \\\\&R+h_{1}=6378.9 \times 10^{3} \mathrm{~m}\end{aligned}[/tex]
In conclusion, acceleration due to gravity at this point will be
[tex]g_{1}=\frac{G M}{\left(\bar{R}+\overline{h_{1}}\right)^{2}}$\\\\$g_{1}=\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{\left(6378.9 \times 10^{3}\right)^{2}}$\\\\$g_{1}=9.789 \mathrm{~m} / \mathrm{s}^{2}$[/tex]
for [tex]$h_{2}=7900 \mathrm{~km}$[/tex]
[tex]R+h_{2}=6371+7900\\\\R+h_{2}=14271 \mathrm{~km}\\\\$g_{2}=\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{\left(14271 \times 10^{3}\right)^{2}}$\\\\$g_2=1.955 \mathrm{~m} / \mathrm{s}^{2}$[/tex]
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Piston 1 in the figure has a diameter of 1.87 cm. Piston 2 has a diameter of 9.46 cm.
In the absence of friction, determine the force F, necessary to support an object with a mass of 991 kg placed on piston 2. (Neglect the height difference between the bottom of the two pistons, and assume that the pistons are massless).
Force necessary to support the object on piston 2 is 24× 10⁴ N.
To find the answer, we need to know about the force and pressure on piston 1 and piston 2.
What's the pressure on piston 2?The force on piston 2= mass × acceleration due to gravity= 991 Kg × 9.8 = 9414.5N
Mathematically, force= pressure/areaPressure= force × area of piston= 9414.5N × π(9.46² cm² /4)
= 9414.5N × π(9.46²× 10^(-4)m²/4)
= 66.2 N/m²
What's the force needed to held the mass on piston 2?Pressure on piston 2 = pressure on piston 1Force on piston 1= pressure on piston 1/area of piston 1= 66.2/ π(1.87² cm² /4)
= 66.2/ π(1.87²×10^(-4)m² /4)
= 24× 10⁴ N
Thus, we can conclude that force necessary to support the object on piston 2 is 24× 10⁴ N.
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A solid cylinder of uniform density of 0.85 g/cm3 floats in a glass of water tinted light blue by food coloring.
Its circular surfaces are horizontal. What effect will the following changes, each made to the initial system, have on X, the height of the upper surface above the water? The liquids added do not mix with the water, and the cylinder never hits the bottom.
1. The cylinder is replaced with one that has the same density and diameter, but with half the height.
2. Some of the water is removed from the glass.
3. A liquid with a density of 1.06 g/cm3 is poured into the glass.
4. The cylinder is replaced with one that has the same height and diameter, but with density of 0.83 g/cm3.
5. A liquid with a density of 0.76 g/cm3 is poured into the glass.
6. The cylinder is replaced with one that has the same density and height, but 1.5× the diameter.
Options are: Increase, Decrease, No change
The buoyant force acting on the cylinder is, [tex]Fb = \rho Ahg[/tex]. Here A is the cross-sectional area of the cylinder, h is the height of the cylinder, ρ is the density of the cylinder, and g is the acceleration due to gravity.
This buoyant force is also equal to the volume of the fluid displaced. [tex]Fb = \sigma h(A-x)g[/tex]. Here σ is the density of the fluid.
Equate the above two equations and solve for x.
[tex]\rho Ahg = \sigma A(h-x)g\\\rho h = \sigma h - \sigma x\\x = \frac{(\sigma - \rho)h}{\sigma}[/tex]
So, the distance x depends on the density of the fluid, density of the cylinder and the height of the cylinder.
1. The density of the cylinder is same and distance x is independent of the diameter of the cylinder. Therefor, there will be no change in the distance x. Hence, the correct answer is No change.
2. Now the height is changing keeping the density same. As the distance x is directly proportional to the height, the distance x will increase.
3. The density of the added liquid is greater than of the water and it does not mix with the water. So, the liquid will settle down and there will be no change in the distance x.
4. The density of the added liquid is less than that of the water and it does not mix with the water. So, the liquid will not settle down and the distance x will change. The change in distance x can be determined as follow:
[tex]\rho Agh = \sigma' Axg + \sigma A(h-x)g\\\rho h=\sigma' x + \sigma h - \sigma x\\x=(\frac{\sigma - \rho}{\sigma - \sigma'})h[/tex]
Here, σ' is the density of the added liquid.
From the above relation it is clear, that on adding the liquid of the density less than that of water, the denominator term become small ando so the value of x will increase.
5. On removing some of the water inside the glass, the height of the water column will decrease, but the value of x does not depend on the height of the water column. So, there will be no change in the distance x.
6. The density of the new cylinder is smaller than that of the earlier one. So, the numerator term will increase. Therefore, the value of x will increase.
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Find the speed of a satellite in a circular orbit around the Earth with a radius 3.57 times the mean radius of the Earth. (Radius of Earth = 6.37×10^3 km, mass of Earth = 5.98×10^24 kg, G = 6.67×10^-11 Nm^2/kg^2.)
Answer: The speed of a satellite in a circular orbit around the Earth with a radius 3.57 times the mean radius of the Earth is 4188.11 m/s.
Explanation: To find the answer, we need to know about the equation of motion of a satellite around earth.
What is the equation of motion of a satellite around earth?We have gravitational force of attraction between the satellite of mass m and earth of mass M as,[tex]F_g=\frac{GMm}{r^2}[/tex]
The expression for centripetal force of,[tex]F_c=\frac{mv^2}{r} \\[/tex]
These two forces are equal for a satellite around earth.[tex]\frac{GMm}{r^2} =\frac{mv^2}{r} \\thus,\\v=\sqrt{\frac{GM}{r} }[/tex]
How to solve the problem?Given that,[tex]r=3.57 R_E=3.57*6.37*10^3=22.74*10^3 km\\M=5.98*10^24kg\\G=6.67*10^{-11}Nm^2/kg^2[/tex]
Thus, the speed of the satellite will be,[tex]v=\sqrt{\frac{6.67*10^{-11}*5.98*10^{24}}{22.74*10^6m} } =4188.11 m/s[/tex]
Thus, we can conclude that the speed of satellite will be 4188.11 m/s.
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In a circular orbit with a radius that is 3.57 times the mean radius of the Earth, a satellite moves at a speed of 132.43km/s.
In order to get the solution, we must understand the satellite's planetary motion equation.
What is the satellite's orbital motion equation?The earth's mass M and the satellite's mass M are attracted to one another by gravity.[tex]F_g=\frac{GMm}{r^2}[/tex]
The term used to describe centripetal force of,[tex]F_c=\frac{MV^2}{r}[/tex]
When a satellite orbits the earth, these two forces are equivalent. Thus, the velocity will be,[tex]\frac{GMm}{r^2}=\frac{mV^2}{r}\\V=\sqrt{\frac{GM}{r} } =\sqrt{\frac{6.67*10{-11}*5.98*10^{24}}{22.74*10^3} } \\V=132.43*10^3m/s[/tex]
As a result, we may estimate that the satellite will move at a speed of 132.43 km/s.
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What is grandfather Paradox?
A grandfather paradox is a situation where individual travels to the past and then introduces a change which affects or contradicts the present.
What is a grandfather paradox?A paradox is a situation or statement which involves two contradictions.
A grandfather paradox is a situation which is defined by the ability of an individual to travel to a time in the past usually before the birth of their grandfather and then introduces a change which affects or contradicts the present. For example, killing the grandfather to prevent their birth.
In conclusion, a grandfather paradox is is an event which contradicts the present as a result of a change done to the past.
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An object 10mm in height is located 20cm from a crown glass spherical surface whose power is +10.00DS. Locate the image
The image is present at 20cm from the crown glass spherical surface.
To find the answer, we need to know about the lens formula.
What's the lens formula?It's (1/V)-(1/U)= (1/f)V= image distance from the lens, U= object distance, f= focal length of the lensWhat's the image distance, if object is present at 20cm from crown glass of power 10DS?Focal length (f)= 1/ power = 1/10 = 0.1 m U= -20cm = -0.2m (-ve sign due to sign convection)(1/V)-(-1/0.2)= (1/0.1)=> (1/V)+5=10
=> 1/V= 5
=> V=0.2m = 20cm
Thus, we can conclude that the image is present at 20cm.
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Hi I have a question it’s not about the subject but is at the same time what is Physics?
Answer:
the branch of science that is concerned with nature and properties of matter and energy.
Explanation:
a study of the basis of what does what in science.
If you speed through a construction zone while workers are present, your fines could be:.
If you speed through a construction zone while workers are present, your fines could be as much as one thousand dollars.
What is a Fine?
This is referred to as the amount which is instructed by a court or authority to be paid as a result of it being a penalty for various types of offences. each crime has its specific fine which helps to serve as deterrent for unlawful behavior in the community.
it is always best not to speed when within a construction zone which has workers present in the area. This is ideal as it helps to prevent incidences of accidents or death.
it is therefore the reason why a fine of 1000 dollars is to be paid so that people can think of such high amount before performing certain types of activities when driving and makes it the most appropriate choice.
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A man pushing a crate of mass
m = 92.0 kg
at a speed of
v = 0.845 m/s
encounters a rough horizontal surface of length
ℓ = 0.65 m
as in the figure below. If the coefficient of kinetic friction between the crate and rough surface is 0.351 and he exerts a constant horizontal force of 280 N on the crate.
(a) Find the magnitude and direction of the net force on the crate while it is on the rough surface.
magnitude_____N
What is the direction?
1. Opposite as the motion of the crate
2. Same as the motion of the crate
(b) Find the net work done on the crate while it is on the rough surface.
______J
(c) Find the speed of the crate when it reaches the end of the rough surface.
_______m/s
(a) The magnitude and direction of the net force on the crate while it is on the rough surface is 36.46 N, opposite as the motion of the crate.
(b) The net work done on the crate while it is on the rough surface is 23.7 J.
(c) The speed of the crate when it reaches the end of the rough surface is 0.45 m/s.
Magnitude of net force on the crateF(net) = F - μFf
F(net) = 280 - 0.351(92 x 9.8)
F(net) = -36.46 N
Net work done on the crateW = F(net) x L
W = -36.46 x 0.65
W = - 23.7 J
Acceleration of the cratea = F(net)/m
a = -36.46/92
a = - 0.396 m/s²
Speed of the cratev² = u² + 2as
v² = 0.845² + 2(-0.396)(0.65)
v² = 0.199
v = √0.199
v = 0.45 m/s
Thus, the magnitude and direction of the net force on the crate while it is on the rough surface is 36.46 N, opposite as the motion of the crate.
The net work done on the crate while it is on the rough surface is 23.7 J.
The speed of the crate when it reaches the end of the rough surface is 0.45 m/s.
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A rocket takes off from Earth's surface, accelerating straight up at 63.2 m/s2. Calculate the normal force (in N) acting on an astronaut of mass 84.4 kg, including her space suit. (Assume the rocket's initial motion parallel to the +y-direction. Indicate the direction with the sign of your answer.)
______N
From the calculation, the normal force is 6161.2 N.
What is the normal force?The normal force is given by the expression;
N - mg = ma
Then;
N = mg + ma
m = 84.4 kg
g = 9.8 m/s^2
a = 63.2 m/s2
Now we have;
N = m(g + a)
N = 84.4 (9.8 + 63.2)
N = 6161.2 N
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The small spherical planet called "Glob" has a mass of 7.88×1018 kg and a radius of 6.32×104 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.44×103 m, above the surface of the planet, before it falls back down.
1. What was the initial speed of the rock as it left the astronaut's hand? (Glob has no atmosphere, so no energy is lost to air friction. G = 6.67×10-11 Nm2/kg2.)
2. A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Calculate the speed of the satellite.
The initial speed of the rock as it left the astronaut's hand is 168 m/s.
The speed of the satellite is 60.2 m/s.
Acceleration due to gravity of the satellite
g = GM/R²
where;
M is mass of the satelliteR is radius of the satelliteg = (6.67 x 10⁻¹¹ x 7.88 x 10¹⁸)/(6.32 x 10⁴)²
g = 0.132 m/s²
initial speed of the rock when it reaches maximum heightv² = u² - 2gh
0 = u² - 2gh
u² = 2gh
u = √2gh
u = √(2 x 9.8 x 1440)
u = 168 m/s
Speed of the satellitev = √GM/r
v = √[(6.67 x 10⁻¹¹ x 7.88 x 10¹⁸)/(1.45 x 10⁵)]
v = 60.2 m/s
Thus, the initial speed of the rock as it left the astronaut's hand is 168 m/s.
The speed of the satellite is 60.2 m/s.
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a body has a mass of 2kg.it accelerats from 20m/s to 40m/s in 4 seconds.the resultant force is
The resultant force is 8N
Given that mass is 2kg , v= 40m/s, u =20m/s and we need to calculate resultant force
F=ma
m is given
so for a
v-u/t=a { first equation of motion }
40-20/4= 4
so a=4
F = ma =2*4 = 8N
The difference between the forces that are acting on an object as part of a system is known as the resultant force.
v = u + at is the first equation of motion. Here, v denotes the end speed, u the starting speed, an acceleration, and t the passage of time. The first equation of motion is provided by the velocity-time relation, which may be used to calculate acceleration.
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A 500 N force accelerates an object at 20 m s-2. What is its mass?
Answer: The mass of the object is 25kg.
The given question deals with Newton's second law of motion and its applications.
Explanation: Given force, F=500N
acceleration, a=20 m/[tex]s^{2}[/tex]
From Newton's 2nd law of motion , we have
F=ma where m=mass of the object
⇒500=m×20
⇒m=500/20=25
∴ Mass of the object is 25 kg .
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1.A virtual image is formed 20.5 cm from a concave mirror having a radius of curvature of 31.5 cm.
(a) Find the position of the object.
(b) What is the magnification of the mirror?
2.A contact lens is made of plastic with an index of refraction of 1.45. The lens has an outer radius of curvature of +2.04 cm and an inner radius of curvature of +2.48 cm. What is the focal length of the lens?
The position of the object is = -68cm
The magnification of the mirror= 0.3
Calculation of object distanceThe image distance = 20.5cm
The focal length= R/2 = 31.5/2= 15.75
The object distance= ?
Using the lens formula,1/f = 1/v-1/u
1/u = 1/v- 1/f
1/u = 1/20.5 - 1/15.75
1/u = 0.0489- 0.0635
1/u = -0.0146
u = -68cm
The magnification of the mirror is image size/object size
= 20.5cm/-68cm
= 0.3
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Find the speed of a satellite in a circular orbit around the Earth with a radius 3.57 times the mean radius of the Earth. (Radius of Earth = 6.37×103 km, mass of Earth = 5.98×1024 kg, G = 6.67×10-11 Nm2/kg2.)
The orbiting velocity of the satellite is 4.2km/s.
To find the answer, we need to know about the orbital velocity of a satellite.
What's the expression of orbital velocity of a satellite?Mathematically, orbital velocity= √(GM/r)r = radius of the orbital, M = mass of earthWhat's the orbital velocity of a satellite orbiting earth with a radius 3.57 times the earth radius?M= 5.98×10²⁴ kg, r= 3.57× 6.37×10³ km = 22.7×10⁶mOrbital velocity= √(6.67×10^(-11)×5.98×10²⁴/22.7×10⁶)=4.2km/s
Thus, we can conclude that the orbiting velocity of the satellite is 4.2km/s.
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At t=0s a small "upward" (positive y) pulse centered at x = 4.0 m is moving to the right on a string with fixed ends at x=0.0m and x = 13.0 m . The wave speed on the string is 3.5 m/s . At what time will the string next have the same appearance that it did at t=0s?
The next time the string will have the same appearance that it did at t=0s is 2.29 s.
Frequency of the wave
v = fλ
f = v/λ
where;
λ is wavelengthhalf of the upward pulse is a quarter of wavelength = ¹/₄ x 4 m = 1 m
f = 3.5/1
f = 3.5 Hz
Time of motion when the pulse is at 4 mt1 = 4/3.5 = 1.143 s
The next time the string will have the same appearance that it did at t=0s.
d = 4 m x 2 = 8 m
t2 = 8/3.5
t2 = 2.29 s
Thus, the next time the string will have the same appearance that it did at t=0s is 2.29 s.
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A 45-kg pole vaulter running at 10 m/s vaults over the bar. Her speed when she is above the bar is 1.1 m/s. Neglect air resistance, as well as any energy absorbed by the pole, and determine her altitude as she crosses the bar.
_______m
The altitude or height of the pole vaulter as she crosses the bar is 4.04 m.
What is the height of the pole vaulter?The height of the pole vaulter is determined from the change in kinetic energy which is equal to the potential energy at that height.
Potential energy = Change in kinetic energymgh = m(v - u)²/2h = (v - u)²/2g
h = (10 - 1.1)²/2 * 9.8
h = 4.04 m.
In conclusion, the height is determined from the potential energy at that height.
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A plank of length L=2.200 m and mass M=4.00 kg is suspended horizontally by a thin cable at one end and to a pivot on a wall at the other end as shown. The cable is attached at a height H=1.70 m above the pivot and the plank's CM is located a distance d=0.700 m from the pivot.
Calculate the tension in the cable.
Hello!
Let's begin by doing a summation of torques, placing the pivot point at the attachment point of the rod to the wall.
[tex]\Sigma \tau = 0[/tex]
We have two torques acting on the rod:
- Force of gravity at the center of mass (d = 0.700 m)
- VERTICAL component of the tension at a distance of 'L' (L = 2.200 m)
Both of these act in opposite directions. Let's use the equation for torque:
[tex]\tau = r \times F[/tex]
Doing the summation using their respective lever arms:
[tex]0 = L Tsin\theta - dF_g[/tex]
[tex]dF_g = LTsin\theta[/tex]
Our unknown is 'theta' - the angle the string forms with the rod. Let's use right triangle trig to solve:
[tex]tan\theta = \frac{H}{L}\\\\tan^{-1}(\frac{H}{L}) = \theta\\\\tan^{-1}(\frac{1.70}{2.200}) = 37.69^o[/tex]
Now, let's solve for 'T'.
[tex]T = \frac{dMg}{Lsin\theta}[/tex]
Plugging in the values:
[tex]T = \frac{(0.700)(4.00)(9.8)}{(2.200)sin(37.69)} = \boxed{20.399 N}[/tex]
A constant force F is applied on a body for a time interval of delta t.This force changes the velocity of the body from V1 to V2.then change in momentum in time interval will be
The change in momentum in time interval, given the data will be F × Δt
What is momentum?Momentum is defined as the product of mass and velocity. It is expressed as
Momentum = mass × velocity
What is impulse?This is defined as the change in momentum of an object.
Impulse = change in momentum
But
Impulse = force × time
Therefore
Force × time = change in momentum
How to determine the change in momentumInitial velocity = v₁ Finalvelocity = v₂Force = FChange in time = ΔtChange in momentum = ?Force × time = change in momentum
F × Δt = change in momentum
Change in momentum = F × Δt
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Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.17 times a second. A tack is stuck in the tire at a distance of 0.351 m from the rotation axis. Noting that for every rotation the tack travels one circumference, find the tack's tangential speed
The tangential speed of the wheel is determined as 4.786 m/s.
Tangential speed of the wheel
The tangential speed of the wheel is calculated as follows;
v = ωr
where;
ω is angular speed in rad/sr is radius of the circular pathv = (2.17 x 2π rad)/s x 0.351 m
v = 4.786 m/s
Thus, the tangential speed of the wheel is determined as 4.786 m/s.
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1. A block of mass 0.4kg resting on the top of an inclined plane of height 20m starts to slide down on the surface of the incline to its foot, and then continues its slide horizontally. At a distance of 5m from the foot of the incline there is another block of the same mass resting on the horizontal surface to undergo an elastic collision. Next to the second block, there is a light spring of constant k = 4000N/m fixed freely against a wall. The spring is supposed to make a head-on collision with the second block. See the arrangements as in 1. Assuming all surfaces being frictionless, (a) calculate the kinetic energy of the firs block just at the foot of the incline; (b) calculate the kinetic and gravitational potential energies of the first block halfway down the incline; (c) calculate the speeds of the two blocks just after their collision; (d) compute the maximum compression of the spring resulted from its collision with the second block; (e) determine the maximum work done by the spring on the second block.
The kinetic energy of the first block just at the foot of the incline is 78.4J, the kinetic and gravitational potential energies of the first block halfway down the incline are same, and which is equal to 39.2J. The speeds of the two blocks just after their collision interchange with the values before collision.
To find the answer, we need to know about the concept of collision and kinetic energy.
How to find the kinetic energy of the first block just at the foot of the incline?Given that, the block of mass 0.4kg resting on the top of an inclined plane of height 20m.Thus, at the top of the incline it has a potential energy, and the kinetic energy will be equal to zero, or we can say that the total energy of the system is equal to the potential energy at topmost point.[tex]TE=KE+PE\\T=PE=m_1gh=(0.4*9.8*20)=78.4 J[/tex]
We have to find the kinetic energy of the first block just at the foot of the incline, and at the bottom point the PE=0, or we can say that the total energy or the potential energy is converted into kinetic energy.[tex]TE=KE=78.4J[/tex]
What is the kinetic and gravitational potential energies of the first block halfway down the incline?At the halfway, the PE will be,[tex]U'=m_1gh'=mg\frac{h}{2} \\U'=39.2J[/tex]
As we know that, the energy is conserved at each point of the motion.[tex]TE=78.4 J\\KE'+PE'=78.4J\\KE'=78-U'=78.4-39.2=39.2J[/tex]
How to find the speeds of the two blocks just after their collision?We have the KE at bottom point as, 78.4J. Thus, the velocity of first block at the bottom before collision will be,[tex]KE=\frac{1}{2} mv^2=78.4J\\v=\sqrt{\frac{2KE}{m} } =4m/s[/tex]
This is the velocity of the block 1 of mass m1 before collision, we can say, u1.As we know that, the 2 nd block of mass m2 is at rest, thus, u2=0.Given that, the collision is elastic. Thus, both the KE and the momentum will be conserved.[tex]\frac{1}{2}m_1u_1^2+ \frac{1}{2}m_2u_2^2=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2[/tex]
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]
We have,[tex]m_1=m_2=m\\u_1=4m/s\\u_2=0\\v_1=?\\v_2=?[/tex]
Substituting this in both the equations, we get,[tex]\frac{1}{2}m*4^2=\frac{1}{2}m(v_1^2+v_2^2)\\(v_1^2+v_2^2)=16[/tex] from resolving KE equation.
[tex]4m=m(v_2+v_1)\\4=v_2+v_1\\v_1=4-v_2[/tex] From resolving momentum conservation.
solving both, we get,[tex]v_2=4m/s\\v_1=0[/tex]
Thus, we can conclude that, the kinetic energy of the first block just at the foot of the incline is 78.4J, the kinetic and gravitational potential energies of the first block halfway down the incline are same, and which is equal to 39.2J. The speeds of the two blocks just after their collision interchange with the values before collision.
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A crate with a mass of 175.5 kg is suspended from the end of a uniform boom with a mass of 94.7 kg. The upper end of the boom is supported by a cable attached to the wall and the lower end by a pivot (marked X) on the same wall.
Calculate the tension in the cable.
The tension in the cable is equal to 323.5 N.
What is the tension in the cable?The tension, T in the cable is determined by taking moments about the pivot marked X.
The angles of the boom and the cable with the horizontal are first calculated.
Angle of the boom with horizontal, θ = tan⁻¹(5/10) = 26.56°
The angle of cable with horizontal, B = tan⁻¹(4/10) = 21.80
Taking moments about the pivot:
175.5 * cos 26.56 + 94.7 * cos 26.56 * 0.5 = T (sin(26.56 + 21.80) * 1
Tension = 241.68/0.747
Tension = 323.5 N
In conclusion, the tension in the cable helps to suspend the crate.
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6.
A swimmer bounces straight up from a diving board and falls feet first into a pool. She
starts with a velocity of 4.00 m/s, and her takeoff point is 1.80 m above the pool. (3pt)
a) How long are her feet in the air?
b) What is her highest point above the board?
c) What is her velocity when her feet hit the water?
a) Her feet are in the air for 0.73+0.41 = 1.14 seconds
b) Her highest height above the board is 0.82 m
c) Her velocity when her feet hit the water is 7.16 m/s
Given,t = Time taken
u = Initial velocity = 4 m/s
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s²
a) Her feet are in the air for 0.73+0.41 = 1.14 seconds
s = ut + [tex]\frac{1}{2}[/tex]at²
2.62 = 0t + [tex]\frac{1}{2}[/tex] ₓ 9.8 ₓ [tex]t^{2}[/tex]
t = 0.73 s
b) Her highest height above the board is 0.82 m
The total height she would fall is 0.82+1.8 = 2.62 m
v = u + at
0 = 4 ₋ 9.8 ₓ t
t = 0.41 s
s = ut +[tex]\frac{1}{2}[/tex] at²
s = 4 ₓ 0.41 ₊ [tex]\frac{1}{2}[/tex] ₓ ₋9.8 ₓ 0.41 [tex]t^{2}[/tex]
c) Her velocity when her feet hit the water is 7.16 m/s
[tex]v = u + at \\v = 0 + 9.8[/tex] ₓ [tex]0.73[/tex]
v = 7.16 m/s
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