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AleksandrR [38]
3 years ago
13

The image formed by the pinhole camera is smaller than that of the object true or false

Physics
1 answer:
Zina [86]3 years ago
4 0

Answer: True? since it has a small hole it would depend on how far away the camera is to the object.

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1) Andrea and Chuck are riding on a merry-go-round. Andrea rides on a horse at the outer rim of the circular platform, twice as
telo118 [61]

Explanation:

The tangential speed of Andrea is given by :

v=r\omega

Where

r is radius of the circular path

ω is angular speed

The merry-go-round is rotating at a constant angular speed. Let the new distance from the center of the circular platform is r'

r' = 2r

New angular speed,

v'=r'\omega'\\\\v'=(2r)\omega\\\\v'=2r\omega\\\\v'=2v

New angular speed is twice that of the Chuck's speed.

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Leni [432]

C is the correct answer. Opposites attract!

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This graph shows velocity vs. time. What does the slope of the line represent? A. speed B. force C. acceleration D. distance
PilotLPTM [1.2K]
I believe the answer is C. Acceleration. (:
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3 years ago
A small meteorite with mass of 1 g strikes the outer wall of a communication satellite with a speed of 2Okm/s (relative to the s
strojnjashka [21]

Answer:

The energy coverted to heat is 200 kilojoules.

Explanation:

GIven the absence of external forces exerted both on the small meteorite and on the communication satellite, the Principle of Linear Momentum is considered and let suppose that collision is completely inelastic and that satellite is initially at rest. Hence, the expression for the satellite-meteorite system:

m_{M}\cdot v_{M} + m_{S}\cdot v_{S} = (m_{M}+m_{S})\cdot v

Where:

m_{M}, m_{S} - Masses of the small meteorite and the communication satellite, measured in kilograms.

v_{M}, v_{S} - Speeds of the small meteorite and the communication satellite, measured in meters per second.

v - Final speed of the satellite-meteorite system, measured in meters per second.

The final speed of the satellite-meteorite system is cleared:

v = \frac{m_{M}\cdot v_{M}+m_{S}\cdot v_{S}}{m_{M}+m_{S}}

If m_{M} = 1\times 10^{-3}\,kg, m_{S} = 200\,kg, v_{M} = 20000\,\frac{m}{s} and v_{S} = 0\,\frac{m}{s}, the final speed is now calculated:

v = \frac{(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)}{1\times 10^{-3}\,kg+200\,kg}

v = 0.1\,\frac{m}{s}

Which means that the new system remains stationary and all mechanical energy from meteorite is dissipated in the form of heat. According to the Principle of Energy Conservation and the Work-Energy Theorem, the change in the kinetic energy is equal to the dissipated energy in the form of heat:

K_{S} + K_{M} - K - Q_{disp} = 0

Q_{disp} = K_{S}+K_{M}-K

Where:

K_{S}, K_{M} - Initial translational kinetic energies of the communication satellite and small meteorite, measured in joules.

K - Kinetic energy of the satellite-meteorite system, measured in joules.

Q_{disp} - Dissipated heat, measured in joules.

The previous expression is expanded by using the definition for the translational kinetic energy:

Q_{disp} = \frac{1}{2}\cdot [m_{M}\cdot v_{M}^{2}+m_{S}\cdot v_{S}^{2}-(m_{M}+m_{S})\cdot v^{2}]

Given that m_{M} = 1\times 10^{-3}\,kg, m_{S} = 200\,kg, v_{M} = 20000\,\frac{m}{s}, v_{S} = 0\,\frac{m}{s} and v = 0.1\,\frac{m}{s}, the dissipated heat is:

Q_{disp} = \frac{1}{2}\cdot \left[(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)^{2}+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)^{2}-(200.001\,kg)\cdot \left(0.001\,\frac{m}{s} \right)^{2}\right]Q_{disp} = 200000\,J

Q_{disp} = 200\,kJ

The energy coverted to heat is 200 kilojoules.

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3 years ago
The Rigidbody component adds collision to a GameObject
kifflom [539]

Rigidbodies are components that allow a GameObject<u> to react to real-time physics. </u>

Explanation:

  • Rigidbodies are components that allow a GameObject to react to real-time physics. This includes reactions to forces and gravity, mass, drag and momentum. You can attach a Rigidbody to your GameObject by simply clicking on Add Component and typing in Rigidbody2D in the search field.
  • A rigidbody is a property, which, when added to any object, allows it to interact with a lot of fundamental physics behaviour, like forces and acceleration. You use rigidbodies on anything that you want to have mass in your game.
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  • If you had a game with only two objects in it, and both move kinematically, in theory you would only need a rigidbody on one of them, even though they both move.
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2 years ago
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