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3 years ago

A force of 500 N acts horizontally on a 10,000 g body. What is its horizontal acceleration

Physics

1 answer:

Paladinen [302]3 years ago

8 0

200n because it's 2×5=10so maybe try solving the problem like that ok does that help

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If Jerome is swinging on a rope and transferring energy from gravitational potential energy to kinetic energy, A)compression B)w

Gnesinka [82]

Salutations!

If Jerome is swinging on a rope and transferring energy from gravitational potential energy to kinetic energy, _______________ is being done.

<span>If Jerome is swinging on a rope and transferring energy from gravitational potential energy to kinetic energy, work is being done. Energy being transferred and the object begins to move is called work.

Thus, your answer is option B.

Hope I helped (:

Have a great day!</span>

7 0

3 years ago

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A sled of mass 10 kg slides along the ice. it has an initial speed of 2 m/s but stops because of friction. How much work is done

NeX [460]

Answer: The correct answer is option B.

Explanation:

Mass of the sled = 10 kg

Initial speed of the sled = 2 m/s

Kinetic energy of the sled = $\frac{1}{2}mv^2$

$\frac{1}{2}\times 10 kg\times (2 m/s)^2=20 Joules$

Work done by the sled = 20 joules

The work done by the friction will be in opposite direction and equal to the magnitude of the work done of the sled that - 20 J.

Hence, correct answer is option B.

6 0

3 years ago

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A car speeding down the highway honks its horn, which has a frequency 392 Hz, but a resting bystander hears the frequency 440 Hz

Natali [406]

Answer:

37.42 m/s

Explanation:

We know that apparent frequency, $\bar f$ is given by

$\bar f=f\frac {V}{V-V_s}$ where f is the given frequency in this case 392, V is the speed of sound in air which is given as 343 and $V_s$ is the speed of car which is unknown, \bar f is given as 440 Hz

$440=392\times \frac {343}{343-V_s}\\343-V_s=392\times \frac {343}{440}=305.5818182\\V_s=343-305.5818182=37.41818182\approx 37.42 m/s$

8 0

3 years ago

Part A A uniform solid disk of radius 1.60 m and mass 2.30 kg rolls without slipping to the bottom of an inclined plane. If the

kifflom [539]

Answer:

Height will be 3.8971 m

Explanation:

We have given that radius of the solid r = 1.60 m

Mass of the solid disk m = 2.30 kg

Angular velocity $\omega =4.46rad/sec$

Moment of inertia is given by $I=\frac{1}{2}mr^2$

Transnational Kinetic energy is given by $KE=\frac{1}{2}mv^2$ as we know that v = $v=\omega r$

So $KE=\frac{1}{2}m(\omega r)^2$

Rotational kinetic energy is given by $KE_{ROTATIONAL}=\frac{1}{2}I\omega ^2=\frac{1}{2}\left ( \frac{1}{2}mr^2 \right )\omega ^2=\frac{1}{4}m(r\omega )^2$