All categories

3 years ago

Which of the following is required for work to be done on an object?

Physics

1 answer:

ad-work [718]3 years ago

7 0

I think that in order for work to be done, the object must move in the direction of the force and move over a distance.

You might be interested in

Can someone please explain number 8?

guajiro [1.7K]

Answer: Ax=(Vx-Vox)/(T)

Vx=Vox+Ax*T

Solving for Ax in terms of Vx, Vox, T

Vx-Vox=Ax*t

Ax=(Vx-Vox)/(T)

This is saying the acceleration in the x-direction can be found by taking the difference between the finial and initial Velocity in x-direction and dividing it by the Total Time.

Any questions please feel free to ask. Thanks

8 0

3 years ago

What is energy? it's from my physics book

s344n2d4d5 [400]

Energy is the capacity of doing work

4 0

3 years ago

Read 2 more answers

A force that is doing work on a ball when that ball is falling through the air.

Ronch [10]

Answer:

i think the answer is gravity

3 0

3 years ago

Define s.i unit of force explanation

kvv77 [185]

<h3><u>Answer</u> :</h3>

◈ As per newton's second law of motion, Force is defined as the product of mass and acceleration $.$

Mathematically,

$\dag\:\boxed{\bf{\purple{F=ma}}}$

Unit of mass : kg

Unit of acceleration : m/s²

Therefore,

Unit of force ➠ <u>kg m/s²</u>

SI unit : <u>N (newton)</u> or <u>kg m/s²</u>

3 0

4 years ago

Your grandmother enjoys creating pottery as a hobby. She uses a potter's wheel, which is a stone disk of radius R-0.520 m and ma

Lesechka [4]

Answer:

0.54454

104.00902 N

Explanation:

m = Mass of wheel = 100 kg

r = Radius = 0.52 m

t = Time taken = 6 seconds

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity

$\alpha$ = Angular acceleration

Mass of inertia is given by

$I=\dfrac{mr^2}{2}\\\Rightarrow I=\dfrac{100\times 0.52^2}{2}\\\Rightarrow I=13.52\ kgm^2$

Angular acceleration is given by

$\alpha=\dfrac{\tau}{I}\\\Rightarrow \alpha=\dfrac{\mu fr}{I}\\\Rightarrow \alpha=\dfrac{\mu 50\times 0.52}{13.52}$

Equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=\omega_i+\dfrac{\mu (-50)\times 0.52}{13.52}t\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{\mu (-50)\times 0.52}{13.52}\times 6\\\Rightarrow 0=6.28318-11.53846\mu\\\Rightarrow \mu=\dfrac{6.28318}{11.53846}\\\Rightarrow \mu=0.54454$

The coefficient of friction is 0.54454

At r = 0.25 m

$\omega_f=\omega_i+\dfrac{0.54454 (-50)\times 0.52}{13.52}6\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{0.54454 f\times 0.25}{13.52}6\\\Rightarrow 2\pi=0.06041f\\\Rightarrow f=\dfrac{2\pi}{0.06041}\\\Rightarrow f=104.00902\ N$

The force needed to stop the wheel is 104.00902 N

5 0

3 years ago