Compare and Contrast Potential Energy with Thermal Energy

ASAP FIRST TO ANSWER CORRECTLY WILL BE BRAINLIEST

ioda

Answer:

1.Divergent

2.Diverge

3.Himalayan

Explanation:

4 0

3 years ago

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The momentum of a car 54,400 kg * m/s. If the car weighs 16,680N what is the velocity of the car

wariber [46]

G=mg=>m=G/g=16680/9.8=1702 kg

p=mv=>v=p/m=54400/1702=32 m/s

6 0

3 years ago

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"Sarah is out playing fetch with her dog, and throws a tennis ball as far as she can. At the moment the ball reaches its maximum

e-lub [12.9K]

Answer:

<h2>a.The ball's vertical acceleration is downwards.</h2><h2>e.The ball's horizontal acceleration is zero</h2>

Explanation:

We are given that Sarah throws a tennis ball as far as she can.

At the moment the ball reaches its maximum height.

We have to find the true statement if air resistance is neglect.

When air resistance is negligible then the force act on the ball is force due to gravity.

The ball throw vertically then the acceleration act on the ball is acceleration due to gravity.

The value of g=-9.8 m/square sec

It acts on the ball in downward direction .

Therefore, the ball's vertical acceleration is downwards.

The horizontal acceleration is zero because the ball reaches at maximum height then there is no force which act in horizontal direction on the ball.

Therefore, horizontal acceleration of the ball is zero.

Hence, option a and e are true.

3 0

3 years ago

The mass of the hammer is 0.454 kg. Calculate the weight of the hammer.

Taya2010 [7]

Explanation:

weight =0.454 × 9.8=4.4492N

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3 years ago

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A luggage handler pulls a suitcase of mass 19.6 kg up a ramp inclined at an angle 24.0 ∘ above the horizontal by a force F⃗ of m

Dvinal [7]

(a) 638.4 J

The work done by a force is given by

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the object

$\theta$ is the angle between the direction of the force and the displacement

Here we want to calculate the work done by the force F, of magnitude

F = 152 N

The displacement of the suitcase is

d = 4.20 m along the ramp

And the force is parallel to the displacement, so $\theta=0^{\circ}$ . Therefore, the work done by this force is

$W_F=(152)(4.2)(cos 0)=638.4 J$

b) -328.2 J

The magnitude of the gravitational force is

W = mg

where

m = 19.6 kg is the mass of the suitcase

$g=9.8 m/s^2$ is the acceleration of gravity

Substituting,

$W=(19.6)(9.8)=192.1 N$

Again, the displacement is

d = 4.20 m

The gravitational force acts vertically downward, so the angle between the displacement and the force is

$\theta= 90^{\circ} - \alpha = 90+24=114^{\circ}$

Where $\alpha = 24^{\circ}$ is the angle between the incline and the horizontal.

Therefore, the work done by gravity is

$W_g=(192.1)(4.20)(cos 114^{\circ})=-328.2 J$

c) 0

The magnitude of the normal force is equal to the component of the weight perpendicular to the ramp, therefore:

$R=mg cos \alpha$

And substituting

m = 19.6 kg

g = 9.8 m/s^2

$\alpha=24^{\circ}$

We find

$R=(19.6)(9.8)(cos 24)=175.5 N$

Now: the angle between the direction of the normal force and the displacement of the suitcase is 90 degrees:

$\theta=90^{\circ}$

Therefore, the work done by the normal force is

$W_R=R d cos \theta =(175.4)(4.20)(cos 90)=0$

d) -194.5 J

The magnitude of the force of friction is

$F_f = \mu R$

where

$\mu = 0.264$ is the coefficient of kinetic friction

R = 175.5 N is the normal force

Substituting,

$F_f = (0.264)(175.5)=46.3 N$

The displacement is still

d = 4.20 m

And the friction force points down along the slope, so the angle between the friction and the displacement is

$\theta=180^{\circ}$

Therefore, the work done by friction is

$W_f = F_f d cos \theta =(46.3)(4.20)(cos 180)=-194.5 J$

e) 115.7 J

The total work done on the suitcase is simply equal to the sum of the work done by each force,therefore:

$W=W_F + W_g + W_R +W_f = 638.4 +(-328.2)+0+(-194.5)=115.7 J$

f) 3.3 m/s

First of all, we have to find the work done by each force on the suitcase while it has travelled a distance of

d = 3.80 m

Using the same procedure as in part a-d, we find:

$W_F=(152)(3.80)(cos 0)=577.6 J$

$W_g=(192.1)(3.80)(cos 114^{\circ})=-296.9 J$

$W_R=(175.4)(3.80)(cos 90)=0$

$W_f =(46.3)(3.80)(cos 180)=-175.9 J$

So the total work done is

$W=577.6+(-296.9)+0+(-175.9)=104.8 J$

Now we can use the work-energy theorem to find the final speed of the suitcase: in fact, the total work done is equal to the gain in kinetic energy of the suitcase, therefore

$W=\Delta K = K_f - K_i\\W=\frac{1}{2}mv^2\\v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(104.8)}{19.6}}=3.3 m/s$

6 0

3 years ago