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3 years ago

HELP ASAP Miguel claims that as a ball falls, it loses potential energy. Izzy claims that as a ball falls, it

Physics

1 answer:

Bogdan [553]3 years ago

7 0

Answer and Explanation:

They are both correct.

This is because potential energy is when there is potential for the object to have energy, while kinetic energy is energy that is moving already.

So, as the ball falls, it loses its potential energy, or "still/Stalled" energy, and gains kinetic energy, or moving energy.

<em><u>#teamtrees #PAW (Plant And Water)</u></em>

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A skydiver, with a mass of 60 kg is falling at a velocity of 30 m/s. Calculate his kinetic energy.

ddd [48]

Answer:

27000 J

Explanation:

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3 years ago

A ski jumper starts with a horizontal take-off velocity of 25 m/s and lands on a straight landing hill inclined at 30o. Determin

melamori03 [73]

Answer:

0.34s, 8.5m,31.89m

Explanation:

The above motion defines a projectile motion.

Now the athletes lands on a cliff 30° to the horizontal this means the velocity at that point would be 25m/s cos30°

Now from Newton's law of motion.

The body would be decelerating so,

V = u - gt

Where u is initial velocity and v is final velocity. g is acceleration of free fall due to gravity.

Hence,

V-U/ -g = t

Hence 25cos30 - 25/ -9.8 = 0.34s.

2.Now the length of the jump is defined as the total horizontal distance which is marked off by the horizontal velocity and time taken for take off and landing.

Hence Distance,S = u × t

25 ×0.34 =8.5m.

3. The maximum height is defined that at that point the Final velocity is 0m/s

Now the initial velocity is 25m/s

From Newton's law that;

V2= U2 -2gH; where U and V are initial and final velocity and H is height.

Hence H = V2-U2/-2g

=(0)^2- (25)^2/ -2×9.8

= -625/-19.6 =31.89m

8 0

3 years ago

The outer layer of cable on a cable reel is 16.2 cm from the center of the reel. The reel is initially stationary and can rotate

ahrayia [7]

Answer:

B. w=12.68rad/s

C. α=3.52rad/s^2

Explanation:

We can solve this problem by taking into account that (as in the uniformly accelerated motion)

$\theta=\omega_{0}t+\frac{1}{2}\alpha t^{2}\\\theta = \frac{s}{r}$ ( 1 )

where w0 is the initial angular speed, α is the angular acceleration, s is the arc length and r is the radius.

In this case s=3.7m, r=16.2cm=0.162m, t=3.6s and w0=0. Hence, by using the equations (1) we have

$\theta=\frac{3.7m}{0.162m}=22.83rad$

$22.83rad=\frac{1}{2}\alpha (3.6s)^2\\\\\alpha=2\frac{(22.83rad)}{3.6^2s}=3.52\frac{rad}{s^2}$

to calculate the angular speed w we can use $\alpha=\frac{\omega _{f}-\omega _{i}}{t _{f}-t _{i}}\\\\\omega_{f}=\alpha t_{f}=(3.52\frac{rad}{s^2})(3.6)=12.68\frac{rad}{s}$

Thus, wf=12.68rad/s

C) We can use our result in B)

$\alpha=3.52\frac{rad}{s^2}$

I hope this is useful for you

regards

3 0

3 years ago

Read 2 more answers

An ideal spring has spring constant ks (to distinguish it from the electrostatic constant k) and equilibrium length l. Then, you

nexus9112 [7]

Answer:

q = square root (4KsL³/k)

The force of extension of the spring is equal to the force of repulsion between the two like charges. Two like charges(positive or negative) would always repel each other and two unlike charges would always attract each other. This electric force between the charges is what is responsible for the stretching of the spring. The electric force causes the spring to increase in length from L to 2L. Equating these forces, that is the electric force between the charges and the elastic force of the spring and rearranging the variables gives the expression to obtain q.

Explanation:

See the attachment below for full solution.

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4 years ago

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vazorg [7]

Is there a question in here somewhere?

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