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zimovet [89]
3 years ago
10

The same force is applied to two skateboards. One rolls across the room and the other moves a few feet and comes to a stop. Wher

e was there more work done?
a. The skateboard that traveled the shorter distance shows more work because there was more resistance.
b. The skateboard that traveled the longer distance shows more work because it was lighter.
c. The skateboard that traveled further shows more work because the distance was greater.
d. The skateboard that traveled the shorter distance shows more work because the force was greater.
Physics
2 answers:
podryga [215]3 years ago
8 0

The question and all of the choices are rather messy and misleading.

Here's a simple statement that you can use to make the best match:

If the two forces are equal, then more work is done by the force that
continues to be applied through the greater distance.

mestny [16]3 years ago
6 0

Answer:

<h2>C</h2>

Explanation:

<h2>THIS IS TRUE BECAUSE I TOOK THE TEST, and because THE EQUATION FOR WORK IS WORK = FORCE * DISTANCE AND EVEN THO THEY BOTH HAVE THE SAME FORCE THE OTHER ONE HAS MORE DISTANCE, ITS LIKE SAYING WHICH EQUAION HAS A BIGGER ANSWER 1*1 OR 1*2 WELL ITS OBVIOUSLY THE SECOUND EQUATION SO THE ANSWER IS C LIKE I SAID BEFORE</h2><h2 /><h2>BTW  AL2006 IS WRONG AND I DONT EVEN UNDERSTAND WHAT HE EVEN THO HE IS "TRUSTED"</h2>
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What is the name of the process that plants use to remove carbon dioxide from the atmosphere?
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It removes carbon naturally.

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A 2.0 kg block is released from rest at the top of a curved incline in the shape of a quarter of a circle of radius R = 3.0 m. T
Zigmanuir [339]

The block has maximum kinetic energy at the bottom of the curved incline. Since its radius is 3.0 m, this is also the block's starting height. Find the block's potential energy <em>PE</em> :

<em>PE</em> = <em>m g h</em>

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Energy is conserved throughout the block's descent, so that <em>PE</em> at the top of the curve is equal to kinetic energy <em>KE</em> at the bottom. Solve for the velocity <em>v</em> :

<em>PE</em> = <em>KE</em>

58.8 J = 1/2 <em>m v</em> ²

117.6 J = (2.0 kg) <em>v</em> ²

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3 0
3 years ago
A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 67 m/s*2. If he reaches the gr
KIM [24]

Answer:

3.49 seconds

3.75 seconds

-43200 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

s=ut+\frac{1}{2}at^2\\\Rightarrow 50=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{50\times 2}{9.81}}\\\Rightarrow t=3.19\ s

Time the parachutist falls without friction is 3.19 seconds

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 50+0^2}\\\Rightarrow v=31.32\ m/s

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

v=u+at\\\Rightarrow 11=31.32+9.81t\\\Rightarrow t=\frac{11-31.32}{-67}=0.3\ s

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=0t+\frac{1}{2}\times a\times t^2\\\Rightarrow \frac{s}{2}=\frac{1}{2}at^2

s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=u1.1+\frac{1}{2}\times a\times 1.1^2

Now the initial velocity of the last half height will be the final velocity of the first half height.

v=u+at\\\Rightarrow v=at

Since the height are equal

\frac{1}{2}at^2=u1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow \frac{1}{2}at^2=at1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow 0.5t^2-1.1t-0.605=0\\\Rightarrow 500t^2-1100t-605=0

t=\frac{11\left(1+\sqrt{2}\right)}{10},\:t=\frac{11\left(1-\sqrt{2}\right)}{10}\\\Rightarrow t=2.65, -0.45

Time taken to fall the first half is 2.65 seconds

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v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-240^2}{2\times \frac{8}{12}}\\\Rightarrow a=-43200\ ft/s^2

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5 0
3 years ago
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brilliants [131]
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