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2 years ago

In Hooke's law, Fspring=kΔx, what does the ∆x stand for

Physics

1 answer:

monitta2 years ago

7 0

Answer:

Change in Displacement

Explanation:

delta/triangle = change

x = displacement

formula (if needed): final x - initial x

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Olivia bought new gym shoes to play volleyball because she kept slipping when she ran in her old shoes .how will the new soles h

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3 years ago

Read 2 more answers

During which two time intervals does the particle undergo equal displacement?

san4es73 [151]

Answer:

BC and DE

Explanation:

In the given figure, the velocity time graph is shown. We know that the area under v-t curve gives the displacement of the particle.

Area under AB, $d_1=\dfrac{1}{2}\times 2\times 10=10\ m$

Area under BC, $d_2=\dfrac{1}{2}\times 2\times 4=4\ m$

Area under CD, $d_3=\dfrac{1}{2}\times 2\times 7=7\ m$

Area under DE, $d_4=\dfrac{1}{2}\times 2\times 4=4\ m$

Area under EF, $d_5=\dfrac{1}{2}\times 2\times 3=3\ m$

So, form above calculations it is clear that, during BC and DE undergo equal displacement. Hence, the correct option is (c) "BC and DE = 4 meters".

4 0

2 years ago

1.imagine you were able to throw a ball in a frictionless environment such as outer space.once you let go of the ball,what will

Scorpion4ik [409]

Imagine you were able to throw a ball in a frictionless environment
such as outer space. Once you let go of the ball, it will travel forever
in a straight line, and at a constant speed. (At least until it bumps into
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A car accelerates down the road. The reaction to the tires pushing
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3 years ago

Calculate a pendulum's frequency of oscillation (in Hz) if the pendulum completes one cycle in 0.5 s.

Marina86 [1]

Time taken to complete one oscillation for a pendulum is Time Period, T = 0.5 s
Frequency of the pendulum oscillation = 1 / Time Period => f = 1 / T = 1 / 0.5
Frequency f = 2 Hz

3 0

3 years ago

How would a spinning disk's kinetic energy change if its moment of inertia was five times larger but its angular speed was five

Keith_Richards [23]

Answer:

The kinetic energy of a spinning disk will be reduced to a tenth of its initial kinetic energy if its moment of inertia is made five times larger, but its angular speed is made five times smaller.

Explanation:

Let us first consider the initial characteristics of the angular motion of the disk

moment of inertia = $I$

angular speed = ω

For the second case, we consider the characteristics to now be

moment of inertia = $5I$ (five times larger)

angular speed = ω/5 (five times smaller)

Recall that the kinetic energy of a spinning body is given as

$KE = \frac{1}{2}Iw^{2}$

therefore,

for the first case, the K.E. is given as

$KE = \frac{1}{2}Iw^{2}$

and for the second case, the K.E. is given as

$KE = \frac{1}{2}(5I)(\frac{w}{5} )^{2} = \frac{5}{50}Iw^{2}$

$KE = \frac{1}{10}Iw^{2}$

this is one-tenth the kinetic energy before its spinning characteristics were changed.

This implies that the kinetic energy of the spinning disk will be reduced to a tenth of its initial kinetic energy if its moment of inertia is made five times larger, but its angular speed is made five times smaller.

6 0

2 years ago