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3 years ago

A rock that has deformed ____ under stress keeps its new shape when the stress is released.

Physics

1 answer:

Anna007 [38]3 years ago

8 0

Answer:

Elastically

Explanation:

A rock that has deformed Elastically under stress keeps its new shape when the stress is released.

In elastic deformation the original shape of the object is regained when the stress is removed. Whereas in plastic deformation the original shape is parmanently deformed with the application of stress.

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A string has its 4th harmonic at 31.5 Hz. What is the fundamental frequency?

seropon [69]

Given data

*The given 4th harmonic frequency is 31.5 Hz

The fundamental frequency is calculated as

$\begin{gathered} f_n=\frac{31.5}{4} \\ =7.875\text{ Hz} \end{gathered}$

Hence, the fundamental frequency is 7.875 Hz

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1 year ago

What type of infant temperament is generally associated with better adjustment in adulthood?

Rashid [163]

Saying no and not throwing fits and manners.

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2 years ago

If two identical trees are cut down, one with a hand saw, and one with an electric saw...

nataly862011 [7]

The hand saw would involve more work because it takes more time and effort.

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3 years ago

The magnitude of the force associated with the gravitational field is constant and has a value FF . A particle is launched from

uysha [10]

Answer:

The kinetic energy of the particle will be 12U₀

Explanation:

Given that,

A particle is launched from point B with an initial velocity and reaches point A having gained U₀ joules of kinetic energy.

Constant force = 12F

According to question,

The kinetic energy is

$U_{0}=Fx$ ....(I)

Constant force = 12F

A resistive force field is now set up ,

Resistive force is given by,

$F_{r}=12F$

When the particle moves from point B to point A then,

We need to calculate the kinetic energy

Using formula for kinetic energy

$U=F_{r}x$

Put the value of $F_{r}$

$U=12Fx$

Now, from equation (I)

$U=12U_{0}$

Hence, The kinetic energy of the particle will be 12U₀.

7 0

3 years ago

Two cars are traveling along a straight line in the same direction, the lead car at 25 m/s and the other car at 35 m/s. At the m

Phoenix [80]

Answer:

a. $t_1=12.5\ s$

b. $a_2=-13.61\ m.s^{-2}$ must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c. $t_2=2.5714\ s$ is the time taken to stop after braking

Explanation:

Given:

speed of leading car, $u_1=25\ m.s^{-1}$

speed of lagging car, $u_{2}=35\ m.s^{-1}$

distance between the cars, $\Delta s=45\ m$

deceleration of the leading car after braking, $a_1=-2\ m.s^{-2}$

Time taken by the car to stop:

$v_1=u_1+a_1.t_1$

where:

$v_1=0$ , final velocity after braking

$t_1=$ time taken

$0=25-2\times t_1$

$t_1=12.5\ s$

using the eq. of motion for the given condition:

$v_2^2=u_2^2+2.a_2.\Delta s$

where:

$v_2=$ final velocity of the chasing car after braking = 0

$a_2=$ acceleration of the chasing car after braking

$0^2=35^2+2\times a_2\times 45$

$a_2=-13.61\ m.s^{-2}$ must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

time taken by the chasing car to stop:

$v_2=u_2+a_2.t_2$

$0=35-13.61\times t_2$

$t_2=2.5714\ s$ is the time taken to stop after braking

7 0

3 years ago