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Alik [6]
3 years ago
11

The ability of solid rock to flow is called _____.

Physics
2 answers:
zysi [14]3 years ago
3 0
The ability of solid rock to flow is called Plasticity.

Plasticity is the deformation of a solid mateial which resulted in non reversible changes of shape in response to applied force.

Another example of plasticity is when you bend metal through a forcer force to create an art or kitchen set
adoni [48]3 years ago
3 0

The ability of solid rock to flow is called Plasticity.

The adaptability of an organism to changes in its environment or differences between its various habitats.

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While riding a bicycle, if you stop pedaling you will still continue to move forward due to ______.
adelina 88 [10]
When riding a bicycle, if you stop pedaling you will still continue to move forward due to inerita.

3 0
3 years ago
Longitudional waves travel through a series of ________ and ___________.
nekit [7.7K]

Answer:

compressions; rarefactions

Explanation:

4 0
3 years ago
A ball is dropped from rest from the top of a building of height h. At the same instant, a second ball is projected vertically u
uranmaximum [27]

Answer:

a) t = \sqrt{\frac{h}{2g}}

b) Ball 1 has a greater speed than ball 2 when they are passing.

c) The height is the same for both balls = 3h/4.

Explanation:

a) We can find the time when the two balls meet by equating the distances as follows:

y = y_{0_{1}} + v_{0_{1}}t - \frac{1}{2}gt^{2}  

Where:

y_{0_{1}}: is the initial height = h

v_{0_{1}}: is the initial speed of ball 1 = 0 (it is dropped from rest)

y = h - \frac{1}{2}gt^{2}     (1)

Now, for ball 2 we have:

y = y_{0_{2}} + v_{0_{2}}t - \frac{1}{2}gt^{2}    

Where:

y_{0_{2}}: is the initial height of ball 2 = 0

y = v_{0_{2}}t - \frac{1}{2}gt^{2}    (2)

By equating equation (1) and (2) we have:

h - \frac{1}{2}gt^{2} = v_{0_{2}}t - \frac{1}{2}gt^{2}

t=\frac{h}{v_{0_{2}}}

Where the initial velocity of the ball 2 is:

v_{f_{2}}^{2} = v_{0_{2}}^{2} - 2gh

Since v_{f_{2}}^{2} = 0 at the maximum height (h):

v_{0_{2}} = \sqrt{2gh}

Hence, the time when they pass each other is:

t = \frac{h}{\sqrt{2gh}} = \sqrt{\frac{h}{2g}}

b) When they are passing the speed of each one is:

For ball 1:

v_{f_{1}} = - gt = -g*\sqrt{\frac{h}{2g}} = - 0.71\sqrt{gh}

The minus sign is because ball 1 is going down.

For ball 2:

v_{f_{2}} = v_{0_{2}} - gt = \sqrt{2gh} - g*\sqrt{\frac{h}{2g}} = (\sqrt{1} - \frac{1}{\sqrt{2}})*\sqrt{gh} = 0.41\sqrt{gh}

Therefore, taking the magnitude of ball 1 we can see that it has a greater speed than ball 2 when they are passing.

c) The height of the ball is:

For ball 1:

y_{1} = h - \frac{1}{2}gt^{2} = h - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h

For ball 2:

y_{2} = v_{0_{2}}t - \frac{1}{2}gt^{2} = \sqrt{2gh}*\sqrt{\frac{h}{2g}} - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h

Then, when they are passing the height is the same for both = 3h/4.

I hope it helps you!                  

7 0
3 years ago
In one contest at the county fair, a spring-loaded plunger launches a ball at a speed of 3.2m/s from one corner of a smooth, fla
lara31 [8.8K]

Answer:

Explanation:

Given

Speed of ball u=3.2\ m/s

Plane is inclined at an angle 20^{\circ}

To win the Game we need to hit the target at x=2.4\ m away

Launch angle of ball \theta

Motion of ball can be considered in two planes i.e. Vertical to the plane and horizontal to the plane

So Net acceleration in vertical plane is g\sin 20

Range of Projectile is given by

R=\frac{u^2\sin 2\theta }{g}

for R=2.4\ m

2.4=\frac{3.2^2\times sin 2\theta }{g\sin 20}

\sin 2\theta =\frac{2.4\times 9.8\times \sin 20}{3.2^2}

\sin 2\theta =0.7855

2\theta =51.77

\theta =25.88^{\circ}

so ball must be launched at an angle of 25.88^{\circ}

4 0
3 years ago
A diverging lens can be defined as a lens hat causes parallel light rays to___
r-ruslan [8.4K]
I'm pretty sure it's B

I hope that helps 
6 0
3 years ago
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