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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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What is the source of the radioactivity?
Nataly_w [17]
The Earth itself is a source of terrestrial radiation. Radioactive materials including uranium, thorium, and radium exist naturally in soil and rock. Essentially all air contains radon<span> , which is responsible for most of the dose that Americans receive each year from natural background sources.</span>
4 0
3 years ago
Humans have three types of cone cells in their eyes, which are responsible for color vision. Each type absorbs a certain part of
xxMikexx [17]

Answer:

Frequency of the light will be equal to 6.97\times 10^{1}Hz

Explanation:

We have given wavelength of the light \lambda =430nm=430\times 10^{-9}m

Velocity of light is equal to v=3\times 10^8m/sec

We have to find the frequency of light

We know that velocity is equal to v=\lambda f, here \lambda is wavelength and f is frequency of light

So frequency of light will be equal to f=\frac{v}{\lambda }=\frac{3\times 10^8}{430\times 10^{-9}}=6.97\times 10^{1}Hz

So frequency of the light will be equal to 6.97\times 10^{1}Hz

5 0
3 years ago
A gang of robbers is escaping across city roofs at night. They come to the edge of one building and need to drop down to their g
REY [17]

Answer:

a) They will hit the ground with a speed of 19.6 m/s.

b) They are at a height of 20 m.

c) It is not a safe jump.

Explanation:

Hi there!

a) The equations of height and velocity in function of time of a free falling body are the following:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = height of the object at time t.

h0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.8 m/s² considering downward as negative direction).

v = velocity of the object at time t.

Using the equation of velocity, let's find the velocity at which they will hit the ground. The pebble is dropped (initial velocity = 0) and it takes 2 s to reach the ground:

v = v0 + g · t     (v0 = 0)

v = g · t

v = -9.8 m/s² · 2.0 s

v = -19.6 m/s

They will hit the ground with a speed of 19.6 m/s.

b)Now, we have to use the equation of height:

h = h0 + v0 · t + 1/2 · g · t²

If we place the origin of the frame of reference on the ground, we have to find the initial height (h0) knowing that at t = 2.0 s, h = 0 m

0 m = h0 - 1/2 · 9.8 m/s² · (2.0 s)²

h0 = 1/2 · 9.8 m/s² · (2.0 s)²

h0 = 20 m

They are at a height of 20 m.

c)According to a NASA paper (Issues on Human Acceleration Tolerance After Long-Duration Space Flights, figure 10), if you fall with a vertical velocity greater than 17 m/s it is unlikely that you will survive. So, it is not a safe jump.  

3 0
3 years ago
If
timama [110]

Answer:

Workdone = 1960 Joules.

Explanation:

Given the following data;

Mass = 5kg

Force = 49N

Height (distance) = 40m

To find the workdone;

Workdone = force * distance

Substituting into the equation, we have;

Workdone = 49*40

Workdone = 1960 Joules.

Therefore, the amount of work done on the bowling ball to lift it is 1960 Joules.

7 0
3 years ago
A weight of 30.0 N is suspended from a spring that has a force constant of 220 N/m. The system is undamped and is subjected to a
Nimfa-mama [501]

Answer:

F_0 = 393 N

Explanation:

As we know that amplitude of forced oscillation is given as

A = \frac{F_0}{ m(\omega^2 - \omega_0^2)}

here we know that natural frequency of the oscillation is given as

\omega_0 = \sqrt{\frac{k}{m}}

here mass of the object is given as

m = \frac{W}{g}

\omega_0 = \sqrt{\frac{220}{\frac{30}{9.81}}}

\omega_0 = 8.48 rad/s

angular frequency of applied force is given as

\omega = 2\pi f

\omega = 2\pi(10.5) = 65.97 rad/s

now we have

0.03 = \frac{F_0}{3.06(65.97^2 - 8.48^2)}

F_0 = 393 N

6 0
3 years ago
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