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3 years ago

Recrystallization occurs in which type of rock

Physics

2 answers:

djverab [1.8K]3 years ago

5 0

Only in metamorphic rocks does this occur.

Dmitry_Shevchenko [17]3 years ago

4 0

igneous rocks i hope i helped

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What is the net force on a 4000 kg car that doubles its speed from 15 m/s to 30 m/s over 10 seconds

kozerog [31]

The net force of the car is 6000 N.

Why?

To calculate the net force of the car during the given time, we first need to calculate its acceleration, and then, use Newton's Second Law equation to calculate the net force.

We can calculate the acceleration of the car during the given tieme using the following equation:

$a=\frac{v-v_{o}}{time}$

Substituting the given information, we have:

$a=\frac{30\frac{m}{s} -15\frac{m}{s} }{10s}=\frac{15\frac{m}{s} }{10s}=1.5\frac{m}{s^{2} }$

Now that we have calculated the acceleration, we can calculate the net force:

$Force=mass*acceleration\\\\Force=4000kg*1.5\frac{m}{s^{2}}=6000\frac{kg.m}{s^{2}}=6000N$

Hence, we have that the net force is equal to 6000N.

Have a nice day!

3 0

3 years ago

A block of mass 3.1 kg, sliding on a horizontal plane, is released with a velocity of 2.3 m/s. The blocks slides and stops at a

jeka57 [31]

To solve this problem we will apply the concepts given by the kinematic equations of motion. For this purpose it will be necessary with the given data to obtain the deceleration. With this it will be possible again to apply one of the kinematic equations of motion that does not depend on time, but on distance, to find how far the block would slide with the quadruplicate velocity

Our values are given as,

$\text{Initial speed} =V_i = 2.3 m/s$

$\text{Final speed}= V_f = 0 m/s$

$\text{Stopping distance = }d = 1.9 m$

$a = acceleration$

$\text{mass} = m = 3.1kg$

Using the kinematic equation of motion we have

$V_f^2 = V_i^2 + 2 a d$

$0^2 = 2.3^2 + 2 a (1.9)$

$a = -1.39211 m/s^2$

Now if the initial velocity is quadrupled we have that,

$\text{Initial speed} =V_i' = 2.3*4 m/s = 9.2m/s$

$\text{Final speed}= V_f' = 0 m/s$

$\text{Stopping distance = }d'$

$V_f^2' = V_i^2' + 2 a d'$

Replacing the values

$0^2 = 9.2^2+ 2 (-1.39211) d'$

$d' = 30.44m$

Therefore the block would have slipped around 30.44 if its initial velocity quadrupled.

3 0

3 years ago

An atom has an atomic number 17 and atomic mass number 35. How many neutrons are present in its nucleus?

Anna [14]

The number of neutrons that are present in its nucleus is 18. The correct option among all the options that are given in the question is the second option or option "B". Mass number minus the atomic number gives the number of neutrons. I hope that this is the answer that has come to your help.

7 0

4 years ago

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A red 120 kg bumper car moving at 4 m/s collides with a green 100 kg bumper car moving at 3 m/s. The red bumper car bounces off

Mariulka [41]

We can utilize the equation pi=pf (p=momentum). because p=mv, we can sub in the initial masses and velocities as well as the finals in order to form an equation. this results in 120(4)+100(3)=120(2)+100x. this can be simplified to 540=100x, or x=5.4 m/s.

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3 years ago

A 8.0-g bullet with an initial velocity of 115.0 m/s lodges in a block of wood and comes to rest at a

Gnesinka [82]

A. When your fire a bullet into the air, it typically takes between 20 and 90 second for it to come down,depending on the angle it was fired at,its muzzle velocity and caliber.

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3 years ago