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vekshin1
3 years ago
13

Why are my rift zones common places for igneous rocks to form

Physics
1 answer:
andrew-mc [135]3 years ago
6 0
The thin crust of a rift zone causes melting in the upper mantle, resulting in volcanic activity.
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Mrrafil [7]

Answer:

A) T.

Explanation:

Kepler's third law states that the orbital period (T) of a satellite is related with the radius (R) and the mass of the object (M) it orbits:

T=\frac{2\pi R^{\frac{3}{2}}}{\sqrt{GM}}  

So the orbital period is independent of the mass of the satellite, that means no matter the mass every satellite at a radius R around the earth have an orbital period A.

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3 years ago
Initilal velocity of projectile at an angle of 22.5° is 10m/s. Then what is the magnitude of range of projectile?
crimeas [40]

Answer:

0.707m

Explanation:

from formula of range i.e R=Usin2Q/g

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3 years ago
Explain why filing cabinets are<br>designed so that only one draw<br>can be pulled out at a time.​
natta225 [31]

Answer:

Cabinets are made to open one at a time so that there's not much of space taken up when all the drawers are open. Hope this helps.

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3 years ago
How do noise and vibration affect you when operating a boat?
kogti [31]
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3 years ago
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During an adiabatic process an object does 100 J of work and its temperature decreases by 5 K. During another process it does 25
Liono4ka [1.6K]

Answer:

The heat capacity for the second process is 15 J/K.

Explanation:

Given that,

Work = 100 J

Change temperature = 5 k

For adiabatic process,

The heat energy always same.

dQ=0

dU=-dW

We need to calculate the number of moles and specific heat

Using formula of heat

dU=nC_{v}dT

nC_{v}=\dfrac{dU}{dT}

Put the value into the formula

nC_{v}=\dfrac{-100}{5}

nC_{v}=-20\ J/K

We need to calculate the heat

Using formula of heat

dQ=nC_{v}(dT_{1})+dW_{1}

Put the value into the formula

dQ=-20\times5+25

dQ=-75\ J

We need to calculate the heat capacity for the second process

Using formula of heat

dQ=nC_{v}(dT_{1})

Put the value into the formula

-75=nC_{v}\times(-5)

nC_{v}=\dfrac{-75}{-5}

nC_{v}=15\ J/K

Hence, The heat capacity for the second process is 15 J/K.

5 0
4 years ago
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