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3 years ago

A push or pull that causes an object's motion to change is called _____

Physics

1 answer:

12345 [234]3 years ago

6 0

Answer:

Force

Explanation:

Force is a pull or push that causes An objects motion to change . When force is applied to an object , it either pushes the object to or pulls it from something thereby changing the motion of the object. A good example is the earth’s gravitational force. This force pulls or attracts a body to the center of the earth . Another example Is experienced when kicking a football from one part of a football field to another. The ball is kicked with force and this causes it to move from one place another .

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Which labels correctly identify the layers most closely associated with gamma rays and visible light? Z: Gamma rays X: Visible l

Marianna [84]

The answer is Z: Gamma rays X: Visible light. Gamma ray is the shortest and has the greatest frequency and power. That is why it is in letter Z.

4 0

3 years ago

A typical car has 16 L of liquid coolant circulating at a temperature of 95 ∘C through the engine's cooling system. Assume that,

amm1812

Answer:

$\Delta V=0.0592\ L$

Explanation:

Given:

Initial temperature of the coolant, $T_f=95^{\circ}C$

final temperature of the coolant, $T_f=105^{\circ}C$

total volume of the coolant, $V_c=16\ L$

coefficient of volume expansion for coolant, $\beta_c=410\times 10^{-6}\ ^{\circ}C^{-1}$

volume of Al radiator, $V_a=2\ L$

volume of steel radiator, $V_s=14\ L$

We have:

coefficient of volume expansion for Aluminium, $\beta_a=75\times 10^{-6}\ ^{\circ}C^{-1}$

coefficient of volume expansion for steel, $\beta_a=35\times 10^{-6}\ ^{\circ}C^{-1}$

Now, change in volume of the coolant after temperature rises:

$\Delta V_c=V_c.\beta_c.\Delta T$

$\Delta V_c=16\times 410\times 10^{-6}\times (105-95)$

$\Delta V_c=0.0656\ L$

Now, volumetric expansion in Aluminium radiant:

$\Delta V_a=V_a.\beta_a.\Delta T$

$\Delta V_a=2\times 75\times 10^{-6}\times (105-95)$

$\Delta V_a=0.0015\ L$

Now, volumetric expansion in steel radiant:

$\Delta V_s=V_s.\beta_s.\Delta T$

$\Delta V_s=14\times 35\times 10^{-6}\times (105-95)$

$\Delta V_s=0.0049\ L$

∴Total extra accommodation volume created after the expansion:

$V_X=\Delta V_s+\Delta V_a$

$V_X=0.0049+0.0015$

$V_X=0.0064\ L$

Hence, the volume that will overflow into the small reservoir will be the volume of coolant that will be extra after the expanded accommodation in the radiator.

$\Delta V=\Delta V_c-\Delta V_X$