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3 years ago

12

__________ is the total kinetic energy of all atoms in a substance while temperature is a measure of the average kinetic energy

of all atoms in a body.

1 answer:

MAXImum [283]3 years ago

5 0

The answer is heat :)

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Can an element be a molecule

Serga [27]

an element can make a molecule. so technically yes.

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3 years ago

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Any fracture or system of fractures along which Earth moves is known as a

bogdanovich [222]

Any fracture or system of fractures along which Earth moves is known as a D.fault.

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3 0

3 years ago

A block is projected up a frictionless inclined plane with initial speed v0 = 7.14 m/s. The angle of incline is θ = 36.5°. (a) H

Wewaii [24]

Answer:

$(a)x=4.37m\\\\(b)t=1.225s\\\\(c) v_{f}=7.14m/s$

Explanation:

Given data

$v_{o}=7.14m/s\\\alpha =36.5^o$

For Part (a)

Starting with the -ve acceleration of the body (opposite to the gravitational force)

$a=-gSin\alpha \\a=-(9.8m/s^2)Sin(36.5)\\a=-5.83m/s^2$

Using equation of motion

$v_{f}^2=v_{o}^2+2ax\\(0m/s)^2=(7.14m/s)^2+2(-5.83m/s^2)x\\-(7.14m/s)^2=2(-5.83m/s^2)x\\x=\frac{-(7.14m/s)^2}{2(-5.83m/s^2}\\ x=4.37m$

For Part (b)

Using the result in Part (a) we can substitute in other equation of motion to get time t:

$x=\frac{1}{2}vt\\ 4.37m=\frac{1}{2}(7.14m/s)t\\ (7.14m/s)t=2*(4.37)\\t=8.744/7.14\\t=1.225s$

For Part (c)

At state 2 where vo=0m/s and the acceleration is positive (same direction as the gravitational force)

$a=gSin\alpha \\a=(9.8m/s^2)Sin(36.5)\\a=5.83m/s^2\\\\\\v_{f}^2=v_{o}^2+2ax\\v_{f}^2=(0m/s)^2+2(5.83m/s^2)(4.37m)\\v_{f}^2=50.95\\v_{f}=\sqrt{50.95}\\ v_{f}=7.14m/s$

4 0

3 years ago

Please help I’m so confused

Ilya [14]

<em>1.wavelength</em>

<em>2.trough</em>

<em>3.amplitude</em>

<em>4.crest</em>

5 0

3 years ago

two cars are moving at constant speeds in a straight line along a major highway. The first is travelling at 20ms^-1 and the seco

Reika [66]

Answer:

$t=750s$

Explanation:

The two cars are under an uniform linear motion. So, the distance traveled by them is given by:

$\Delta x=vt\\x_f-x_0=vt\\x_f=x_0+vt$

$x_f$ is the same for both cars when the second one catches up with the first. If we take as reference point the initial position of the second car, we have:

$x_0_1=6km\\x_0_2=0$

We have $x_f_1=x_f_2$ . Thus, solving for t:

$x_0_1+v_1t=x_0_2+v_2t\\x_0_1=t(v_2-v_1)\\t=\frac{x_0_1}{v_2-v_1}\\t=\frac{6*10^3m}{28\frac{m}{s}-20\frac{m}{s}}\\t=750s$

8 0

3 years ago