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3 years ago

12

Temperature is a measure of the average speed of the ____ of a thing.

2 answers:

Pie3 years ago

7 0

Answer:

it is atoms and molecules

Luba_88 [7]3 years ago

3 0

The answer is atoms and molecules

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How much power will it take to move a 50Kg box 10m across a floor that has -50N of Frictional forces in 3 seconds?

likoan [24]

Answer: 83.3 W

Explanation: I think, I’m not sure. If I’m wrong correct me ;)

8 0

3 years ago

What happens when an object with a lower density is placed in a container with an

maksim [4K]

The only thing that definitely happens in every such case is:
The container becomes heavier.

5 0

3 years ago

Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m

Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

$A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}$ .....(I)

$tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}$ ....(II)

Put the value of $\omega=0.628\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}$

$A=0.0198$

From equation (II)

$tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}$

$d=0.0023$

Put the value of $\omega=3.14\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}$

$A=0.0203$

From equation (II)

$tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}$

$d=0.0120$

Put the value of $\omega=6.28\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}$

$A=0.0209$

From equation (II)

$tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}$

$d=0.0257$

Put the value of $\omega=9.42\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}$

$A=0.0217$

From equation (II)

$tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}$

$d=0.0413$

Hence, This is the required solution.

5 0

3 years ago

falling objects drop with an average acceleration of 9.8m/sec/sec. if an object falls from a tall building how long will it take

sveta [45]

Answer:

5 seconds

Explanation:

<em>Acceleration = (final velocity - initial velocity) ÷ time</em>

<em> $a = \frac{v - u}{t}$ </em>

<em> $9.8 = \frac{49 - 0}{t}$ </em>

<em> $9.8 = \frac{49}{t}$ </em>

<em> $9.8t = 49$ </em>

<em> $t = \frac{49}{9.8}$ </em>

<em> $t = 5$ </em>

8 0

3 years ago

A block of mass, m, sits on the ground. A student pulls up on

kakasveta [241]

Answer a

Explanation: a

3 0

3 years ago