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2 years ago

11

Can an object's average velocity equal zero when object's speed is greater than zero.

1 answer:

notsponge [240]2 years ago

6 0

No, because the magnitude of the velocity is always greater than speed.

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Solve the science problem

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What is the problem I cant help unless you have the problem.

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¿Cuál es el lado positivo sobre el auge de la <br><br>industrialización<br>

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El lado positivo que tuvo el auge de la industrialización es principalmente que: Permitió el desarrollo económico de una gran cantidad de países. Marco un nuevo estilo de vida con mayor índice de globalización y producción. Creó una gran cantidad de fuentes de empleo.

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3 years ago

A 9V battery produces a 1.5A current in a piece of wire. What is the resistance of the wire?

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Using Ohm's Law and a bit of substitution, we can use voltage divided by current to solve for resistance. Doing that, we'll get 6 Ohm.

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2 years ago

At a race car driving event, a staff member notices that the skid marks left by the race car are

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A car that experiences a deceleration of -41.62 m/s² and comes to a stop after 10.99 m has an initial velocity of 30.60 m/s.

A car experiences a deceleration (a) of -41.62 m/s² and comes to a stop (final velocity = v = 0 m/s) after 10.99 m (s).

We can calculate the initial velocity of the car (u) using the following kinematic equation.

$v^{2} = u^{2} + 2as\\\\u = \sqrt[]{v^{2}-2as} = \sqrt[]{(0m/s)^{2}-2(-42.61m/s^{2} )(10.99m)} = 30.60m/s$

A car that experiences a deceleration of -41.62 m/s² and comes to a stop after 10.99 m has an initial velocity of 30.60 m/s.

Learn more: brainly.com/question/14851168

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2 years ago

A sample of an ideal gas is in a tank of constant volume. The sample absorbs heat energy so that its temperature changes from 33

Anuta_ua [19.1K]

Answer:

$\frac{v_{2}}{v_{1}}=2$ .

Explanation:

The average kinetic energy per molecule of a ideal gas is given by:

$\bar{K}=\frac{3k_{B}T}{2}$

Now, we know that $\bar{K} = (1/2)m\bar{v}^{2}$

Before the absorption we have:

$(1/2)m\bar{v_{1}}^{2}=\frac{3k_{B}T_{1}}{2}$ (1)

After the absorption,

$(1/2)m\bar{v_{2}}^{2}=\frac{3k_{B}T_{2}}{2}$ (2)

If we want the ratio of v2/v1, let's divide the equation (2) by the equation (1)

$\frac{v_{2}^{2}}{v_{1}^{2}}=\frac{T_{2}}{T_{1}}$

$\frac{v_{2}}{v_{1}}=\sqrt{\frac{T_{2}}{T_{1}}}$

$\frac{v_{2}}{v_{1}}=\sqrt{\frac{1340}{335}}$

$\frac{v_{2}}{v_{1}}=\sqrt{4}$

Therefore the ratio will be $\frac{v_{2}}{v_{1}}=2$

I hope it helps you!

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