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2 years ago

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At constant temperature, the volume of the container that a sample of nitrogen gas is in is doubled. As a result the pressure of

the nitrogen gas is halved. The amount of nitrogen gas is unchanged in this process. This is an example of:
A) Boyle's Law
B) Avogadro's Law
C) Charle's Law
D) Dalton's Law
E) Gay-Lussac's Law

1 answer:

In-s [12.5K]2 years ago

6 0

Answer:

A) Boyle's Law

Explanation:

Boyle's Law states that pressure is inversely proportional to volume if the temperature and volume remain constant in closed system

${P}=\frac{1}{V}$

${P_1}\times {V_1}={P_2}\times {V_2}$

As it is seen here the volume and temperature remain constant but the volume of the container is doubled.

So, this is an example of Boyle's Law.

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Explanation:

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2 years ago

You must determine the length of a long, thin wire that is suspended from the ceiling in the atrium of a tall building. A 2.00-c

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Answer:

Explanation:

Let L be the length of the wire.

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2 years ago

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Explanation:

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2 years ago

1)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it r

DENIUS [597]

Answer:

1)

$v_{oy}=11.29\ m/s$

2)

$y=7.39\ m$

Explanation:

<u>Projectile Motion</u>

When an object is launched near the Earth's surface forming an angle $\theta$ with the horizontal plane, it describes a well-known path called a parabola. The only force acting (neglecting the effects of the wind) is the gravity, which acts on the vertical axis.

The heigh of an object can be computed as

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

Where $y_o$ is the initial height above the ground level, $v_{oy}$ is the vertical component of the initial velocity and t is the time

The y-component of the speed is

$v_y=v_{oy}-gt$

1) We'll find the vertical component of the initial speed since we have not enough data to compute the magnitude of $v_o$

The object will reach the maximum height when $v_y=0$ . It allows us to compute the time to reach that point

$v_{oy}-gt_m=0$

Solving for $t_m$

$\displaystyle t_m=\frac{v_{oy}}{g}$

Thus, the maximum heigh is

$\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}$

We know this value is 8 meters

$\displaystyle y_o+\frac{v_{oy}^2}{2g}=8$

Solving for $v_{oy}$

$\displaystyle v_{oy}=\sqrt{2g(8-y_o)}$

Replacing the known values

$\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}$

$\displaystyle v_{oy}=11.29\ m/s$

2) We know at t=1.505 sec the ball is above Julie's head, we can compute

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}$

$\displaystyle y=1.5\ m+16,991\ m-11.098\ m$

$y=7.39\ m$

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2 years ago

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2 years ago

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