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Oksi-84 [34.3K]
3 years ago
12

A gas is stored in a sealed container of constant volume. The temperature of the gas increases.

Physics
1 answer:
Alexus [3.1K]3 years ago
6 0

Answer:

If the gas volume is decreased, the container wall area decreases and the molecule-wall collision frequency increases, both of which increase the pressure exerted by the gas. Avogadro's law. At constant pressure and temperature, the frequency and force of molecule-wall collisions are constant.

Explanation:

As the temperature increases, the average kinetic energy increases as does the velocity of the gas particles hitting the walls of the container. The force exerted by the particles per unit of area on the container is the pressure, so as the temperature increases the pressure must also increase.

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The voltage across the terminals of an ac power supply varies with time according to V=V0cos(t). The voltage amplitude is V0 = 4
kirza4 [7]

Answer:

A) V_rms = 29 V

B) Vav = 0 V

Explanation:

A) We are told that;

V = V_o cos ωt

voltage amplitude; V = V_o = 41.0V

Now, the formula for the root-mean-square potential difference Vrms is given as;

V_rms = V/√2

Thus plugging in relevant values, we have;

V_rms = 41/√2

V_rms = 29 V

B) Due to the fact that the voltage is sinusoidal from the given V = V_o cos ωt, we can say that the average potential difference Vav between the two terminals of the power supply would be zero.

Thus; Vav = 0 V

5 0
3 years ago
Help pleaseee!! I’ll give brainliest! And points!
zmey [24]

Answer:

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Explanation:

7 0
2 years ago
To water the yard you use a hose with a diameter of 3.2 cm. Water flows from the hose with a speed of 1.1 m/s. If you partially
cupoosta [38]

Answer:

The speed of water flow inside the pipe at point - 2 = 34.67 m / sec

Explanation:

Given data

Diameter at point - 1 = 3.2 cm

Velocity at point - 1 = 1.1 m / sec = 110 cm / sec

Diameter at point - 2 = 0.57 cm

Velocity at point - 2 = ??

We know that from the continuity equation the rate of flow is constant inside  a pipe between two points.

Thus

⇒ A_{1} × V_{1} = A_{2} × V_{2}

⇒  \frac{\pi }{4} × d_{1} ^{2} × V_{1} =

⇒  d_{1} ^{2} × V_{1} =  d_{2} ^{2}  × V_{2}

⇒  (3.2)^{2} × 110 = (0.57)^{2} × V_{2}

⇒ V_{2} = 3467 cm / sec

⇒ V_{2} = 34.67 m / sec  

Thus the speed of water flow inside the pipe at point - 2 = 34.67 m / sec

3 0
3 years ago
A 0.0450-kg golf ball initially at rest is given a speed of 25.0 m/s when a club strikes. If the club and ball are in contact fo
LuckyWell [14K]

Answer:

Average force, F = 562.5 N

Explanation:

Mass of the golf ball, m = 0.045 kg

Initially, it is at rest, u = 0

Final speed of the ball, v = 25 m/s

The club and the ball are in contact for, t=2\ ms=2\times 10^{-3}\ s

We need to find the average force acting on the ball. It can be calculated using the formula as :

F\times t=m(v-u)

F=m\dfrac{v-u}{t}

F=0.045\times \dfrac{25}{2\times 10^{-3}}  

F = 562.5 N

So, the average force acting on the ball is 562.5 N. Hence, this is the required solution.

5 0
3 years ago
The given function represents the position of a particle traveling along a horizontal line. s(t) = 2t3 − 3t2 − 12t + 6 for t ≥ 0
avanturin [10]

Answer:

(a) v(t) = 6t^2 - 6t - 12, a(t) = 12t - 6

(b) When 0 \leq t < 0.5, object is slowing down, when t > 0.5 object is speeding up.

Explanation:

(a) To get the velocity function, we need to take the derivative of the position function.

v(t) = \frac{ds(t)}{dt}  = (2t^{3})^{'} - (3t^{2})^{'} - (12t)^{'} + 6^{'} = 6t^{2} - 6t - 12

To get the acceleration function, we need to take the derivative of the velocity function.

a(t) = \frac{dv(t)}{dt} = (6t^{2})^{'} - (6t)^{'} - (12)^{'} = 12t - 6

(b) The object is slowing down when velocity is decreasing by time (decelerating) hence a < 0

12t - 6 < 0 \\12t < 6 \\t < 0.5

On the other hand, object is speeding up when a > 0

12t - 6 > 0 \\12t > 6 \\t > 0.5

Therefore, when 0 \leq t < 0.5, object is slowing down, when t > 0.5 object is speeding up.

6 0
3 years ago
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