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Rudiy27
2 years ago
11

When are you most aware of your motion in a moving vehicle: when it is moving steadily in a straight line or when it is accelera

ting? If you were in a car that moved with absolutely constant velocity (no bumps at all), would you be aware of the motion?
(A) Accelerating. You would not be aware of the motion if you did not look outside the car.

(B) Accelerating. You would be aware of the motion because humans can sense speed.

(C) Moving steadily in a straight line. You would be aware of the motion because you can feel velocity even in a closed car.

(D) Moving steadily in a straight line. You would be aware of the motion because you can feel the speed.
Physics
1 answer:
Ghella [55]2 years ago
6 0

Answer:

Acceleration is percieved, not constant velocity.

Explanation:

You are most aware when the vehicle is accelerating. At constant velocity you would not be aware of the motion. Only if the system is accelerated the dynamics must be solved considering a pseudo-force (of inertial origin) acting.

It's because of this that:

(A) False. The acceleration can be detected from the inside of a closed car.

(B) False. You would be aware of the motion, but not because humans can sense speed but acceleration.

(C) False. Constant velocity cannot be felt in a closed car.

(D) False. Again, you can't feel constant speed.

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harmful effects

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3 years ago
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Use the dimensional analysis and check the correctness of given equation:-<br> PV= nRT
Harman [31]

PV=nRT

Here

P=Pressure

V=Volume

n=Molarity

R=universal gas constant

T=Temperature.

LHS

\\ \tt\bull\leadsto PV

\\ \tt\bull\leadsto [ML^2T^{-2}][M^0L^3T^0]

\\ \tt\bull\leadsto [ML^5T^{-2}]

RHS

\\ \tt\bull\leadsto nRT

\\ \tt\bull\leadsto [M^0L^{3}T^0][M^1 L^2 T^{-2}K^{-1}][M^0L^0T^0K^1]

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LHS=RHS

hence verified

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3 years ago
What value is closest to the mass of the atom?....
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6 0
3 years ago
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A traves de una manguera de 1 in de diámetro fluye gasolina con una velocidad media de 5ft/s ¿cuál es el gasto?
jonny [76]

Answer:

El gasto de gasto es de aproximadamente 0.0273 pies cúbicos por segundo.

Explanation:

El gasto es el flujo volumétrico de gasolina (Q), medido en pies cúbicos por segundo, que sale de la manguera. Asumiendo que la velocidad de salida es constante, tenemos que el gasto a través de la manguera es:

Q = \frac{\pi}{4}\cdot D^{2}\cdot v (1)

Donde:

D - Diámetro de la manguera, medido en pies.

v - Velocidad medida de salida, medida en pies por segundo.

Si sabemos que D = \frac{1}{12}\,ft y v = 5\,\frac{ft}{s }, entonces el gasto de gasolina es:

Q = \frac{\pi}{4}\cdot \left(\frac{1}{12}\,ft \right)^{2} \cdot \left(5\,\frac{ft}{s} \right)

Q \approx 0.0273\,\frac{ft^{3}}{s}

El gasto de gasto es de aproximadamente 0.0273 pies cúbicos por segundo.

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