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3 years ago

If you have just combed your hair and you hold the comb near a small, uncharged object, what will most likely happen?

Physics

2 answers:

AnnyKZ [126]3 years ago

8 0

It won't happen anything i think it will just be a something like a small thing

mihalych1998 [28]3 years ago

7 0

Nothing will happen, because if you comb your hair the comb will have an electric charge and it needs an opposite electric charge for anything to happen.

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Un objeto, que se encuentra a nivel del suelo, es lanzado verticalmente hacia arriba con una velocidad de 160 km/hr. ¿Qué altura

LiRa [457]

I don’t speak Spanish?

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2 years ago

A ball has a mass of 1.5kg and is thrown straight up with a speed of 60m/s, what is the ball’s momentum:

madam [21]

Answer:

Assumption: the air resistance on this ball is negligible. Take $g = 10\; \rm m \cdot s^{-2}$ .

a. The momentum of the ball would be approximately $60\;\rm kg \cdot m \cdot s^{-1}$ two seconds after it is tossed into the air.

b. The momentum of the ball would be approximately $\rm \left(-45\; \rm kg \cdot m \cdot s^{-1}\right)$ three seconds after it reaches the highest point (assuming that it didn't hit the ground.) This momentum is smaller than zero because it points downwards.

Explanation:

The momentum $p$ of an object is equal its mass $m$ times its velocity $v$ . That is: $\vec{p} = m \cdot \vec{v}$ .

Assume that the air resistance on this ball is negligible. If that's the case, then the ball would accelerate downwards towards the ground at a constant $g \approx -10\; \rm m \cdot s^{-2}$ . In other words, its velocity would become approximately $10\; \rm m \cdot s^{-1}$ more negative every second.

The initial velocity of the ball is $60\; \rm m \cdot s^{-1}$ . After two seconds, its velocity would have become $60\;\rm m \cdot s^{-1} + 2\; \rm s \times \left(-10\;\rm m \cdot s^{-1}\right) = 40\; \rm m \cdot s^{-1}$ . The momentum of the ball at that time would be around $p = m \cdot v \approx 60\; \rm kg \cdot m \cdot s^{-1}$ .

When the ball is at the highest point of its trajectory, the velocity of the ball would be zero. However, the ball would continue to accelerate downwards towards the ground at a constant $g \approx -10\; \rm m \cdot s^{-2}$ . That's how the ball's velocity becomes negative.

After three more seconds, the velocity of the ball would be $0\; \rm m \cdot s^{-1} + 3\; \rm s \times \left(-10\; \rm m \cdot s^{-2}\right) = -30 \; \rm m \cdot s^{-1}$ . Accordingly, the ball's momentum at that moment would be $p = m \cdot v \approx \left(-45\; \rm kg \cdot m \cdot s^{-1}\right)$ .

3 0

2 years ago

A bike, a truck, and a train—all without passengers, motors, or engines—roll down the same hill. Put the vehicles in order from

Pepsi [2]

Answer: WAIT WHATTTT i have that same test due today and the answer is in explanation

Explanation:

Bike, truck train. we are in the same school i think. Its imma say the incisal JMES Im Lusi i used to help in the library

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2 years ago

Are the objects described here in static equilibrium, dynamic equilibrium, or not equilibrium at all? Explain.

Alexandra [31]

Let us examine the given situations one at a time.

Case a. A 200-pound barbell is held over your head.
The barbell is in static equilibrium because it is not moving.
Answer: STATIC EQUILIBRIUM

Case b. A girder is being lifted at a constant speed by a crane.
The girder is moving, but not accelerating. It is in dynamic equilibrium.
Answer: DYNAMIC EQUILIBRIUM

Case c: A jet plane has reached its cruising speed at an altitude.
The plane is moving at cruising speed, but not accelerating. It is in dynamic equilibrium.
Answer: DYNAMIC EQUILIBRIUM

Case d: A box in the back of a truck doesn't slide as the truck stops.
The box does not slide because the frictional force between the box and the floor of the truck balances out the inertial force. The box is in static equilibrium.
Answer: STATIC EQUILIBRIUM

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2 years ago

Read 2 more answers

A 7.0 kg bowling ball has a moment of inertia of 2.8x10-2 kg m2, and a radius of 0.10 m. If it rolls down the lane at an angular

slega [8]

Hi there!

Angular momentum is equivalent to:

$\large\boxed{L = I\omega}$