Without friction, what force is needed to maintain a 1,000 kg car in uniform motion for 30 minutes?

1 answer:

6 0

<span>The answer is none. According to the first law of Newton, an object stays at the same speed in the same direction if there are not forces unbalancing the object. Without friction, the car would be moving forever, unless there is another force accelerating or stopping the car.</span>

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If a car is traveling at an average speed of 60 kilometers per hour, how long does it take to travel 12 kilometers?

Otrada [13]

Answer: 0.2 hour

Explanation:

Speed = Distance / Time

Time = Distance / Speed

= 12/60

= 0.2 hour

8 0

3 years ago

Assign a positive velocity to the red box and negative velocity to the blue box. Are they moving in opposite direction?

attashe74 [19]

Answer:

<em>Yes, they are moving in opposite direction one to the other.</em>

Explanation:

Velocity is a vector quantity, which means that it has both magnitude and direction. The magnitude shows the size of the velocity, and the direction shows which way it is moving in reference to a chosen reference direction. If the red box is assigned a positive velocity, and the blue box is assigned a negative velocity, as indicated in the question, then it means that the red box, and the blue box, both move in opposite direction to the other.

4 0

3 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 $p = m \cdot v \approx \left(-45\; \rm kg \cdot m \cdot s^{-1}\right)$ .