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Temka [501]
3 years ago
5

To demonstrate the tremendous acceleration of a top fuel drag racer, you attempt to run your car into the back of a dragster tha

t is "burning out" at the red light before the start of a race. (Burning out means spinning the tires at high speed to heat the tread and make the rubber sticky.) You drive at a constant speed of v0 toward the stopped dragster, not slowing down in the face of the imminent collision. The dragster driver sees you coming but waits until the last instant to put down the hammer, accelerating from the starting line at constant acceleration, a. Let the time at which the dragster starts to accelerate be t = 0.
(A) What is tmax, the longest time after the dragster begins to accelerate that you can possibly run into the back of the dragster if you continue at your initial velocity?
Physics
1 answer:
Daniel [21]3 years ago
5 0

Answer:

Explanation:

If the dragster  attains the speed equal to that of the car which is moving with constant velocity of v₀ , before the two close in contact with each othe , there will not be collision .

So the dragster starting from rest , must attain the velocity v₀ in the maximum time given that is tmax .

v = u + a t

v₀ = 0 + a tmax

tmax = v₀ / a

The value of tmax is v₀ / a .

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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).

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Which things determine an element's electronegativity? Check all that apply. A. The size of the atom of that element B. Whether
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D. The number of electrons
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A 0.320 kg ball approaches a bat horizontally with a speed of 14.0 m/s and after getting hit by the bat, the ball moves in the o
katen-ka-za [31]

Answer:

<h2>42.67N</h2>

Explanation:

Step one:

<u>Given </u>

mass m= 0.32kg

intital velocity, u= 14m/s

final velocity v= 22m/s

time= 0.06s

Step two:

<u>Required</u>

Force F

the expression for the force is

F=mΔv/t

F=0.32*(22-14)/0.06

F=(0.32*8)/0.06

F=2.56/0.06

F=42.67N

The average force exerted on the bat 42.67N

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2 years ago
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