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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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An empty graduated cylinder weighs 55.26 g. When filled with 48.1 mL of an unknown liquid, it weighs 92.39 g. The density of the
katrin [286]
<h2>Density of the unknown liquid is 771.93 kg/m³</h2>

Explanation:

An empty graduated cylinder weighs 55.26 g

Weight of empty cylinder = 55.26 g = 0.05526 kg

Volume of liquid filled = 48.1 mL = 48.1 x 10⁻⁶ m³

Weight of cylinder plus liquid = 92.39 g = 0.09239 kg

Weight of liquid = 0.09239 - 0.05526

Weight of liquid = 0.03713 kg

We have

        Mass = Volume x Density

        0.03713 = 48.1 x 10⁻⁶ x Density

        Density = 771.93 kg/m³

Density of the unknown liquid is 771.93 kg/m³

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3 years ago
A 1000 kg elevator accelerates upward at 1.0 m/s2 for 10 m, starting from rest. a. How much work does gravity do on the elevator
cricket20 [7]

Answer:

a)= 98kJ

b)=108kJ

c) = 10kJ

Explanation:

a. The work that is done by gravity on the elevator is:

Work = force * distance  

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= 1000 * 9.81 * 10  

= 98,000 J

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b)The net force equation in the cable

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T = m(g+a)

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T = 10800N

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c) PE at 10m = 1000 * 9.81 * 10 = 98,100 J  

Work done by cable = PE +KE  

108,100 J = KE + 98,100 J  

KE = 10,000 J

= 10kJ

=

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