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mixer [17]
4 years ago
13

A car accelerates uniformly from rest to 20 m/sec in 5.6 sec along a level stretch of road. Ignoring friction, determine the ave

rage power required to accelerate the car if (a) the weight of the car is 9,000 N, and (b) the weight of the car is 14,000 N.
Physics
1 answer:
bonufazy [111]4 years ago
3 0

Answer:

(a) P=33000W

(b) P=51000W

Explanation:

The average power is defined as the amount of work done during a time interval:

P=\frac{W}{t}(1)

According to work-energy theorem, the work done is equal to the change in kinetic energy. So, we have:

W=\Delta K\\W=K_f-K_0\\W=\frac{mv_f^2}{2}-\frac{mv_0^2}{2}\\(2)

Recall that the weight is given by:

w=mg\\m=\frac{w}{g}(3)

The car accelerates uniformly from rest (v_0=0). Replacing (3) in (2), we have:

W=\frac{wv_f^2}{2g}

(a) Finally, we replace this in (1):

P=\frac{wv_f^2}{2gt}\\P=\frac{9000N(20\frac{m}{s})^2}{2(9.8\frac{m}{s^2})(5.6s)}\\P=33000W

(b)

P=\frac{14000N(20\frac{m}{s})^2}{2(9.8\frac{m}{s^2})(5.6s)}\\P=51000W

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alexandr402 [8]

9,000,000 ÷ 60,000,000 = 0.15m/y

5 0
3 years ago
A student exerts a force of 500 n pushing a box 10 m across the floor in 4 s. how much work does the student perform?
gregori [183]
Work = Force * distance
W = Fd

Given F = 500 N, d = 10 m
W = (500)(10)
W = 5000 J

The work done is 500 Joules. The time of 4 s is irrelevant in this case.
6 0
3 years ago
A small drop of water is suspended motionless in air by a uniform electric field that is directed upward and has a magnitude of
uysha [10]

Answer:

q = 4.04×10^-12C

n = 2.53×10^7

Explanation:

1. Electrostatic force on charge due to electrical field E is F = qE

Gravitational force due to weight of drop of mass M is F = mg, when net force is zero we have to find the charges on the drop.

2. Charge on drop: since gravitational force is acting downward, electric field is acting upward and we equate both equations.

Therefore, Fg = Fe

qe = mg

q = mg/E

q = 3.50×10^-9kg×9.8m/s²/8480N/c

q = 4.04 × 10^-12

3. Number of protons by quantitization law

n = q/e

n = 4.04× 10^-12/8480

n = 2.53× 10^7

8 0
3 years ago
If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?
vovikov84 [41]

Complete Question

In an action movie, the villain is rescued from the ocean by grabbing onto the ladder hanging from a helicopter. He is so intent on gripping the ladder that he lets go of his briefcase of counterfeit money when he is 130 m above the water. If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?

Answer:

The speed of the helicopter is u  =  7.73 \  m/s

Explanation:

From the question we are told that

   The height at which he let go of the brief case is  h =  130 m  

    The  time taken before the the brief case hits the water is  t =  6 s

Generally the initial speed of the  briefcase (Which also the speed of the helicopter )before the man let go of it is  mathematically evaluated using kinematic equation as

      s = h+  u t +  0.5 gt^2

Here s  is the distance covered by the bag at sea level which is zero

      0 = 130+  u * (6) +  0.5  *  (-9.8) * (6)^2

=>    0 = 130+  u * (6) +  0.5  *  (-9.8) * (6)^2

=>   u  =  \frac{-130 +  (0.5 * 9.8 *  6^2) }{6}

=>   u  =  7.73 \  m/s

     

7 0
3 years ago
At a weightlifting competition two competitors list of the same way to the same height the second competitor accomplished to liv
UNO [17]
One more what ? I’m confused ?
4 0
3 years ago
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