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Wittaler [7]
4 years ago
6

a horizontal force of 2500 N is applied for a time of 30 seconds to the car mass of 1000 kg if the car is at rest and friction i

s negligible determine the distance travelled by the car​
Physics
1 answer:
Digiron [165]4 years ago
6 0

Newton’s second law: F=ma. 2500=1000a. a=2.5m/s*s. Final velocity is 2.5*30=75m/s. Average speed is (Vi+Vf)/2, which is 37.5m/s. D=vt. 37.5*30=1125m. The distance traveled is 1125 meters

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By chromatography it's the answer
8 0
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A car travels straight for 20 miles on a road that is 30° north of east. What is the east component of the car’s displacement to
Virty [35]
17.3 would be the correct answer.

4 0
4 years ago
Read 2 more answers
a tennis ball is thrown straight up with an initial velocity of 22.5 m/s how much total time is the ball in the air
lutik1710 [3]

Answer:

4.6s

Explanation:

v=u+at

0=22.5+(-9.8)t

-22.5=-9.8t

t=-22.5/-9.8

t=2.295 s

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6 0
3 years ago
1. If a car travels East 60 km for 4 hours. What was the velocity of the car?
True [87]

Answer:

15km/h East (15m/s East option)

Explanation:

Velocity = (change in) Distance/(change in) Time

The distance here is 60km, and the time is 4h, as given by the question. Therefore the velocity is 60km/4h = 15km/h.

To convert km/h to m/s, we just divide the value by 3.6, 15/3.6= 4.17m/s (2dp), which isn't actually an option here, so I'm assuming maybe a mistake in unit for the question?

'Velocity' is a vector quantity, meaning it has a size<em> </em>and a direction, as opposed to speed, a scalar quantity, which only has size. Therefore we need to add a direction for it to be velocity. The given direction here is east, so the velocity of the car is 15km/h East. (I would choose the 15m/s East as the question likely has a unit error and is closest.)

Hope this helped!

4 0
3 years ago
A 591 μF capacitor is discharged through a resistor, whereby its potential difference decreases from its initial value of 88.5 V
Stels [109]

Answer:

2.6 kilo Ohm

Explanation:

Capacitance, C = 591 μF = 591 x 10^-6 F

Vo = 88.5 V

V = 11.9 V

t = 3.09 s

Let the resistance is R.

V = V_{0}e^{\frac{-t}{RC}}

\frac{11.9}{88.5} = e^{\frac{-t}{RC}}

0.135 = e^{\frac{-t}{RC}}

Take natural log on oth the sides

ln 0.135 = - 3.09 / RC

RC = 1.545

R = 1.545 / ( 591 x 10^-6)

R = 2614.2 ohm

R = 2.6 kilo Ohm

Thus the resistance is  2.6 kilo Ohm.

8 0
3 years ago
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