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3 years ago

A cheetah accelerates from rest to 30m/s in 3 seconds. Calculate the acceleration of the cheetah.

Physics

1 answer:

fiasKO [112]3 years ago

4 0

30m/s times 3s= 90 m/s

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Explanation:

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A 4000-kg car bumps into a stationary 6000kg truck. The Velocity of the car before the collision was +4m/s and -1m/s after the c

Fynjy0 [20]

Answer:

-4.33m/s

Explanation:

Using the law of conservation of momentum

m1u1+m2u2 = (m1+m2)v

m1 and m2 are the masses

u1 and u2 are their respective velocities

v is the final velocity

From the question;

m1 = 4000kg

m2 = 6000kg

u1 = 4m/s

u2 = ?

v =-1m/s

Substitute into the formula and get u2

4000(4)+6000u2 = (4000+6000)(-1)

16000 + 6000u2 = -10000

6000u2 = -10000-16000

6000u2 = -26000

6u2 = -26

u2 = -26/6

u2 = -13/3

u2 = -4.33m/s

Hence the velocity of the truck is -4.33m/s

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3 years ago

Two capacitors of 36 µF and 3.8 µF are connected parallel and charged with a 51 V power supply. Calculate the total energy store

olga nikolaevna [1]

Answer:

U=0.0517 J

Explanation:

Given that

The values of capacitance

C₁ = 36 µF

C₂ = 3.8 µF

We know that when capacitor are connected in the parallel then the equivalent value of the capacitance is given as

C=C₁ +C₂

Now by putting the values

C= 36 + 3.8 µF

C = 39.8 µF

The voltage difference

ΔV= 51 V

The total stored energy in capacitor is given as

$U=\dfrac{1}{2}C\Delta V^2$

Now by putting the values in the above equation we get

$U=\dfrac{1}{2}\times 39.8\times 10^{-6}\times 51^2\ J$

U=0.0517 J

Therefore the stored energy in the capacitors will be 0.0517 J.

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3 years ago