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crimeas [40]
3 years ago
8

An object in space is initially stationary relative to the Earth. Then, a force begins acting on the object, starting with a for

ce of 0 N, and increasing at a uniform rate until the magnitude of the force is 36 N after the force has acted for 35 s. The body has a mass of 10 kg. What is the speed of the object at 35 s?1. 4.4 m/s2. 8.8 m/s3. 11458.3 m/s4. 2750 m/s5. Insufficient information to answer6. 22916.7 m/s7. 110 m/s8. 5500 m/s9. 220 m/s
Physics
1 answer:
11111nata11111 [884]3 years ago
8 0

Answer:

v_f =63 m/s

Explanation:

given,

starting force = 0 N

uniform rate increase to 36 N

time of action of Force = 35 s

mass of the body = 10 Kg

Speed of the object = ?

From the given data

if we plot F-t curve  we will get a triangular shape

we know,

Impulse = Area between F-t curve

            = (1/2) x base x height

             = 0.5 x 35 x 36

             = 630 N.s

now use Impulse-momentum theorem

Impulse = change in momentum

630 = 10 x (v_f - vi)

630 = 10 x (v_f - 0)

v_f =63 m/s

Speed of the object at 35 sec is equal to v_f =63 m/s

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Answer:

x=2.4365\ m

and

x=-1.4365\ m

Explanation:

Given:

  • first charge, q_1=5\times 10^{-3}\ C
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\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}

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