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umka21 [38]
3 years ago
12

If It took 40 Seconds for a vehicle of weight 40,000 Newton to move round a 7 metres, What time will it achieve same feat for an

other circular path Of radius 21 metres given the Condition of the vehicle remains the same.​
Physics
1 answer:
Svetllana [295]3 years ago
3 0

Answer:

  t = 376.99 s

Explanation:

We must solve this problem with the equations and kinematics, let's start by looking for the speed of the vehicle,

         v = d / t

         v = 7/40

         v = 0.175 m / s

Since the speed e remains constant, we must find the length of the circle is

          L = 2π r

          L = 2π 21

          L = 131.95 m

In the problem it does not specify clearly, but in general the curves of the road correspond to half a circle, so the length of the road is

         L ’= L / 2

         L ’= 131.95 / 2 = 65.97 m

as the speed is constant

          t = L ’/ v

          t = 65.97 / 0.175

          t = 376.99 s

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Answer:

A) m2 = 98.71g

B) v_f2 = 1.86 m/s

Explanation:

We are given;

Mass of cart; m1 = 340g

Initial speed; v_i1 = 1.2 m/s

Final speed; v_f1 = 0.66 m/s

A)Since the collision is elastic, we can simply apply the conservation of momentum to get;

m1•(v_i1) = m1•(v_f1) + m2•(v_f2) - - - - - (eq1)

From conservation of kinetic energy, we have;

(1/2)m1•(v_i1)² = (1/2)m1•(v_f1)² + (1/2)m2•(v_f2)² - - - - eq(2)

Let's make v_f2 the subject in eq 2;

Thus,

v_f2 = √([m1•(v_i1)² - m1•(v_f1)²]/m2)

v_f2 = √([m1((v_i1)² - (v_f1)²)]/m2)

Let's put this for v_f2 in eq1 to obtain;

m2 = {m1((v_i1) - (v_f1))}/√([m1((v_i1)² - (v_f1)²)]/m2)

Let's square both sides to give;

(m2)² = {m1•m2((v_i1) - (v_f1))²}/([(v_i1)² - (v_f1)²]

This gives;

m2 = {m1((v_i1) - (v_f1))²}/([(v_i1)² - (v_f1)²]

Plugging in the relevant values to get;

m2 = {340((1.2) - (0.66))²}/([(1.2)² - (0.66)²]

m2 = 98.71g

B) from equation 1, we have;

m1•(v_i1) = m1•(v_f1) + m2•(v_f2)

Making v_f2 the subject, we have;

v_f2 = m1[(v_i1) - (v_f1)]/m2

Plugging in the relevant values to get;

v_f2 = 340[(1.2) - (0.66)]/98.71

v_f2 = 1.86 m/s

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