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bulgar [2K]
3 years ago
12

A van of mass 1200kg is moving with speed of 90km/hr. It is brought in 3 seconds by applying brakes. Calculate the force applied

to bring van to rest. ​
Physics
1 answer:
Dimas [21]3 years ago
4 0

Answer:

Explanation:

The main equation to solve this is F = ma, where F is the force applied to the brakes with respect to its acceleration. We have the mass that we need, but we do not have the acceleration. That's the first thing we have to find. However, our velocity needs to be stated in m/s and right now it's in km/h. Converting that:

90\frac{km}{hr}*\frac{1000m}{1km}*\frac{1hr}{3600s} =25\frac{m}{s} Now we're ready to find the acceleration:

a=\frac{v_f-v_0}{t} where the top line there translates to the final velocity minus the initial velocity.

a=\frac{0-25}{3.0} so the acceleration is -8.3 m/s/s

We can use that now in the force equation above:

F = 1200(-8.3) and

F = -1.0 × 10⁴ N (that's 10,000 N to the correct number of sig dig's; the negative sign means that the force is being applied in the direction opposite to that which the van is currently moving)

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Electric energy is generated at a nuclear power plant using which type<br> of nuclear reaction
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Read 2 more answers
A 97.3 kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.67 r
Yuliya22 [10]

Answer:

w = 1.14 rad / s

Explanation:

This is an angular momentum exercise. Let's define a system formed by the three bodies, the platform, the bananas and the monkey, in such a way that the torques during the collision have been internal and the angular momentum is preserved.

Initial instant. The platform alone

        L₀ = I w₀

Final moment. When the bananas are on the shelf

we approximate the bananas as a point load and the distance is indicated

x = 0.45m

        L_f = (m x² + I ) w₁

angular momentum is conserved

         L₀ = L_f

         I w₀ = (m x² + I) w₁

         w₁ = \frac{I}{m x^2 + I}  \ w_o

Let's repeat for the platform with the bananas and the monkey is the one that falls for x₂ = 1.73 m

initial instant. The platform and bananas alone

        L₀ = I₁ w₁

         I₁ = (m x² + I)

           

final instant. After the crash

        L_f = I w

        L_f = (I₁ + M x₂²) w

the moment is preserved

        L₀ = L_f

        (m x² + I) w₁ = ((m x² + I) + M x₂²) w

         (m x² + I) w₁ = (I + m x² + M x₂²) w

we substitute

         w =  \frac{m x^2 +I}{I + m x^2 + M x_2^2} \ \frac{I}{m x^2 + I} \ w_o

         w =  \frac{I}{I + m x^2 + M x_2^2} \ w_o

the moment of inertia of a circular disk is

         I = ½ m_p x₂²

we substitute

         w =  \frac{ \frac{1}{2} m_p x_2^2 }{ \frac{1}{2} m_p x_2^2 + M x_2^2 + m x^2} \ \ w_o

let's calculate

          w =\frac{ \frac{1}{2} \ 97.3 \ 1.73^2 }{ \frac{1}{2} \ 97.3 \ 1.73^2 + 21.9 \ 1.73^2 + 9.67 \ 0.45^2 } \ \ 1.67

          w =  \frac{145.60 }{145.60 \ + 65.54 \ + 1.958} \ \ 1.67

          w = 1.14 rad / s

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Answer:

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