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Natali [406]
4 years ago
11

A 1200-kg car initially at rest undergoes constant acceleration for 9.4 s, reaching a speed of 11 m/s. It then collides with a s

tationary car that has a perfectly elastic spring bumper. What is the final kinetic energy of the two car system?
Physics
1 answer:
Natalka [10]4 years ago
8 0

To solve this problem we will apply the principle of conservation of energy and the definition of kinematic energy as half the product between mass and squared velocity. So,

KE_i = KE_f

KE_f = \frac{1}{2} mv^2

Here,

m = Mass

V = Velocity

Replacing,

KE_f = \frac{1}{2} (12000)(11)^2

KE_f = 72600J

Therefore the  final kinetic energy of the two car system is 72.6kJ

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A crow is flying horizontally with a constant speed of 2.70m/s when it releases a claim from its beak. The clan lands on the roc
jenyasd209 [6]

Given:

Speed = 2.70 m

Time, t = 2.10 seconds

Let's solve for the following:

• (a) The horizontal component of the velocity.

To find the horizontal component, apply the formula:

V_{ox}=V_o\cos \theta

Where:

Vo is the initial speed = 2.70 m

θ = 0 degrees

Hence, we have:

\begin{gathered} V_{ox}=2.70\cos 0 \\  \\ V_{ox}=2.7\text{ m/s} \end{gathered}

The horizontal component of the velocity just before it lands is 2.70 m/s.

• (b) The vertical component of the velocity.

To find the vertical component, apply the formula:

V_{oy}=V_{0y}-gt=\text{V}_{oy}\text{ sin}\Theta-gt

Where:

g is the acceleration due to gravity = 9.8 m/s²

t is the time = 2.10 s

Hence, we have:

\begin{gathered} V_{oy}=V_{oy}\sin \theta-gt \\  \\ V_{oy}=2.70\sin 0-9.8(2.10) \\  \\ V_{oy}=0-20.58 \\  \\ V_{oy}=-20.58\text{ m/s} \end{gathered}

The vertical component of the velocity just before it lands is -20.58 m/s.

(c) Here, the initial speed is equal to the constant horizontal speed.

Therefore, in part (a) the horizontal component will increase in the x-direction if the speed of the crow is increased.

The initial vertical velocity is 0 m/s in both cases.

Therefore, in part (b) the vertical component will remain constant.

ANSWER:

(a) 2.70 m/s

(b) -20.58

(c) In part (a) the horizontal component will increase, while in part (b) the vertical component will remain constant.

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