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3 years ago

A car with mass 450 kg has a kinetic energy of 16,256 j. What is the speed of the car?

Physics

2 answers:

Lana71 [14]3 years ago

7 0

Answer:

Speed of the car is 8.49 m/s

Explanation:

It is given that,

Mass of the car, m = 450 kg

Kinetic energy of the car, E = 16256 J

We need to find the speed of the car. The kinetic energy of the car is given by :

$E=\dfrac{1}{2}mv^2$

$v=\sqrt{\dfrac{2E}{m}}$

$v=\sqrt{\dfrac{2\times 16256\ J}{450\ kg}}$

v = 8.49 m/s

So, the speed of the car is 8.49 m/s. Hence, this is the required solution.

Likurg_2 [28]3 years ago

5 0

E = 1/2 * m * v²

v = √(2 * E / m)

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3 years ago

A baseball is batted from a height of 1.09 m with a speed of

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(a) The horizontal and vertical components of the ball’s initial velocity is 37.8 m/s and 12.14 m/s respectively.

(b) The maximum height above the ground reached by the ball is 8.6 m.

(d) The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

<h3>Horizontal and vertical components of the ball's velocity</h3>

Vx = Vcosθ

Vx = 39.7 x cos(17.8)

Vx = 37.8 m/s

Vy = Vsin(θ)

Vy = 39.7 x sin(17.8)

Vy = 12.14 m/s

<h3>Maximum height reached by the ball</h3>

$H = \frac{v^2 sin^2(\theta)}{2g} \\\\H = \frac{(39.7)^2 \times (sin17.8)^2}{2(9.8)} \\\\H = 7.51 \ m$

Maximum height above ground = 7.51 + 1.09 = 8.6 m

<h3>Distance off course after 2 second </h3>

Upward speed of the ball after 2 seconds, V = V₀y - gt

Vy = 12.14 - (2x 9.8)

Vy = - 7.46 m/s

Horizontal velocity will be constant = 37.8 m/s

Resultant speed of the ball after 2 seconds = √(Vy² + Vx²)

$V = \sqrt{(-7.46)^2 + (37.8)^2} \\\\V = 38.53 \ m/s$

<h3>Resultant speed of the ball and crosswind</h3>

$V = \sqrt{38.52^2 + 4^2} \\\\V = 38.72 \ m/s$

<h3>Distance off course the ball would be carried</h3>

d = Δvt = (38.72 - 38.53) x 2

d = 0.38 m

The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

Learn more about projectiles here: brainly.com/question/11049671

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