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Leona [35]
4 years ago
10

A 3-kg ball is thrown with a speed of 8 m/s at an unknown angle above the horizontal. The ball attains a maximum height of 2.8 m

before striking the ground.
If air resistance is negligible, what is the value of the kinetic energy of the ball at its highest point?
Physics
1 answer:
Alina [70]4 years ago
8 0

Answer:

13.5 J

Explanation:

mass of ball, m = 3 kg

maximum height, h = 2.8 m

initial speed, u = 8 m/ s

Angle of projection, θ

use the formula of maximum height

H = \frac{u^{2}Sin^{2}\theta }{2g}

2.8 = \frac{8^{2}Sin^{2}\theta }{2\times 9.8}

Sin θ = 0.926

θ = 67.8°

The velocity at maximum height is u Cosθ = 8 Cos 67.8 = 3 m/s

So, kinetic energy at maximum height

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

K = 0.5 x 3 x 3 x 3

K = 13.5 J

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beks73 [17]

Answer:

I for a rod about its center is I = M L^2 / 12

A = L (1/2 - 1/3) = L / 6     distance from middle to location L/3

A^2 = L^2 / 36

I = M L^2 (1/12 + 1/36) = M L^2 (4 / 36) = M L^2 / 9

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Alex17521 [72]

Answer:

E=1824.81 V/m

Explanation:

Given that

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As we know that electric filed given as

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Given that D is 24° with respect to the perpendicular to the electrodes.So we have to take cos component of D.

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d= 3 cos24°

d = 2.74 mm

So

E=\dfrac{V}{d}

E=\dfrac{5}{2.74\times 10^{-3}}

E=1824.81 V/m

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3 years ago
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Answer:

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blondinia [14]

Answer:

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<h2><u><em>Pls Mark Brainliest</em></u></h2>
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