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Assoli18 [71]
2 years ago
13

The acceleration of gravity is -9.8 m/s2. A sling shot fires a rock straight up into the air with a speed of +39.2 m/s. 1. what

is its velocity after 2 seconds?
Physics
1 answer:
Vanyuwa [196]2 years ago
4 0

Given that,

The acceleration of gravity is -9.8 m/s²

Initial velocity, u = 39.2 m/s

Time, t = 2 s

To find,

The final velocity of the shot.

Solution,

Let v is the final velocity of sling shot. Using first equation of motion to find it.

v = u +at

Here, a = -g

v = u-gt

v = (39.2)-(9.8)(2)

v = 19.6 m/s

So, its velocity after 2 seconds is 19.6 m/s.

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3 years ago
A room with 3.1-m-high ceilings has a metal plate on the floor with V = 0V and a separate metal plate on the ceiling. A 1.1g gla
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The ball traveled 0.827 m

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charge on the glass, q = 4.7 nC

speed of the glass ball, v = 4.8 m/s

voltage of the ceiling, V = +3.0 x 10⁶ V

The repulsive force experienced by the ball when shot to the ceiling with positive voltage, can be calculated using Coulomb's law;

F = qV/d

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|F| = 4.548 x 10⁻³ N

F = - 4.548 x 10⁻³ N

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F_{net} = F_c - mg\\\\F_{net} = -4.548 *10^{-3} - (1.1*10^{-3} * 9.8)\\\\F_{net} = -15.328*10^{-3} \ N

The work done between the ends of the plate is equal to product of the  magnitude of net force on the ball and the distance traveled by the ball.

W = F_{net} *h\\\\W = 15.328 *10^{-3} *  h

W = K.E

15.328*10^{-3} *h = \frac{1}{2}mv^2\\\\ 15.328*10^{-3} *h = \frac{1}{2}(1.1*10^{-3})(4.8)^2\\\\ 15.328*10^{-3} *h =0.0127\\\\h = \frac{0.0127}{15.328*10^{-3}}\\\\ h = 0.827 \ m

Therefore, the ball traveled 0.827 m

4 0
3 years ago
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