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2 years ago

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A small object is placed at the top of an incline that is essentially frictionless. The object slides down the incline onto a ro

ugh horizontal surface, where it stops in 5.0 s after traveling 60 m. a) What is the speed of the object at the bottom of the incline and its acceleration along the horizontal surface? b) What is the height of the incline?

1 answer:

Georgia [21]2 years ago

4 0

Answer

given,

time taken to stop by the object = 5 s

distance travel before stopping = 60 m

final velocity = 0

using equation of motion

v = u + at

0 = u - 5 a

$a_x = \dfrac{u}{5}$

$s = u t + \dfrac{1}{2}at^2$

$s = 5u + \dfrac{1}{2}\times \dfrac{4}{5}\times u^2$

$60 = 2.5 u$

u = 24 m/s

$a_x = \dfrac{24}{5}$

a_x = 4.8 m/s²

b) using energy conservation

$\dfrac{1}{2}mv_i^2 + \dfrac{1}{2}mv_f^2 = mg(\Delta h)$

$\dfrac{1}{2}v_i^2= g(\Delta h)$

$(\Delta h) = \dfrac{0.5 \times 24^2}{9.8}$

Δh = 29.38 m

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What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1

notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is $F_{net}= 3.22*10^{-5} \ J$

Explanation:

From the question we are told that

The third charge is $q_3 = 55 nC = 55 *10^{-9} C$

The position of the third charge is $x = -1.220 \ m$

The first charge is $q_1 = -16 nC = -16 *10^{-9} \ C$

The position of the first charge is $x_1 = -1.650m$

The second charge is $q_2 = 32 nC = 32 *10^{-9} C$

The position of the second charge is $x_2 = 0 \ m$

The distance between the first and the third charge is

$d_{1-3} = -1.650 -(-1.220)$

$d_{1-3} = -0.43 \ m$

The force exerted on the third charge by the first is

$F_{1-3} = \frac{k q_1 q_3}{d_{1-3}^2}$

Where k is the coulomb's constant with a value $9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

$F_{1-3} = \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}$

$F_{1-3} = 4.28 *10^{-5} \ N$

The distance between the second and the third charge is

$d_{2-3} = 0- (-1.22)$

$d_{2-3} =1.220 \ m$

The force exerted on the third charge by the first is mathematically evaluated as

$F_{2-3} = \frac{k q_2 q_3}{d_{2-3}^2}$

substituting values

$F_{2-3} = \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}$

$F_{2-3} = 1.06*10^{-5} N$

The net force is

$F_{net} = F_{1-3} -F_{2-3}$

substituting values

$F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}$

$F_{net}= 3.22*10^{-5} \ J$

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Answer:

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