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2 years ago

5

calculate the magnitude of impulse applied to a 0.75 kilogram cart to change its velocity from 0.50 meters per second east to 2.

00 meters per second east

1 answer:

ella [17]2 years ago

8 0

Answer:

1.125 N s towards East

Explanation:

since both velocities are in same direction hence change in velocity is

Δ V = final - initial

= 2.00 - 0.50

= 1.50 towards East

impulse = change of linear momentum

= mass × change in velocity

= 0.75 ×1.50

= 1.125 N s towards East

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two people, each with a mass of 70 kg, are wearing inline skates and are holding opposite ends of a 15m rope. One person pulls f

Zina [86]

Answer:

7.75 s

Explanation:

Newton's second law:

∑F = ma

35 N = (70 kg) a

a = 0.5 m/s²

Given v₀ = 0 m/s and Δx = 15 m:

Δx = v₀ t + ½ at²

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t = 7.75 s

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A penny is placed on a rotating turntable. Where on the turntable does the penny require the largest centripetal force to remain

Artyom0805 [142]

Answer:

m = mass of the penny

r = distance of the penny from the center of the turntable or axis of rotation

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F = m r w²

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Explanation:

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2 years ago

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2 years ago

Read 2 more answers

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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2 years ago

A woman is applying 300N/m2 of pressure on to door with her hand. Her hand has area of 0.02m2. Work out the force being applied

never [62]

Answer:

6N

Explanation:

Given parameters:

Pressure applied by the woman = 300N/m²

Area = 0.02m²

Unknown:

Force applied = ?

Solution:

Pressure is the force per unit area on a body

Pressure = $\frac{force}{area}$

Force = Pressure x area

Force = 300 x 0.02 = 6N

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2 years ago