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polet [3.4K]
3 years ago
10

A package is pushed across the floor a distance of 90 feet by exerting a force of 32 lbs downward at an angle of 24 ∘ with the h

orizontal. How much work is done? Round to the nearest foot pound.
Physics
1 answer:
Crazy boy [7]3 years ago
4 0

Answer:

2632 foot-pound

Explanation:

Work done: Work is said to be done when ever a force moves a body through a given distance. The S.I unit of force is Newton (N).

From the question,

The expression for work done is given as,

W = Fdcos∅......................... Equation 1

Where W = work done, F = force, d = distance, ∅ = angle between the force and the horizontal.

Given: F = 32 lbs, d = 90 feet, ∅ = 24°

substitute into equation 1

W = 32×90×cos24

W = 2880(0.914)

W = 2632.32

W = 2632 foot-pound

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Answer:

At an angle of 30^{\circ}

Explanation:

Assume the river flows from East to West so for the swimmer to cross across it, assume he crosses it from West to East.

The resultant speed will be given by

R= \sqrt {10^{2}-5^{2}=\sqrt {75}\approx 8.66 km/h\\Direction=sin^{-1}\frac {5}{10}\approx 30^{\circ}

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3 years ago
Two cars are moving with velocities 70km/hr and 50km/hr in east and west direction respectively.
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70 + 50 = 120 km/hr

Explanation:

The driver of either car would see the other car approaching or departing at 120 kph

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2 years ago
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Answer:

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2 years ago
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Alborosie

Answer:

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7 0
2 years ago
Chapter 21, Problem 009 Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.12
PilotLPTM [1.2K]

Answer:

a) -1.325 μC

b) 4.17μC

Explanation:

First, you need to know that charge is conserved. So, the adition of the charges, as there is no lost in charge, should always be the same. Also, after the wire is removed, both spheres will have the same charge, trying to find equilibrium. In summary:

q_1 + q_2 = constant\\q_1_f = q_2_f |Then\\q_1_f + q_2_f = 2q_1_f = q_1_o+q_2_o\\q_1_f = q_2_f = \frac{q_1_o+q_2_o}{2}

We know both q1f and q2f must be positive, because the negative charge at the beginning was the the smaller.

The electrostatic force is equal to:

F_e = k\frac{q_1q_2}{r^2}

K is the Coulomb constant, equal to 9*10^9 Nm^2/C^2

Now, we are told that the electrostatic force after the wire is equal to 0.0443 N:

F_e_f = k \frac{q_1_fq_2_f}{r^2} = k\frac{\frac{q_1_o+q_2_o}{2}\frac{q_1_o+q_2_o}{2}}{r^2} = k\frac{(q_1_o+q_2_o)^2}{4r^2}  \\(q_1_o+q_2_o) = \sqrt{\frac{F_e_f*4r^2}{k}} = \sqrt{\frac{0.0443N *4(0.641m)^2}{9*10^9Nm^2/C^2} } = 2.844 *10^{-6}C \\ q_1_o = 2.844*10^{-6}C - q_2_o

Originally, the force is negative because it was an attraction force, therefore, its direction was opposite to the direction of the repulsive force after the wire:

F_e_o = k\frac{q_1_oq_2_o}{r^2}\\ q_1_oq_2_o = \frac{F_e_o*r^2}{k} = \frac{-0.121N(0.641m)^2}{9*10^9Nm^2/C^2} = -5.524*10^{-12}

(2.844*10^{-6}C - q_2_o)q_2_o = -5.524*10^{-12}\\0 = q_2_o^2 - 2.844*10^{-6}q_2_o - 5.524*10^{-12}

Solving the quadratic equation:

q_2_o = 4.17*10^{-6}C | -1.325 * 10^{-6}C

for this values q_1 wil be:

q_1_o =  -1.325 *10^{-6}C | 4.17*10^{-6}C

So as you can see, the negative charge will always be -1.325 μC and the positive 4.17μC

5 0
3 years ago
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