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user100 [1]
3 years ago
15

A bicyclist rides 1.59 km due east, while the resistive force from the air has a magnitude of 7.00 N and points due west. The ri

der then turns around and rides 1.59 km due west, back to her starting point. The resistive force from the air on the return trip has a magnitude of 7.00 N and points due east. Find the work done by the resistive force during the round trip. Number Units
Physics
1 answer:
kramer3 years ago
8 0

Answer:

The required work done by the resistive force = -22.3 kJ

Explanation:

We understand that the amount of work done for a force F and displacements is provided as the product of force multiplied by displacement, which can be represented as W = F.S

Now, when the cyclist travels east and returns west, the force and displacement directions will be opposite and they make an angle 180 degrees apart.

∴

W = -2 × 7.00 × 1590

W = -22260 J

W = -22.3 kJ

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How much force is needed to accelerate a 1,100 kg car at a rate of 1.5 m/s2?
Anna [14]
Assuming there is no force of friction...

F = ma
F = (1300kg)(1.5m/s^2)
F = 1950N
Just multiply mass by acceleration.
1300 x 1.5 = 1950N.
7 0
3 years ago
How many cubic meters of habitable space was available in skylab?
loris [4]

361 cubic meters of habitable space was available in the skylab. None of the given options are correct.

<h3>Who launched skylab?</h3>

NASA launched the first American space station, called Skylab.

The fuel tank of the Saturn SIV-B rocket stage served as the Skylab's residential quarters.

Skylab's livable capacity was large in comparison to the spacecraft that American astronauts had previously used 361 cubic meters.

Skylab offered 361 cubic meters of livable area.

Hence, none of the given options is correct.

To learn more about the skylab refer;

brainly.com/question/23389881

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4 0
2 years ago
A 13.3 kg box sliding across the ground
adelina 88 [10]

Answer:

<em>0.25</em>

Explanation:

According to newtons law of motion

\sum F_x = ma

F_f =  ma

nR = ma

nmg = ma

ng = a

n = a/g

g is the acceleration due to gravity

Given

a = 2.42m/s²

g = 9.8m/s²

Substitute into the formula;

n = 2.42/9.8

n = 0.25

<em>Hence the coefficient of kinetic friction is 0.25</em>

<em></em>

5 0
3 years ago
NASA has asked your team of rocket scientists about the feasibility of a new satellite launcher that will save rocket fuel. NASA
kkurt [141]

Answer:

The answer is "q=0.0945\,C".

Explanation:

Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).

U_i=\frac{1}{4\pi\epsilon_0} \times \frac{4\times4q^2}{\sqrt{(15)^2+(5/\sqrt2)^2}}

It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:

U_f=\frac{1}{4\pi\epsilon_0}\ \times \frac{4\times4q^2}{( \frac{5}{\sqrt{2}})}

Potential energy shifts:

= U_f -U_i \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{\sqrt{(15)^2+( \frac{5}{\sqrt{2})^2)}}\right ) \\\\   =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ 15 +( \frac{5}{2})}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{30+5}{2})}}\right )\\\\

=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{35}{2})}}\right )\\\\=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{17.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 24.74- 5 }{87.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 19.74- 5 }{87.5}}\right )\\\\ =\frac{4q^2}{\pi\epsilon_0}\left ( 0.2256 }\right )\\\\= \frac{0.28 \times q^2}{ \epsilon_0}\\\\=q^2\times31.35 \times10^9\,J

Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.

=\frac{GMm}{R}-\frac{GMm}{R+h} \\\\=(6.67\times10^{-11}\times6.0\times10^{24}\times100)\left(\frac{1}{6400\times1000}-\frac{1}{6700\times1000} \right ) \\\\ =(6.67\times10^{-11}\times6.0\times10^{26})\left(\frac{1}{64\times10^{5}}-\frac{1}{67\times10^{5}} \right ) \\\\=(6.67\times6.0\times10^{15})\left(\frac{67 \times 10^{5} - 64 \times 10^{5}  }{ 4,228 \times10^{5}} \right ) \\\\

=( 40.02\times10^{15})\left(\frac{3 \times 10^{5}}{ 4,228 \times10^{5}} \right ) \\\\ =40.02 \times10^{15} \times 0.0007 \\\\

\\\\ =0.02799\times10^{10}\,J \\\\= q^2\times31.35\times10^{9} \\\\ =0.02799\times10^{10} \\\\q=0.0945\,C

This satellite is transmitted by it system at a height of 300 km and not in orbit, any other mechanism is required to bring the satellite into space.

6 0
3 years ago
What is transferred when a wave hits the beach
Degger [83]
A tsunami or a high tide wave
hoep this helped u!!
8 0
4 years ago
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