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grandymaker [24]
2 years ago
11

A force of 385 N is applied in pushing a stalled automobile at a constant speed for a distance of 150 m. How much work (in J) wa

s done on the car by the force?
Physics
1 answer:
Helga [31]2 years ago
3 0

Answer:57,750J

Explanation:

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il63 [147K]

Explanation:

There are many ways to achieve a healthier lifestyle and increase fitness. Healthy eating and constant exercise are essential for maintaining good physical and mental health. Through effective nutrition the human body is able to function better, there is an increase in disposition, improvement in immunity, improvement in blood rates, etc. Physical exercise, on the other hand, assists in strengthening muscles, controlling blood pressure, preventing cardiovascular disease, greater satisfaction, less stress, etc.

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6 0
3 years ago
Kon'nichiwa~<br>please help me with this question!!​
andrezito [222]

Let's check the relationship

\\ \rm\hookrightarrow g=\dfrac{GM}{r^2}

\\ \rm\hookrightarrow g\propto G

So

  • Raindrops will fall faster . .
  • Also walking on ground would become more difficult as g increases.

Option C is wrong by now .Let's check D once

\\ \rm\hookrightarrow T\propto \dfrac{1}{\sqrt{g}}

  • So time period of simple pendulum would decrease.

4 0
2 years ago
(a) According to Hooke's Law, the force required to hold any spring stretched x meters beyond its natural length is f(x)=kx. Sup
KengaRu [80]

Answer:

a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve r(\theta) = 2\cdot \sin 5\theta is 4\pi.

Explanation:

a) The work, measured in joules, is a physical variable represented by the following integral:

W = \int\limits^{x_{f}}_{x_{o}} {F(x)} \, dx

Where

x_{o}, x_{f} - Initial and final position, respectively, measured in meters.

F(x) - Force as a function of position, measured in newtons.

Given that F = k\cdot x and the fact that F = 25\,N when x = 0.3\,m - 0.2\,m, the spring constant (k), measured in newtons per meter, is:

k = \frac{F}{x}

k = \frac{25\,N}{0.3\,m-0.2\,m}

k = 250\,\frac{N}{m}

Now, the work function is obtained:

W = \left(250\,\frac{N}{m} \right)\int\limits^{0.05\,m}_{0\,m} {x} \, dx

W = \frac{1}{2}\cdot \left(250\,\frac{N}{m} \right)\cdot [(0.05\,m)^{2}-(0.00\,m)^{2}]

W = 0.313\,J

The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.

b) Let be r(\theta) = 2\cdot \sin 5\theta. The area of the region enclosed by one loop of the curve is given by the following integral:

A = \int\limits^{2\pi}_0 {[r(\theta)]^{2}} \, d\theta

A = 4\int\limits^{2\pi}_{0} {\sin^{2}5\theta} \, d\theta

By using trigonometrical identities, the integral is further simplified:

A = 4\int\limits^{2\pi}_{0} {\frac{1-\cos 10\theta}{2} } \, d\theta

A = 2 \int\limits^{2\pi}_{0} {(1-\cos 10\theta)} \, d\theta

A = 2\int\limits^{2\pi}_{0}\, d\theta - 2\int\limits^{2\pi}_{0} {\cos10\theta} \, d\theta

A = 2\cdot (2\pi - 0) - \frac{1}{5}\cdot (\sin 20\pi-\sin 0)

A = 4\pi

The area of the region enclosed by one loop of the curve r(\theta) = 2\cdot \sin 5\theta is 4\pi.

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3 years ago
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NISA [10]
The person should start to slow down but if close enough or in the intersection go threw. Otherwise come to a complete stop until the light turns green again
4 0
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What evidence did wegener have to support his theory of plate tectonics?
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Answer:

Wegener first thought of this idea by noticing that the different large landmasses of the Earth almost fit together like a jigsaw puzzle. The continental shelf of the Americas fits closely to Africa and Europe, and other continents showed the same trend. Wegner also analyzed both sides of the Atlantic Ocean for rock type, geological structures and fossils and noticed that there was a significant similarity between matching sides of the continents, especially in fossil plants.

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