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Wittaler [7]
3 years ago
9

a body weights 28N at a height of 3200km from the earth surface.What will be the gravitational force on that body if its lies on

the earth surface.
Physics
1 answer:
alekssr [168]3 years ago
4 0

Answer:

The object would weight 63 N on the Earth surface

Explanation:

We can use the general expression for the gravitational force between two objects to solve this problem, considering that in both cases, the mass of the Earth is the same. Notice as well that we know the gravitational force (weight) of the object at 3200 km from the Earth surface, which is (3200 + 6400 = 9600 km) from the center of the Earth:

F_G=G\,\frac{M_E\,m}{d^2} \\28\,\,N=G\,\frac{M_E\,m}{9600000^2}

Now, if the body is on the surface of the Earth, its weight (w) would be:

F_G=G\,\frac{M_E\,m}{d^2} \\w=G\,\frac{M_E\,m}{6400000^2}

Now we can divide term by term the two equations above, to cancel out common factors and end up with a simple proportion:

\frac{w}{28} =\frac{9600000^2}{6400000^2} \\\frac{w}{28} =\frac{9}{4} \\\\ \\w=\frac{9\,*\,28}{4}\,\,\,N\\w=63\,\,N \\

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Flauer [41]
<h3>It takes 60 seconds to do the work</h3>

<em><u>Solution:</u></em>

Given that,

Force = 100 newtons

Distance = 15 meters

Power = 25 watts

To find: time it takes to do the work

<em><u>Find the work done:</u></em>

work = force \times displacement\\\\work = 100\ newtons \times 15\ meters\\\\work = 1500\ joule

<em><u>Find the time taken</u></em>

power = \frac{work}{time}\\\\25\ watts = \frac{1500\ joule}{time}\\\\time = \frac{1500\ joule}{25\ watt}\\\\time = 60\ second

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3 years ago
Water is flowing in a pipe with a varying cross-sectional area, and at all points the water completely fills the pipe. At point
Salsk061 [2.6K]

Answer:

The fluids speed at a) 0.105\,m^{2}  and b) 0.047\,m^{2} are 2.33\,\frac{m}{s^{2}}  and 5.21\,\frac{m}{s^{2}} respectively

c) Th volume of water the pipe discharges is: 882\,m^{3}  

Explanation:

To solve a) and b) we should use flow continuity for ideal fluids:

\Delta Q=0(1)

With Q the flux of water, but Q is Av using this on (1) we have:

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a) Here, we use that point 2 has a cross-sectional area equal to A_{2}=0.105\,m^{2}, so now we can solve (2) for v_{2}:

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b) Here we use point 2 as A_{2}=0.047\,m^{2}:

v_{2}=\frac{A_{1}v_{1}}{A_{2}}=\frac{(0.070)(3.5)}{0.047}\approx5.21\,\frac{m}{s^{2}}

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A_{1}v_{1}=\frac{V}{t}, solving for V:

V=A_{1}v_{1}t=(0.070m^{2})(3.5\frac{m}{s})(3600s)=882\,m^{3}

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A jogger jogs around a circular track with a diameter of 275 m in 14.0 minutes. what was the jogger's average speed in m/s?
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The jogger's average speed 1.03 m/s

<h3>The Speed and the Velocity of a Particle in a Circle</h3>

The speed of a particle is a circle will always be constant while the velocity will not. That is, velocity varies.

Given that a jogger jogs around a circular track with a diameter of 275 m in 14.0 minutes. First convert the minutes to seconds

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Therefore, the jogger's average speed 1.03 m/s approximately

Learn more about Circular Motion here: brainly.com/question/20905151

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