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ZanzabumX [31]
3 years ago
5

A planet in elliptical orbit around a star moves from the point in its orbit furthest from the star (A) to the closest point (P)

. Choose the planet to be your system. The work done by the force of gravity during this movement is:
Physics
1 answer:
Virty [35]3 years ago
6 0

Answer:

Zero work done,since the body isn't acting against  or by gravity.

Explanation:

Gravitational force is usually  considered as work done against gravity (-ve) and work by gravity ( +ve ) and also When work isn't done by or against gravity work done in this case is zero.

Gravitational force can be define as that force that attracts a body to any other phyical body or system that have mass.

The planet been considered as our system in this case is assumed to have mass, and ought to demonstrate such properties associated with gravitational force in such system. Such properties include the return of every object been thrown up as a result of gravity acting downwards. The orbiting nature of object along an elliptical part when gravitational force isn't acting on the body and it is assumed to be zero.

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In a 100 mm diameter horizontal pipe, a venturimeter of 0.5 contraction ratio has been fitted. The head of water on the meter wh
horrorfan [7]

Answer:

the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

Explanation:

Given:

Diameter of the pipe = 100mm = 0.1m

Contraction ratio = 0.5

thus, diameter at the throat of venturimeter = 0.5×0.1m = 0.05m

The formula for discharge through a venturimeter is given as:

Q=C_d\frac{A_1A_2}{\sqrt{A_1^2-A_2^2}}\sqrt{2gh}

Where,

C_d is the coefficient of discharge = 0.97 (given)

A₁ = Area of the pipe

A₁ = \frac{\pi}{4}0.1^2 = 7.85\times 10^{-3}m^2

A₂ = Area at the throat

A₂ = \frac{\pi}{4}0.05^2 = 1.96\times 10^{-3}m^2

g = acceleration due to gravity = 9.8m/s²

Now,

The gauge pressure at throat = Absolute pressure - The atmospheric pressure

⇒The gauge pressure at throat = 2 - 10.3 = -8.3 m (Atmosphric pressure = 10.3 m of water)

Thus, the pressure difference at the throat and the pipe = 3- (-8.3) = 11.3m

Substituting the values in the discharge formula we get

Q=0.97\frac{7.85\times 10^{-3}\times 1.96\times 10^{-3}}{\sqrt{7.85\times 10^{-3}^2-1.96\times 10^{-3}^2}}\sqrt{2\times 9.8\times 11.3}

or

Q=\frac{0.97\times15.42\times 10^{-6}\times 14.88}{7.605\times 10^{-3}}

or

Q = 29.28 ×10⁻³ m³/s

Hence, the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

5 0
3 years ago
if a gas has a gage pressure of 156 kpa its absolute pressure is approximately .....A... 56 kPa ....B....100 kPa.......C......25
NeX [460]

256 kPa because p-guage + p-absolute + p-atmospheric = 256

7 0
3 years ago
The spans of time that make up the Earth's geologic timescale are generally defined by the different types of
Trava [24]
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5 0
3 years ago
A box is moving along the x-axis and its position varies in time according to the expression:
Colt1911 [192]

Answer:

38.4 m/s

Explanation:

a) at t = 3.2s. x = 6 * 3.2^2 = 61.44 m

b) at t = 3.2 + Δt. x = 6*(3.2 + \Delta t)^2

c) As Δt approaches 0. We can find the velocity by the ratio of Δx/Δt

v = \frac{\Delta x}{\Delta t} = \frac{x_2 - x_1}{\Delta t}

v = \frac{6*(3.2 + \Delta t)^2 - 61.44}{\Delta t}

v = \frac{6(3.2^2 + 6.4\Delta t + \Delta t^2) - 61.44}{\Delta t}

v = \frac{61.44 + 38.4\Delta t + \Delta t^2 - 61.44}{\Delta t}

v = \frac{\Delta t(38.4 + \Delta t)}{\Delta t}

v = 38.4 + \Delta t

As Δt approaches 0, v = 38.4 + 0 = 38.4 m/s

3 0
3 years ago
During the fission reaction shown, how did the target nucleus change ?
Zarrin [17]

Answer:

A. The target nucleus split into two nuclei, each with fewer nucleons than the original.

Explanation:

5 0
3 years ago
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