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mars1129 [50]
4 years ago
13

5. Each of five satellites makes a circular orbit about an object that is much more massive than any of the satellites. The mass

and orbital radius of each satellite is given below. Which satellite has the greatest speed?

Physics
1 answer:
Vikentia [17]4 years ago
4 0

The options of the question are missing. I have attached it.

Answer:

Satellite in option B will have the greatest speed.

Explanation:

From kepplers third law, we know that

V² = GM/R

Thus, v = √(GM/R)

Where;

v is velocity

G is gravitational constant

M is mass

R is radius

Looking at the options, let's start from the first one;

Option A

Here, mass = (1/2)m and radius = R

So, v = √(GM/R) thus v = √(G(m/2)/R) = √(Gm/2R)

Option B

Here, mass = m and radius = (1/2)R

Thus v = √(Gm/(R/2)) = √(2Gm/R)

Option C

Here, mass = m and radius = R

Thus v = √(Gm/R)

Option D

Here, mass = m and radius = 2R

Thus v = √(Gm/(2R))

Now, inspecting all the options, it's clear that option B will have the greatest velocity because it's numerator is the biggest and will in turn lead to higher velocity.

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Taking the resistivity of platinoid as 3.3 x 10-7 m, find the resistance of 7.0 m of platinoid wire of average diameter 0.14 cm.
mr_godi [17]

Answer:

1.5 \Omega

Explanation:

The resistance of a wire is given by the equation:

R=\rho \frac{L}{A}

where

\rho is the resistivity of the material

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A is the cross-sectional area of the wire

In this problem, we have a wire of platinoid, whose resistivity is

\rho = 3.3\cdot 10^{-7} \Omega m

The length of the wire is

L = 7.0 m

And its radius is

r=\frac{0.14 cm}{2}=0.07 cm = 7\cdot 10^{-4} m, so the cross-sectional area is

A=\pi r^2=\pi(7\cdot 10^{-4})^2=1.54\cdot 10^{-6}m^2

Solving for R, we find the resistance of the wire:

R=(3.3\cdot 10^{-7})\frac{7.0}{1.54\cdot 10^{-6}}=1.5 \Omega

3 0
3 years ago
A spacecraft on its way to Mars has small rocket engines mounted on its hull; one on its left surface and one on its back surfac
lozanna [386]

Answer:

v=545.41 \frac{m}{s}

β=-25.93

Explanation:

Give the acceleration in 'x' and 'y' also the time can find the initial velocity using equation of uniform acceleration motion

For axis x

a_{x}=5.10\frac{m}{s^{2}} \\v_{fx}=3780\frac{m}{s} \\v_{fx}=v_{ox}+a_{x}*t\\v_{ox}=v_{fx}-a_{x}*t\\v_{ox}=3780 \frac{m}{s}-5.1\frac{m}{s^{2} }*645s\\v_{ox}=490.5\frac{m}{s}

For axis y

a_{y}=7.30\frac{m}{s^{2}} \\v_{fy}=4470\frac{m}{s} \\v_{fy}=v_{oy}+a_{y}*t\\v_{oy}=v_{fy}-a_{y}*t\\v_{oy}=4470\frac{m}{s}-7.30\frac{m}{s^{2} }*645s\\v_{oy}=-238.5\frac{m}{s}

Maginuted

v=\sqrt{v_{xo}^{2} +v_{yo}^{2} } \\v=\sqrt{490.5x^{2} +(-238.5)^{2} }\\v=545.41 \frac{m}{s}

The  direction is knowing when find the angle so

\beta =tan^{-1}*\frac{v_{yo} }{v_{xo}}\\\beta =tan^{-1}*\frac{-238.5}{490.5}\\\beta =tan^{-1}*-0.48\\\beta =-25.93

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3 years ago
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Answer:

in the downward movement of the movement when the constant is lost

Explanation:

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the piston position is

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The point where you lose contact is

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