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gregori [183]
3 years ago
5

State the law of universal gravitation, and use examples to explain how changes in mass and changes in distance affect gravitati

onal force.
Physics
1 answer:
juin [17]3 years ago
6 0
F(g)= Gm1m2/ r^2 If mass is increased, so will the force of gravity because it is in direct relationship with the gravitational force, but if distance is increased, the force of gravity will decrease because it is indirectly related ( since it is on the bottom of the equation)
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requires more power

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Can enter through any cavity lined with mucus membranes
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mucus

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A single loop of nickel wire, lying flat in a plane, has an area of 7.40 cm^2 and a resistance of 2.40 Ω. A uniform magnetic fie
ale4655 [162]

Explanation:

It is given that,

Area of nickel wire, A=7.4\ cm^2=7.4\times 10^{-4}\ m^2

Resistance of the wire, R = 2.4 ohms

Initial value of magnetic field, B_1=0.5\ T

Final magnetic field, B_2=3\ T

Time, t = 1.12 s

Let I is the induced current in the loop of wire over this time. Te emf induced in the wire is given by Faraday's law as :

\epsilon=-\dfrac{d\phi}{dt}

\epsilon=-\dfrac{d(BA)}{dt}

\epsilon=-A\dfrac{d(B)}{dt}

\epsilon=-A\dfrac{B_2-B_1}{t}

\epsilon=-7.4\times 10^{-4}\times \dfrac{3-0.5}{1.12}

\epsilon=1.65\times 10^{-3}\ V

Induced current in the loop of wire is given by :

I=\dfrac{\epsilon}{R}

I=\dfrac{1.65\times 10^{-3}}{2.4}

I=6.87\times 10^{-4}\ A

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7 0
3 years ago
A satellite in outer space is moving at a constant velocity of 21.4 m/s in the y direction when one of its onboard thruster turn
kumpel [21]

Answer:

a) The magnitude of the satellite's velocity when the thruster turns off is approximately 24.177 meters per second.

b) The direction of the satellite's velocity when the thruster turns off is approximately 62.266º.

Explanation:

Statement is incomplete. The complete description is now described below:

<em>A satellite in outer space is moving at a constant velocity of 21.4 m/s in the y direction when one of its onboard thruster turns on, causing an acceleration of 0.250 m/s2 in the x direction. The acceleration lasts for 45.0 s, at which point the thruster turns off. </em>

<em>(a) What is the magnitude of the satellite's velocity when the thruster turns off</em>

<em>(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis. ° counterclockwise from the +x-axis</em>

Let be x and y-directions orthogonal to each other and the satellite is accelerated uniformly from rest in the +x direction and moves at constant velocity in the +y direction. The velocity vector of the satellite (\vec{v}_{S}), measured in meters per second, is:

\vec{v}_{S} = (v_{o,x}+a_{x}\cdot t)\,\hat{i}+v_{y}\,\hat{j}

Where:

v_{o,x} - Initial velocity in +x direction, measured in meters per second.

a_{x} - Acceleration in +x direction, measured in meter per square second.

t - Time, measured in seconds.

v_{y} - Velocity in +y direction, measured in meters per second.

If we know that v_{o,x} = 0\,\frac{m}{s}, a_{x} = 0.250\,\frac{m}{s^{2}}, t = 45\,s and v_{y} = 21.4\,\frac{m}{s}, the final velocity of the satellite is:

\vec{v}_{S} = \left[0\,\frac{m}{s}+\left(0.250\,\frac{m}{s^{2}} \right)\cdot (45\,s) \right]\,\hat{i}+\left(21.4\,\frac{m}{s} \right)\,\hat{j}

\vec{v_{S}} = 11.25\,\hat{i}+21.4\,\hat{j}\,\,\left[\frac{m}{s} \right]

a) The magnitud of the satellite's velocity can be found by the resource of the Pythagorean Theorem:

\|\vec {v}_{S}\| = \sqrt{\left(11.25\,\frac{m}{s} \right)^{2}+\left(21.4\,\frac{m}{s} \right)^{2}}

\|\vec{v}_{S}\| \approx 24.177\,\frac{m}{s}

The magnitude of the satellite's velocity when the thruster turns off is approximately 24.177 meters per second.

b) The direction of the satellite's velocity when the thruster turns off is determined with the help of trigonometric functions:

\tan \alpha = \frac{v_{y}}{v_{x}} = \frac{21.4\,\frac{m}{s} }{11.25\,\frac{m}{s} }

\tan \alpha = 1.902

\alpha = \tan^{-1}1.902

\alpha \approx 62.266^{\circ}

The direction of the satellite's velocity when the thruster turns off is approximately 62.266º.

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3 years ago
To avoid using the variable "a"
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Answer:

option c .

Explanation:

..............................

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