Parallel Circuits:

Which form of energy is due to the movement of electrons?

Kruka [31]

Electrical Energy : b

3 0

3 years ago

Gravity on the moon is about 1/6 th the gravity felt on the earth. This is because A) the moon is so far away from the earth. B)

ankoles [38]

So we want to know why is there a difference between the force of gravity on the Moon and the force of gravity of the Earth. So the gravitational force between two objects depends on the masses of both objects. That can be seen from Newtons universal law of gravity. F=G*m1*m2*(1/r^2). So lets say we are holding an object of mass m=1kg on a height r=1m on the Moon and we are holding the same object on the Earth also on the same height of r=1m. The Gravitational force on the Earth will be Fg=G*M*m*(r^2) where M is the mass of the Earth. The force between the moon and that object will be Fg=G*n*m*(r^2), where n is the mass of the moon. Since mass of the Moon is much smaller than mass of the Earth, The gravitational force between the Moon and that body will be almost 6 times smaller than the gravitational force between the Earth and that body. So the correct answer is B.

4 0

3 years ago

Read 2 more answers

A race car has a mass of 820 kg. It starts from rest and travels 50.0m in 3.0s. The car is uniformly accelerated during the enti

Contact [7]

A=DELTAv/DELTAt=50/3
f=ma=820.50/3

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:

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.

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

(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

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.

4 0

3 years ago

When we experience positive "g forces", it is as if we have become...

zhenek [66]

Answer:

heavier

Explanation:

7 0

3 years ago