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Eddi Din [679]
3 years ago
10

Use the right-hand rule for magnetic force to determine the charge on the moving particle. This is a charge.

Physics
2 answers:
swat323 years ago
6 0

Answer:

The answer is negative

Explanation:

german3 years ago
4 0

Answer:

Use the right-hand rule for magnetic force to determine the charge on the moving particle.

This is a  

negative

charge

Explanation:

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The distance and displacement of a object in motion can be the same (true or false)
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Aswer:

False, the values ​​of the distance traveled and the displacement only coincide when the trayectorie is a straight line. Otherwise, the distance will always be greater than the offset.

Although these terms are used synonymously in other cases, they are totally different. Since the distance that a mobile travels is the equivalent of the length of its trajectory. Whereas, the displacement will be a vector magnitude.

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You measure the velocity of a drag racer that accelerates with constant acceleration. You want to plot the data and determine th
Ivenika [448]

Answer:

a) Linear equation

Explanation:

Definition of acceleration

a=\frac{dv}{dt}\\

if a=constant and we integrate the last equation

v(t)=v_{o}+a*t

So the relation between the time and the velocity is linear. If we plot the velocity in function of time, the plot is a line, and the acceleration is the slope of this line.

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3 years ago
What type of heat transfer occurs in conduction
Veseljchak [2.6K]

Answer:

when two object at different temperatures are in contact with each other.

Explanation:

6 0
3 years ago
the record for a ski jump is 180 m set in 1989. Assume the jumper comes off the end of the ski jump horizontally and falls 90 m
Mashutka [201]
Let's write the equations of motion on both x- (horizontal) and y- (vertical) axis. On the x-axis, it's a uniform motion with constant velocity vx. On the y-axis, it is a uniformly accelerated motion with initial height h=90 m and acceleration of g=9.81 m/s^2 pointing down (so with a negative sign):
S_x(t)=v_x t
S_y(t)=h- \frac{1}{2} gt^2

First, let's find the time at which the jumper reaches the ground. This happens when Sy(t)=0:
0=h- \frac{1}{2} g t^2
and so 
t= \sqrt{ \frac{2h}{g} }= \sqrt{ \frac{2\cdot 90m}{9.81 m/s^2} }=4.28 s

Then, we can find the horizontal speed. In fact, we know that at the time t=4.28 s, when the jumper reached the ground, he covered exactly 180 m, so Sx=180 m. Using this into the law of motion in x, we find
v_x= \frac{S_x}{t} = \frac{180 m}{4.28 s} =42 m/s

8 0
3 years ago
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