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Aleks04 [339]
3 years ago
5

What are the magnitude and direction of the acceleration of an electron at a point where the electric field has magnitude 6100 n

/c and is directed due north?
Physics
1 answer:
Hoochie [10]3 years ago
4 0

Force on electron due to electric field is given by

F = eE

F = 1.6 * 10^{-19}* 6100

F = 9.76 * 10^{-16} N

now the acceleration is given by

a = \frac{F}{m}

a = \frac{9.76 * 10^{-16}}{9.1 * 10^{-31}}

a = 1.07 * 10^{15} m/s^2

so above is the magnitude of acceleration and its direction is opposite to field as electron is negatively charged so direction is towards SOUTH

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1. Do you think that hand signals are important in officiating games? why?​
alex41 [277]

Yeah, hand signals are important in officiating games. The hand signals are given by the referee who first signals the fault and then indicates which team has won the point. A point is indicated by one finger at the side of the court to indicate the winner of the rally.

5 0
3 years ago
A student throws a 130 g snowball at 6.5 m/s at the side of the schoolhouse, where it hits and sticks. What is the magnitude of
Alex73 [517]

Answer:

4.7 N

Explanation:

130 g = 0.13 kg

The momentum of the snowball when it's thrown at the wall is

p = mv = 0.13*6.5 = 0.845 kgm/s

Which is also the impulse. From here we can calculate the magnitude of the average force F knowing the duration of the collision is 0.18 s

p = F\Delta t

F*0.18 = 0.845

F = 0.845 / 0.18 = 4.7 N

8 0
3 years ago
Based on the data given, in what direction will the car accelerate?
skad [1K]
Vertical forces:
There is a force of 579N acting upward, and a force of 579N
acting downward.
The vertical forces are balanced ... they add up to zero ...
so there's no vertical acceleration. 
Not up, not down.

Horizontal forces:
There is a force of 487N acting to the left, and a force of 632N
acting to the right.
The net horizontal force is

        (487-left + 632-right)  -  (632-right - 487-right) =  145N to the right.

The net force on the car is all to the right.
The car accelerates to the right.
7 0
3 years ago
reasons. 5. Why is the unit of temperature called a fundamental unit? Give reasons. ring derived unit.​
Furkat [3]

Explanation:

It doesn't depends upon other.

It have it's own identity.

It's a lot easier to measure temperature than to measure the motion of component particles.

8 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.

4 0
3 years ago
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