Answer:
The magnitude of the electric field be 171.76 N/C so that the electron misses the plate.
Explanation:
As data is incomplete here, so by seeing the complete question from the search the data is
vx_0=1.1 x 10^6
ax=0 As acceleration is zero in the horizontal axis so
Equation of motion in horizontal direction is given as


Now for the vertical distance
vy_o=0
than the equation of motion becomes

Now using this acceleration the value of electric field is calculated as

Here a is calculated above, m is the mass of electron while q is the charge of electron, substituting values in the equation

So the magnitude of the electric field be 171.76 N/C so that the electron misses the plate.


Now


- Lower mass=Higher acceleration
- Lower Force=Lower Acceleration
Option B has lowest mass and highest force hence its correct
Answer:
Acceleration of the crate is 0.362 m/s^2.
Explanation:
Given:
Mass of the box, m = 40 kg
Applied force, F = 15 N
Angle at which the force is applied,
= 15°
We have to find the magnitude of the acceleration.
Let the acceleration be "a".
FBD is attached with where we can see the horizontal and vertical component of force.
⇒
and ⇒ 
⇒
⇒ 
⇒ Applying concept of forces.
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⇒ 
⇒
<em> ...Newtons second law Fnet = ma</em>
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⇒ Plugging the values.
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<em>...f is the friction which is zero here.</em>
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⇒ 
Magnitude of the acceleration of the crate is 0.362 m/s^2.
DOES GOD EXIST? That is one of the best.
Answer:
During charging by conduction, both objects acquire the same type of charge. If a negative object is used to charge a neutral object, then both objects become charged negatively. ... In this case, electrons are transferred from the neutral object to the positively charged rod and the sphere becomes charged positively.