A 2.8-kg physics cart is moving forward with a speed of 45 cm/s. A 1.9-kg brick is dropped from rest and lands on the cart. The
1 answer:
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Answer:
The magnitude of the electric field is 5.75 N/C towards positive x- axis.
Explanation:
Given that,
Point charge at origin = 2 nC
Second charge = 5 nC
Distance at x axis = 8 m
We need to calculate the electric field at the point x = 2 m
Using formula of electric field
Put the value into the formula
The direction is toward positive x- axis.
Hence, The magnitude of the electric field is 5.75 N/C towards positive x- axis.
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.
