Answer:
They are the same (assuming there is no air friction)
Explanation:
Take a look at the picture.
When the first ball (the one thrown upward) gets to the point marked as A, the speed will has the exact same value V but the velocity will now point downward (just like the second ball).
So if you think about it, the first ball, from point A to the ground, will behave exactly like the second ball (same initial speed, same height).
That is why the speeds will be the same when they reach the ground.
F = k(Q₁Q₂)/r²
where k is Coulomb's constant 8.99e9
F = 8.99e9(2.8e-6)²/0.5² = 0.2819N
Answer:
speed of the proton is 6.286 ×
m/s
Explanation:
given data
charge q= −60.0 nC
inner radius a = 20.0 cm
outer radius b = 24.0 cm
charge density ρ = −2.05 µC/m³
to find out
What is the speed of the proton
solution
we know that force on the proton due to this electric field is express as
F = q × E ...................1
here F is force and q is charge and E is electric filed so
if v be the speed of the proton in circular orbit than force will be
F = 
....................2
from equation 1 and 2
q × E = .......................3
so
here total charge Q on shell is
Q = ρ × V
here ρ is density and V is volume
Q = 
put here value
Q = 
Q = 50.01 × 
C
and
total charge enclosed by Gaussian surface is
qin = q + Q
qin = −60 × C - 50.01 × C
qin = - 110.01 × C
and
from Gauss law
E ×4×π×b² = 
E = ![\frac{110.01*10^{-9} }{\epsilon *4 *\pi *b^2[tex]}](https://tex.z-dn.net/?f=%5Cfrac%7B110.01%2A10%5E%7B-9%7D%20%7D%7B%5Cepsilon%20%2A4%20%2A%5Cpi%20%2Ab%5E2%5Btex%5D%7D)
E = 
E = 17189.06 N/C
so
from equation 3
q × E =
v = 
v = 
v = 6.286 × m/s
so speed of the proton is 6.286 × m/s
Answer:
The car would travel after applying brakes is, d = 14.53 m
Explanation:
Given that,
The time taken to apply brakes fully is, t = 0.5 s
The velocity of the car, v = 29.06 m/s
The distance traveled by the car in 0.5 s, d = ?
The relation between the velocity, displacement, and time is given by the formula
d = v x t m
Substituting the values in the above equation,
d = 29.06 m/s x 0.5 s
= 14.53 m
Therefore, the car would travel after applying brakes is, d = 14.53 m
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