4. No motion
5. Threre will be movement in one direction or another
Answer:
1.5 m/s/s
Explanation:
This problem can be solved using equation of motion given below
where v is the final velocity
u is the initial velocity
t is the time
given
final velocity = 9 m/s
initial velocity = 0 m/s
t = 6 seconds
lets substitute these value in v = u + at
9 = 0 + a*6
=> 9 = 6a
=> a = 9/6 = 3/2 = 1.5
Thus, acceleration is 1.5 m/s/s
Answer:
1039.23 m/s
Explanation:
Given:
Height, H = 3 x 10⁵m
acceleration of gravity on lo, a = 1.80 m/s² toward moon's surface i.e -1.80m/s² with respect to the earth
Now, at its highest point the velocity i.e the final velocity 'v' = 0m/s.
Now using the Newton's equation of motion
v² - u² =2as
where,
v = final velocity
u = initial velocity
a = acceleration
s = distance
now,
substituting the values we get
0²-u² = 2 × (-1.80) × 3 x 10⁵
or
Hence, the <u>initial velocity</u><u> with which the material is ejected is </u><u>1039.23 m/s</u>
Answer:
A) 0 for r < R B) Q/4πε₀r² for r > R
Explanation:
Here is the complete question
Part A Find the electric field inside a hollow plastic ball of radius R that has charge Q uniformly distributed on its outer surface. Give your answer as a multiple of Q/ε0.
Part B Find the electric field outside this ball. Give your answer as a multiple of Q/ε0. Express your answer in terms of some or all of the variables R, r and the constant π.
Solution
Using Gauss' law ∫E.dA = q/ε₀. Where E is the electric field, dA is the area vector and q is the charge enclosed.
A For r < R The direction of the electric field is directed radially inward and r is outward and the angle between them is 180°. So E.dA = EdAcos180 = -EdA
∫-EdA = q/ε₀
-E∫dA = q/ε₀
-E4πr² = q/ε₀ (∫dA = 4πr² since it is a sphere)
E = -q/4πr²ε₀
But for r < R q = 0. So,
E = -q/4πr²ε₀ = -0/4πr²ε₀ = 0
B For r > R The direction of the electric field is directed radially outward and r is outward and the angle between them is 0°. So E.dA = EdAcos0 = EdA
∫EdA = q/ε₀ where Q is the charge on the hollow plastic ball
E∫dA = q/ε₀
E4πr² = q/ε₀ (∫dA = 4πr² since it is a sphere)
E = q/4πr²ε₀
But for r > R q = Q. So,
E = Q/4πr²ε₀ = Q/4πε₀r²
light can travel 5,874,601,670,040 in a year around the earth