Answer:
the correct answer is C, E’= 4E
Explanation:
In this exercise you are asked to calculate the electric field at a given point
E = 
indicates that the field is E for r = 2m
E = 
(1)
the field is requested for a distance r = 1 m
E ’= k \frac{q}{r'^2}
E ’= k q / 1
from equation 1
4E = k q
we substitute
E’= 4E
so the correct answer is C
The final speed when the mass is 40,000kg height is 2.5km and 500,000N of force is 176.8 m / s
According to Newton's second law of motion,
F = m a
F = Force
m = Mass
a = Acceleration
m = 40000 kg
F = 500000 N
a = F / m
a = 500000 / 40000
a = 12.5 m / s²
a = v / t
v = d / t
v = Velocity
t = Time
d = Distance
d = 2.5 km = 2500 m
a = d / t²
12.5 = 2500 / t²
t² = 200
t = 14.14 s
v = 2500 / 14.14
v = 176.8 m / s
Therefore, the final speed is 176.8 m / s
To know more about Newton's second law of motion
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Answer: q=5.70 x 10^13 C
Explanation:
gravitational attraction = electrostatic repulsion GMm/d^2 = kQ^2/d^2 as you can see the d^2 cancel out. that is why lunar distance is irrelevant. G is the universal gravitational constant = 6.67 x 10^-11 m^3 / kgs^2 M is earth's mass = 5.972 × 10^24 kg m is moon's mass = 7.342×10^22 kg Q is charge on earth and moon. k is coulomb's constant = 9 x10^9 N m^2 /C^2 On solving equation for Q. Q = sqrt (GMm/k) = sqrt ( 6.67 x 10^-11 x 5.972 x 10^24 * 7.342×10^22 / 9 x10^9) = 5.70 x 10^13 C
Explanation:
The pressure of an object is calculated as follows: 
The area of test tube stopper is A= π
Since radius is half the diameter we have the following:
R = 
= 0.60 cm
Thus the area of the test tube becomes:
A = π 
= π 
= 1.14 * 
Now the force from the outside of the tube.
= 
= 11.55N
The net upward force on the stopper is equal to the difference of force exerted by pressure from inside of the tube and that from the outside.
= 23.1 N
