Answer:
q1 = q₂= -3
therefore each sphere has the same charge of -3 untis
Explanation:
The metallic spheres have mobile charge, so when the two spheres come into contact the total charge
Q_total = q₁ + q₂
Q_total = -2 -4
Q_total = -6 units
it is distributed in between the two spheres evenly since the charges of the same sign repel each other.
When the spheres separate each one has
q₁ = -6/2
q1 = q₂= -3
therefore each sphere has the same charge of -3 untis
Given that the space station is free of gravitational force, it is required that it spins an certain speed to acquire centripetal acceleration.
In this case, you want that the centripetal acceleration, Ac, equals g (gravitational acceleration on the earth), becasue this will cause a centripetal force equal to the weight on earth.
The formula for centripetal acceleration is Ac = [angular velocity]^2*R
where R = [1/2]50.0m = 25.0 m
Ac = 9.81 m/s^2
=> [angular velocity]^2 = Ac/R = 9.81m/s^2v/ 25.0m = 0.3924 (rad/s)^2
[angular velocity] = √(0.3924) rad/s = 0.63 rad/s
Answer: 0.63 rad/s
Answer:
ΔD = 2.29 10⁻⁵ m
Explanation:
This is a problem of thermal expansion, if the temperature changes are not very large we can use the relation
ΔA = 2α A ΔT
the area is
A = π r² = π D² / 4
we substitute
ΔA = 2α π D² ΔT/4
as they do not indicate the initial temperature, we assume that ΔT = 75ºC
α = 1.7 10⁻⁵ ºC⁻¹
we calculate
ΔA = 2 1.7 10⁻⁵ pi (1.8 10⁻²) ² 75/4
ΔA = 6.49 10⁻⁷ m²
by definition
ΔA = A_f- A₀
A_f = ΔA + A₀
A_f = 6.49 10⁻⁷ + π (1.8 10⁻²)² / 4
A_f = 6.49 10⁻⁷ + 2.544 10⁻⁴
A_f = 2,551 10⁻⁴ m²
the area is
A_f = π D_f² / 4
A_f =
D_f =
D_f = 1.80229 10⁻² m
the change in diameter is
ΔD = D_f - D₀
ΔD = (1.80229 - 1.8) 10⁻² m
ΔD = 0.00229 10⁻² m
ΔD = 2.29 10⁻⁵ m
Answer:

Explanation:
Given:
- mass of bullet,
- initial velocity of bullet,

- displacement of the bullet in the target,

Here as given in the question the bullet penetrates the target by the given displacement of the bullet into it. During this process it faces deceleration and hence it comes to rest.
- so, final velocity of the bullet,

Now using the equation of motion:

where:
acceleration of the bullet


<u>Now the force of resistance offered by the target in stopping it:</u>



Answer:
it depends if it is alpha ,beta or gamma
Explanation:
if alpha use paper
if beta use some clothing or aluminium
if it is gamma use lead