Answer:
An acceleration of 5m/s^2 means that the velocity of a body is increasing by 5m/s per second in a certain direction
Explanation:
Answer:
The current in the circuit at a time interval of τ seconds after the switch has been closed is 0.123 A
Explanation:
The time constant for an R and C in series circuit is given by τ = RC.
R = 3000 ohms, C = 0.5 × 10⁻⁶ F = 5.0 × 10⁻⁷ F
τ = 3000 × 5 × 10⁻⁷ = 0.015 s
The voltage across a capacitor as it charges is given be
V(t) = Vs (1 - e⁻ᵏᵗ)
where k = 1/τ
At the point when t = τ, the expassion becomes
V(t = τ) = 1000 (1 - e⁻¹) = 0.632 × 1000 = 632 V
Current flows as a result of potential difference,.
Current in the circuit at this time t = τ is given by
I = (Vs - Vc)/R
Vs = source voltage = 1000 V
Vc = Voltage across the capacitor = 632 V
R = 3000 ohms
I = (1000 - 632)/3000 = 0.123 A
Answer:
15.7m/s
Explanation:
To solve this problem, we use the right motion equation.
Here, we have been given the height through which the ball drops;
Height of drop = 14.5m - 1.9m = 12.6m
The right motion equation is;
V² = U² + 2gh
V is the final velocity
U is the initial velocity = 0
g is the acceleration due to gravity = 9.8m/s²
h is the height
Now insert the parameters and solve;
V² = 0² + 2 x 9.8 x 12.6
V² = 246.96
V = √246.96 = 15.7m/s
Joules is a unit for work which may decomposed into N.m. Work is a quantity which is a product of force (in this case, the woman's weight) and the distance she has traveled.
W = F x d ; d = W / F
Substituting the given,
d = (3.5 x 10^4 J) / (55 kg x 9.8 m/s²) = 64.94 m
Thus, the woman can climb up to 64.94 meters.
Answer:
It continue to move forward at a constant velocity which will be slower than before the front thruster was fired.
Explanation:
Before the front thruster was fired, the spacecraft was already moving at a particular velocity.
After the malfunction, the front thruster is fired and then the force exerted by that front thruster slows the spacecraft down, as we are told.
By using the rear thruster to exert a force equal to that from the front thrusters, a force equal in magnitude to that of the front thrusters is added, cancelling out the effect of the front thrusters. Because the spacecraft is already moving at a slower speed at this point compared to the beginning, it continues to move at that speed.
