Answer:
3×10⁹ W
Explanation:
Power = work / time
Power = force × distance / time
Power = force × velocity
P = Fv
P = (10⁶ N) (3000 m/s)
P = 3×10⁹ W
Answer:
b
Explanation:
Given:
- The ball is fired at a upward initial speed v_yi = 2*v
- The ball in first experiment was fired at upward initial speed v_yi = v
- The ball in first experiment was as at position behind cart = x_1
Find:
How far behind the cart will the ball land, compared to the distance in the original experiment?
Solution:
- Assuming the ball fired follows a projectile path. We will calculate the time it takes for the ball to reach maximum height y. Using first equation of motion:
v_yf = v_yi + a*t
Where, a = -9.81 m/s^2 acceleration due to gravity
v_y,f = 0 m/s max height for both cases:
For experiment 1 case:
0 = v - 9.81*t_1
t_1 = v / 9.81
For experiment 2 case:
0 = 2*v - 9.81*t_2
t_2 = 2*v / 9.81
The total time for the journey is twice that of t for both cases:
For experiment 1 case:
T_1 = 2*t_1
T_1 = 2*v / 9.81
For experiment 2 case:
T_2 = 2*t_2
T_2 = 4*v / 9.81
- Now use 2nd equation of motion in horizontal direction for both cases:
x = v_xi*T
For experiment 1 case:
x_1 = v_x1*T_1
x_1 = v_x1*2*v / 9.81
For experiment 2 case:
x_2 = v_x2*T_2
x_2 = v_x2*4*v / 9.81
- Now the x component of the velocity for each case depends on the horizontal speed of the cart just before launching the ball. Using conservation of momentum we see that both v_x2 = v_x1 after launch. Since the masses of both ball and cart remains the same.
- Hence; take ratio of two distances x_1 and x_2:
x_2 / x_2 = v_x2*4*v / 9.81 * 9.81 / v_x1*2*v
Simplify:
x_1 / x_2 = 2
- Hence, the amount of distance traveled behind the cart in experiment 2 would be twice that of that in experiment 1.
Equilibrium implies that the net force on the body is 0, therefore the body undergoes no acceleration. No acceleration means the velocity doesn't change.
The body's velocity stays constant over time.
Answer:
11.6 g/m³
Explanation:
The complete question is,
The device shows the relative humidity at 22 degrees celsius. What's the water vapor density if the maximum water vapor in air at this temperature is 20 gram/cubic meter? a device showing that at 22 degrees celsius the relative humidity is 58%.
Solution:
saturation of water in air is showed by the relative humidity. As it is 58% and not 100%. Lets scale density that we have, if humidity is 100%.
density is 20 gram/m³ when the humidity is 100%
When humidity is 58%
0.58 x 20 = 11.6 g/m³
Therefore, density is 11.6g/m³
It will most likely break. The frame is 5 newton's and the string can only hold 2.5 newton's.
