Answer:
50000 N
Explanation:
From the question given above, the following data were obtained:
Mass (m) of bullet = 0.050 kg
velocity (v) = 400 m/s
Distance (s) = 0.080 m
Force (F) =?
Next, we shall determine the acceleration of the bullet. This can be obtained as follow:
Initial velocity (u) = 0 m/s
Final velocity (v) = 400 m/s
Distance (s) = 0.080 m
Acceleration (a) =?
v² = u² + 2as
400² = 0 + (2 × a × 0.08)
160000 = 0 + 0.16a
160000 = 0.16a
Divide both side by 0.16
a = 160000 / 0.16
a = 1×10⁶ m/s²
Finally, we shall determine the force exerted by the bullet on the target. This can be obtained as follow:
Mass (m) of bullet = 0.050 kg
Acceleration (a) of bullet = 1×10⁶ m/s²
Force (F) =?
F = ma
F = 0.050 × 1×10⁶
F = 50000 N
Thus, the bullet exerted a force of 50000 N on the target.
Answer:
Amplitude, A = 0.049 meters
Explanation:
Given that,
A harmonic wave travels in the positive x direction at 6 m/s along a taught string. A fixed point on the string oscillates as a function of time according to the equation :
.......(1)
The general equation of a wave is given by :
.......(2)
A is amplitude of wave
On comparing equation (1) and (2) we get :
A = 0.049 meters
So, the amplitude of the wave is 0.049 meters.
Answer:
the average velocity of car A between t1 and t2greater is greater than the average velocity of B berween t1 and t2
Explanation:
Velocity is displacement over time,
Displacement is the distance covered relative to the initial starting position
For A:
at time ti, A moved from Xo to 2Xo, displacement is 2Xo.
at time t2 a moves with speed 3V, hence, his new position will be 3Xo from 2Xo which will be at 5Xo. A's displacement is 5Xo from starting point.
For B:
at time ti, B moved from Xo to 2Xo, displacement is 2Xo.
at time t2 a moves with speed V in the opposite position so he'll be back to his starting point, hence, his new position will be at Xo. A's displacement is 0 from his starting point.
