Answer:
The speed of the man is 4.54 m/s.
Explanation:
Given that,
Mass of man=8100 g
Mass of stone = 79 g
Speed = 4.5 m/s
We need to calculate the speed of the man
Using momentum of conservation
![(m_{1}+m_{2})V=m_{1}v_{1}+m_{2}v_{2}](https://tex.z-dn.net/?f=%28m_%7B1%7D%2Bm_%7B2%7D%29V%3Dm_%7B1%7Dv_%7B1%7D%2Bm_%7B2%7Dv_%7B2%7D)
Where,
=mass of man
=mass of stone
=velocity of man
=velocity of stone
Put the value in the equation
As the stone is away from the man
So, the speed of stone is zero
![8179\times4.5=8100\times v+0](https://tex.z-dn.net/?f=8179%5Ctimes4.5%3D8100%5Ctimes%20v%2B0)
![v=\dfrac{8179\times4.5}{8100}](https://tex.z-dn.net/?f=v%3D%5Cdfrac%7B8179%5Ctimes4.5%7D%7B8100%7D)
![v = 4.54\ m/s](https://tex.z-dn.net/?f=v%20%3D%204.54%5C%20m%2Fs)
Hence, The speed of the man is 4.54 m/s.
See the attached figure.
The black arrows represent the two given vectors. The dashed black arrows are these same vectors, but translated so that the end of one vector is aligned with the start of another.
The red vector is their sum.
In case you also need to find the magnitude and direction of the sum, we have
A = (120 N) (cos(30°) i + sin(30°) j) = (60√3 i + 60 j) N
B = (-100 N) (cos(90°) i + sin(90°) j) = (-100 j) N
⇒ A + B = (60√3 i - 40 j) N
⇒ ||A + B|| = √((60√3)² + (-40)²) N = 20√31 N
and its direction relative to the positive horizontal axis (rightward) is θ such that
tan(θ) = (-40) / (60√3) = -2/(3√3)
⇒ θ = arctan(-2/(3√3)) ≈ -21.05°
Potential energy (Ep) is equal to the mass (m) of the subject multiplied by the height (h) from sea level that its positioned multiplied by the earth acceleration (g).
Ep = mgh
In your case m = 70 kg, g = 9.8m/s^2 , h = 3m , all the units are in SI, so simple multiplication gives Ep = 2060 J or Nm^2
Answer:
The speed of the ball relative to the ground if the ball is thrown forward is 58.1 m/s
Explanation:
Given;
speed of the ball thrown inside boxcar,
= 23.2 m/s
speed of the boxcar moving along the tracks,
= 34.9 m/s
Determine the speed of the ball relative to the ground if the ball is thrown forward.
If the ball is thrown forward, the speed of the ball relative to the ground will be sum of the ball's speed plus speed of the boxcar.
![V_{relative \ speed} = V_B + V_T\\\\V_{relative \ speed} = 23.2 + 34.9\\\\V_{relative \ speed} = 58.1 \ m/s](https://tex.z-dn.net/?f=V_%7Brelative%20%5C%20speed%7D%20%3D%20V_B%20%2B%20V_T%5C%5C%5C%5CV_%7Brelative%20%5C%20speed%7D%20%3D%2023.2%20%2B%2034.9%5C%5C%5C%5CV_%7Brelative%20%5C%20speed%7D%20%3D%2058.1%20%5C%20m%2Fs)
Therefore, the speed of the ball relative to the ground if the ball is thrown forward is 58.1 m/s.