Answer:
The speed of the car when load is dropped in it is 10.90 m/s.
Explanation:
Given that,
Mass of the railroad car, m₁ = 11600kg
Mass of the load, m₂ = 5420kg
It can be assumed as the speed of the car, u₁ = 16 m/s
Initially, it is at rest, u₂ = 0
The additional load is dropped onto the car.
Let v is the speed of the car. It can be calculated using the conservation of momentum as :
![m_1u_1+m_2u_2=(m_1+m_2)vv=\dfrac{m_1u_1}{m_1+m_2}v=\dfrac{11600\times 16}{11600+5420}v = 10.90 m/s](https://tex.z-dn.net/?f=m_1u_1%2Bm_2u_2%3D%28m_1%2Bm_2%29vv%3D%5Cdfrac%7Bm_1u_1%7D%7Bm_1%2Bm_2%7Dv%3D%5Cdfrac%7B11600%5Ctimes%2016%7D%7B11600%2B5420%7Dv%20%3D%2010.90%20m%2Fs)
So, the speed of the car when load is dropped in it is 10.90 m/s.
Centripetal acceleration is the motion inwards towards the center of a circle. It is given by the square of velocity, divided by the radius of the circular path.
ac = v²/r, where ac is the centripetal acceleration in m/s²
Therefore,
Centripetal acceleration = 3²/20m
= 9/20m
= 0.45 m/s²
Thus, the centripetal acceleration is 0.45 m/s²
Answer:
![\mu_s=0.47](https://tex.z-dn.net/?f=%5Cmu_s%3D0.47)
Explanation:
Just before the box starts moving, according to Newton's first law, we have:
![\sum F_x:F_f-W_x=0(1)\\\sum F_y:N-W_y=0(2)](https://tex.z-dn.net/?f=%5Csum%20F_x%3AF_f-W_x%3D0%281%29%5C%5C%5Csum%20F_y%3AN-W_y%3D0%282%29)
The sine of the angle (25°) of a right triangle is defined as the opposite cathetus (
) to that angle divided into the hypotenuse (W):
![sin25^\circ=\frac{W_x}{W}\\W_x=Wsin25^\circ\\W_x=mgsin25^\circ(3)](https://tex.z-dn.net/?f=sin25%5E%5Ccirc%3D%5Cfrac%7BW_x%7D%7BW%7D%5C%5CW_x%3DWsin25%5E%5Ccirc%5C%5CW_x%3Dmgsin25%5E%5Ccirc%283%29)
The cosine of the angle (25°) of a right triangle is defined as the adjacent cathetus (
) to that angle divided into the hypotenuse (W):
![cos25^\circ=\frac{W_y}{W}\\W_y=Wcos25^\circ\\W_y=mgcos25^\circ(4)](https://tex.z-dn.net/?f=cos25%5E%5Ccirc%3D%5Cfrac%7BW_y%7D%7BW%7D%5C%5CW_y%3DWcos25%5E%5Ccirc%5C%5CW_y%3Dmgcos25%5E%5Ccirc%284%29)
Recall that the maximum frictional force is defined as:
![F_f=\mu_s N(5)](https://tex.z-dn.net/?f=F_f%3D%5Cmu_s%20N%285%29)
Replacing (3) in (1) and (4) in (2):
![F_f=mgsin25^\circ(6)\\N=mgcos25^\circ(7)](https://tex.z-dn.net/?f=F_f%3Dmgsin25%5E%5Ccirc%286%29%5C%5CN%3Dmgcos25%5E%5Ccirc%287%29)
Replacing (7) and (6) in (5) and solvinf for
:
![\mu_s=\frac{mgsin25^\circ}{mgcos25^\circ}\\\mu_s=\frac{sin25^\circ}{cos25^\circ}\\\mu_s=0.47](https://tex.z-dn.net/?f=%5Cmu_s%3D%5Cfrac%7Bmgsin25%5E%5Ccirc%7D%7Bmgcos25%5E%5Ccirc%7D%5C%5C%5Cmu_s%3D%5Cfrac%7Bsin25%5E%5Ccirc%7D%7Bcos25%5E%5Ccirc%7D%5C%5C%5Cmu_s%3D0.47)
Answer:
C. 960.8°C
Explanation:
When freezing, silver changes state from liquid to solid. Also, when melting, silver changes state from solid to liquid. Therefore, the freezing point and melting point share the same temperature. In this case, 960.8°C
The force experienced by the foot is 1000 N
Explanation:
We can solve the question by applying Newton's third law, which states that:
<em>"When an object A exerts a force (called action) on an object B, then object B exerts an equal and opposite force (called reaction) on object A"
</em>
In this situation, we can identify the foot as object A, and the ball as object B.
We are told that the ball experiences a 1000 N force, so the foot exerts a force of 1000 N on the ball (action). As a consequence of Newton's third law, therefore, the ball also exerts an equal and opposite force of 1000 N (reaction) on the foot.
It is important to remember that action and reaction do not act on the same object, so they never appear at the same time in the same free-body diagram (which shows only the forces acting on one object).
Learn more about Newton's third law:
brainly.com/question/11411375
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