Answer:
velocity and displacement answer
Explanation:
thanks me
The question is incomplete. Here is the complete question.
Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.
(a) What is the total mass of the three boxes?
(b) What is the mass of each box?
Answer: (a) Total mass = 2384.5kg;
(b) m1 = 915kg;
m2 = 605kg;
m3 = 864.5kg;
Explanation: The image of the boxes is described in the picture below.
(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:
![F_{pull}=m_{T}.a](https://tex.z-dn.net/?f=F_%7Bpull%7D%3Dm_%7BT%7D.a)
![m_{T}=\frac{F_{pull}}{a}](https://tex.z-dn.net/?f=m_%7BT%7D%3D%5Cfrac%7BF_%7Bpull%7D%7D%7Ba%7D)
![m_{T}=\frac{3615}{1.516}](https://tex.z-dn.net/?f=m_%7BT%7D%3D%5Cfrac%7B3615%7D%7B1.516%7D)
![m_{T}=2384.5](https://tex.z-dn.net/?f=m_%7BT%7D%3D2384.5)
Total mass of the system of boxes is 2384.5kg.
(b) For each mass, analyse each box and make them each a free-body diagram.
<u>For </u>
<u>:</u>
The only force acting On the
box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.
![m_{1} = \frac{F_{12}}{a}](https://tex.z-dn.net/?f=m_%7B1%7D%20%3D%20%5Cfrac%7BF_%7B12%7D%7D%7Ba%7D)
![m_{1} = \frac{1387}{1.516}](https://tex.z-dn.net/?f=m_%7B1%7D%20%3D%20%5Cfrac%7B1387%7D%7B1.516%7D)
= 915kg
<u>For </u>
<u>:</u>
There are two forces acting on
: tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:
![m_{2} = \frac{F_{23}-F_{12}}{a}](https://tex.z-dn.net/?f=m_%7B2%7D%20%3D%20%5Cfrac%7BF_%7B23%7D-F_%7B12%7D%7D%7Ba%7D)
![m_{2} = \frac{2304-1387}{1.516}](https://tex.z-dn.net/?f=m_%7B2%7D%20%3D%20%5Cfrac%7B2304-1387%7D%7B1.516%7D)
= 605kg
<u>For </u>
<u>:</u>
![m_{3} = m_{T} - (m_{1}+m_{2})](https://tex.z-dn.net/?f=m_%7B3%7D%20%3D%20m_%7BT%7D%20-%20%28m_%7B1%7D%2Bm_%7B2%7D%29)
![m_{3} = 2384.5-1520.0](https://tex.z-dn.net/?f=m_%7B3%7D%20%3D%202384.5-1520.0)
= 864.5kg
Answer:
translucent
Explanation:
you can see light coming thru but u cant see thru the glass.
Answer:
(A) L = 115.3kgm²/s
(B) dL/dt = 94.1kgm²/s²
Explanation:
The magnitude of the angular momentum of the rock is given by the foemula
L = mvrSinθ
We have been given θ = 36.9°, m = 2.0kg, v = 12.0m/s and r = 8.0m.
Therefore L = 2.00 × 12 × 8.0 × Sin 36.9° =
115.3 kgm²/s
(B) The magnitude of the rate of angular change in momentum is given by
dL /dt = d(mvrSinθ)/dt = mgrSinθ = 2.00 × 9.8 × 8.0× Sin36.9 = 94.1kgm²/s²
Answer:
The centripetal force acting on the car is proportional to the mass of the car.
Explanation:
Let,
The mass of the car be 'm'
The velocity of the car moving in the curved path be 'v'
The radius of the curved path be 'r'
According to physics, a body moving ion circular path experience a force directed along the radius of the path. This force is called centripetal force.
The formula for centripetal force is,
<em>F = mv²/r</em>
Where,
a = v²/r
So, if the mass of the car changes, the centripetal force also changes proportionally according to the above equation.