The mass would be the same
47kg on the moon as well
Answer:
<h3>62.5N</h3>
Explanation:
The pressure at one end of the piston is equal to the pressure on the second piston.
Pressure = Force/Area
F1/A1 = F2/A2
Given
F1 = 250N
A1 = 2.0m²
A2 = 0.5m²
F2 = ?
Substituting the given values in the formula;
250/2 = F2/0.5
cross multiply
250*0.5 = 2F2
125 = 2F2
F2 = 125/2
F2 = 62.5N
Hence the force needed to lift this piston if the area of the second piston is 0.5 m^2 is 62.5N
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:




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.


= 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:


= 605kg
<u>For </u>
<u>:</u>


= 864.5kg
The correct answer as the first one above !
Answer:
e.26m/s
Explanation:
Vf=Vi+at (1)
Vf=9j+(2i-4j)t
X= X₀+at
now, in the i direction
15=O+2t or t=7.5 when x position is 15
Lets put that into the (1) equation, solve for Vf.
Vf=9j+(2i-4j)7.5
Vf= 15i - 21j
Speed=
Vf= 25.8 m/s