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
Answer:
Newton's third law of motion states that for every action, there is equal and opposite reaction.
While space walking, when the astronaut gets detached from the space ship, she floats in space holding a wrench. In order to get back to the spaceship, she should throw the wrench in the opposite direction of the spaceship. This action would cause a reaction on her own body and she would be pushed away from the wrench and towards the spaceship. Thus, she can return back to the spaceship in this way.
Hi,
The answer is D, Gravitational.
Hope this helps.
r3t40
Answer:
ma+mgsinh0+f=F∴(25)(0.75)+(25)(10)sinh0+μkN=F∴18.75+(250)(0.6h)+μk(mgcosh0=F⟹18.75+150+μk((25)(10)(0.76))=500∴168.75+μk(190)=500⟹μk(190)=331.25⟹μk=1.74
Explanation:
Answer:
i think that the answer might be be C. one day. because it last about 4 to 6 hours