Answer:
<em>The answer is B</em>
Explanation:
<em>I got this from study island</em>
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:
Ratio of magnetic field will be 
Explanation:
We have given radius of the loop r = 30 mm = 0.03 m
We know that magnetic field at the center of the loop is given by
---------eqn 1
Number of turns in the solenoid is given as n = 3 turn per mm = 3000 turn per meter
We know that magnetic field due to solenoid is given by
-------------eqn 2
Now dividing eqn 1 by eqn 2

By definition we have that the final speed is:
Vf² = Vo² + 2 * a * d
Where,
Vo: Final speed
a: acceleration
d: distance.
We cleared this expression the acceleration:
a = (Vf²-Vo²) / (2 * d)
Substituting the values:
a = ((0) ^ 2- (60) ^ 2) / ((2) * (123) * (1/5280))
a = -77268 mi / h ^ 2
its stopping distance on a roadway sloping downward at an angle of 17.0 ° is:
First you must make a free body diagram and see the acceleration of the car:
g = 32.2 feet / sec ^ 2
a = -77268 (mi / h ^ 2) * (5280/1) (feet / mi) * (1/3600) ^ 2 (h / s) ^ 2
a = -31.48 feet / sec ^ 2
A = a + g * sin (θ) = -31.48 + 32.2 * sin17.0
A = -22.07 feet / sec ^ 2
Clearing the braking distance:
Vf² = Vo² + 2 * a * d
d = (Vf²-Vo²) / (2 * a)
Substituting the values:
d = ((0) ^ 2- (60 * (5280/3600)) ^ 2) / (2 * (- 22.07))
d = 175.44 feet
answer:
its stopping distance on a roadway sloping downward at an angle of 17.0 ° is 175.44 feet