If the crate shown here is moving at a constant speed in a straight line and the force applied is 310 N, what is the magnitude o
f the force of friction?
A.310 N
B.0 N
C.620 N
A.310 N
B.0 N
C.620 N
1 answer:
Answer:
the answer is A.
Explanation:
Using the laws of newton:
∑F = ma
where ∑F is the sumatory of forces acting in the system, m the mass and a the acelertion of the system.
Then, if the block is moving with constant velocity, its aceleration is equal to 0, so:
∑F = m(0)
∑F = 0
It means that:
F - = 0
where F is the force applied and is the friction force. Replacing the value of F, we get:
310N - = 0
Finally, solving for :
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Answer:
Final kinetic energy,
Explanation:
It is given that,
Kinetic energy of toy car,
Frictional force, F = 0.5 N
Distance, d = 6 m
We need to find the kinetic energy after this frictional force has acted on it. We know that frictional force is an opposing force. The work done by the toy car is given by :
W = -3 J
We know that the work done is equal to the change in kinetic energy. Let is the final kinetic energy.
So, the final kinetic energy is the toy car is 5 J. Hence, this is the required solution.
Answer:
Magnetic field, B = 0.0196 T
Explanation:
A current of 0.6 amps flows through a solenoid 6500 turns and 25 cm long. What is the value of the magnetic field inside it?
We have,
Current, I = 0.6 A
Number of turns in the solenoid, N = 6500
Length of solenoid, l = 25 cm = 0.25 m
The magnetic field inside the solenoid is given by :
So, the value of the magnetic field inside the solenoid is 0.0196 T.
Answer:
(a) 19.62 N
(b) Box moves down the slope
(c) 24.43 N
Explanation:
(a)
2 Kg box causes tension
hence
where m is mass and g is gravitational force
T'=4*9.81 sin 35= 22.5071 N
Since T' is greater than , then the box moves down the slope
(c)
Acceleration a=
When moving, the box will exert force T"=
T"= 4*9.81 sin 35 +(4*0.48)= 24.43 N