Answer:
m = 9795.9 kg
Explanation:
v = 35 m/s
KE = 6,000,000 J
Plug those values into the following equation:
6,000,000 J = (1/2)(35^2)m
---> m = 9795.9 kg
Answer:
The force pulling the roller along the ground is 128.55 N
Explanation:
A force of 200 N acting at an angle of 50° with the ground level
This force is pulled a garden roller
We need to find the force pulling the roller along the ground
The force that pulling the roller along the ground is the horizontal
component of the force acting
→ The force acting is 200 N at direction 50° with ground (horizontal)
→ The horizontal component = F cosФ
→ F = 200 N , Ф = 50
→ The horizontal component = 200 cos(50) = 128.55 N
128.55 N is the horizontal component of the force that pulling the
roller along the ground
<em>The force pulling the roller along the ground is 128.55 N</em>
Before going to answer this question first we have to know the fundamental principle of magnetism.
A magnet have two poles .The important characteristic of a magnet is that like poles will repel each other while unlike poles will attract each other.
Through this concept the question can be answered as explained below-
A-As per first option the side of magnet A is repelled by the south pole of magnet B. Hence the pole of a must be south .It can't be north as it will lead to attraction.
B-The side of magnet A is repelled by the north pole of magnet B. Hence the side of A must be north pole.It can't be a south pole.
C-The side of magnet A is attracted by the south pole of magnet B .Hence the side of magnet A must be north.Hence this is right
D-The side of magnet A is attracted by the north pole of magnet B. Hence the side of A must south.It can't be north as it will lead to repulsion.
Hence the option C is right.
The work done by the force is 47.1 J
Explanation:
The work done by a force in moving an object is given by
(1)
where
F is the magnitude of the force
d is the distance covered by the object
is the angle between the direction of the force and the motion of the object
In this problem, the force applied to the object is
F = 3.0 N
This force is always tangential to the track: this means that at every instant, the force is parallel to the motion of the object, so
And the distance covered is equal to the circumference of the circle, which is:
where r = 2.5 m is the radius.
Now we can substitute into eq.(1) to find the work done:
Learn more about work:
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Answer:
No, it is not proper to use an infinitely long cylinder model when finding the temperatures near the bottom or top surfaces of a cylinder.
Explanation:
A cylinder is said to be infinitely long when is of a sufficient length. Also, when the diameter of the cylinder is relatively small compared to the length, it is called infinitely long cylinder.
Cylindrical rods can also be treated as infinitely long when dealing with heat transfers at locations far from the top or bottom surfaces. However, it not proper to treat the cylinder as being infinitely long when:
* When the diameter and length are comparable (i.e have the same measurement)
When finding the temperatures near the bottom or top of a cylinder, it is NOT PROPER TO USE AN INFINITELY LONG CYLINDER because heat transfer at those locations can be two-dimensional.
Therefore, the answer to the question is NO, since it is not proper to use an infinitely long cylinder when finding temperatures near the bottom or top of a cylinder.
