Answer:
The magnitude of the emf induced in the coil is 60 mV.
Explanation:
We have,
Side of the square coil, a = 24 cm = 0.24 m
Number of turns in the coil, N = 2
It is placed in a uniform magnetic field that makes an angle of 60 degrees with the plane of the coil. If the magnitude of this field increases by 6.0 mT every 10 ms, we need to find the magnitude of the emf induced in the coil.
We know that the induced emf is given by the rate of change of magnetic flux throughout the coil. So,
is the angle between magnetic field and the normal to area vector.
But in this case, a uniform magnetic field that makes an angle of 60 degrees with the plane of the coil. So, the angle between magnetic field and the normal to area vector is 90-60=30 degrees.
Now, induced emf becomes :
or
So, the magnitude of the emf induced in the coil is 60 mV. Hence, this is the required solution.
Heat the tea because the heat make the tea's particles move around more dissolving the sugar quicker
We want to study the impact of a sledgehammer and a wall.
Before the sledgehammer hits the wall, it has a given velocity and a given mass, so it has momentum and it has kinetic energy.
When it hits the wall, the velocity of the hammer disappears, this means that the energy is transferred to the wall, this "transfer of energy" can be thought of a force applied for a really short time on the wall, which for the third law of Newton, the force is also applied on the hammer.
This is why you feel the impact on the handle when you hit something with a hammer, this also means that some of the energy is dissipated on your arms.
Now, because the wall is made of a material usually not as strong as the head of the sledgehammer, we will see that in this interaction the wall seems more affected than the hammer, but the forces that each one experiences are exactly equal in magnitude.
If you want to learn more, you can read:
brainly.com/question/13952508
Answer:
200 , 0 , 133.33333
Explanation:
velocity = change of X / change of T
so
400/2 = 200
0/2 = 0
400/3 = 133.33333
The work done by a constant force in a rectilinear motion is given by:
where F is the magnitude of the force, d is the distance and θ is the angle between the force and the displacement vector.
In this case we have two forces then we need to add the work done by each of them; for the first force we have a magnitude of 17 N, a displacement of 12 m and and angle of 0° (since both the displacement and the force point right); for the second force we have a magnitude of 36 N, a displacement of 12 m and an angle of 30°. Plugging these values we have that the total work is:
Therefore, the total work done is 578.123 J and the answer is option E
