Explanation:
If a metal rod of length L moves with velocity v is moving perpendicular to its length, in a magnetic field B, the induced emf is given by :
The electric field in the conductor is given by :
It is clear that the electric field is independent of the length of the rod. If the length of the rod is doubled, the electric field in the rod remains the same.
The horizontal range of the missile is b) 176 m
Explanation:
The motion of the missile is a projectile motion, so it consists of two independent motions:
- A uniform motion with constant velocity along the horizontal direction
- A uniformly accelerated motion with constant acceleration (equal to the acceleration of gravity) in the vertical-downward direction
To find the time of flight of the missile, we study the vertical motion. We can use the following suvat equation:
where:
s = 75 m is the vertical displacement of the missile (the height of the building)
is the initial vertical velocity (the missile is thrown horizontally)
t is the time of flight
is the acceleration of gravity
Solving for t, we find the time of flight:
This means that the missile takes 3.91 s to reach the ground.
Now we study the horizontal motion: the missile moves with a constant horizontal velocity of
Therefore, the distance covered in a time t is
and by substituting t = 3.91 s, we find the horizontal range of the missile:
Learn more about projectile motion:
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A) 11.1 cm
We can find the focal length of the lens by using the lens equation:
where
f is the focal length
p = 16.0 cm is the distance of the object from the lens
q = 36.0 cm is the distance of the image from the lens (taken with positive sign since it is on the opposide side to the image, so it is a real image)
Solving the equation for f:
B) Converging
The focal length is:
- Positive for a converging lens
- Negative for a diverging lens
In this case, the focal length is positive, so it is a converging lens.
C) 18.0 mm
The magnification equation states that:
where
is the heigth of the image
is the height of the object
Solving the formula for , we find
So the image is 18 mm high.
D) Inverted
From the magnification equation we have that:
- When the sign of is positive, the image is erect
- When the sign of is negative, the image is inverted
In this case, is negative, so the image is inverted.