Answer:
e = 30 V
Explanation:
given,
N = 30 turns
Area = 0.25 m²
angular speed = ω = 100 rad/s
Magnetic field = 0.04 T
maximum induced emf in the loop = ?
e = N B A ω
e = 30 x 0.04 x 0.25 x 100
e = 30 V
hence, the maximum emf induced in the loop is equal to e = 30 V
Hi,
A material that is malleable and conducts electricity is most likely a metal.
Hope this helps!
Well, there's a lot of friction going on there, so the snowball gradually
loses kinetic energy just from bouncing and plowing through the snow
on the ground.
But I don't think you're asking about that. I think you're ignoring that
for the moment, and asking how its kinetic energy changes as its
mass increases. We know that
Kinetic Energy = (1/2) (mass) (speed²)
and THAT seems to say that more mass means more kinetic energy.
So maybe the snowball's kinetic energy increases as it picks up
more mass.
Don't you believe it !
Remember: Energy always has to come from somewhere ... a motor,
a jet, a push, gravity ... something ! It doesn't just appear out of thin air.
If the snowball were rolling down hill, then it could get more kinetic energy
from gravity. But if it's rolling on level ground, then it can never have any
more kinetic energy than you gave it when you pushed it and let it go.
If snow or leaves stick to it and its mass increases, then its speed must
decrease, in order to keep the same kinetic energy.
Answer:
a= g = - 9.81 m/s2.
The following equations will be helpful:
a = (vf - vo)/t d = vot + 1/2 at2 vf2 = vo2 + 2ad
When you substitute the specific acceleration due to gravity (g), the equations are as follows:
g = (vf - vo)/t d = vot + 1/2 gt2 vf2 = vo2 + 2gd
If the object is dropped from rest, the initial velocity ("vi") is zero. This further simplifies the equations to these:
g = vf /t d = 1/2 gt2 vf2 = 2gd
The sign convention that we will use for direction is this: "down" is the negative direction. If you are given a velocity such as -5.0 m/s, we will assume that the direction of the velocity vector is down. Also if you are told that an object falls with a velocity of 5.0 m/s, you would substitute -5.0 m/s in your equations. The sign convention would also apply to the acceleration due to gravity as shown above. The direction of the acceleration vector is down (-9.81 m/s2) because the gravitational force causing the acceleration is directed downward.
hope this info helps you out!
Answer:
v_oy = 16.33 m/s
Explanation:
To find the vertical velocity of the tiger, you use the information about the horizontal velocity and maximum horizontal distance traveled.
You use the following formula for the range of the trajectory:
( 1 )
v_ox: horizontal initial velocity = 4.5m/s
v_oy: vertical initial velocity = ?
g: gravitational acceleration = 9.8m/s^2
x_max: range of the trajectory = 15 m
You do v_oy the subject of the formula ( 1 ) and you replace the values of the other parameters in order to calculate v_oy:
hence, the initial vertical velocity of the tiger is 16.33m/s
