Answer:
b. The side the boy is sitting on will tilt downward
Explanation:
Initially, the seesaw is balanced because the torque exerted by the boy is equal to the torque exerted by the girl:
where
Wb is the weight of the boy
db is the distance of the boy from the pivot
Wg is the weight of the girl
dg is the distance of the girl from the pivot
When the boy moves backward, the distance of the boy from the pivot (
increases, therefore the torques are no longer balanced: the torque exerted by the boy will be larger, and therefore the side of the boy will tilt downward.
Answer:
The relation between frequency and time period is given by:
f = 1/T
Explanation:
In a wave motion, the particle move about the mean position with the passage of time. The particles rise to reach the highest point which is crest, and similarly falls to reach the lowest point which is trough. The cycle keeps on repeating.
The time period of the wave can be defined as the time taken to complete one such cycle. Time period is given by:
T = 2π/ω
Frequency can be defined as the number of cycles completed in unit time, which can be taken as the inverse of time period. frequency is given by
f = ω/2π
or
f = 1/T
Answer:
5.565 V
Explanation:
Radius of coil of generator=r=0.14 m
Length of wire=l=10 m
Magnetic field,B=0.24 T
Angular speed,
We have to find the peak emf of the generator.
Peak(maximum) induced emf of generator=
Using the formula
Answer:
i dont know
Explanation:im sorry to do this to you but you dont have to watch ads if you answer questions
To solve the problem it is necessary to use Newton's second law and statistical equilibrium equations.
According to Newton's second law we have to
where,
m= mass
g = gravitational acceleration
For the balance to break, there must be a mass M located at the right end.
We will define the mass m as the mass of the body, located in an equidistant center of the corners equal to 4m.
In this way, applying the static equilibrium equations, we have to sum up torques at point B,
Regarding the forces we have,
Re-arrange to find M,
Therefore the maximum additional mass you could place on the right hand end of the plank and have the plank still be at rest is 16.67Kg
