Frank gives a box of books a push to the right, which makes it slide across his
1 answer:
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Answer:
Explanation:
For sound waves we have v=d/t where v is the speed of sound and d the distance between the astronauts, while for electromagnetic waves we have c=D/t where c is the speed of light and D the distance between the spaceship and Earth. <em>We have written both times as the same</em> because is what is imposed by the problem, so we have t=d/v=D/c, which means:
And for our values:
Answer:
the length of the simple pendulum is 0.25 m.
Explanation:
Given;
mass of the air-track glider, m = 0.25 kg
spring constant, k = 9.75 N/m
let the length of the simple pendulum = L
let the frequency of the air-track glider which is equal to frequency of simple pendulum = F
The oscillation frequency of air-track glider is calculated as;
The frequency of the simple pendulum is given as;
Thus, the length of the simple pendulum is 0.25 m.
The electric field of a very large (essentially infinitely large) plane of charge is given by:
E = σ/(2ε₀)
E is the electric field, σ is the surface charge density, and ε₀ is the electric constant.
To determine σ:
σ = Q/A
Where Q is the total charge of the sheet and A is the sheet's area. The sheet is a square with a side length d, so A = d²:
σ = Q/d²
Make this substitution in the equation for E:
E = Q/(2ε₀d²)
We see that E is inversely proportional to the square of d:
E ∝ 1/d²
The electric field at P has some magnitude E. Now we double the side length of the sheet while keeping the same amount of charge Q distributed over the sheet. By the relationship of E with d, the electric field at P must now have a quarter of its original magnitude: