5.6 g/ml. That is the density.
Answer:
Explanation:
Given that,
B(t) = B0 cos(ωt) • k
Radius r = a
Inner radius r' = a/2 and resistance R.
Current in the loop as a function of time I(t) =?
Magnetic flux is given as
Φ = BA
And the Area is given as
A = πr², where r = a/2
A = πa²/4
Then,
Φ = ¼ Bπa²
Φ(t) = ¼πa²Bo•Cos(ωt)
Then, the EMF is given as
ε(t) = -dΦ/dt
ε(t) = -¼πa²Bo • -ωSin(ωt)
ε(t) = ¼ωπa²Bo•Sin(ωt)
From ohms law,
ε = iR
Then, i = ε/R
I(t) = ¼ωπa²Bo•Sin(ωt) /R
This is the current induced in the loop.
Check attachment for better understanding
Answer:
<u>We are given: </u>
initial velocity (u) = 0 m/s
final velocity (v) = 10 m/s
displacement (s) = 20 m
acceleration (a) = a m/s/s
<u>Solving for 'a'</u>
From the third equation of motion:
v² - u² = 2as
replacing the variables
(10)² - (0)² = 2(a)(20)
100 = 40a
a = 100 / 40
a = 2.5 m/s²
Answer:
5.01 J
Explanation:
Info given:
mass (m) = 0.0780kg
height (h) = 5.36m
velocity (v) = 4.84 m/s
gravity (g) = 9.81m/s^2
1. First, solve for Kinetic energy (KE)
KE = 1/2mv^2
1/2(0.0780kg)(4.84m/s)^2 = 0.91 J
so KE = 0.91 J
2. Next, solve for Potential energy (PE)
PE = mgh
(0.0780kg)(9.81m/s^2)(5.36m) = 4.10 J
so PE = 4.10 J
3. Mechanical Energy , E = KE + PE
Plug in values for KE and PE
KE + PE = 0.91J + 4.10 J = 5.01 J
There is not enough information given to answer with. The force of gravity at the planet's surface depends on the planet's radius as well as its mass. The planet could have exactly the same mass as Earth has. But if it's radius is only 71% of Earth's radius, then gravity on its surface will be twice as strong as gravity on Earth.
