Few moons
I just took the test and got it right
Angular velocity = (75x2pie)/60
=2.5pie ras^-1
linear velocity(or speed) at end of string, v = radius x angular velocity
v= 0.5 x 2.5pie
v=3.93 ms^-1
tension of string (I beleve is centeral force aplied by string), F= (mv^2)/r
F= (0.2 x 3.93^2)/0.5
F=6.18 N
(sorry if wrong)
The gravitational force between the Earth and the satellite (its "weight") is inversely proportional to the distance between the centers of both objects.
On the surface, their centers are separated by 1 Earth radius.
12,000 miles above the surface, they're separated by 4 Earth radiii.
(4/1) = 4
So after the move, the satellite's weight is (1/4²) = 1/16 of its surface weight.
(321 lb) / (16) = (20 and a hair) lb
The correct choice from the given list is " <em>>20 lb "</em> .
Answer:
2090 J
Explanation:
the work done to move the cart is equal to the product between the force applied and the distance traveled:
In this case, the force applied is F=209 N, while the distance covered is d=10 m, therefore the work done is
Answer:
1.925 μC
Explanation:
Charge: This can be defined as the product of the capacitance of a capacitor and the voltage. The S.I unit of charge is Coulombs (C)
The formula for the charge stored in a capacitor is given as,
Q = CV ................... Equation 1
Where Q = charge, C = Capacitor, V = Voltage.
Note: 1 μF = 10⁻⁶ F
Given: C = 0.55 μF = 0.55×10⁻⁶ F, V = 3.5 V.
Substitute into equation 1
Q = 0.55×10⁻⁶×3.5
Q = 1.925×10⁻⁶ C.
Q = 1.925 μC
Hence the charge on the plate = 1.925 μC
