The resistance R of a piece of wire is given by

where

is the resistivity of the material, L is the length of the wire and A is its cross-sectional area.
Using this formula, and labeling with A the aluminum and with T the tungsten wire, we can write the ratio between

(the resistance of the tungsten wire) and

(the resistance of the aluminum wire):

the two wires are identical, so L and A are the same for the two wires and simplify in the ratio, and we get:

By using the resistivity of the aluminum:

and the resistivity of the tungsten:

m we can get the resistance of the tungsten wire:
Answer:
(a) A = 0.650 m
(b) f = 1.3368 Hz
(c) E = 17.1416 J
(d) K = 11.8835 J
U = 5.2581 J
Explanation:
Given
m = 1.15 kg
x = 0.650 cos (8.40t)
(a) the amplitude,
A = 0.650 m
(b) the frequency,
if we know that
ω = 2πf = 8.40 ⇒ f = 8.40 / (2π)
⇒ f = 1.3368 Hz
(c) the total energy,
we use the formula
E = m*ω²*A² / 2
⇒ E = (1.15)(8.40)²(0.650)² / 2
⇒ E = 17.1416 J
(d) the kinetic energy and potential energy when x = 0.360 m.
We use the formulas
K = (1/2)*m*ω²*(A² - x²) (the kinetic energy)
and
U = (1/2)*m*ω²*x² (the potential energy)
then
K = (1/2)*(1.15)*(8.40)²*((0.650)² - (0.360)²)
⇒ K = 11.8835 J
U = (1/2)*(1.15)*(8.40)²*(0.360)²
⇒ U = 5.2581 J
Answer:
This refers when a spiker quickly strides towards the net before they jump in the air for an attack.
1) velocity: 98 m/s downward
Explanation:
The marble moves by uniformly accelerated motion, with constant acceleration a=g=-9.8 m/s^2 directed towards the ground. Therefore, its velocity at time t is given by:

where
is the initial velocity
is the acceleration
is the time
Substituting t = 10 s, we find:

And the negative sign means the direction of the velocity is downward.
2) Distance covered: 490 m
The distance covered in an uniformly accelerated motion can be found with the formula:

where
is the acceleration
t is the time
Substituting t=10 s, we find

And the negative sign means the displacement is below the dropping point.
Answer:
12 volts
Explanation:
4 + 4 + 4 = 12 volts all volts on parallel circuit are the same