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zheka24 [161]
3 years ago
5

The intensity at a distance of 6.0 m from a source that is radiating equally in all directions is 6.0 × 10-10 w/m2 . what is the

power emitted by the source?
Physics
1 answer:
satela [25.4K]3 years ago
3 0
The intensity is defined as the ratio between the power emitted by the source and the area through which the power is calculated:
I= \frac{P}{A} (1)
where
P is the power
A is the area

In our problem, the intensity is I=6.0 \cdot 10^{-10} W/m^2. At a distance of r=6.0 m from the source, the area intercepted by the radiation (which propagates in all directions) is equal to the area of a sphere of radius r, so:
A=4 \pi r^2 = 4 \pi (6.0 m)^2 = 452.2 m^2

And so if we re-arrange (1) we find the power emitted by the source:
P=IA = (6.0 \cdot 10^{-10}W/m^2)(452.2 m^2)=2.7 \cdot 10^{-7} W
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A swift blow with the hand can break a pine board. As the hand hits the board, the kinetic energy of the hand is transformed int
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Answer:

A. The hand must move with a velocity of 6.98 m/s to break the board.

B. Average force on the hand = 1025 N

Explanation:

A.To determine the speed the hand must move with to break the board, the force workdone in breaking the board is found first.

Workdone = force × distance

Minimum force required = 870 N;

Distance moved by board/Deflection in order to break = 1.4 cm = 0.014 M

Workd done = 870 N × 0.014 m = 12.18 Nm or 12.18 J

This work done = Kinetic energy of the hand

Kinetic energy = mv²/2 ; where m is mass and v is velocity

Mass of hand = 0.50 Kg, velocity = ?, K.E. = 12.18 J

v² = 2 KE/m

v = √2KE/m

v = √(2 × 12.18/0.50)

v = 6.98 m/s

Therefore, the hand must move with a velocity of 6.98 m/s to break the board.

B. Average force on the hand

This can be determined using the equation of motion, v² = u² + 2as to find acceleration, since force = mass × acceleration

From the equation of motion, a = v² - u²/2s

At rest, v = 0, u = 6.98, s = 1.2 cm = 0.012 m

a = 0² - 6.98²/ 2 × 0.012

a = -2030 m/s²

Force = 2030 m/s² × 0. 50 kg = 1015 N

Therefore, Average force on the hand = 1025 N

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A closely wound circular coil has a radius of 6.00 cmand carries a current of 2.65 A. How many turns must it have if the magneti
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Answer:

Given:

radius of the coil, R = 6 cm = 0.06 m

current in the coil, I = 2.65 A

Magnetic field at the center, B = 6.31\times 10^{4} T

Solution:

To find the number of turns, N, we use the given formula:

B = \frac{\mu_{o}NI}{2R}

Therefore,

N = \frac{2BR}{\mu_{o}I}

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