A smoke alarm beeps twice per minute. What is the period and frequency 1 po

If the current through a 20-Ω resistor is 8.0 A , how much energy is dissipated by the resistor in 1.0 h ? Express your answer w

marshall27 [118]

Answer:

$P(3600)=593.247W$

Explanation:

First, let's find the voltage through the resistor using ohm's law:

$V=IR=20*8=160V$

AC power as function of time can be calculated as:

$P(t)= V*I*cos(\phi)-V*I*cos(2 \omega t-\phi)$ (1)

Where:

$\phi=Phase\hspace{3}angle\\\omega= Angular\hspace{3}frequency$

Because of the problem doesn't give us additional information, let's assume:

$\phi=0\\\omega=2 \pi f=2*\pi *(60)=120\pi$

Evaluating the equation (1) in t=3600 (Because 1h equal to 3600s):

$P(3600)=160*8*cos(0)-160*8*cos(2*120\pi*3600-0)\\P(3600)=1280-1280*cos(2714336.053)\\P(3600)=1280-1280*0.5365255751\\P(3600)=1280-686.7527361=593.2472639\approx=593.247W$

5 0

4 years ago

5. If the momentum of a body is increased by 100% then the percentage increase in kinetic

Mnenie [13.5K]

The answer is the letter ( d )

5 0

3 years ago

Elliptical galaxies are frequently found a) Inside the Milky Way b) In galaxy clusters c) In the Galactic bulge d) In the Local

sergejj [24]

Answer:

The correct option is b) In galaxy clusters

Explanation:

A type of galaxy that appear elliptical in shape and have an almost featureless and smooth image is known as the elliptical galaxy.

An elliptical galaxy is three dimensional and consists of more than one hundred trillion stars which are present in random orbits around the centre.

Elliptical galaxy is generally found in the galaxy clusters.

5 0

3 years ago

A motorcycle starts at rest and accelerates at a rate of 3 meters per second squared (m/s2) over a time period of 5 seconds (s).

Alex787 [66]

Final velocity = 3 x 5 = 15 m/s

5 0

3 years ago

An abstract sculpture consists of a ball (radius R = 76 cm) resting on top of a cube (each side L = 200 cm long). The ball and t

scZoUnD [109]

Answer:

132.9 cm

Explanation:

Data provided in the question:

Radius of the basll = 76 cm = 0.76 m

Side of the box = 200 cm = 2 m

Density of the ball and cube are equal

let the density be 'D'

Now,

Mass of ball, M = Volume × Density

= $\frac{4}{3}\pi r^3$ × D

= $\frac{4}{3}\pi (0.76)^3$ × D

= 1.838D

Mass of cube, m = L³ × D

= 2³ × D

= 8D

Thus,

center of mass, y = [ My₁ + my₂ ] ÷ [M + m]

here,

y₁ = center of mass of ball with respect to floor

as the center mass of sphere lies at the center of the sphere

= Length of cube + radius of sphere

= 2 + 0.76

= 2.76 m

y₂ = Center of mass of cube = $\frac{L}{2}=\frac{2}{2}$ = 1 m

Thus,

y = [ ( 1.838D × 2.76 ) + (8D × 1 ) ] ÷ [1.838D + 8D]

= 13.07288D ÷ 9.838D

= 1.329 m

or

= 132.9 cm

3 0

3 years ago