All categories

2 years ago

A circuit breaker is tripped when the

Physics

1 answer:

SCORPION-xisa [38]2 years ago

7 0

Answer:

P= 15A x 120 V P= 1800 W

I believe, I hope this helps

Explanation:

You might be interested in

a car has a mass of 1.00 x 10^3 kilograms and it has an acceleration of 4.5 meters/second what is the net force on the car

Reptile [31]

The answer is 4.5 x10^3 newtons.

6 0

3 years ago

Read 2 more answers

Help ASAP<br> Answer question 3 please!!!

PolarNik [594]

Answer:

the answer is ture

Explanation:

because the energy substances must absorb in order to change from liquid to gas

8 0

3 years ago

What happens to the speed of the particles if the size of the particle is increased

bearhunter [10]

The particle slows down.

6 0

3 years ago

A satellite orbits a planet of unknown mass in a circular orbit of radius 2.3 x 104 km. The gravitational force on the satellite

sladkih [1.3K]

Answer:

The kinetic energy is $KE = 7.59 *10^{10} \ J$

Explanation:

From the question we are told that

The radius of the orbit is $r = 2.3 *10^{4} \ km = 2.3 *10^{7} \ m$

The gravitational force is $F_g = 6600 \ N$

The kinetic energy of the satellite is mathematically represented as

$KE = \frac{1}{2} * mv^2$

where v is the speed of the satellite which is mathematically represented as

$v = \sqrt{\frac{G M}{r^2} }$

=> $v^2 = \frac{GM }{r}$

substituting this into the equation

$KE = \frac{ 1}{2} *\frac{GMm}{r}$

Now the gravitational force of the planet is mathematically represented as

$F_g = \frac{GMm}{r^2}$

Where M is the mass of the planet and m is the mass of the satellite

Now looking at the formula for KE we see that we can represent it as

$KE = \frac{ 1}{2} *[\frac{GMm}{r^2}] * r$

=> $KE = \frac{ 1}{2} *F_g * r$

substituting values

$KE = \frac{ 1}{2} *6600 * 2.3*10^{7}$

$KE = 7.59 *10^{10} \ J$

7 0

3 years ago

A physics professor demonstrates the Doppler effect by tying a 600 Hz sound generator to a 1.0-m-long rope and whirling it aroun

nalin [4]

Answer:

Therefore the the highest frequency is 620Hz and lowest frequency is 580Hz

Explanation:

Given data

Source Frequency fs=600Hz

Length r=1.0m

RPM=100 rpm

The speed of the generator is calculated as:

$v_{s}=rw\\v_{s}=r(2\pi f)$

Substitute the given values

$v_{s}=(1.0m)2\pi (\frac{100}{60}rev/s )\\v_{s}=10.47m/s$

For approaching generator the frequency is calculated as:

$f_{+}=\frac{f_{s}}{1-\frac{v_{s}}{v} }\\f_{+}=\frac{600Hz}{1-\frac{10.47m/s}{343m/s} } \\f_{+}=620Hz$

On the other hand,for the receding generator,Doppler's effect is expressed as:

$f_{-}=\frac{f_{s}}{1+\frac{v_{s}}{v} }\\f_{-}=\frac{600Hz}{1+\frac{10.47m/s}{343m/s} } \\f_{-}=580Hz$

Therefore the the highest frequency is 620Hz and lowest frequency is 580Hz

8 0

3 years ago