All categories

3 years ago

A circuit with a resistance of 25 ohms is carrying a current of 2

1 answer:

7 0

The heat developed can be calculated using the power rules as follows:

P = I x V = I^2 x R

First solution is to use the given values of I and R as follows:
P = I^2 x R = (2)^2 x 25 = 4 x 25 = 100 watt

Second solution is to use ohm's law to get V, then substitute to get the power as follows:
V = I x R = 2 x 25 = 50 volts
Power = V x I = 50 x 2 = 100 watt

You might be interested in

A constant net force acts on an object. which of the following best describes the object's motion?

Anton [14]

The object is moving with a decreasing acceleration. The object is moving with a constant velocity.

4 0

4 years ago

It has been suggested, and not facetiously, that life might have originated on Mars and been carried to Earth when a meteor hit

damaskus [11]

Answer:

$a=3125000 m/s^2\\a=3.125*10^6 m/s^2$

Acceleration, in m/s, of such a rock fragment = $3.125*10^6m/s^2$

Explanation:

According to Newton's Third Equation of motion

$V_f^2-V_i^2=2as$

Where:

$V_f$ is the final velocity

$V_i$ is the initial velocity

a is the acceleration

s is the distance

In our case:

$V_f=V_{escape}, V_i=0,s=4 m$

So Equation will become:

$V_{escape}^2-V_i^2=2as\\V_{escape}^2-0=2as\\V_{escape}^2=2as\\a=\frac{V_{escape}^2}{2s}\\a=\frac{(5*10^3m)^2}{2*4}\\a=3125000 m/s^2\\a=3.125*10^6 m/s^2$

Acceleration, in m/s, of such a rock fragment = $3.125*10^6m/s^2$

5 0

3 years ago

A 100-kg spacecraft is in a circular orbit about Earth at a height h = 2RE .

maria [59]

To solve this problem it is necessary to apply the concepts related to the conservation of the Gravitational Force and the centripetal force by equilibrium,

$F_g = F_c$

$\frac{GmM}{r^2} = \frac{mv^2}{r}$

Where,

m = Mass of spacecraft

M = Mass of Earth

r = Radius (Orbit)

G = Gravitational Universal Music

v = Velocity

Re-arrange to find the velocity

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

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

$v = \sqrt{\frac{GM}{r}}$

PART A ) The radius of the spacecraft's orbit is 2 times the radius of the earth, that is, considering the center of the earth, the spacecraft is 3 times at that distance. Replacing then,

$v = \sqrt{\frac{(6.67*10^{-11})(5.97*10^{24})}{3*(6.371*10^6)}}$