Light travels in a straight line at a constant speed of 3,0 x 108 m/s for 4,1

What is the voltage across each resistor?

kogti [31]

Answer:

see solution below

Explanation:

The given resistors are connected in series.

Equivalent resistance in series = 30 + 55 + 15

Equivalent resistance in series Rt = 100 ohms

Since the potential difference in the circuit = 36V

Get the current in the circuit first

I = V/Rt

I = 36/100

I = 0.36A

Get the voltage across 30ohms resistor;

V30 = 0.36 * 30

V30 = 10.8volts

Hence the voltage across the 30ohms resistor is 10.8volts

Get the voltage across 55ohms resistor;

V55 = 0.36 * 55

V55 = 19.8volts

Hence the voltage across the 55ohms resistor is 19.8volts

Get the voltage across 15ohms resistor;

V15 = 0.36 * 15

V15 = 5.4volts

Hence the voltage across the 15ohms resistor is 5.4volts

4 0

3 years ago

Will give you brainliest pls help

ExtremeBDS [4]

1. All the relevant resistors are in series, so the total (or equivalent) resistance is the sum of the resistances of the resistors: 20 Ω + 80 Ω + 50 Ω = 150 Ω [choice A].

2. The ammeter will read the current flowing through this circuit. We can find the ammeter reading using Ohm's law in terms of the electromotive force provided by the battery: I = ℰ/R = (30 V)(150 Ω) = 0.20 A [choice C].

3. The voltmeter will measure the potential drop across the 50 Ω resistor, i.e., the voltage at that resistor. We know from question 2 that the current flowing through the resistor is 0.20 A. So, from Ohm's law, V = IR = (0.20 A)(50 Ω) = 10. V, which will be the voltmeter reading [choice F].

4. Trick question? If the circuit becomes open, then no current will flow. Moreover, even if the voltmeter were kept as element of the circuit, voltmeters generally have a very high resistance (an ideal voltmeter has infinite resistance), so the current moving through the circuit will be negligible if not nil. In any case, the ammeter reading would be 0 A [choice B].

4 0

2 years ago

Which of the following is not used in calculating acceleration?

Leno4ka [110]

Answer:

l think C am not pretty show

4 0

2 years ago

An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km fr

mestny [16]

An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km from its center. The black hole is 5.00 times the mass of the sun and has a Schwarzschild radius of 15.0 km. The astronaut is positioned inside the spaceship such that one of her 0.030 kg ears is 6.0 cm farther from the black hole than the center of mass of the spacecraft and the other ear is 6.0 cm closer.

What is the tension between her ears?

Would the astronaut find it difficult to keep from being torn apart by the gravitational forces?

Answer:

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

Explanation:

Given that:

Orbital radius of the spacecraft (R) = 120 Km = 120 × 10³ m

Mass of the black hole (m) = $5 \ * (M \ _{sun})$

where : $M_{sun} = 1.99*10^{33} \ kg$

Then; we have:

$m = 5*(1.99*10^{30} \ kg ) \\ = 9.95*10^{30} kg$

Schwarzchild radius of the black hole

r - 15.0 km

Mass of each ear $m_{ear} = 0.030 \ kg$

Farther distance between one ear and the black hole (d) = 6.0 cm

= 0.06 m

Closer distance between the other ear and the black home is (d) 6.0 cm

= 0.6 cm

NOW, If we assume that the tension force should be T; then definitely the two ears will posses the same angular velocity .

The net force on the ear closer to the black hole will be:

$\frac{GMm_{ear} }{(R-d)}- T = m_{ear} (R - d) \omega^2$

$\frac{GMm_{ear} }{(R-d)^2}- \frac{T}{(R-d)} = m_{ear} \omega^2 \ ----> \ (1)$

The net force on the ear farther to the black hole is :

$\frac{GMm_{ear} }{(R+d)}- T = m_{ear} (R + d) \omega^2$

$\frac{GMm_{ear} }{(R+d)^2}- \frac{T}{(R+d)} = m_{ear} \omega^2 \ ----> \ (2)$

Equating equation (1) and (2) & therefore making (T) the subject of the formula; we have:

$T = \frac{3GMm_{ear}d}{R^3}$

$T = \frac{3(6.67*10^{-11}N.m^2/kg^2)(1.95*10^{30}kg)(0.03kg)(0.06m)}{(120*10^3m)^3}$

$T = 2073.9 N\\T = 2.07 KN$

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

3 0

3 years ago

A police car chases a speeder along a straight road towards a cliff both vehicles move at 160km/h the siren on the police car pr

natta225 [31]

Answer:

f ’= 97.0 Hz

Explanation:

This is an exercise of the doppler effect use the frequency change due to the relative movement of the fort and the observer

in this case the source is the police cases that go to vs = 160 km / h

and the observer is vo = 120 km / h

the relationship of the doppler effect is

f ’= f₀ (v + v₀ / v- $v_{s}$ )

let's reduce the magnitude to the SI system

v_{s} = 160 km / h (1000 m / 1km) (1h / 3600s) = 44.44 m / s

v₀ = 120 km / h (1000m / 1km) (1h / 3600s) = 33.33 m / s

we substitute in the equation of the Doppler effect

f ‘= 100 (330+ 33.33 / 330-44.44)

f ’= 97.0 Hz

4 0

3 years ago