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3 years ago

If the current in the circuit above is 10 amps and the resistance is 3 ohms, what is the voltage?

1 answer:

4 0

It's really difficult to make out the circuit above. Quite frankly, your
question leaves me to wonder how far 'above' it may be.
The best I can do will be to try and fabricate an answer based on the
information given in the text of the question, augmented only by my own
training, chutzpah, and life experiences.

If the circuit ... wherever it is ... consists entirely of the single 3-ohm
resistance and no other components, and the current through the
resistance is 10 Amperes, then

Voltage = (Current) x (resistance) = 30 volts .

You might be interested in

Light could be thought of as a stream of tiny particles discharged by _______ objects that travel in straight paths. *

Gelneren [198K]

Answer: Light could be thought of as a stream of tiny particles discharged by luminous objects that travel in straight paths.

Explanation:

We can define "radiation" as the transmision of energy trough waves or particles.

Particularly, light is a form of electromagnetic radiation, so the "tiny particles" of light are discharged by a radiating object, particularly we can be more explicit and call it a luminous object, in this way we are being specific about the nature of the radiation of the object.

5 0

3 years ago

An astronaut goes out for a space walk. Her mass (including space suit, oxygen tank, etc.) is 100 kg. Suddenly, disaster strikes

Marina CMI [18]

Answer:

Unknown variables:

velocity of the astronaut after throwing the tank.

maximum distance the astronaut can be away from the spacecraft to make it back before she runs out of oxygen.

Known variables:

velocity and mass of the tank.

mass of the astronaut after and before throwing the tank.

maximum time it can take the astronaut to return to the spacecraft.

To obtain the velocity of the astronaut we use this equation:

-(momentum of the oxygen tank) = momentum of the astronaut

-mt · vt = ma · vt

Where:

mt = mass of the tank

vt = velocity of the tank

ma = mass of the astronaut

va = velocity of the astronaut

To obtain the maximum distance the astronaut can be away from the spacecraft we use this equation:

x = x0 + v · t

Where:

x = position of the astronaut at time t.

x0 = initial position.

v = velocity.

t = time.

The maximum distance the astronaut can be away from the spacecraft is 162 m.

Explanation:

Hi there!

Due to conservation of momentum, the momentum of the oxygen tank when it is thrown away must be equal to the momentum of the astronaut but in opposite direction. In other words, the momentum of the system astronaut-oxygen tank is the same before and after throwing the tank.

The momentum of the system before throwing the tank is zero because the astronaut is at rest:

Initial momentum = m · v

Where m is the mass of the astronaut plus the equipment (100 kg) and v is its velocity (0 m/s).

Then:

initial momentum = 0

After throwing the tank, the momentum of the system is the sum of the momentums of the astronaut plus the momentum of the tank.

final momentum = mt · vt + ma · va

Where:

mt = mass of the tank

vt = velocity of the tank

ma = mass of the astronaut

va = velocity of the astronaut

Since the initial momentum is equal to final momentum:

initial momentum = final momentum

0 = mt · vt + ma · va

- mt · vt = ma · va

Now, we have proved that the momentum of the tank must be equal to the momentum of the astronaut but in opposite direction.

Solving that equation for the velocity of the astronaut (va):

- (mt · vt)/ma = va

mt = 15 kg

vt = 10 m/s

ma = 100 kg - 15 kg = 85 kg

-(15 kg · 10 m/s)/ 85 kg = -1.8 m/s

The velocity of the astronaut is 1.8 m/s in direction to the spacecraft.

Let´s place the origin of the frame of reference at the spacecraft. The equation of position for an object moving in a straight line at constant velocity is the following:

x = x0 + v · t

where:

x = position of the object at time t.

x0 = initial position.

v = velocity.

t = time.

Initially, the astronaut is at a distance x away from the spacecraft so that

the initial position of the astronaut, x0, is equal to x.

Since the origin of the frame of reference is located at the spacecraft, the position of the spacecraft will be 0 m.

The velocity of the astronaut is directed towards the spacecraft (the origin of the frame of reference), then, v = -1.8 m/s

The maximum time it can take the astronaut to reach the position of the spacecraft is 1.5 min = 90 s.

Then:

x = x0 + v · t

0 m = x - 1.8 m/s · 90 s

Solving for x:

1.8 m/s · 90 s = x

x = 162 m

The maximum distance the astronaut can be away from the spacecraft is 162 m.

6 0

3 years ago

Calculate how far ahead of the drop zone a pilot would have needed to drop humanitarian aid packages if the delivery occurred at

ozzi

Answer:

S = 11488.42 m

Explanation:

Given,

The speed of the jet, v = 0.74 Mach

= 253.82 m/s

The altitude of the jet, h = 10000 m

Since the jet is travelling horizontally, the vertical component of velocity Vy = 0

The equation for a projectile projected from a height h is given by,

S = Vx [Vy + √(Vy² +2gh)] / g

Since Vy = 0

S = Vx √(Vy² +2gh) / g

= 253.82 x √(2 x 9.8 x 10000) / 9.8

= 11488.42 m

Hence, the humanitarian aid package should be delivered ahead of distance, S = 11488.42 m

6 0

3 years ago

A simple series circuit consists of a 130 ? resistor, a 30.0V battery, a switch, and a 2.10 pF parallel-plate capacitor (initial

V125BC [204]

Answer with Explanation:

We are given that

Resistance,R=130 ohm

Potential difference, V=30 V

Capacitor,C=2.1pF= $2.1\times 10^{-12} F$

$1pF=10^{-12} F$

$d=5.0 mm=5\times 10^{-3} m$

$1mm=10^{-3} m$

a.Maximum flux

$\phi=EA=\frac{V}{d}\times \frac{Cd}\epsilon_0}=\frac{CV}{\epsilon_0}$

$\epsilon_0=8.85\times 10^{-12}$

$\phi=\frac{2.1\times 10^{-12}\times 30}{8.85\times 10^{-12}}=7.12Vm$

b.Maximum displacement current, $I=\frac{V}{R}=\frac{30}{130}=0.23 A$

c.We have to find electric flux at t=0.5 ns

$t=0.5ns=0.5\times 10^{-9} s$

$1ns=10^{-9}s$

$q=CV(1-e^{-\frac{t}{RC}})$

$q=30\times 2.1\times 10^{-12}(1-e^{-\frac{0.5\times 10^{-9}}{130\times 2.1\times 10^{-9}})$

$q=52.9\times 10^{-12} C$

$\phi=\frac{q}{\epsilon_0}=\frac{52.9\times 10^{-12}}{8.85\times 10^{-12}}=5.98Vm$

d.Displacement current at t=0.5ns

$I=(\frac{V}{R})e^{-\frac{t}{RC}}=\frac{30}{130}e^{-\frac{0.5\times 10^{-9}}{130\times 2.1\times 10^{-12}}}$

$I=0.037 A$

5 0

3 years ago

Nancy ran a distance of 5km in 30 minutes. What is her speed in meters(m) per hour)h)

Virty [35]

The speed of Nancy who ran a distance of 5 km in 30 minutes is 10000 meters(m) per hour(h).

<h3>What is the rate of speed?</h3>

The rate of speed is the rate at which the total distance is travelled in the time taken. Rate of speed can be given as,

$s=\dfrac{d}{t}$

Here, (d) is the distance travelled by the object and (t) is the time taken but the object to cover that distance.

Nancy ran a distance of 5 km in 30 minutes. There are 60 minutes in one hour. Thus, the time in hour for which Nancy ran is,

$t=\dfrac{30}{60}\\t=0.5\rm\; hr$

The meters in 5 kilometers is,

$d=5\times1000\\d=5000\rm\; m$

She ran 5000 kilometers in 0.5 hours. Thus, the speed of her is,

$s=\dfrac{5000}{0.5}\\s=10000\rm\; km/hr$

Thus, the speed of Nancy who ran a distance of 5 km in 30 minutes is 10000 meters(m) per hour(h).

Learn more about the rate of speed here:

brainly.com/question/359790

#SPJ4

7 0

3 years ago