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konstantin123 [22]
4 years ago
13

When a resistor is connected to a 12-V source, it draws a 185-mA current. The same resistor connected to a 90-V source draws a 1

.25-A current. (a) Is the resistor ohmic? Justify your answer mathematically. (b) What is the rate of Joule heating in this resistor in both cases?
Physics
2 answers:
Svetach [21]4 years ago
8 0

Answer:

The resistor is not Ohmic

Explanation:

In the first case, from ohm's law: V=IR

R=V/I= 12/0.185=64.9 ohm

In the second case:

R=V/I= 90/1.25= 72ohm

The relationship between voltage and current is linear according to ohms law but that is not the case here.

Joule heating = I^2R

For case1: (0.185)^2 × 64.9 = 2.2W

For case2: (1.25)^2 ×72 =112.5W

topjm [15]4 years ago
7 0

Answer:

Explanation:

Ohm's law is define by the formula

V = IR where I is current in A, R is resistance in ohms

to calculate the resistance we use the formula above

I = 185 mA / 1000 = 0.185 A

case 1, R = V / I = 12 V / 0.185 A = 64.86 ohms

case 2, R = V / I = 90 / 1.25 A = 72 ohms

a) the resistor is not ohmic since there is no linear relationship between R and V ( as voltage increases, resistance ought to also increase)

b) rate of Joule heating in the case 1, P, power = IV = 0.185 A × 12 = 2.22 W

case 2, P = IV = 1.25 A × 90 V = 112.5 W

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<u>Metal detectors work by transmitting an electromagnetic field from the search coil into the ground. Any metal objects (targets) within the electromagnetic field will become energised and retransmit an electromagnetic field of their own. The detector’s search coil receives the retransmitted field and alerts the user by producing a target response. metal detectors are capable of discriminating between different target types and can be set to ignore unwanted targets. </u>

1. Search Coil

The detector’s search coil transmits the electromagnetic field into the ground and receives the return electromagnetic field from a target.

2. Transmit Electromagnetic Field (visual representation only - blue)

The transmit electromagnetic field energises targets to enable them to be detected.

3. Target

A target is any metal object that can be detected by a metal detector. In this example, the detected target is treasure, which is a good (accepted) target.

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3 years ago
Dr. John Paul Stapp was U.S. Air Force officer who studied the effects of extreme deceleration on the human body. On December 10
Mars2501 [29]

Answer:

a) a=5.7551 \times g

b) d=20.5539\times g

Explanation:

Given:

  • speed of rocket initially, v_i=0\ m.s^{-1}
  • top speed of rocket after acceleration, v=282\ m.s^{-1}
  • time taken to get to the top speed, t_i=5\ m.s^{-1}
  • final speed of the rocket, v_f=0\ m.s^{-1}
  • time taken to get to the final speed after reaching the top speed, t_f=1.4\ s

Now the acceleration:

a=\frac{v-v_i}{t_i}

a=\frac{282-0}{5}

a=56.4\ m.s^{-2}

Now as a fraction of gravity:

a=\frac{56.4}{9.8}\times g

a=5.7551 \times g

Now, the deceleration:

d=\frac{0-282}{1.4}

d=201.4285\ m.s^{-2}

Now as a fraction of gravity:

d=\frac{201.4285}{9.8}\times g

d=20.5539\times g

6 0
4 years ago
A small metal ball is given a negative charge, then brought near to end a of the rod (figure 1). What happens to end a of the ro
erma4kov [3.2K]

What happens to end a of the rod when the ball approaches it closely this first time is; It is strongly attracted.

<h3>Electrostatics</h3>

I have attached the image of the rod.

We are told that the ball is much closer to the end of the rod than the length of the rod. Thus, if we point down the rod several times, the distance of approach will experience no electric field and as such the charge on end point A of the rod must be comparable in magnitude to the charge on the ball.

This means that their fields will cancel.

Finally, we can conclude that when a charge is brought close to a conductor, the opposite charges will all navigate to the point that is closest to the charge and as a result, a strong attraction will be created.

This also applies to a strong conducting rod and therefore it is strongly attracted.

Read more about Electrostatics at; brainly.com/question/18108470

7 0
2 years ago
A spaceship, at rest in some inertial frame in space, suddenly needs to accelerate. The ship forcibly expels 103 kg of fuel from
Cerrena [4.2K]

Answer:

V_s = 1.8*10^5m/s

Explanation:

There is no external force applied, therefore there is a moment's preservation throughout the trajectory.

<em>Initial momentum  = Final momentum. </em>

The total mass is equal to

m_T= m_1 +m_2

Where,

m_1 = mass of ship

m_2 = mass of fuell expeled.

As the moment is conserved we have,

0=V_fm_2+V_sm_1

Where,

V_f = Velocity of fuel

V_s =Velocity of Space Ship

Solving and re-arrange to V_swe have,

V_s = \frac{V_f m_2 }{m_1}

V_s = \frac{3/5c}{10^6}

V_s = 3.5*10^{-3}c

Where c is the speed of light.

Therefore the ship be moving with speed

V_s = \frac{3}{5}*10^{-3}*3*10^8m/s

V_s = 1.8*10^5m/s

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3 years ago
Christopher drives into the city to buy new new hockey equipment. Because of traffic conditions, he averages only 15 mph. On the
alina1380 [7]

Answer:

36 minutes

Explanation:

Distance= Time*Speed

Let time taken to go to city be x hours, therefore, time to travel back is (2-x) hours

The distance to and from the city are equal hence

15x=35(2-x)

15x=70-35x

50x=70 hence x=70/50=1.4 hours

Time to drive home will be 2-x=2-1.4=0.6 hours

0.6*60=36 minutes

Therefore, time to travel back home is 36 minutes

8 0
3 years ago
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