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Nimfa-mama [501]
3 years ago
7

A current of 15.0 A flows through an electric heater operating on 120 V. Compute the power consumption of the electric heater.

Physics
1 answer:
Rzqust [24]3 years ago
5 0

Answer:

There will be 1800 W power consumption in heater

Explanation:

We have given current flowing in the heater I = 15 A

Voltage on which heater is operating V = 120 volt

We have to find the power consumption in the heater

We know that power consumption is given by P = VI

So power consumption in heater = 120 × 15 = 1800 W

So there will be 1800 W power consumption in heater

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A 400-kg object has a 100-Newton rightward net force being applied to it. What is the magnitude of the rightward acceleration on
aliya0001 [1]

Answer:

The answer to your question is a = 0.25 m/s²

Explanation:

Data

mass = m = 400 kg

Force = F = 100 N

acceleration = a = ? m/s²

Process

To solve this problem use Newton's second law that states that the force applied to an object is directly proportional to the mass of the body times its acceleration.

Formula

                       F = ma

solve for a

                       a = \frac{F}{m}

Substitution

                       a = \frac{100}{400}

Simplification and result

                              a = 0.25 m/s²

5 0
3 years ago
Through what potential difference would you need to accelerate an alpha particle, starting from rest, so that it will just reach
Svetlanka [38]

Answer:

\Delta V    = 1.8 \times 10^7 V

Explanation:

GIVEN

diameter = 15 fm  =15 \times 10^{-15}m

we use here energy conservation

K_{i}+U_{i} =K_{f}+U_{f}

there will be some initial kinetic  energy but after collision kinetic energy will zero

K_{i} + 0 = 0 + \frac{1}{4 \pi \epsilon _{0}} \frac{(2e)(92e)}{7.5 \times 10^{-15}}

on solving these equations we get kinetic energy initial

KE_{i} = 5.65\times 10 ^{-12} \times \frac  {1 eV}{1.6 \times 10^{-19}}

KE_{i} = 35.33 J ..............(i)

That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2  e

and gains kinetic energy K  =e∆V  ..........(ii)

 by accelerating through a potential difference ∆V

Thus the alpha particle will

just reach the {238}_U nucleus after being accelerated through a potential difference  ∆V

equating (i) and second equation we get

e∆V  = 35.33 Me V

\Delta V = \frac{35.33}{2}  MV\\\Delta V    = 1.8 \times 10^7 V

7 0
3 years ago
3. How do you represent the strength of a<br> force in a free-body diagram?
ivann1987 [24]
Generally, the length of the line will indicate how strong the force is. If you have two opposing forces and one is higher than the other, you would draw the line of the higher force visibly longer.
3 0
3 years ago
1. A spring extends by 10 cm when a mass of 100 g is attached to it. What is the spring constant? (Calculate your answer in N/m)
cupoosta [38]

Answer:

1) k = 10 [N/m]

2) a-) x = 0.4 [m]

b)  x = 0.075 [m]

Explanation:

To be able to solve this type of problems that include springs we must use Hooke's law, which relates the force to the deformed length of the spring and in the same way to the spring coefficient.

F = k*x

where:

F = force [N] (units of Newtons]

k = spring constant  [N/m]

x = distance = 10 [cm] = 0.1 [m]

Now, the weight is equal to the product of the mass by the gravity

W = m*g = F

where:

m = mass = 100 [g] = 0.1 [kg]

g = gravity acceleration = 10 [m/s²]

F = 0.1*10 = 1 [N]

Now clearing k

k = F/x

k = 1/0.1

k = 10 [N/m]

2)

a ) if the force is 4 [N]

clearing x

x = F/k

x = 4/10

x = 0.4 [m]

m = 75 [g] = 0.075 [kg]

W = m*g = F

F = 0.075*10 = 0.75 [N]

x = .75/10

x = 0.075 [m]

5 0
2 years ago
Each of the following statements is arguably true of thermometers. Which of them is most helpful to keep in mind if you are cond
Lera25 [3.4K]

Answer:

The temperature reported by a thermometer is never precisely the same as its surroundings

Explanation:

In this experiment to determine the specific heat of a material the theory explains that when a heat interchange takes place between two bodies that were having different temperatures at the start, the quantity of heat the warmer body looses is equal to that gained by the cooler body to reach the equilibrium temperature. <u>This is true only if no heat is lost or gained from the surrounding.</u> If heat is gained or lost from the surrounding environment, the temperature readings by the thermometer will be incorrect. The experimenter should therefore keep in mind that for accurate results, the temperature recorded by the thermometer is similar to that of the surrounding at the start of the experiment and if it differs then note that there is either heat gained or lost to the environment.

3 0
3 years ago
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