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4 years ago

How do you calculate resistance with coulombs, time and J's?

Physics

1 answer:

Mars2501 [29]4 years ago

4 0

Hey user

The energy E in joules (J) is equal to the voltage V in volts (V), times the electrical charge Q in coulombs (C):

E(J) = V(V) × Q(C)

joule = volt × coulomb

J = V × C

Example

What is the energy in joules that is consumed in an electrical circuit with voltage supply of 15V and charge flow of 4 coulombs?

E = 15V × 4C = 60J

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What are the contrast of adaptation vs natural selection?

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The difference between adaptation and natural selection is that adaptation is the characteristic while natural selection is the mechanism that increases the probability that an advantageous characteristic is passed on and becomes commonplace.

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3 years ago

(b) During one day, 250 kg of water is pumped through

ser-zykov [4K]

Answer: The energy incident on the solar panel during that day is $9.24 \times 10^{7} J$ .

Explanation:

Given: Mass = 250 kg

Initial temperature = $16^{o}C$

Final temperature = $38^{o}C$

Specific heat capacity = 4200 $J/kg^{o}C$

Formula used to calculate the energy is as follows.

$q = m \times C \times (T_{2} - T_{1})$

where,

q = heat energy

m = mass of substance

C = specific heat capacity

$T_{1}$ = initial temperature

$T_{2}$ = final temperature

Substitute the values into above formula as follows.

$q = 250 kg \times 4200 J/kg^{o}C \times (38 - 16)^{o}C\\= 250 kg \times 4200 J/kg^{o}C \times 22^{o}C$

As it is given that water absorbs 25% of the energy incident on the solar panel. Hence, energy incident on the solar panel can be calculated as follows.

$\frac{25}{100} \times q = 250 kg \times 4200 J/kg^{o}C \times 22^{o}C\\q = 9.24 \times 10^{7} J$

Thus, we can conclude that the energy incident on the solar panel during that day is $9.24 \times 10^{7} J$ .

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3 years ago

What is a hypothesis?

Likurg_2 [28]

Answer:

Explanation:

a statement that can be tested through the scientific method

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3 years ago

If an object has a mass of 10 kilograms, how much does it weigh in newtons?

schepotkina [342]

10 kilograms of mass weighs 98.1 newtons on Earth,
16.2 newtons on the Moon, 37.1 newtons on Mars,
and other weights in other places.

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4 years ago

Two ropes are attached to either side of a 100.0 kg wagon as shown below. The rope on the right is pulled at an angle 40.0° rela

NikAS [45]

The acceleration of the wagon is found by applying Newton's Second Law of motion.

1. The responses for question 1 are;

x-component of the tension in the rope on the right is approximately 91.93 N

y-component of the tension in the rope on the right is approximately 71.135 N

x-component of the tension in the rope on the left is -80.0 N

y-component of the tension in the rope on the left is 0

2. The net force in the x-direction is approximately 11.93 N

3. The net acceleration of the wagon in the horizontal direction is approximately 0.1193 m/s².

Reasons:

The given parameters are;

Mass of the wagon, m = 100.0 kg

Angle of inclination to the horizontal of the rope to the right, θ = 40.0°

Tension in the rope on the right = 120.0 N

Direction in which the rope on the left is pulled = To the west

Tension in the rope on the left = 80.0 N

1. The x and y component of the tension in the rope on the right are;

x-component = 120.0 N × cos(40.0°) ≈ 91.93 N

y-component = 120.0 N × sin(40.0°) ≈ 77.135 N

The x and y component of the tension in the rope on the left are;

x-component = 80.0 N × cos(180°) = -80.0 N

y-component = 80.0 N × sin(180°) = 0.0 N

2. The net force in the horizontal direction, Fₓ, is found as follows;

Fₓ = The x-component of the rope on the left + The x-component of the rope on the right

Which gives;

Fₓ = 91.93 N - 80.0 N = 11.93 N

3. The net acceleration of the block is given as follows;

According to Newton's Second Law of motion, we have;

Force in the horizontal direction, Fₓ = Mass of wagon, m × Acceleration of the wagon in the horizontal direction, aₓ

Fₓ = m × aₓ

Therefore;

$\displaystyle a_x = \frac{F_x}{m} \approx \frac{11.93 \, N}{100.0 \, kg} = \mathbf{0.1193 \ m/s^2}$

The acceleration of the wagon in the horizontal direction, aₓ ≈ 0.1193 m/s².

Learn more here:

brainly.com/question/20357188

8 0

3 years ago