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1 year ago

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Physics

1 answer:

Alona [7]1 year ago

7 0

In the circuit, the total current is 60 A and the voltage is 240 V. If R1

resistance is 20 Ω, then the R2 resistance would be 5 Ω

<h3>What is resistance?</h3>

Resistance is the obstruction of electrons in an electrically conducting material. The SI unit of the resistance is Ohm

The mathematical relation for resistance can be understood with the help of the empirical relation provided by Ohm's law.

V=IR

where V is the voltage

I is the current

R is the resistance

For the given problem V= 240 V and I = 60 A

R = V/I

R =240/60

R =4 Ω

For calculating equivalent resistance in parallel combination.

1/Re = 1/R1 + 1/R2

1/4 = 1/20 + 1/R2

R2 = 5 Ω

Thus , the R2 resistance would be 5 Ω, therefore the correct answer is option B.

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If a battery is running down, a light bulb in the circuit will become dimmer because when voltage drops

lana [24]

A. There is less current flowing through the bulb.

Ohms law dictates that Electrical voltage = current x resistance. To understand current you would divide voltage by resistance. The less voltage the less current will run through the circuit.

Mark as brainliest if satisfied.

6 0

2 years ago

A box is placed on a conveyor belt that moves with a constant speed of 1.05 m/s. The coefficient of kinetic friction between the

sveticcg [70]

Answer:

The box stops in 0.139 seconds, after moving 7.29cm (0.0729m) backwards relative to the belt.

Explanation:

As the box is initially at rest relative to the earth, it is moving backwards with a speed of 1.05m/s relative to the belt. Then, the frictional force acts on the box to make it stop relative to the belt. So, we first have to write the equations of motion of the box in each axis:

$x: f_k=ma\implies a=\frac{f_k}{m} \\\\y: N-mg=0\implies N=mg$

Since the frictional force $f_k$ is equal to $f_k=\mu_k N=\mu_k mg$ , then we have that the acceleration is:

$a=\frac{\mu_k mg}{m}=\mu_k g$

Now, from the definition of acceleration we get:

$a=\frac{v-v_0}{t}\implies t=\frac{v-v_0}{a}$

And, as the final velocity is zero because the box gets to a stop, we have:

$t=-\frac{v_0}{a}=-\frac{v_0}{\mu_k g}$

(Don't worry about the negative sign. It will disappear because the initial velocity is also negative, since we take the box initially moving backwards)

Then, plugging in the given values, we calculate the time:

$t=-\frac{(-1.05m/s)}{0.770(9.81m/s^{2})}=0.139s$

In words, the time the box takes to stop sliding relative to the belt is 0.139s.

The displacement of the box in this time, is given by the kinematics formula:

$v^{2}=v_0^{2}+2ax\implies x=-\frac{v_0^{2}}{2\mu_kg}$

Finally, we calculate the displacement:

$x=-\frac{(1.05m/s)^{2} }{2(0.770)(9.81m/s^2)}=-0.0729m=-7.29cm$

This means that the box moves 7.29cm backwards relative to the belt.

4 0

2 years ago

Find the moment of inertia about each of the following axes for a rod that is 0.360 {cm} in diameter and 1.70 {m} long, with a m

mamaluj [8]

The complete question is;

Find the moment of inertia about each of the following axes for a rod that is 0.36 cm in diameter and 1.70m long, with a mass of 5.00 × 10 ^(−2) kg.

A) About an axis perpendicular to the rod and passing through its center in kg.m²

B) About an axis perpendicular to the rod and passing through one end in kg.m²

C) About an axis along the length of the rod in kg.m²

Answer:

A) I = 0.012 kg.m²

B) I = 0.048 kg.m²

C) I = 8.1 × 10^(-8) kg.m²

Explanation:

We are given;

Diameter = 0.36 cm = 0.36 × 10^(−2) m

Length; L = 1.7m

Mass;m = 5 × 10^(−2) kg

A) For an axis perpendicular to the rod and passing through its center, the formula for the moment of inertia is;

I = mL²/12

I = (5 × 10^(−2) × 1.7²)/12

I = 0.012 kg.m²

B) For an axis perpendicular to the rod and passing through one end, the formula for the moment of inertia is;

I = mL²/3

So,

I = (5 × 10^(−2) × 1.7²)/3

I = 0.048 kg.m²

C) For an axis along the length of the rod, the formula for the moment of inertia is; I = mr²/2

We have diameter = 0.36 × 10^(−2) m, thus radius;r = (0.36 × 10^(−2))/2 = 0.18 × 10^(−2) m

I = (5 × 10^(−2) × (0.18 × 10^(−2))^2)/2

I = 8.1 × 10^(-8) kg.m²

3 0

3 years ago

Are small molecular units joined together in large molecules

inysia [295]

<span>Polymers are small molecular units joined together in large molecules.

hope this helps!</span>

7 0

3 years ago

How is modeling useful in studying the structure of the atom?

Mandarinka [93]

Modelling the structure of the atom is important because modeling replaces the real system with something similar but easier to examine. Option B

<h3>What is modeling?</h3>

A model is a representation of reality. We know that a model could help us to recreate reality in a manner that we could be able to relate fully with it. A model could be used also a means of explanation.

The atomic models that we have usually help us to understand more abut the atom. Therefore, modelling the structure of the atom is important because modeling replaces the real system with something similar but easier to examine. Option B

Learn more about modelling the atom:brainly.com/question/1596638

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6 0

1 year ago