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3 years ago

What is power when 3.2 A flows through an 18 ohm resistor?

Physics

1 answer:

vichka [17]3 years ago

4 0

To find the power formula on the electrical side you should have to know ohm's law of voltage that is V=I*R.

Here power formula is P=V*I.

Interms of resistance P=(IR)*I ==>I² R.

Here the power dissipated when 3.2 ampere of current flows through an 18-ohm resistor is

P=I² R ==>(3.2)² (18).

====>10.24(18)=184.32W.

W= Watts.

Watts is a unit of power.

Hope this helps

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You pour 250 g of tea into a Styrofoam cup, initially at 80∘C and stir in a little sugar using a 100-g aluminum 20∘C spoon and l

jok3333 [9.3K]

To solve this problem it is necessary to apply the concepts related to the conservation of energy and heat transferred in a body.

By definition we know that the heat lost must be equal to the heat gained, ie

$Q_g = Q_l$

Where,

Q = Heat exchange

The heat exchange is defined as

$Q = c_p m \Delta T$

Where,

$c_p =$ Specific heat

m = mass

$\Delta T=$ Change in Temperature

Therefore replacing we have that

$Q_g = Q_l$

$c_{p-tea} m \Delta T = c_{p-al} m \Delta T$

Replacing with our values we have that

$0.25*4180*(80-T) = 0.1*900*(T-20)$

$11.61*(80-T) = T-20$

$T= \frac{948.8}{11.61}$

$T = 75.24\°C$

Therefore the highest possible temperature of the spoon when you finally take it out of the cup is 75.24°C

3 0

3 years ago

A 425 g block is released from rest at height h0 above a vertical spring with spring constant k = 460 N/m and negligible mass. T

Morgarella [4.7K]

Answer:

(a) = +5.38m (b) = -5.38m (c) = 1.246m (d) = +0.3771m.

Explanation:

Initially the spring is at equilibrium,

Work done by all forces = change in kinetic energy

Work = ∇K.E

Work = Kf -Ki =0

Since the work done = 0 since the body is at rest.

W(spring) + W(gravity) = 0

W(spring) = -W(gravity)

Work done by the block on the spring = W(block/spring)

W(block/spring) + W(spring) = 0

W(spring) = -∫kx.dx

W(spring) = ½k(X²i - X²f) ; Xi =0, Xf = 15.3cm = 0.153m

W(spring) = -½* 460 * (0.153)²

W(spring) = - 5.38NM

Work done by block on spring = + 5.38NM

(b). Workdone by spring on the block = -5.38NM.

Note: This is so because the displacement of the force is in the opposite direction to the previous one since they counter each other to maintain equilibrium.

(C). W(spring) +W(gravity) = 0

½kx² + mg(h + x) = 0

-5.83 + mg(h + 0.153) =0

5.83 = 0.425*9.8 (h + 0.153)

5.83 = 4.165(h + 0.153)

H = 1.399 - 0.153

H = 1.246m

(D).

If the release height was 6ho

H = 6* 1.246m = 7.476m

W(spring) = W(gravity)

½kx² = mg(7.476 + x)

Note: At maximum compression, the blocks would be at rest.

½Kx² = mg(h + x)

½ * 460 * x² = 0.425 * 9.8 * (7.476 + x)

230x² = 4.165 (7.476 + x)

230x² = 31.137 + 4.165x

230x² - 4.165x - 31.137 = 0

Solving the quadratic equation ( i would suggest you use formula method for easy navigation of the variables)

X = + 0.3771m or -0.3589m

But we can't have a negative compression value,

X = + 0.3771m

7 0

4 years ago

Read 2 more answers

A current-carrying wire is bent into a circular loop of radius R and lies in an xy plane. A uniform external magnetic field B in

Ilya [14]

Answer:

F = 2π I R B

Explanation:

The magnetic force is described by the equation.

F = q v x B = i L x B

Where i is the current, L is a vector that points in the direction of the current (length) and B is the magnetic field.

This equation can be used in scalar form and the direction of the force found by the right hand ruler, the thumb goes in the direction of L, the fingers extended in the direction of B and the palm of the hand indicates the direction of the force if the load is positive

F = i L B sin θ

In this case the wire is in the xy plane and the z-axis field whereby they are perpendicular, θ = 90º and sin 90 = 1

F = i L B

The loop length is

L = 2π R

F = i 2π R B

F = 2π I R B

The force is in the loop

8 0

3 years ago

A 1100 kg car rounds a curve of radius 68 m banked at an angle of 16 degrees. If the car is traveling at 95 km/h, will a frictio

Mariulka [41]

Answer:

Yes. Towards the center. 8210 N.

Explanation:

Let's first investigate the free-body diagram of the car. The weight of the car has two components: x-direction: towards the center of the curve and y-direction: towards the ground. Note that the ground is not perpendicular to the surface of the Earth is inclined 16 degrees.

In order to find whether the car slides off the road, we should use Newton's Second Law in the direction of x: F = ma.

The net force is equal to $F = \frac{mv^2}{R} = \frac{1100\times (26.3)^2}{68} = 1.1\times 10^4~N$

Note that 95 km/h is equal to 26.3 m/s.

This is the centripetal force and equal to the x-component of the applied force.

$F = mg\sin(16) = 1100(9.8)\sin(16) = 2.97\times10^3$

As can be seen from above, the two forces are not equal to each other. This means that a friction force is needed towards the center of the curve.

The amount of the friction force should be $8.21\times 10^3~N$

Qualitatively, on a banked curve, a car is thrown off the road if it is moving fast. However, if the road has enough friction, then the car stays on the road and move safely. Since the car intends to slide off the road, then the static friction between the tires and the road must be towards the center in order to keep the car in the road.

5 0

4 years ago

A tall tube is evacuated and its stopcock is closed. The open end of the tube is immersed in a container of water (density 10^3

LuckyWell [14K]

Answer:

10.19 m

Explanation:

Water will rise to equalize the pressure inside and outside the tube.

The equation of pressure is given by

$p=\rho gh$

Where,

p = Pressure of air = 10⁵ N/m²

ρ = Density of water = 10³ kg/m³

g = Acceleration due to gravity = 9.81 m/s²

h = Height the water will rise

$h=\frac{p}{gh}\\\Rightarrow h=\frac{10^5}{9.81\times 10^3}\\\Rightarrow h=10.19\ m$

∴ The water will rise by 10.19 m.

7 0

3 years ago