All categories

2 years ago

A

Physics

1 answer:

OleMash [197]2 years ago

7 0

Answer:

R= 5.4 ohms

Explanation:

Given that

V= 9 V

Power ,P= 15 W

Lets take resistor resistance = R

We know that Power given as

$P=\dfrac{V^2}{R}$

V=Voltage

P=Power

R=Resistance

$R=\dfrac{V^2}{P}$

Now by putting the all the values in the above equation

$R=\dfrac{9^2}{15}$

R= 5.4 ohms

Therefore the resistance of the resister will be 5.4 ohms.

You might be interested in

Calculate the change in heat of the aluminum; show all calculations. Calculate the change in heat of the water; show all calcula

tatyana61 [14]

Calculate the change in heat of the aluminum; show all calculations. Calculate the change in heat of the water; show all calculations. Are the two values the same? Why or why not? See the attached picture for the numbers.

I got -3443.14 J for the aluminum and 3443.595 for the water

6 0

2 years ago

Water is flowing in a pipe with a circular cross section but with varying cross-sectional area, and at all points the water comp

slamgirl [31]

(a) 5.66 m/s

The flow rate of the water in the pipe is given by

$Q=Av$

where

Q is the flow rate

A is the cross-sectional area of the pipe

v is the speed of the water

Here we have

$Q=1.20 m^3/s$

the radius of the pipe is

r = 0.260 m

So the cross-sectional area is

$A=\pi r^2 = \pi (0.260 m)^2=0.212 m^2$

So we can re-arrange the equation to find the speed of the water:

$v=\frac{Q}{A}=\frac{1.20 m^3/s}{0.212 m^2}=5.66 m/s$

(b) 0.326 m

The flow rate along the pipe is conserved, so we can write:

$Q_1 = Q_2\\A_1 v_1 = A_2 v_2$

where we have

$A_1 = 0.212 m^2\\v_1 = 5.66 m/s\\v_2 = 3.60 m/s$

and where $A_2$ is the cross-sectional area of the pipe at the second point.

Solving for A2,

$A_2 = \frac{A_1 v_1}{v_2}=\frac{(0.212 m^2)(5.66 m/s)}{3.60 m/s}=0.333 m^2$

And finally we can find the radius of the pipe at that point:

$A_2 = \pi r_2^2\\r_2 = \sqrt{\frac{A_2}{\pi}}=\sqrt{\frac{0.333 m^2}{\pi}}=0.326 m$

6 0

2 years ago

The latent heat of fusion for Aluminium is 3.97 x 105. How much energy would be required to melt 0.75 kg of it?

RoseWind [281]

Answer:

$E = 2.9775\times10^5 J$

Explanation:

Given: The latent heat of fusion for Aluminum is $L = 3.97\times10^5 J/Kg$

mass to be malted m = 0.75 Kg

Energy require to melt E = mL

$E = 3.97\times10^5\times0.75 = 2.9775\times10^5 J$

Therefore, energy required to melt 0.75 Kg aluminum

$E = 2.9775\times10^5 J$

5 0

2 years ago

The wavelengths of radio waves are much _______________ than the wavelength of microwaves. therefore, radio waves carry much ___

nydimaria [60]

The wavelengths of radio waves are much "Longer" than the wavelength of microwaves therefore, radio waves carry much "Lower" <span>energy than a microwave.

Hope this helps!</span>

7 0

2 years ago

Read 2 more answers

A rocket sled accelerates from rest for a distance of 645 m at 16.0 m/s2. A parachute is then used to slow it down to a stop. If

inessss [21]

Answer:

the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake

Explanation:

This problem can be solved using the kinematics relations, let's start by finding the final velocity of the acceleration period

v² = v₀² + 2 a₁ x

indicate that the initial velocity is zero

v² = 2 a₁ x

let's calculate

v = $\sqrt {2 \ 15.0 \ 645}$

v = 143.666 m / s

now for the second interval let's find the distance it takes to stop

v₂² = v² - 2 a₂ x₂

in this part the final velocity is zero (v₂ = 0)

0 = v² - 2 a₂ x₂

x₂ = v² / 2a₂

let's calculate

x₂ = $\frac{ 143.666^2 }{2 \ 18.2}$

x₂ = 573 m

as the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake

3 0

2 years ago