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2 years ago

Please help!! Look at pic

Physics

1 answer:

kkurt [141]2 years ago

7 0

Answer:

Explanation:

Voltage and Current relation is given by V=IR
I=9A

R=4Ohm

V=IR
V=9x4=36Volts

Resistance = 4 Ohm

Current = 9 Ampere

Voltage = 36 Volts

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A proton with a speed of 3.2 x 106 m/s is shot into a region between two plates that are separated by a distance of 0.23 m. As t

harina [27]

Answer:

<h2>The magnitude of the magnetic is 0.145 T</h2>

Explanation:

Given :

Speed of proton $v = 3.2 \times 10^{6}\frac{m}{s}$

Mass of proton $m = 1.67 \times 10^{-27}$ Kg

The force on the proton in magnetic field is given by,

$F = q (v \times B)$

$F = qvB \sin \theta$

But $\sin 90 = 1$ (∵ Force is perpendicular to the velocity so $\theta = 90$ )

$F = qvB$

When particle enter in magnetic field at the angle of 90° so particle moves in circle

So force is given by,

$F = \frac{mv^{2} }{r}$

Where $r =$ radius but in our case 0.23 m, $q = 1.6 \times 10^{-19}$ C

By comparing above two equation,

$B = \frac{mv}{qr}$

$B = \frac{1.67 \times 10^{-27} \times 3.2 \times 10^{6} }{1.6 \times 10^{-19} \times 0.23 }$

$B = 0.145$ T

5 0

3 years ago

If force remains constant and acceleration decreases what must happen to the mass

blondinia [14]

That can only be happening if the mass mysteriously increased somehow. I'd like to know how in the world THAT happened.

4 0

3 years ago

An emf is induced in response to a change in magnetic field inside a loop of wire. Which of the following changes would increase

goldenfox [79]

Answer:

changing the magnetic field more rapidly

Explanation:

According to Faraday's law, whenever there is a change in the magnetic lines of force, it leads the production of induced emf. The magnitude of induced emf is proportional to to the rate of change of flux.

Hence if the magnetic field inside a loop of wire is changed rapidly, the magnitude of induced emf increases in accordance with Faraday's law of electromagnetic induction stated above when the magnetic field is changed more rapidly, hence the answer.

8 0

3 years ago

I would love to stretch a wire from our house to the Shop so I can 'call' my husband in for meals. The wire could be tightened t

dezoksy [38]

Note: I'm not sure what do you mean by "weight 0.05 kg/L". I assume it means the mass per unit of length, so it should be "0.05 kg/m".

Solution:
The fundamental frequency in a standing wave is given by
$f= \frac{1}{2L} \sqrt{ \frac{T}{m/L} }$
where L is the length of the string, T the tension and m its mass. If we plug the data of the problem into the equation, we find
$f= \frac{1}{2 \cdot 24 m} \sqrt{ \frac{240 N}{0.05 kg/m} }=1.44 Hz$

The wavelength of the standing wave is instead twice the length of the string:
$\lambda=2 L= 2 \cdot 24 m=48 m$

So the speed of the wave is
$v=\lambda f = (48 m)(1.44 Hz)=69.1 m/s$

And the time the pulse takes to reach the shop is the distance covered divided by the speed:
$t= \frac{L}{v}= \frac{24 m}{69.1 m/s}=0.35 s$

7 0

4 years ago

A sound is recorded at 19 decibels. What is the intensity of the sound?

sp2606 [1]

$1 \times 10^{-10.1} \mathrm{Wm}^{-2}$ is the intensity of the sound.

Answer: Option B

<u>Explanation:</u>

The range of sound intensity that people can recognize is so large (including 13 magnitude levels). The intensity of the weakest audible noise is called the hearing threshold. (intensity about $1 \times 10^{-12} \mathrm{Wm}^{-2}$ ). Because it is difficult to imagine numbers in such a large range, it is advisable to use a scale from 0 to 100.

This is the goal of the decibel scale (dB). Because logarithm has the property of recording a large number and returning a small number, the dB scale is based on a logarithmic scale. The scale is defined so that the hearing threshold has intensity level of sound as 0.

$\text { Intensity }(d B)=(10 d B) \times \log _{10}\left(\frac{I}{I_{0}}\right)$

Where,

I = Intensity of the sound produced

$I_{0}$ = Standard Intensity of sound of 60 decibels = $1 \times 10^{-12} \mathrm{Wm}^{-2}$

So for 19 decibels, determine I as follows,

$19 d B=(10 d B) \times \log _{10}\left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)$

$\log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=\frac{19}{10}$

$\log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=1.9$

When log goes to other side, express in 10 to the power of that side value,

$\left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)=10^{1.9}$

$I=1 \times 10^{-12} \mathrm{Wm}^{-2} \times 10^{1.9}=1 \times 10^{-12-1.9}=1 \times 10^{-10.1} \mathrm{Wm}^{-2}$

5 0

3 years ago