Ask question

Ask question

All categories

bearhunter [10]

2 years ago

12

a square loop of wire side 3.0 cm carries 3.0 A of current. A uniform magnetic field of magnitude 0.67 T makes an angle of 37 de

grees with the plane of the loop. What is the magnitude of the torque on the loop

1 answer:

Varvara68 [4.7K]2 years ago

8 0

Explanation:

Below is an attachment containing the solution.

You might be interested in

An oven uses 4500 watts of power in 2 minutes, how much work was done?

jolli1 [7]

First of all we state the formula
Power=work done/time
we can rearrange this formula as well
work done=power x time

Since the SI unit of time is in seconds we change the minutes to seconds
2mins= 60x2 = 120 seconds

Using our formula (work done=power x time) we simply put in the values
work done = 4500 x 120
work done = 540,000J

4 0

2 years ago

The specification limits for a product are 8 cm and 10 cm. A process that produces the product has a mean of 9.5 cm and a standa

My name is Ann [436]

Answer:

The value of Cpk is 0.83.

Explanation:

Given that,

Upper specification limits = 10 cm

lower specification limits = 8 cm

Mean = 9.5

Standard deviation = 0.2 cm

We need to calculate the process capability

Using formula of Cpk

$Cpk=min(\dfrac{USL-mean}{3\times SD}, \dfrac{mean-LSL}{3\times SD})$

Put the value into the formula

$Cpk=min(\dfrac{10-9.5}{3\times0.2}, \dfrac{9.5-8}{3\times0.2})$

$Cpk=min(0.83,2.5)$

$Cpk=0.83$

Hence, The value of Cpk is 0.83.

4 0

2 years ago

The real advantage to hydrostatic weighing is that it __________.

katovenus [111]

The real advantage to hydrostatic weighing is that it gives one of the most accurate measurements of body fat.

7 0

2 years ago

Read 2 more answers

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu

vlada-n [284]

Answer:

$\frac{1}{10}$ M

Explanation:

To apply the concept of <u>angular momentum conservation</u>, there should be no external torque before and after

As the <u>asteroid is travelling directly towards the center of the Earth</u>, after impact ,it <u>does not impose any torque on earth's rotation,</u> So angular momentum of earth is conserved

⇒ $I_{1} \times W_{1} =I_{2} \times W_{2}$

$I_{1}$ is the moment of interia of earth before impact
$W_{1}$ is the angular velocity of earth about an axis passing through the center of earth before impact
$I_{2}$ is moment of interia of earth and asteroid system
$W_{2}$ is the angular velocity of earth and asteroid system about the same axis

let $W_{1}=W$

since $\text{Time period of rotation}∝$ $\frac{1}{\text{Angular velocity}}$

⇒ if time period is to increase by 25%, which is $\frac{5}{4}$ times, the angular velocity decreases 25% which is $\frac{4}{5}$ times

therefore $W_{1}$ $=$ $\frac{4}{5} \times W_{1}$

$I_{1}=\frac{2}{5} \times M\times R^{2}$ (moment of inertia of solid sphere)

where M is mass of earth

R is radius of earth

$I_{2}=\frac{2}{5} \times M\times R^{2}+M_{1}\times R^{2}$

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)

where $M_{1}$ is mass of asteroid

⇒ $\frac{2}{5} \times M\times R^{2} \times W_{1}=}(\frac{2}{5} \times M\times R^{2}$ $+$ $M_{1}\times R^{2})\times(\frac{4}{5} \times W_{1})$

$\frac{1}{2} \times M\times R^{2}$ = $(\frac{2}{5} \times M\times R^{2}$ + $M_{1}\times R^{2})$

$M_{1}\times R^{2}=$ $\frac{1}{10} \times M\times R^{2}$

⇒ $M_{1}=}$ $\frac{1}{10} \times M$

3 0

2 years ago

The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz. Find the possible range of wavelengths in ai

taurus [48]

Answer:

The possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.

Explanation:

Given that,

The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz.

The speed of sound in air is 343 m/s.

To find,

The wavelength range for the corresponding frequency.

Solution,

The speed of sound is given by the following relation as :

$v=f_1\lambda_1$

Wavelength for f = 45 Hz is,

$\lambda_1=\dfrac{v}{f_1}$

$\lambda_1=\dfrac{343}{45}=7.62\ m$

Wavelength for f = 375 Hz is,

$\lambda_2=\dfrac{v}{f_2}$

$\lambda_2=\dfrac{343}{375}=0.914\ m/s$

So, the possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.

6 0

3 years ago