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2 years ago

Eeeee help pls pls pls

Physics

1 answer:

Dima020 [189]2 years ago

5 0

Answer:

hey dream I m your big fan for u I have answered this question The third one is correct.

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A cylindrical region of radius R contains a uniform magnetic field parallel to its axis. The field is zero outside the cylinder.

PtichkaEL [24]

The magnitude of the induced electric field is (RdB/dt)/4

The induced electric field is gotten from

-∫E.dl = dФ/dt where E = induced electric field, dl = path length vector, Ф = magnetic flux through cylindrical region = AB where A = area of magnetic flux = πR² where R = radius of cylindrical region and B = magnetic field.

So, -∫E.dl = dФ/dt

-∫E.dl = dAB/dt

-∫Edlcos0 = AdB/dt (where E.dl = Edlcos0 = Edl since E and dl are parallel to each other.)

So -∫Edl = πR²dB/dt

-E∫dl = πR²dB/dt (∫dl = 2πr since the integral is the circumference of the path)

-E(2πr) = πR²dB/dt (we integrate dl from r = 0 to 2R)

-E2π(2R - 0) = πR²dB/dt

-E4πR= πR²dB/dt

E = πR²dB/dt ÷ 4πR

E = -(RdB/dt)/4

So, the magnitude of the induced electric field is (RdB/dt)/4

Learn more about induced electric field here:

brainly.com/question/15730392

3 0

2 years ago

A 50 mm diameter steel shaft and a 100 mm long steel cylindrical bushing with an outer diameter of 70 mm have been incorrectly s

BARSIC [14]

Answer:

The axial force is $P = 15.93 k N$

Explanation:

From the question we are told that

The diameter of the shaft steel is $d = 50mm$

The length of the cylindrical bushing $L =100mm$

The outer diameter of the cylindrical bushing is $D = 70 \ mm$

The diametral interference is $\delta _d = 0.005 mm$

The coefficient of friction is $\mu = 0.2$

The Young modulus of steel is $207 *10^{3} MPa (N/mm^2)$

The diametral interference is mathematically represented as

$\delta_d = \frac{2 *d * P_B * D^2}{E (D^2 -d^2)}$

Where $P_B$ is the pressure (stress) on the two object held together

So making $P_B$ the subject

$P_B = \frac{\delta _d E (D^2 - d^2)}{2 * d* D^2}$

Substituting values

$P_B = \frac{(0.005) (207 *10^{3} ) * (70^2 - 50^2))}{2 * (50) (70) ^2 }$

$P_B = 5.069 MPa$

Now he axial force required is

$P = \mu * P_B * A$

Where A is the area which is mathematically evaluated as

$\pi d l$

So $P = \mu P_B \pi d l$

Substituting values

$P = 0.2 * 5.069 * 3.142 * 50 * 100$

$P = 15.93 k N$

8 0

3 years ago

Define 1 unit electricity

Tomtit [17]

Answer:

A unit is represented in kWH or Kilowatt Hour. This is the actual electricity or energy used. If you use 1000 Watts or 1 Kilowatt of power for 1 hour then you consume 1 unit or 1 Kilowatt-Hour (kWh) of electricity.

4 0

2 years ago

If the average distance between bumps on a road is about 10 m and the natural frequency of the suspension system in the car is a

Mnenie [13.5K]

When you hit a bump every 0.9 seconds.

3 0

3 years ago

Observe yourself breathing and count the number of times you inhale per second. During each breath you probably inhale 0.66 L of

Pavel [41]

To solve this exercise it is necessary to apply the concepts related to Robert Boyle's law where:

$PV=nRT$