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krok68 [10]
2 years ago
12

A circular loop of wire of radius 18 cm moves at a speed of 6.3 m/s in the direction of a 0.16 T uniform magnetic field, perpend

icular to the plane of the loop. What is the emf induced in the loop
Physics
1 answer:
Degger [83]2 years ago
3 0

Answer:

0

Explanation:

Since every value is constant, it shows us that there is no induced emf in the loop. This is according to Faraday's Law [ ε = -N(Δф/Δτ) ].

Best of Luck!

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Energy that flows from an objective with a higher temperature to an object with a lower temperature
dalvyx [7]
That's THERMAL energy, often referred to as "heat".
7 0
3 years ago
A soccer player kicks a rock horizontally off a 37 m high cliff into a pool of water. If the player hears the sound of the splas
Alchen [17]
<h2>Answer:</h2>

<em><u>(a). 16.741 m/s</u></em>

<em><u>(b). 15.75 m/s</u></em>

<h2>Explanation:</h2>

In the question,

Height of the cliff, h = 37 m

Time taken to reach the sound to us = 2.92 s

Speed of the sound in air at room temperature = 343 m/s

Now,

Let us say the speed of the ball = u m/s

So,

Time taken by the ball to reach at the bottom, t is given by,

t=\sqrt{\frac{2h}{g}}\\t=\sqrt{\frac{2\times 37}{9.8}}\\t=2.747\,s

So,

Splash is heard after = 2.92 s

So,

<u>Time taken by sound to travel the shortest distance along the hypotenuse of the triangle</u> thus formed is,

t = 2.92 - 2.747

t = 0.172 s

Now,

Distance traveled by sound is given by,

Distance=343\times 0.172\\Distance=59.02\,m

So,

In the triangle using the Pythagoras Theorem,

Horizontal distance traveled is,

D=\sqrt{59.02^{2}-37^{2}}\\D=45.99\,m

So,

Speed of throwing of ball is given by,

Distance=Speed\times Time\\45.99=u\times 2.747\\u=16.741\,m/s

<em><u>Therefore, the speed of the ball = 16.741 m/s.</u></em>

(b).

If,

Speed of sound = 331 m/s

So,

<u>Distance traveled by sound</u> is,

Distance=331\times 0.172\\Distance=56.932\,m

So,

Distance traveled in the horizontal by ball is,

Distance=\sqrt{56.932^{2}-37^{2}}\\Distance=43.269\,m

So,

Speed of the ball thrown is given by,

Speed=\frac{Distance}{Time}\\Speed=\frac{43.269}{2.747}\\Speed=15.75\,m/s

<em><u>Therefore, the speed of the ball = 15.75 m/s.</u></em>

8 0
3 years ago
The volume of liquid flowing per second is called the volume flow rate Q and has the dimensions of [L]3/[T]. The flow rate of li
melamori03 [73]

Answer: n=4

Explanation:

We have the following expression for the volume flow rate Q of a hypodermic needle:

Q=\frac{\pi R^{n}(P_{2}-P_{1})}{8\eta L}  (1)

Where the dimensions of each one is:

Volume flow rate Q=\frac{L^{3}}{T}

Radius of the needle R=L

Length of the needle L=L

Pressures at opposite ends of the needle P_{2} and P_{1}=\frac{M}{LT^{2}}

Viscosity of the liquid \eta=\frac{M}{LT}

We need to find the value of n whicha has no dimensions, and in order to do this, we have to rewritte (1) with its dimensions:

\frac{L^{3}}{T}=\frac{\pi L^{n}(\frac{M}{LT^{2}})}{8(\frac{M}{LT}) L}  (2)

We need the right side of the equation to be equal to the left side of the equation (in dimensions):

\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{n}}{LT}  (3)

\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{n-1}}{T}  (4)

As we can see n must be 4 if we want the exponent to be 3:

\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{4-1}}{T}  (5)

Finally:

\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{3}}{T}  (6)

8 0
3 years ago
What type of stress is shown in the following diagram?
leva [86]
Here stress is parallel to the surface of the body. So it's a Shear stress.
7 0
3 years ago
Read 2 more answers
Determine the thrust that a boat with a volume of 1.2m³ receives when it is stranded at sea. The density of seawater is 1020kg /
puteri [66]

Answer:

The maximum possible up-thrust on the boat is 11,995.2 N

Explanation:

According to Archimedes' principle, the thrust received by an object immersed a fluid is equal to the weight of the fluid displaced;

The given parameter of the boat in sea water are;

The volume of the boat = 1.2 m³

The density of seawater = 1020 kg/m³

Density = Mass/Volume

Therefore, Mass = Density × Volume

The maximum volume of water that the boat displaces = 1.2 m³

The mass of the water displaced by the boat = (Density of seawater) × (Volume of seawater displaced)

∴ The maximum possible mass of the water displaced by the boat = 1.2 m³ × 1020 kg/m³ = 1224 kg

The maximum possible mass of the water displaced by the boat, m = 1224 kg

Weight = Mass, m × g

Where;

g = The acceleration due to gravity = 9.8 m/s²

The up-thrust on the boat = The weight of the seawater displaced

∴ The maximum possible up-thrust on the boat = m × g = 1224 kg × 9.8 m/s² = 11,995.2 N

The maximum possible up-thrust on the boat = 11,995.2 N.

3 0
2 years ago
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