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elena-s [515]
2 years ago
9

A 20-turn coil of area 0.32 m2 is placed in a uniform magnetic field of 0.055 T so that the perpendicular to the plane of the co

il makes an angle of 30∘ with respect to the magnetic field.
The flux through the coil is
Physics
1 answer:
makvit [3.9K]2 years ago
5 0

Answer:

1.5 * 10^-2 Tm^2

Explanation:

Electric Flux = B.A cos(theta)

B = 0.055 T

A = 0.32 m^2

theta = 30

Electric Flux = (0.055 T).(0.32 m^2).Cos(30) = 0.0152 = 1.5 * 10^-2 Tm^2

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Find the kinetic energy of a 0.1-kilogram toy truck moving at the speed of 1.1 meters per second.
Viktor [21]

Answer:

0.061 J

Explanation:

The kinetic energy of an object is given by

K=\frac{1}{2}mv^2

where

m is the mass of the object

v is its speed

For the toy truck in the problem, we have

m = 0.1 kg is its mass

v = 1.1 m/s is its speed

Putting the numbers into the equation, we find

K=\frac{1}{2}(0.1 kg)(1.1 m/s)^2=0.061 J

6 0
3 years ago
A runner has a speed of 5m/s and a mass of 130 kg what is his kinetic energy?
arlik [135]
M= 130 kg

v= 5 m/s

kinetik energy = ½• m•v²


= ½ • 130• 5²
= ½•130•25
= ½•3250
= 1625.
4 0
3 years ago
Read 2 more answers
magnitude of velocity does not change but the direction changes. Can we say that it is an accelerated motion?
Olenka [21]

Answer:

MRCORRECT has answered the question

Explanation:

Since velocity is a vector, it can change either in magnitude or in direction. Acceleration is therefore a change in either speed ordirection, or both. Keep in mind that althoughacceleration is in the direction of the changein velocity, it is not always in the direction ofmotion.

4 0
3 years ago
A 17.6-kg block rests on a horizontal table and is attached to one end of a massless, horizontal spring. By pulling horizontally
leonid [27]

In order get the block up to a speed of 3.58 m/s in 1.77 s, it must undergo an acceleration <em>a</em> of

<em>a</em> = (3.58 m/s) / (1.77 s) ≈ 2.02 m/s²

When the spring is getting pulled, Newton's second law tells us

• the net vertical force is

∑ <em>F</em> = <em>n</em> - <em>mg</em> = 0

where ∑ <em>F</em> is the net force, <em>n</em> is the magnitude of the normal force, and <em>mg</em> is the weight of the block - it follows that <em>n</em> = <em>mg</em> ; and

• the net horizontal force is

∑ <em>F</em> = <em>F</em> - <em>f</em> = <em>ma</em>

where <em>F</em> is the applied force, <em>f</em> is kinetic friction, <em>m</em> is the block's mass, and <em>a</em> is the acceleration found earlier. <em>F</em> stretches the spring by <em>x</em> = 0.250 m, so we have

<em>F</em> - <em>f</em> = <em>kx</em> - <em>µn</em> = <em>kx</em> - <em>µmg</em> = <em>ma</em>

where <em>k</em> is the spring constant and <em>µ</em> is the coefficient of kinetic friction.

When the block is being pulled at a constant speed, Newton's second law says

• the net vertical force is still

∑ <em>F</em> = <em>n</em> - <em>mg</em> = 0

so that <em>n</em> = <em>mg</em> again; and

• the net horizontal force is

∑ <em>F</em> = <em>F</em> - <em>f</em> = 0

This time, <em>F</em> stretches the spring by <em>y</em> = 0.0544 m, so we have

<em>F</em> - <em>f</em> = <em>ky</em> - <em>µmg</em> = 0

Solve the equations in boldface for <em>k</em> and <em>µ</em> :

<em>kx</em> - <em>µmg</em> = <em>ma</em>

<em>ky</em> - <em>µmg</em> = 0

==>   <em>k</em> (<em>x</em> - <em>y</em>) = <em>ma</em>

==>   <em>k</em> = <em>ma</em> / (<em>x</em> - <em>y</em>)

==>   <em>k</em> = (17.6 kg) (2.02 m/s²) / (0.250 m - 0.0544 m) ≈ 182 N/m

Then

<em>ky</em> - <em>µmg</em> = 0

==>   <em>µ</em> = <em>ky </em>/ (<em>mg</em>)

==>   <em>µ</em> = (182 N/m) (0.0544 m) / ((17.6 kg) <em>g</em>) ≈ 0.0574

3 0
3 years ago
A hydrogen fuel cell supplies power for a small motor. the fuel cell delivers a current of 0.5 a and a voltage of 0.43 v. what i
gogolik [260]
We want to know what is the power supplied by the power cell if the current I=0.5 A and the voltage V=0.43 V. The equation for power P is P= I*V, so:

P=I*V=0.5*0.43=0.215 W

So the correct answer is that the power cell is supplying the motor with P=0.215 W of power. 
8 0
3 years ago
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