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Ostrovityanka [42]
3 years ago
10

A rectangular loop of wire that can rotate about an axis through its center is placed between the poles of a magnet in a magneti

c field with a strength of 0.40 T. The length of the loop L is 0.16 m and its width w is 0.040 m. What is the magnetic flux through the loop when the plane of the loop is perpendicular to the magnetic field?

Physics
1 answer:
Afina-wow [57]3 years ago
3 0

Explanation:

hope this helps you...........

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1. The Moon's mass is 7.34 x 1022 kg, and it is 3.8 x 105 km away from Earth. Earth's
sineoko [7]

Answer:

2.03 x 10²⁴N

Explanation:

Given parameters:

Mass of moon = 7.34 x 10²²kg

Mass of the earth  = 5.97 x 10²⁴kg

Distance  = 3.8 x 10⁵km

Unknown:

Gravitational force of attraction  = ?

Solution:

To find the gravitational force of attraction between the masses, we use the expression below;

   F = \frac{Gm_{1} m_{2}  }{r^{2} }

G is the universal gravitation constant

m is the mass

1 and 2 represents moon and earth

r is  the distance

  F = \frac{6.67 x 10^{-11}  x 7.34 x 10^{22} x 5.97 x 10^{24}  }{(3.8 x 10^{5})^{2}  }

 F = \frac{2.92 x 10^{35} }{1.44 x 10^{11} }  = 2.03 x 10²⁴N

8 0
2 years ago
A box slides down a ramp inclined at 24◦ to the horizontal with an acceleration of 1.7 m/s 2 . The acceleration of gravity is 9.
dsp73

Answer:

<h3>0.445</h3>

Explanation:

In friction, the coefficient of friction formula is expressed as;

\mu = \frac{F_f}{R}

Ff is the frictional force = Wsinθ

R is the reaction = Wcosθ

Substitute inti the equation;

\mu = \frac{Wsin \theta}{W cos\theta} \\\mu = \frac{sin \theta}{cos\theta} \\\mu = tan \theta

Given

θ = 24°

\mu = tan 24^0\\\mu = 0.445\\

Hence the coefficient of kinetic friction between the box and the ramp is 0.445

3 0
3 years ago
How far does a runner run if he runs for 60 seconds at 5 m/s
mylen [45]

Answer: 20 miles

Explanation:

4 0
2 years ago
A wire of radius R has a current I uniformly distributed across its cross-sectional area. Ampere's law is used with a concentric
MrMuchimi

Answer:

Please refer to the figure.

Explanation:

The crucial point here is to calculate the enclosed current. If the current I is flowing through the whole cross-sectional area of the wire, the current density is

J = \frac{I}{\pi R^2}

The current density is constant for different parts of the wire. This idea is similar to that of the density of a glass of water is equal to the density of a whole bucket of water.

So,

J = \frac{I}{\pi R^2} = \frac{I_{enc}}{\pi r^2}\\I_{enc} = \frac{Ir^2}{R^2}

This enclosed current is now to be used in Ampere’s Law.

\mu_o I_{enc} = \int {B} \, dl

Here, \int \, dl represents the circular path of radius r. So we can replace the integral with the circumference of the path, 2\pi r.

As a result, the magnetic field is

B = \frac{\mu_0}{2\pi}\frac{Ir}{R^2}

5 0
3 years ago
A robot standing on a cliff shoots a ball upwards with an initial speed of 30 m/s. What is the height of the cliff given that th
valina [46]

Answer:

C 80 m

Explanation:

Given:

v₀ = 30 m/s

a = -10 m/s²

t = 8 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (30 m/s) (8 s) + ½ (-10 m/s²) (8 s)²

Δy = -80 m

The ball lands 80 m below where it started.  So the height of the cliff is 80 m.

5 0
2 years ago
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