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nadezda [96]
3 years ago
5

A 0.50-T magnetic field is directed perpendicular to the plane of a circular loop of radius 0.25 m. What is the magnitude of the

magnetic flux through the loop
Physics
1 answer:
Cloud [144]3 years ago
6 0

Answer:

The magnitude of the magnetic flux through the loop is 0.0982 T.m²

Explanation:

Given;

magnitude of magnetic field, B = 0.5 T

radius of the loop, r = 0.25 m

Area of the loop is given by;

A = πr²

A = 3.142 x (0.25)²

A = 0.1964 m²

The magnitude of the magnetic flux through the loop is given by;

Ф = BA

Where;

B is the magnitude of the magnetic field

A is area of the field

Ф = 0.5 x 0.1964

Ф = 0.0982 T.m²

Therefore, the magnitude of the magnetic flux through the loop is 0.0982 T.m²

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the pygmy shrew has an average mass of 2.0 g if 49 of these shrew are placed on a spring scale with a spring constant of 24 N/m
olga_2 [115]

Answer:

Spring's displacement, x = -0.04 meters.

Explanation:

Let the spring's displacement be x.

Given the following data;

Mass of each shrew, m = 2.0 g to kilograms = 2/1000 = 0.002 kg

Number of shrews, n = 49

Spring constant, k = 24 N/m

We know that acceleration due to gravity, g is equal to 9.8 m/s².

To find the spring's displacement;

At equilibrium position:

Fnet = Felastic + Fg = 0

But, Felastic = -kx

Total mass, Mt = nm

Fg = -Mt = -nmg

-kx -nmg = 0

Rearranging, we have;

kx = -nmg

Making x the subject of formula, we have;

x = \frac {-nmg}{k}

Substituting into the formula, we have;

x = \frac {-49*0.002*9.8}{24}

x = \frac {-0.9604}{24}

x = -0.04 m

Therefore, the spring's displacement is -0.04 meters.

3 0
3 years ago
Hooke's Law Practice
Oxana [17]

Answer:sheeExplanation:

8 0
3 years ago
An ac series circuit has an impedance of 60 Ohm and
Iteru [2.4K]

Answer:

Power factor of the AC series circuit is cos\phi=0.5

Explanation:

It is given that,

Impedance of the AC series circuit, Z = 60 ohms

Resistance of the AC series circuit, R = 30 ohms

We need to find the power factor of the circuit. It is given by :

cos\phi=\dfrac{R}{Z}

cos\phi=\dfrac{30}{60}

cos\phi=\dfrac{1}{2}

cos\phi=0.5

So, the power factor of the ac series circuit is cos\phi=0.5. Hence, this is the required solution.

6 0
3 years ago
Objects with masses of 165 kg and a 465 kg are separated by 0.340 m. (a) Find the net gravitational force exerted by these objec
Dmitrij [34]

Answer:

(a) F_{net} = 4.19 x 10^{-5} N, and its direction is towards m_{2}.

(b) It must be placed inside a hollow shell.

Explanation:

Let, m_{1} = 165 kg, m_{2} = 465 kg, m_{3} = 60 kg, and the distance between m_{1} and m_{2} is 0.340 m.

(a) Since m_{3} is placed midway between m_{1} and m_{2}, then its distance to both masses is 0.170 m.

From the Newton's law of universal gravitation,

F = \frac{Gm_{1}m_{2}  }{r^{2} }

Where all variables have their usual meaning.

Then,

a. F_{net} = F_{23} - F_{13}

F_{13} = \frac{6.67*10^{-11}*165*60 }{(0.17)^{2} }

     = 2.25 x 10^{-5} N

F_{23} = \frac{6.67*10^{-11}*465*60 }{(0.17)^{2} }

     = 6.44 x 10^{-5} N

∴ F_{net} =  = 6.44 x 10^{-5} - 2.25 x 10^{-5}

              = 4.19 x 10^{-5} N

The net force exerted by the two masses on the 60 kg object is 4.19 x 10^{-5}  N.

(ii) /F_{net}/ = /F_{23}/ - /F_{13}/

              = 6.44 x 10^{-5} - 2.25 x 10^{-5}

              = 4.19 x 10^{-5} N

(iii) The direction of the net force is to the right i.e towards m_{2}.

(b) For the net force experienced by the 60 kg object to be zero, it must be placed inside a hollow shell.

7 0
3 years ago
Pls help will give brainliest
max2010maxim [7]

Explanation:

the correct answer is Option 5.√½gh.

hope this helps you.

5 0
3 years ago
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