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miskamm [114]
2 years ago
13

What is average velocity

Physics
1 answer:
Sphinxa [80]2 years ago
6 0

Answer:

Average velocity is the displacement of an object over time that also in a specified direction...

Explanation:

brainliest plz

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A circular loop of radius 11.7 cm is placed in a uniform magnetic field. (a) If the field is directed perpendicular to the plane
Mama L [17]

Answer:

Magnetic field, B = 0.199 T

Explanation:

It is given that,

Radius of circular loop, r = 11.7 cm = 0.117 m

Magnetic flux through the loop, \phi=8.6\times 10^{-3}\ T/m^2

The magnetic flux linked through the loop is :

\phi=B.A

\phi=BA\ cos\theta

Here, \theta=0

B=\dfrac{\phi}{A}

or

B=\dfrac{\phi}{\pi r^2}

B=\dfrac{8.6\times 10^{-3}}{\pi (0.117 )^2}

B = 0.199 T

So, the strength of the magnetic field is 0.199 T. Hence, this is the required solution.

4 0
3 years ago
Power of sunlight on Earth The Sun emits about 3.9 * 1026 J of electromagnetic radiation each second. (a) Estimate the power tha
ratelena [41]

Answer:

a)6.34 x  10^{7}W/m²

b)1.37 x 10^{3 W/m²

c) see explanation.

Explanation:

a)The relation of intensity'I' of the radiation and area 'A' is given by:

I= P/A

where P= power of sunlight i.e 3.9 x 10^{26} J

and the area of the sun is given by,

A= 4πR_{sun} => 4π(\frac{1.4*10^{9} }{2} )^{2}

A=6.15 x 10^{18}m²

I_{sun} =  3.9 x 10^{26} / 6.15 x 10^{18} =><u> 6.34 x  </u>10^{7}<u>W/m²</u>

b) First determine the are of the sphere in order to determine intensity at the surface of the virtual space

A= 4πR

Now R= 1.5 x 10^{11m

A=  4π x 1.5 x 10^{11 =>2.83 x 10^{23 m²

The power that each square meter of Earths surface receives

I_{earth =   3.9 x 10^{26}/2.83 x 10^{23 =><u>1.37 x </u>10^{3<u> W/m²</u>

<u />

c) in part (b), by assuming the shape of the wavefront of the light emitted by the Sun is a spherical shape so each point has the same distance from the source i.e sun on the wavefront.

6 0
3 years ago
The kinetic energy, k, of an object varies jointly with the mass of the object and the square of the velocity of the object. a 5
o-na [289]

Answer:

KE = 1/2 M V^2 = 1/2 * 25 * 10^2 = 1250 J

Check

M2 = 1/2 M1

V2 = V1 / 2

E2 = 1/2 * 1/4 E1 = E1 / 8 = 10000 / 8 = 1250 J

8 0
2 years ago
Read 2 more answers
A 22-turn circular coil of radius 3.00 cm and resistance 1.00 Ω is placed in a magnetic field directed perpendicular to the plan
zloy xaker [14]

Answer:

23.5 mV

Explanation:

number of turn coil  'N' =22

radius 'r' =3.00 cm=> 0.03m

resistance = 1.00 Ω

B= 0.0100t + 0.0400t²

Time 't'= 4.60s

Note that Area'A' = πr²

The magnitude of induced EMF is given by,

lƩl =ΔφB/Δt = N (dB/dt)A

    =N[d/dt (0.0100t + 0.0400 t²)A        

    =22(0.0100 + 0.0800(4.60))[π(0.03)²]

     =0.0235

     =23.5 mV

Thus, the induced emf in the coil at t = 4.60 s is 23.5 mV

8 0
3 years ago
Why are denser materials found closer to earths center
ICE Princess25 [194]
Denser materials tend to be closer to earths center due to their mass gravity is shown by the equation mg
Which stands for mass x gravity.
5 0
3 years ago
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