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3 years ago

During nuclear fission, great amounts of energy are produced from

Physics

2 answers:

stiks02 [169]3 years ago

6 0

D.very small amounts of mass.

True [87]3 years ago

6 0

During nuclear fission, great amounts of energy are produced from very small amounts of mass. <em>(d)</em>

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An object that weighs 2.450 N is attached to an ideal massless spring and undergoes simple harmonic oscillations with a period o

inysia [295]

The answer is B. 2.45 N/m.

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3 years ago

It take a car 3hour to travel 300kilometers

rjkz [21]

100 kilometers per hour

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4 years ago

Two coils that are separated by a distance equal to their radius and that carry equal currents such that their axial fields add

jok3333 [9.3K]

Answer:

When x = 2.8 cm, $B_{x1} = 0.0265 T$

When x = 5.5 cm, $B_{x2} = 0.0209 T$

when x = 7.3 cm, $B_{x3} = 0.0169 T$

When x = 11.0 cm, $B_{x4} = 0.0103 T$

Explanation:

According to Biot-Savart law,

$B_{x} = \frac{N \mu_{o}IR^{2} }{2(x^{2} +R^{2} )^{3/2} }\\$ .......................(1)

R = 11.0 cm = 0.11 m

I = 17.0 A

N = 300 turns

$\mu_{o} = 4\pi * 10^{-7} N/A^{2}$

When x₁ = 2.8 cm = 0.028 m

$B_{x1} = \frac{300 *(4\pi * 10^{-7} ) * 17 *0.11^{2} }{2(0.028^{2} +0.11^{2} )^{3/2} }\\B_{x1} = 0.0265 T$

When x₂ = 5.5cm = 0.055 m

$B_{x2} = \frac{300 *(4\pi * 10^{-7} ) * 17 *0.11^{2} }{2(0.055^{2} +0.11^{2} )^{3/2} }\\B_{x2} = 0.0209 T$

When x₃ = 7.3 cm = 0.073 m

$B_{x3} = \frac{300 *(4\pi * 10^{-7} ) * 17 *0.11^{2} }{2(0.073^{2} +0.11^{2} )^{3/2} }\\B_{x3} = 0.0169 T$

When X₄ = 11.0 cm = 0.11 m

$B_{x4} = \frac{300 *(4\pi * 10^{-7} ) * 17 *0.11^{2} }{2(0.11^{2} +0.11^{2} )^{3/2} }\\B_{x4} = 0.0103 T$

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4 years ago

Santa has lots of mixed up socks in his sack if he has 6 green socks 4 gold socks 8 black socks and 2 red socks what is the mini

Sergio039 [100]

There are four (4) different colors of socks in his sack.

So, even if he's in a dark room and does it all without looking,
he will absolutely positively definitely have at least 1 matching
pair in has hand after he pulls out 5 socks.

The first 4 could have all been different colors. But the 5th sock
MUST match one of the other four.

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3 years ago

1. Define the following in the case of a concave mirror.

insens350 [35]

Answer:

Principal focus of a concave mirror. The principal focus of a concave mirror is a point on its principal axis to which all the light rays are parallel and close to the axis converge after reflection from the concave mirror.

Focal length of a concave mirror. The focal length of a concave mirror is the distance between its pole and the principal focus

The reflecting surface of a spherical mirror forms a part of a sphere. The centre of this sphere. This point is called the centre of curvature of the spherical mirror. Center of curvature can also be defined as the point in the centre of the sphere from which the mirror was sliced. It is represented by the letter C. Please note that the centre of curvature lies outside the mirror's reflecting surface. The centre of curvature of a concave mirror lies in front of it. However, it lies behind the mirror in case of a convex mirror.

If a concave mirror were thought of as being a slice of a sphere, then there would be a line passing through the center of the sphere and attaching to the mirror in the exact center of the mirror. This line is known as the principal axis.

5 0

4 years ago