If a girl is running along a straight road with a uniform velocity 1.5 m/s, find her

Physics

1 answer:

Dafna1 [17]3 years ago

7 0

Answer:

Explanation:

The definition of acceleration is the change in velocity over a period of time. If the girl's velocity is constant, that means it's not changing. Therefore, acceleration is 0 m/s/s

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Which of the following is an example of revolution?

Slav-nsk [51]

The example of revolution is movement of the earth
around the sun. the example of revolution is the war fought between the colonial people and great Britain. An example of revolution is the introduction of the automobile into society.

hope that helps because you didn't give no question answer <span />

4 0

3 years ago

Read 2 more answers

A solid conducting sphere of radius 2 cm has a charge of 8microCoulomb. A conducting spherical shell of inner radius 4 cm andout

nika2105 [10]

Answer:

C) 7.35*10⁶ N/C radially outward

Explanation:

If we apply the Gauss'law, to a spherical gaussian surface with radius r=7 cm, due to the symmetry, the electric field must be normal to the surface, and equal at all points along it.
So, we can write the following equation:

$E*A = \frac{Q_{enc} }{\epsilon_{0}} (1)$

As the electric field must be zero inside the conducting spherical shell, this means that the charge enclosed by a spherical gaussian surface of a radius between 4 and 5 cm, must be zero too.
So, the +8 μC charge of the solid conducting sphere of radius 2cm, must be compensated by an equal and opposite charge on the inner surface of the conducting shell of total charge -4 μC.
So, on the outer surface of the shell there must be a charge that be the difference between them:

$Q_{enc} = - 4e-6 C - (-8e-6 C) = + 4 e-6 C$

Replacing in (1) A = 4*π*ε₀, and Qenc = +4 μC, we can find the value of E, as follows:

$E = \frac{1}{4*\pi*\epsilon_{0} } *\frac{Q_{enc} }{r^{2} } = \frac{9e9 N*m2/C2*4e-6C}{(0.07m)^{2} } = 7.35e6 N/C$

As the charge that produces this electric field is positive, and the electric field has the same direction as the one taken by a positive test charge under the influence of this field, the direction of the field is radially outward, away from the positive charge.

6 0

4 years ago

A metal wire 1.50 m long has a circular cross section of radius 0.32 mm and an end-to-end resistance of 90.0 Ohms. The metal wir

elixir [45]

Answer:

So after stretching new resistance will be 0.1823 ohm

Explanation:

We have given initially length of the wire $l_1=150m$