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3 years ago

The force of repulsion between two like-charged particles will increase if

Physics

2 answers:

Andru [333]3 years ago

6 0

If the distance between the charged particles is deceased

Elanso [62]3 years ago

5 0

Answer:

B). the distance between the charged particles is decreased.

Explanation:

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A uniform magnetic field is perpendicular to the plane of a circular loop of diameter 13 cm formed from wire of diameter 2.6 mm

I am Lyosha [343]

Answer:

Rate of change of magnetic field is $3.466\times 10^3T/sec$

Explanation:

We have given diameter of the circular loop is 13 cm = 0.13 m

So radius of the circular loop $r=\frac{0.13}{2}=0.065m$

Length of the circular loop $L=2\pi r=2\times 3.14\times 0.065=0.4082m$

Wire is made up of diameter of 2.6 mm

So radius $r=\frac{2.6}{2}=1.3mm=0.0013m$

Cross sectional area of wire $A=\pi r^2=3.14\times0.0013^2=5.30\times 10^{-6}m^2$

Resistivity of wire $\rho =2.18\times 10^{-8}m$

Resistance of wire $R=\frac{\rho L}{A}=\frac{2.18\times 10^{-8}\times 0.4082}{5.30\times 10^{-6}}=1.67\times 10^{-3}ohm$

Current is given i = 11 A

So emf $e=11\times 1.67\times 10^{-3}=0.0183volt$

Emf induced in the coil is $e=-\frac{d\Phi }{dt}=-A\frac{dB}{dt}$

$0.0183=5.30\times 10^{-6}\times \frac{dB}{dt}$

$\frac{dB}{dt}=3.466\times 10^3=T/sec$

8 0

3 years ago

The length of vector A is 93.8 meters and the length of Ay is 38.4 meters, then the length of A, must be

vaieri [72.5K]

Answer:

85.556metres

Explanation:

Using pythagorean theorem

C²=A²+B²

we have c as the hypotenuse vector A thus:

93.8²=A²+38.4²

93.8²-38.4²=A²

8794.44-1474.56=A²

7319.88=A²

A=85.556

3 0

3 years ago

Which description is evidence of chemical weathering?

Gelneren [198K]

Answer:

Explanation: I do not know much about this but the answer that i think it is was b.

8 0

3 years ago

On the asteroid Cere, the acceleration due to gravity is about 0.27 m/s^2. How long would it take a ball to fall to the ground i

KiRa [710]

Answer:

F. 25.82 s

Explanation:

Given:

Δy = 90 m

v₀ = 0 m/s

a = 0.27 m/s²

Find: t

Δy = v₀ t + ½ at²

90 m = (0 m/s) t + ½ (0.27 m/s²) t²

t = 25.82 s

7 0

3 years ago

How does the Earth release energy back into the atmosphere?

JulsSmile [24]

In general, the Earth releases energy back to the atmosphere through reflection, evaporation, and radiation. The Earth gets energy from the sunlight, part of which it absorbs, while part it reflects backwards, thus giving energy to the atmosphere. Also, the heating up of the Earth by the absorbed sunlight, radiates back in the lower layers of the atmosphere, again giving back energy to it. The water vapor is another way in which the Earth gives back energy tot he atmosphere as through the evaporation, the water vapor gets into the lower parts of the atmosphere and gives energy to it.

8 0

3 years ago