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3 years ago

12

a body of mass 2kg is released from rest on a smooth plane at an angle of 60 degree to the horizontal.calculate the acceleration

of the body down the plane (g=10ms-2

1 answer:

Lelu [443]3 years ago

5 0

Answer:

8.66m/s2

Explanation:

from newton second laws F=ma

the force down the plane is the component force along the plane which is mgsin₩ so there fore

a= gsin₩ = 10sin60= 8.66m/s2

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When is the force on a current-carrying wire in a magnetic field at its strongest?

Hitman42 [59]

The forces on a current-carrying wire in a magnetic field are at their strongest when the current is at a 90-degree angle to the field. Option D is correct.

<h3>What is a magnetic field?</h3>

It is the type of field where the magnetic force is obtained. The magnetic force is obtained by the field felt around a moving electric charge.

The complete question is;

"When is the force on a current-carrying wire in a magnetic field at its strongest?

-when the current is at a 0-degree angle to the field

-when the current is at a 30-degree angle to the field

-when the current is at a 45-degree angle to the field

-when the current is at a 90-degree angle to the field"

The magnetic force is found as;

F=BILSINΘ

Where,

Magnetic Field, B

Length of the wire, L

The angle between field and current, Θ

When Θ=90°

The value of the magnetic force is;

F=BIL

When the current is flowing at a 90-degree angle to the magnetic field, the forces acting on a wire carrying a current are the strongest.

Hence, option D is correct.

To learn more about the magnetic field, refer to the link;

brainly.com/question/19542022

#SPJ1

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2 years ago

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 $e=-\frac{d\Phi }{dt}=-A\frac{dB}{dt}$

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

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

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3 years ago

Winds tend to rotate in a counter clockwise direction in the ___ (northern or southern) Hemisphere as they move into a low press

lana [24]

Answer:

Winds tend to rotate in a counter clockwise direction in the center of northern and southern hemisphere.

Explanation:

The wind blows clockwise around a high pressure area in the northern hemisphere and the wind blows counter - clockwise around low pressure.

In the northern hemisphere High-pressure systems rotate clockwise direction and in the southern hemisphere low-pressure systems rotate clockwise direction.

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3 years ago

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An asteroid orbiting the Sun has a mass of 4.00×1016 kg. At a particular instant, it experiences a gravitational force of 3.14×1

Ksivusya [100]

<h2>The asteroid is 4.11 x 10¹¹ m far from Sun</h2>

Explanation:

We have gravitational force

$F=\frac{GMm}{r^2}$

Where G = 6.67 x 10⁻¹¹ N m²/kg²

M = Mass of body 1

M = Mass of body 2

r = Distance between them

Here we have

M = Mass of Sun = 1.99×10³⁰ kg

m = Mass of asteroid = 4.00×10¹⁶ kg

F = 3.14×10¹³ N

Substituting

$F=\frac{GMm}{r^2}\\\\3.14\times 10^{13}=\frac{6.67\times 10^{-11}\times 1.99\times 10^{30}\times 4\times 10^{16}}{r^2}\\\\r=4.11\times 10^{11}m$

The asteroid is 4.11 x 10¹¹ m far from Sun

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3 years ago

How much does coast to coast membership cost?

CaHeK987 [17]

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