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3 years ago

An object has a mass of 20 kg. Determine its weight, in N, at a location where the acceleration of gravity is 9.78 m/s2.

Physics

1 answer:

pychu [463]3 years ago

3 0

Answer:

Weight of the object will be 195.6 N

Explanation:

We have given mass of the object m = 20 kg

And acceleration due to gravity at any location is given $g=9.78m/sec^2$

We have to find the weight of the object in N at the location where value of acceleration due to gravity is $g=9.78m/sec^2$

Weight of the object is given by W = mg , here W is weight , m is mass and g is acceleration due to gravity

So weight W = 20×9.78 = 195.6 N

So weight of the object will be 195.6 N

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A 25-turn circular coil of wire has diameter 1.00 m. It is placed with its axis along the direction of the Earth’s magnetic fiel

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Answer:

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Explanation:

Given that,

Number of turns = 25 turns

Diameter = 1.00 m

Magnetic field = 50.0μm

Time = 0.200 s

Angle = 180°

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Using formula of area

$A=\pr r^2$

$A=\pi\times(\dfrac{1.00}{2})^2$

$A=0.785\ m^2$

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Using formula of emf

$E=NAB\cos\theta$

$E=NA(\dfrac{B\cos0^{\circ}-B\cos180^{\circ}}{dt}$

$E=\dfrac{2NA}{dt}$

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B = magnetic field

Put the value into the formula

$E=\dfrac{2\times25\times0.785\times50.0\times10^{-6}}{0.200}$

$E=9.813\ mV$

Hence, The magnitude of average emf is 9.813 mV.

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4 years ago

A 6.0 kg box slides down an inclined plane that makes an angle of 39° with the horizontal. If the coefficient of kinetic frictio

Blababa [14]

Answer:

(B) 1.6 m/s^2

Explanation:

The equation of the forces acting on the box in the direction parallel to the slope is:

$mg sin \theta - \mu N = ma$ (1)

where

$mg sin \theta$ is the component of the weight parallel to the slope, with m = 6.0 kg being the mass of the box, g = 9.8 m/s^2 being the acceleration of gravity, $\theta=39^{\circ}$ being the angle of the incline

$\mu N$ is the frictional force, with $\mu = 0.6$ being the coefficient of kinetic friction, N being the normal reaction of the plane

a is the acceleration

The equation of the force along the direction perpendicular to the slope is

$N-mg cos \theta =0$

where $mg cos \theta$ is the component of the weight in the direction perpendicular to the slope. Solving for N,

$N=mg cos \theta$

Substituting into (1), solving for a, we find the acceleration:

$a=gsin \theta- \mu g cos \theta=(9.8)(sin 39^{\circ})-(0.6)(9.8)(cos 39^{\circ})=1.6 m/s^2$

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