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4 years ago

An object of mass m slides down an incline with angle 0.

Physics

2 answers:

juin [17]4 years ago

8 0

Answer:

B). $F_N - mgcos\theta = 0$

Explanation:

As we know that the force on the mass "m" which slides along the inclined plane is along the inclined surface

So it must have accelerated along that surface only

So we can write the equation for force along the inclined plane as

$F_{net} = ma$

now if we wish to find net force perpendicular to the plane then we can say that it must be zero

Because perpendicular to the plane the net force on the given mass must be zero as it is not accelerating along that direction

So here we can say

$F_{net} = 0$

so we have

$F_N - mgcos\theta = 0$

MAVERICK [17]4 years ago

5 0

<h3>Answer</h3>

FN - mgcosФ = 0

<h3>Explanation</h3>

Given the mass of an object, forces that will be acting on it are as following

Force due to gravity (weight) of an object
The Normal force (the perpendicular force from that the inclined surface exerts on the object)

Any frictional forces (due to static friction or kinetic friction, if applicable) along the incline
The corresponding force component along the incline (downhill force due to gravity)

Force due to gravity = -mgcosΔ

Normal force = fN

Frictional force = -f

force applied forward = mgsinΔ

Horizontal force = mgsinΔ - f

vertical force = fN - mgcosΔ

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A 1 mW laser beam is incident onto a detector. Determine the fractional fluctuation in number of photons intercepted by the dete

Ugo [173]

Answer:

(a) $625 photons$

(b) $625\times 10^{6}photons$

Explanation:

Given that , the power of the laser beam is,

$P=1 mW\\P=1\times 10^{-3} W$

and time is given is,

$t=1\mu s\\t=10^{-6}s$

Now the energy formula for the laser beam is,

$E=P\times t$

Now,

$E=10^{-3}\times 10^{-6} \\E=10^{-9}J$

(a) The value of energy is given,

$E_{1}=10MeV\\E_{1}=10\times 10^{6}\times 1.6\times 10^{-19}J\\ E_{1}=16\times 10^{-13}J$

Now the no of photons's fraction fluctuation is,

$n=\frac{E}{E_{1} }\\ n=\frac{10^{-9} }{16\times 10^{-13} }\\ E_{1}=625 photons$

Therefore the no of photons is $625 photons$ .

(b)The value of energy is given,

$E_{z}=10eV\\E_{z}=16\times 10^{-19}J$

Now the no of photons's fraction fluctuation is,

$n=\frac{E}{E_{z} }\\ n=\frac{10^{-9} }{16\times 10^{-19} }\\n=625\times 10^{6}photons$

Therefore the no of photons is $625\times 10^{6}photons$ .

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3 years ago

a technician connects the red lead of a voltmeter to the b+ [output] terminal of an ac generator and the black lead to the batte

stealth61 [152]

Answer:

He is testing for voltage drop (voltage drop test)

Explanation:

Connecting the red lead of a voltmeter to the B+ output to an ac generator and black lead to a battery +ve terminal measures the voltage drop in the circuit when the engine is turned ON and all electrical accessories too.

with this he knows if the alternator is charging the battery.

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4 years ago

10. If the mass of the Earth is... increased by a factor of 2, then the Fgrav is ______________ by a factor of _______. ... incr

12345 [234]

Answer:

If the mass of the Earth is increased by a factor of 2, then the Fgrav is increased by a factor of 2.

If the mass of the earth is increased by a factor of 3, then Fgrav is increased by a factor of 3.

If the mass of the earth is decreased by a factor of 4, then the Fgrav is decreased by a factor of 4

Explanation:

In order to solve this question, we must take into account that the force of gravity is given by the following formula:

$F_{g0}=G \frac{mM_{E0}}{r^{2}}$

So if the mass of the earth is increased by a factor of 2, this means that:

$M_{Ef}=2M_{E0}$

so:

$F_{gf}=G \frac{2mM_{E0}}{r^{2}}$

Therefore:

$\frac{F_{gf}}{F_{g0}}=\frac{G \frac{2mM_{E0}}{r^{2}}}{G \frac{mM_{E0}}{r^{2}}}$

When simplifying we end up with:

$\frac{F_{gf}}{F_{g0}}=2$

so if the mass of the Earth is increased by a factor of 2, then the Fgrav is increased by a factor of 2.

If the mass of the earth is increased by a factor of 3

So if the mass of the earth is increased by a factor of 2, this means that:

$M_{Ef}=3M_{E0}$

so:

$F_{gf}=G \frac{3mM_{E0}}{r^{2}}$

Therefore:

$\frac{F_{gf}}{F_{g0}}=\frac{G \frac{3mM_{E0}}{r^{2}}}{G \frac{mM_{E0}}{r^{2}}}$

When simplifying we end up with:

$\frac{F_{gf}}{F_{g0}}=3$

so if the mass of the Earth is increased by a factor of 3, then the Fgrav is increased by a factor of 3.

If the mass of the earth is decreased by a factor of 4

So if the mass of the earth is decreased by a factor of 4, this means that:

$M_{Ef}=\frac{M_{E0}}{4}$

so:

$F_{gf}=G \frac{mM_{E0}}{4r^{2}}$

Therefore:

$\frac{F_{gf}}{F_{g0}}=\frac{G \frac{mM_{E0}}{4r^{2}}}{G \frac{mM_{E0}}{r^{2}}}$

When simplifying we end up with:

$\frac{F_{gf}}{F_{g0}}=\frac{1}{4}$

so if the mass of the Earth is decreased by a factor of 4, then the Fgrav is decreased by a factor of 4.

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3 years ago

Two objects form a closed system. One object, which is at 400 K, absorbs 25.0 kJ of heat from the other object,which is at 500 K

Damm [24]

Explanation:

The given data is as follows.

$T_{1}$ = 400 K, $T_{2}$ = 500 K,

$\Delta Q$ = 25 kJ = 25000 J (as 1 kJ = 1000 J)

Now, we will calculate the change in entropy as follows.

$\Delta S = \frac{\Delta Q}{\Delta T}$

Putting the given values into the above formula as follows.

$\Delta S = \frac{\Delta Q}{\Delta T}$

= $\frac{25000 J}{(500 K - 400 K)}$

= 250 J/K

Hence, we can conclude that change in entropy is 250 J/K.

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4 years ago

Write the untis of following physical quantities electric current , force, frequency, density

Shtirlitz [24]

Answer:

1) electric current - ampere

2) force - joule

3 frequency - hertz

4) density - kilogram per cubic metre

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3 years ago

Read 2 more answers