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4 years ago

A machine can never be 100% efficient because some work is always lost due to .

Physics

2 answers:

konstantin123 [22]4 years ago

6 0

Efficiency of machine is defined by

$\eta = \frac{W_{output}}{Q_{input}}$

here we know that

$W_{output}$ = output useful work

$Q_{input}$ = input energy

Now here if we obtain output work same as input thermal energy then it is 100% efficient

but this is only possible when we have no energy loss while it is not possible due to many reasons

one of the main cause of energy loss is friction

So when machine working there must be friction between its working parts due to which out input energy is lost in different parts and hence we can never be 100% efficient machine.

So here main reason is FRICTIONAL LOSS

klasskru [66]4 years ago

5 0

Friction
Hope it helped

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(b) Released from rest at the same height, a can of frozen juice rolls to the bottom of the same hill. What is the translational

Oksanka [162]

Answer:

Explanation:

a) using the energy conservation equation

mgh = 0.5mv^2 + 0.5Iω^2

I(moment of inertia) (basket ball) = (2/3)mr^2

mgh = 0.5mv^2 + 0.5( 2/3mr^2) ( v^2/r^2)

gh = 1/2v^2 + 1/3v^2

gh = v^2( 5/6)

v = $\sqrt{\frac{6gh}{5} }$

putting the values we get

$6.6 ^{2} = \frac{6\times9.8h}{5}$

solving for h( height)

h = 3.704 m apprx

b) velocity of solid cylinder

mgh = 0.5mv^2 + 0.5( mr^2/2)( v^2/r^2) where ( I ofcylinder = mr^2/2)

g*h = 1/2v^2 + 1/4v^2

g*h = 3/4v^2

putting the value of h and g we get

v= = 6.957 m/s apprx

5 0

4 years ago

When you are standing on your toes, where do you exact the force of gravity

zloy xaker [14]

uhhhh, I think it depends your height or weigh...?

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3 years ago

A system consists of a disk of mass 2.33 kg and radius 50 cm upon which is mounted an annular cylinder of mass 2.20 kg with inne

Gekata [30.6K]

Answer:

Kinetic energy of the system = 2547.41 Joules.

Explanation:

Given:

Disk:

Mass of the disk (m) = $2.33$ kg

Radius of the disk (r) = $50$ cm = $\frac{50}{100} =0.5$ m

Cylinder:

Mass of the annular cylinder (M) = $2.20$ kg

Inner radius of the cylinder $(R_i)$ = $0.2$ m

Outer radius of the cylinder $(R_o)$ = $0.3$ m

The angular speed of the system $(\omega)$ = $15.1$ rev/s

Angular speed in in terms of Rad/sec = $15.1\times 2\pi =94.876$ rad/sec

Formula to be used:

Rotational Kinetic energy, $(KE)_r$ = $\frac{I\times \omega^2}{2}$

So, before that we have to work with the moment of inertia (MOI) of the system.

⇒ MOI of the system = MOI of the disk + MOI of the cylinder

⇒ MOI (system) = $\frac{mr^2}{2} +\frac{M(R_i+R_o)^2}{2}$

⇒ MOI (system) = $\frac{2.33\times (0.5)^2}{2} + \frac{2.20\times (0.2+0.3)^2}{2}$

⇒ MOI (system) = $0.566$ kg.m^2

Now

The rotational Kinetic energy.

⇒ $(KE)_r =\frac{I\omega^2}{2}$

Plugging the values.

⇒ $(KE)_r=\frac{0.566\times (94.876)^2}{2}$

⇒ $(KE)_r=2547.41$ Joules

Then

The kinetic energy of the rotational system is 2547.41 J.

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3 years ago

A comet is approaching Venus on a parabolic path with perigee distance of 18,200 km . Calculate the total time of travel (in hou

Anna35 [415]

Answer:

<em>The time traveled is 1.39 hrs</em>

Explanation:

Equation of Trajectory of a comet is given as

$r=\frac{h^2}{\mu}\frac{1}{1+cos \theta}$

Here

h is the specific angular momentum given as

$h=v_p r_p$

μ is gravitational parameter whose value is $3.24859 \times 10^{14} \, \, m^3/s^2$ for Venus
r_p is the perigee distance of parabolic which is 18200 km
As the path of comet is parabolic so energy is conserved i.e

$\frac{1}{2}m_c v_p^2-\frac{\mu m_c}{r_p}=0\\v_p=\sqrt{\frac{2 \mu}{r_p}}$

So h is given as

$h=v_pr_p\\\\h=\sqrt{2 \mu r_p}\\h=\sqrt{2 \times 3.24859 \times 10^{14} \times 18200 \times 10^3}\\h=1.087 \times 10^{11}$

So for point a where r=24500 km

$r_1=\frac{h^2}{\mu}\frac{1}{1+cos \theta_1}\\24500 \times 10^3=\frac{1.1824 \times 10^{22}}{3.24859 \times 10^{14}}\frac{1}{1+cos \theta_1}\\0.6731=\frac{1}{1+cos \theta_1}\\1+cos \theta_1=\frac{1}{0.6731}\\cos \theta_1=1.4857-1\\cos \theta_1=0.4857\\\theta_1=cos^{-1}0.4857\\\theta_1=1.0636 rad$

So for point a where r=39000 km

$r_2=\frac{h^2}{\mu}\frac{1}{1+cos \theta_2}\\39000 \times 10^3=\frac{1.1824 \times 10^{22}}{3.24859 \times 10^{14}}\frac{1}{1+cos \theta_2}\\1.0715=\frac{1}{1+cos \theta_2}\\1+cos \theta_2=\frac{1}{1.0715}\\cos \theta_2=0.9332-1\\cos \theta_2=-0.0667\\\theta_2=cos^{-1}(-0.0667)\\\theta_2=1.6375 rad$

So as per the Barkers equation

$t_1-T=\sqrt{\frac{2 r_p^3}{\mu}}(D_1+\frac{D_1^3}{3})$

where

$D_1=tan (\theta_1/2)\\D_1=tan(0.5318)\\D_1=0.5883$

$t_2-T=\sqrt{\frac{2 r_p^3}{\mu}}(D_2+\frac{D_2^3}{3})$

where

$D_2=tan (\theta_2/2)\\D_1=tan(0.8187)\\D_2=1.0690$

$t_2-t_1=\sqrt{\frac{2 r_p^3}{\mu}}(D_2+\frac{D_2^3}{3}-D_1+\frac{D_1^3}{3})\\t_2-t_1=\sqrt{\frac{2 (18200 \times 10^3)^3}{3.24859 \times 10^{14}}}((1.0690)+\frac{(1.0690)^3}{3}-(0.5883)-\frac{(0.5883)^3}{3})\\t_2-t_1=\sqrt{\frac{2 (18200 \times 10^3)^3}{3.24859 \times 10^{14}}}(0.8201)\\t_2-t_1=(6092)(0.8201)\\t_2-t_1=4996.21 s\\t_2-t_1=1.39 hrs\\$

So the time traveled is 1.39 hrs

3 0

3 years ago

According to Dutch scientist Christiaan Huygens, what was light made of? atoms particles waves electrons

nekit [7.7K]

Answer:

The right answer is actually waves.

Please give branlist

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3 years ago