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3 years ago

True of false efficiency compared the output work to the output force

1 answer:

8 0

The statement about "<span>efficiency compared the output work to the output force" is false. Efficiency can be compared from the input work to the output work.</span>

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In the Bohr model of the atom, atomic electrons approximatately 'orbit' the nucleus. The hydrogen atom consists of a proton of m

lara31 [8.8K]

Answer:

$F=3.61\times 10^{-47}\ N$

Explanation:

Mass of a proton, $m_p=1.67\times 10^{-27}\ kg$

Mass of an electron, $m_e=9.11\times 10^{-31}\ kg$

The distance between the electron and the proton is, $r=5.3\times 10^{-11}\ m$

We need to find the mutual attractive gravitational force between the electron and proton. The gravitational force is given by :

$F=G\dfrac{m_em_p}{r^2}$

Where G is the universal Gravitational constant

$F=6.67\times 10^{-11}\times \dfrac{9.11\times 10^{-31}\times 1.67\times 10^{-27}}{(5.3\times 10^{-11})^2}\\\\F=3.61\times 10^{-47}\ N$

So, the force between the electron and proton is $3.61\times 10^{-47}\ N$ .

7 0

2 years ago

In the Millikan oil drop experiment the measured charge of any single droplet was always a whole number multiple of -1.60 x 10-1

BaLLatris [955]

Answer:

Number of electrons, n = 6

Explanation:

Total charge in a single droplet, $q=-9.6\times 10^{-19}\ C$

The measured charge of any single droplet, $e=-1.6\times 10^{-19}\ C$

Let n is the number of excess electrons are contained within the drop. According to the quantization of charge :

$q=ne$

$n=\dfrac{q}{e}$

$n=\dfrac{-9.6\times 10^{-19}}{-1.6\times 10^{-19}}$

n = 6

So, there are 6 electrons contained within the drop. Hence, this is the required solution.

7 0

3 years ago

skater spins over a point at a speed of 3.0 rotations per second then the momentum of inertia is 0.60 kg.M2, what is its angular

laiz [17]

Answer:

$L=11.3\ kg-m^2/s$

Explanation:

Given that,

Angular speed of a skater, $\omega=3\ rot/s=18.84\ rad/s$

The moment of inertia of the skater, I = 0.6 kg-m²

We need to find the angular momentum of the skater. The formula for the angular momentum of the skater is given by :

$L=I\omega$

Substitute all the values,

$L=0.6\times 18.84\\\\L=11.3\ kg-m^2/s$

So, its angular momentum is equal to $11.3\ kg-m^2/s$ .

8 0

2 years ago

How do I solve this step by step? I’m really confused

LekaFEV [45]

Step-#1:

Ignore the wire on the right.

Find the strength and direction of the magnetic field at P,

caused by the wire on the left, 0.04m away, carrying 5.0A

of current upward.

Write it down.

Step #2:

Now, ignore the wire on the left.

Find the strength and direction of the magnetic field at P,

caused by the wire on the right, 0.04m away, carrying 8.0A

of current downward.

Write it down.

Step #3:

Take the two sets of magnitude and direction that you wrote down

and ADD them.

The total magnetic field at P is the SUM of (the field due to the left wire)

PLUS (the field due to the right wire).

So just calculate them separately, then addum up.

4 0

3 years ago

A gyroscope slows from an initial rate of 32.0 rad/s at a rate of 0.700 rad/s2. How long does it take to come to rest? How many

Darya [45]

Answer:

$t=45.7s$

$\alpha =116revolutions$

Explanation:

Since we have given values of ω₀=32.o rad/s ,ω=0 and α=-0.700 rad/s² to find t we use below equation

$w=w_{o}+at\\ 0=(32.0rad/s)+(-0.700rad/s^{2} )t\\t=\frac{-32.0}{-0.700} \\t=45.7s$

To find revolutions we use below equation

$w^{2}=w_{o}^{2}+2a\alpha$

Substitute the given values to find revolutions α

$0=(32.0rad/s)^{2}+2(-0.700rad/s^{2} )\alpha \\\alpha =\frac{(-32.0rad/s)^{2}}{2(-0.700rad/s^{2} )} \\\alpha =731rad$

To convert rad to rev:

$\alpha =(731rad)*(\frac{1rev}{2\pi rad} )\\\alpha =116revolutions$

7 0

3 years ago