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vovikov84 [41]
3 years ago
13

if you push your chair across the floor at a constant velocity how does the force of friction compare with the force you exert?

Physics
2 answers:
Alekssandra [29.7K]3 years ago
7 0
The force of friction is equal but opposite to the force that you exert. This is due to newtons 3rd law 'For every action there is an equal but opposite reaction'
Alik [6]3 years ago
5 0
-- If velocity is constant, then there is no net force
on the chair.

-- If there is no net force on the chair, then friction
must exactly balance out your push.

-- The force of friction is exactly equal in magnitude
to your push, and in exactly the opposite direction.
You might be interested in
Difference between displacement and vector quantity
daser333 [38]

The only 'difference' is that they are different categories.

It's like asking "What's the difference between Susie and girl ?"

Or "What's the difference between Cadillac and car ?"

Displacement <em>IS</em> a vector quantity.

5 0
3 years ago
A +0.2 µC charge is in an electric field. What happens if that charge is replaced by a +0.4 µC charge?
AleksandrR [38]

Explanation:

It is known that electric field is responsible for creating electric potential. As a result, it depends only on the electric field and not on the magnitude of charge.

So, when a charge is increased by a factor of 2 then electric potential will remain the same. Since, expression to calculate the electric potential is as follows.

                 U = qV

Since, the electric potential is directly proportional to the charge. Hence, when 0.2 \mu C tends to replaced by 0.4 \mu C then charge is increased by a factor of 2. Hence, the electric potential energy is doubled.

Thus, we can conclude that if that charge is replaced by a +0.4 µC charge then electric potential stays the same, but the electric potential energy doubles.

4 0
3 years ago
uppose the weights of Farmer Carl's potatoes are normally distributed with a mean of 9 ounces and a standard deviation of 0.8 ou
Brrunno [24]

Answer:

x= 9.53 ounces

Explanation:

Given that

Mean ,μ= 9 ounces

Standard deviation ,σ=0.8 ounces

He wants to sell only those potatoes that are among the heaviest 25%.

P=25% = 0.25

When P= 0.25 then Z=0.674

Lest take x is the the minimum weight required to be brought to the farmer's market.

We know that

x = Z . σ + μ

x= 0.674 ₓ 0.8 + 9 ounces

x= 9.53 ounces

3 0
3 years ago
A photon of wavelength 2.78 pm scatters at an angle of 147° from an initially stationary, unbound electron. What is the de Brogl
Elena-2011 [213]

Answer:

2.07 pm

Explanation:

The problem given here is the very well known Compton effect which is expressed as

\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)

here, \lambda is the initial photon wavelength, \lambda^{'} is the scattered photon wavelength, h is he Planck's constant, m_e is the free electron mass, c is the velocity of light, \theta is the angle of scattering.

Given that, the scattering angle is, \theta=147^{\circ}

Putting the respective values, we get

\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.

Here, the photon's incident wavelength is \lamda=2.78pm

Therefore,

\lambda^{'}=2.78+4.45=7.23 pm

From the conservation of momentum,

\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}

where,\vec{P_\lambda} is the initial photon momentum, \vec{P_{\lambda^{'}}} is the final photon momentum and \vec{P_e} is the scattered electron momentum.

Expanding the vector sum, we get

P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta

Now expressing the momentum in terms of De-Broglie wavelength

P=h/\lambda,

and putting it in the above equation we get,

\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}

Therefore,

\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm

This is the de Broglie wavelength of the electron after scattering.

6 0
3 years ago
As air pressure decreases, what happens
Naya [18.7K]

Answer:

I believe it is B, not 100% sure though

Explanation:

5 0
2 years ago
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