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Aliun [14]
3 years ago
15

According to Newton's third law of motion, if you push against a wall, the wall will __________.

Physics
2 answers:
slavikrds [6]3 years ago
3 0

Answer:

push up agaisnt you with equal force.

egoroff_w [7]3 years ago
3 0

Answer:

c

Explanation:

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A piston raises a weight, then lowers it again to its original height. If the process the system follows as the piston rises is
FinnZ [79.3K]

Answer:

Negative

Explanation:

First law of thermodynamic also known as the  law of conservation of energy states that the total energy of an isolated system is constant; energy can be transformed from one form to another, but can be neither created nor destroyed.

The first law relates  relates changes in internal energy to heat added to a system and the work done by a system by the conservation of energy.

The first law is mathematically given as ΔU  = U_{F} - U_{0} = Q - W

Where Q  = Quantity of heat

            W = Work done

From the first law The internal energy has the symbol U. Q is positive if heat is added to the system, and negative if heat is removed; W is positive if work is done by the system, and negative if work is done on the system.

Analyzing the pistol when it raises in isothermal and when it falls in  isobaric state.The following can be said:

In the Isothermal compression of a gas there is work done on the system to decrease the volume and increase the pressure. For work to be done on the system it is a negative work done then.

In the Isobaric State An isobaric process occurs at constant pressure. Since the pressure is constant, the force exerted is constant and the work done is given as PΔV.If a gas is to expand at a constant pressure, heat should be transferred into the system at a certain rate.Isobaric is a fuction of heat which is Isothermal Provided the pressure is kept constant.

In Isobaric definition above it can be seen that " Heat should be transferred into the system ata certain rate. For heat to be transferred into the system work is deinitely been done on the system thereby favouring the negative work done.

6 0
3 years ago
A photon of wavelength 7.33 pm scatters at an angle of 157° from an initially stationary, unbound electron. What is the de Brogl
Ann [662]

Answer:

4.63 p.m.

Explanation:

The problem given here can be solved by the 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=157^{\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-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42(1-cos157^\circ ) p.m.

Therfore,

\lambda^{'}-\lambda=4.64 p.m.

Here, the photon's incident wavelength is \lamda=7.33pm

So,

\lambda^{'}=7.33+4.64=11.97 p.m

From the conservation of momentum,

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

here, \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}}

Therfore,

\lambda_{e}=\frac{7.33\times 11.97}{\sqrt{7.33^{2}+11.97^{2}-2\times 7.33\times 11.97\times cos157^\circ }} p.m.\\\lambda_{e}=\frac{87.7401}{18.935} = 4.63 p.m.

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

8 0
3 years ago
A 160.-kilogram space vehicle is traveling along a
mario62 [17]

If the object is moving in a straight line at a constant speed, then that's
the definition of zero acceleration.  It can only happen when the sum of
all forces (the 'net' force) on the object is zero.

And it doesn't matter what the object's mass is.  That argument is true
for specks of dust, battleships, rocks, stars, rock-stars, planets, and
everything in between.

4 0
3 years ago
A 7300 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.20 m/s2 and feels no appreci
Simora [160]

Answer:

Explanation:

We shall first calculate the velocity at height h = 575 m .

acceleration a = 2.2 m /s²

v² = u² + 2 a s

u is initial velocity , v is final velocity , s is height achieved

v² = 0 + 2 x 2.2 x 575

v = 50.3 m /s

After 575 m , rocket moves under free fall so g will act on it downwards

If it travels further by height H

from the relation

v² = u² - 2 g H

v = 0 , u = 50.3 m /s

H = ?

0 = 50.3² - 2 x 9.8 H

H = 129.08 m

Total height attained by rocket

= 575 + 129.08

= 704.08 m .

4 0
2 years ago
Which of the following is an example of changing physical capital?
rosijanka [135]
<span>C. switching to cheaper fuel. Physical capital pertains to non-human asset. Under this type were the asset that use to process goods and services like machinery, buildings etc.</span>
3 0
3 years ago
Read 2 more answers
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