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erastova [34]
2 years ago
5

In a real isothermal expansion, the temperature of the surroundings must be________the temperature of the gas.

Physics
1 answer:
Lady_Fox [76]2 years ago
7 0
Must be isolated from the temperature of the gas
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8. When a 100 N bag of nails hangs motionless from a single vertical strand of rope, how many newtons of tension are exerted in
Svetllana [295]

If the bag is motionless, then it's not accelerating up or down.
That fact right there tells you that the net vertical force on it
is zero.  So the sum of any upward forces on it is exactly equal
to the downward gravitational force ... the bag's "weight".

If the bag is suspended from a single rope, then the tension
in the rope must be equal to the 100-N weight of the bag.

And if there are four ropes holding it up, then the sum of
the four tensions is 100N.  If the ropes have been carefully
adjusted to share the load equally, then the tension is 25N
in each rope.

8 0
3 years ago
An x-ray photon is scattered by an originally stationary electron. how does the frequency of the scattered photon compare relati
Viefleur [7K]

The frequency of the scattered photon decreases or it will be lower compare to the frequency of incident photon. An x-ray photon scatters in one direction after a collision and some energy is transferred to the electron as it recoils in another direction resulting to have less energy in the scattered photon. In addition, the frequencies will also depend on the differences of the angle at which the scattered photon leaves the collision and this incident is called Compton Effect.

8 0
3 years ago
John ( body mass= 65 kg) is taking off for a long jump . Horizontal accerleration ax is 5m/s^2 and vertical acceleration ay is 0
cestrela7 [59]

Answer:

(a) The horizontal ground reaction force  F_{g,x}=325\, N

(b)  The vertical ground reaction force F_{g,y}=696\, N

(c)   The resultant ground reaction force F_g=768\, N

Explanation:  

Given

John mass , m = 65 kg

Horizontal acceleration , a_x= 5.0 \frac{m}{s^{2}}

Vertical acceleration , a_y=0.9 \frac{m}{s^{2}}

(a) Using Newton's 2nd law in horizontal direction

F_{g,x}=ma_x

=>F_{g,x}=65\times 5\, N=325\, N

Thus the horizontal ground reaction force  F_{g,x}=325\, N

(b) Using Newton's 2nd law in vertical direction

F_{g,y}-mg=ma_y

=>F_{g,y}=mg+ma_y

=>F_{g,y}=65\times (9.81+0.9)\, N=696\, N

Thus the vertical ground reaction force F_{g,y}=696\, N

(c)  Resultant ground reaction force is

F_g=(F_{g,x}^{2}+F_{g,y}^{2})^{\frac{1}{2}}

=>F_g=(325^{2}+696^{2})^{\frac{1}{2}}\, N=768\, N

=>F_g=768\, N

Thus  the resultant ground reaction force F_g=768\, N

5 0
3 years ago
How will the electrostatic force between two electric charges change if the first charge is doubled and the second charge is onl
Vinvika [58]

Answer:

B) \frac{2}{3}

Explanation:

The electric force between charges can be determined by;

F = \frac{kq_{1} q_{2} }{r^{2} }

Where: F is the force, k is the Coulomb's constant, q_{1} is the value of the first charge, q_{2} is the value of the second charge, r is the distance between the centers of the charges.

Let the original charge be represented by q, so that;

q_{1} = 2q

q_{2} = \frac{q}{3}

So that,

F = q_{1}q_{2} x \frac{k}{r^{2} }

  = 2q x \frac{q}{3} x \frac{k}{r^{2} }

  = \frac{2q^{2} }{3} x \frac{k}{r^{2} }

  = \frac{2}{3} x \frac{kq}{r^{2} }

F = \frac{2}{3} x \frac{kq}{r^{2} }

The electric force between the given charges would change by \frac{2}{3}.

4 0
3 years ago
How is a quarter able to role on its edge?
Olegator [25]
Idk its really weird because the ridges are making the quarter stand and therefore able to roll
5 0
3 years ago
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