Answer:
A. the internal energy stays the same
Explanation:
From the first law of thermodynamics, "energy can neither be created nor destroyed but can be transformed from one form to another.
Based on this first law of thermodynamic, the new internal energy of the gas is the same as the internal energy of the original system.
Therefore, when the partition separating the two halves of the box is removed and the system reaches equilibrium again, the internal energy stays the same.
The force of friction is 19.1 N
Explanation:
According to Newton's second law, the net force acting on the bag is equal to the product between its mass and its acceleration:
where
is the net force
m is the mass
a is the acceleration
The bag is moving at constant speed, so its acceleration is zero:
Therefore the net force is zero as well:
Here we are interested only in the forces acting along the horizontal direction, therefore the net force is given by:
where
is the horizontal component of the applied force, with
F = 22.5 N
is the force of friction
And solving for , we find
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Answer:
E = 1.25×10¹³ N/m²
Explanation:
Young's modulus is defined as:
E = stress / strain
E = (F / A) / (dL / L)
E = (F L) / (A dL)
Given:
F = 10 kg × 9.8 m/s² = 98 N
L = 1 m
dL = 10⁻⁵ m
A = π/4 (0.001 m)² = 7.85×10⁻⁷ m²
Solve:
E = (98 N × 1 m) / (7.85×10⁻⁷ m² × 10⁻⁵ m)
E = 1.25×10¹³ N/m²
Round as needed.
Answer:
The electric field strength is 
Solution:
As per the question:
Area of the electrode, 
Charge, q = 50 nC = ![50\times 10^{- 9} C[/etx]Distance, x = 2 mm = [tex]2\times 10^{- 3} m](https://tex.z-dn.net/?f=50%5Ctimes%2010%5E%7B-%209%7D%20C%5B%2Fetx%5D%3C%2Fp%3E%3Cp%3EDistance%2C%20x%20%3D%202%20mm%20%3D%20%5Btex%5D2%5Ctimes%2010%5E%7B-%203%7D%20m)
Now,
To calculate the electric field strength, we first calculate the surface charge density which is given by:
Now, the electric field strength of the electrode is:
where
Answer:
50000 N
Explanation:
From the question given above, the following data were obtained:
Mass (m) of bullet = 0.050 kg
velocity (v) = 400 m/s
Distance (s) = 0.080 m
Force (F) =?
Next, we shall determine the acceleration of the bullet. This can be obtained as follow:
Initial velocity (u) = 0 m/s
Final velocity (v) = 400 m/s
Distance (s) = 0.080 m
Acceleration (a) =?
v² = u² + 2as
400² = 0 + (2 × a × 0.08)
160000 = 0 + 0.16a
160000 = 0.16a
Divide both side by 0.16
a = 160000 / 0.16
a = 1×10⁶ m/s²
Finally, we shall determine the force exerted by the bullet on the target. This can be obtained as follow:
Mass (m) of bullet = 0.050 kg
Acceleration (a) of bullet = 1×10⁶ m/s²
Force (F) =?
F = ma
F = 0.050 × 1×10⁶
F = 50000 N
Thus, the bullet exerted a force of 50000 N on the target.
