When the mass of one object doubles what happens to its gravitational attraction to another object of constant mass, assuming th
1 answer:
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Explanation:
Mass of bumper cars,
Initial speed of car A,
Initial speed of car Z,
Final speed of car A after the collision,
We need to find the velocity of car Z after the collision. Let it is equal to . Using the conservation of momentum as :
So, the velocity of car Z after the collision is (-12 m/s). Hence, this is the required solution.
Answer:
The tension in each half of the rope, is approximately 4,908.8 N
Explanation:
The mass of the acrobat, m = 60 kg
The length of the rope, l = 10 m
The extent by which the center dropped = 30 cm = 0.3 m
Let, 'T' represent the tension in each half of the rope
Weight, W = Mass, m × The acceleration due to gravity, g
∴ W = m × g
The acceleration due to gravity, g ≈ 9.8 m/s²
∴ The weight of the acrobat, W = 60 kg × 9.8 m/s² ≈ 588 N
The angle the dropped rope makes with the horizontal, θ is given as follows;
θ = arctan((0.3 m)/(5 m)) = arctan(0.06) ≈ 3.434°
At equilibrium, the sum of vertical forces, = 0
The vertical component of the tension, , in each half of the rope is given as follows;
= T × sin(θ)
∴ = W + T × sin(θ) + T × sin(θ) = W + 2 × T × sin(θ)
Plugging in the values, with θ = arctan(0.06) for accuracy, we get;
588 N + 2 × T × sin(arctan(0.06) = 0
∴ 2 × -T × sin(arctan(0.06) = 588 N
-T= 588 N/(2 × sin(arctan(0.06)) = 4,908.81208 N ≈ 4,908.8 N
The tension in each half of the rope, T ≈ 4,908.8 N.
Answer:
1) Final Temperature of the gas in A will be GREATER than the temperature in B
2) Diagram of both processes on a single PV has been uploaded below
3) The Initial pressures in containers A and B is 3039.87 J/liters
4) the final volume of container B is 923.36 cm³
Explanation:
Given that;
Temperature = 20°C = 293 K
mass of piston = 10 kg
Area = 100cm³
Volume V = 800 cm³ = 0.8 L
ideal gas constant R = 8.3 J/K·mol
1)
Final Temperature of the gas in A will b GREATER than the temperature in B
2)
Diagram of both processes on a single PV has been uploaded below,
3)
Initial pressures in containers A and B
PV = nRT
P = RT/V
we substitute
P = (8.3 × 293) / 0.8
P = 2431.9 / 0.8
P = 3039.87 J/liters
Therefore, The Initial pressures in containers A and B is 3039.87 J/liters
4)
Given that;
power = 25 W
time t = 15s
the final volume of container B = ?
we know that;
work done = power × time
work done = 25 × 15 = 375
Also work done = P( V₂ - V₁ )
so we substitute
375 = 3039.87 ( V₂ - 0.8 )
( V₂ - 0.8 ) = 375 / 3039.87
V₂ - 0.8 = 0.12336
V₂ = 0.12336 + 0.8
V₂ = 0.92336 Litres
V₂ = 923.36 cm³
Therefore, the final volume of container B is 923.36 cm³
