Answer:
Explanation:
mass of 1 L water = 1 kg .
200⁰F = (200 - 32) x 5 / 9 = 93.33⁰C .
260.928 K = 260.928 - 273 = - 12.072⁰C .
water is at higher temperature .
Let the equilibrium temperature be t .
Heat lost by water = mass x specific heat x fall of temperature
= 1 x 4.2 x 10³ x ( 93.33 - t )
Heat gained by copper
= .25 x .385 x 10³ x ( t + 12.072 )
Heat lost = heat gained
1 x 4.2 x 10³ x ( 93.33 - t ) = .25 x .385 x 10³ x ( t + 12.072 )
93.33 - t = .0229 ( t + 12.072)
93.33 - t = .0229 t + .276
93.054 = 1.0229 t
t = 90.97⁰C .
Answer: hello your question is incomplete below is the missing part
A spherical cavity is hollowed out of the interior of a neutral conducting sphere. At the center of the cavity is a point charge, of positive charge q.
answer:
- q
Explanation:
Since the spherical cavity was carved out of a neutral conducting sphere hence the electric field inside this conductor = zero
given that there is a point charge +q at the center of the spherical cavity hence for the electric field inside the conductor to be = zero the total surface charge qint on the wall of the cavity will be -q
Answer:
Balanced forces are responsible for unchanging motion. Balanced forces are forces where the effect of one force is cancelled out by another. A tug of war, where each team is pulling equally on the rope, is an example of balanced forces. The forces exerted on the rope are equal in size and opposite in direction.
Explanation:
I think it false. Sorry if i'm wrong.
Given:
Area of pool = 3m×4m
Diameter of orifice = 0.076m
Outlet Velocity = 6.3m/s
Accumulation velocity = 1.5cm/min
Required:
Inlet flowrate
Solution:
The problem can be solved by this general formula.
Accumulation = Inlet flowrate - Outlet flowrate
Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice
First, we need to convert the units of the accumulation velocity into m/s to be consistent.
Accumulation velocity = 1.5cm/min × (1min/60s)×(1m/100cm)
Accumulation velocity = 0.00025 m/s
We then calculate the area of the pool and the area of the orifice by:
Area of pool = 3 × 4 m²
Area of pool = 12m²
Area of orifice = πd²/4 = π(0.076m)²/4
Area of orifice = 0.00454m²
Since we have all we need, we plug in the values to the general equation earlier
Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice
0.00025 m/s × 12m² = Inlet flowrate - 6.3m/s × 0.00454m²
Transposing terms,
Inlet flowrate = 0.316 m³/s
