Answer:
It will be B
Explanation:
Since resisitors in series are added together, 1 + 1 + 1 would = 3kilo ohms. But with resistors in parallel would be (1/1+1/1)^-1. That would equal 0.5 Now you have two resistors in series for B, and because now that they are in series you add them together, so 0.5 + 1 = 1.5 kilo ohms which is what is needed.
Answer:
Explanation:
There are three points in time we need to consider. At point 0, the mango begins to fall from the tree. At point 1, the mango reaches the top of the window. At point 2, the mango reaches the bottom of the window.
We are given the following information:
y₁ = 3 m
y₂ = 3 m − 2.4 m = 0.6 m
t₂ − t₁ = 0.4 s
a = -9.8 m/s²
t₀ = 0 s
v₀ = 0 m/s
We need to find y₀.
Use a constant acceleration equation:
y = y₀ + v₀ t + ½ at²
Evaluated at point 1:
3 = y₀ + (0) t₁ + ½ (-9.8) t₁²
3 = y₀ − 4.9 t₁²
Evaluated at point 2:
0.6 = y₀ + (0) t₂ + ½ (-9.8) t₂²
0.6 = y₀ − 4.9 t₂²
Solve for y₀ in the first equation and substitute into the second:
y₀ = 3 + 4.9 t₁²
0.6 = (3 + 4.9 t₁²) − 4.9 t₂²
0 = 2.4 + 4.9 (t₁² − t₂²)
We know t₂ = t₁ + 0.4:
0 = 2.4 + 4.9 (t₁² − (t₁ + 0.4)²)
0 = 2.4 + 4.9 (t₁² − (t₁² + 0.8 t₁ + 0.16))
0 = 2.4 + 4.9 (t₁² − t₁² − 0.8 t₁ − 0.16)
0 = 2.4 + 4.9 (-0.8 t₁ − 0.16)
0 = 2.4 − 3.92 t₁ − 0.784
0 = 1.616 − 3.92 t₁
t₁ = 0.412
Now we can plug this into the original equation and find y₀:
3 = y₀ − 4.9 t₁²
3 = y₀ − 4.9 (0.412)²
3 = y₀ − 0.83
y₀ = 3.83
Rounded to two significant figures, the height of the tree is 3.8 meters.
Answer:
ΔQ = 1.06 KJ
Explanation:
The amount of heat transfer between the piston-cylinder system and the surrounding can easily be found by using the First Law of Thermodynamics. The first law of thermodynamics can be written as follows:
ΔQ = ΔU + W
ΔQ = mΔu + PΔV
where,
ΔQ = Heat transfer between system and surrounding = ?
Δu = specific change in internal energy of the system = - 175 KJ/kg
m = mass of air = 20 g = 0.02 kg
P = Constant Pressure = 101.3 KPa
ΔV = Change in Volume = 0.05 m³ - 0.005 m³ = 0.045 m³
Therefore,
ΔQ = (0.02 kg)(-175 KJ/kg) + (101.3 KPa)(0.045 m³)
ΔQ = -3.5 KJ + 4.56 KJ
<u>ΔQ = 1.06 KJ</u>
Answer:
ay max = 4.91 m/s²
so here acceleration would be either right or left
Explanation:
given data
wide b = 1 m
long l = 2 m
depth d = 3 m
height of tank sides h = 4 m
solution
here for prevent spilling condition is
≤ 
..........1
≤ - 0.50
and when here
= - 
......2
when az is 0 ay will be
ay = - 
and ay max will be
ay max = -( -0.50) (9.81 )
ay max = 4.91 m/s²
so here acceleration would be either right or left
