Answer:
pressure in the second point is P₂ = 165833.75 Pa
Explanation:
given,
water speed = 1.8 m/s
gauge pressure(P₁) = 51500 Pa
gauge pressure at another point (P₂) = ?
distance of the point at 11.5 m
density of the water = 1000 kg/m³
diameter of the second pipe is double than first.
A₁ V₁ = A₂ V₂
π r² × 1.8 = π × (2r)² × V₂
V₂ = 0.45 m/s
now, gauge pressure at P₂
P₂ = 165833.75 Pa
pressure in the second point is P₂ = 165833.75 Pa
Explanation:
So basically yh first off the number for protons and electrons are the same and to find them is to look at the atomic number because the atomic number is the same as protons and electrons.
To find the neutrons you need to take the actual atomic number from the atomic mass unit.
Hope that helps yh don't stress it.
Answer:
see solution below
Explanation:
The given resistors are connected in series.
Equivalent resistance in series = 30 + 55 + 15
Equivalent resistance in series Rt = 100 ohms
Since the potential difference in the circuit = 36V
Get the current in the circuit first
I = V/Rt
I = 36/100
I = 0.36A
Get the voltage across 30ohms resistor;
V30 = 0.36 * 30
V30 = 10.8volts
Hence the voltage across the 30ohms resistor is 10.8volts
Get the voltage across 55ohms resistor;
V55 = 0.36 * 55
V55 = 19.8volts
Hence the voltage across the 55ohms resistor is 19.8volts
Get the voltage across 15ohms resistor;
V15 = 0.36 * 15
V15 = 5.4volts
Hence the voltage across the 15ohms resistor is 5.4volts
Answer:
+b±√b² - 4ac /2a
0.6t ± √36-36/2a
Explanation:
Work done = 1/2 mv² where v = (1.2)²
Therefore, 1/2m(1.2)ω mgh
1/2m (1.2)² = 0.4 × m ×10 5
s = 1.44 / 2.4 = 1.44 / 8
S = ut - 1/2gt²
Where u = 1.2
g = 0.9 × 10
Therefore,
1.8 = 1.2v-2t²
2t²c-1.2t+1.8 = 0
t² - 0.6t + 0.9 = 0
0.6t ± √36-36/2a
Solving this further, we make use of the formula
+b±√b² - 4ac /2a
Answer:
Written in Python
def energyvector(mass):
c = 2.9979 * 10**8
energy = mass * c ** 2
print(round(energy,2))
Explanation:
This line defines the function
def energyvector(mass):
This line initializes the speed of light
c = 2.9979 * 10**8
This line calculates the corresponding energy
energy = mass * c ** 2
This line prints the calculated energy
print(round(energy,2))