Answer:
4A
Explanation:
From Ohms law ;
R= V/I where R is resistance, V is voltage and I is current
Given that;
R= 3Ω
V= 12
Using the values in the formula as;
R= V/I
3=12/I
I= 12/3
I= 4 A
The equatorial diameter of the earth is 12,756 km.
Therefore the equatorial radius is
r = 12756/2 = 6378 km.
The angular velocity of the earth is
ω = (2π radians)/(24 hours) = 0.2618 rad/h
The tangential velocity of a person at the equator is
v = rω
= (6378 km)*( 0.2618 rad/h)
= 1670 km/h
Answer: 1670 km/h
I hope this helps you.
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Answer:
People would watch it please give me one star so I can be first place
Explanation:
IDK
Answer:
potato<-100
print(potato)
sqrt(potato)
potato<-potato*2
print(potato)
Explanation:
The complete question is as follows
Create a variable called potato whose value corresponds to the number of potatoes you’ve eaten in the last week. Or something equally ridiculous. Print out the value of potato.
Calculate the square root of potato using the sqrt() function. Print out the value of potato again to verify that the value of potato hasn’t changed.
Reassign the value of potato to potato * 2.
Print out the new value of potato to verify that it has changed
The question was answered using R programming language.
At line 1, I assumed that I ate 100 potatoes in the previous week.
So, potato = 100
At line 2, the value of potato is printed as 100.
At line 3, the square root of potato is calculated using sqrt function: Square for of 100 = 10
At line 4,the initial value of potato is doubled and stored in potato variable. 100 * 2 = 200
At line 5, the new value of potato is printed: 200.
Answer:
1.737 kJ
Explanation:
Thinking process:
Step 1
Data:
Area of the shaft = 0.8 cm²
Combined mass of shaft and piston = (24.5 + 0.5) kg
= 25 kg
Piston diameter = 0.1 m
External atmospheric pressure = 1 bar = 101.3 kPa
Pressure inside the gas cylinder = 3 bar = 3 × 101.3 kPa
g = 9.81 m/s²
Step 2
Draw a free body diagram
Step 3: calculations
area of the piston = 0.0314 m²
Change in the elevation of the piston, 
z = 
= 
= 0.82 m
Next, we evaluate the work done by the shaft:
= (1668) ( 0.082)
= 1. 37 kJ
Net area for work done = A (piston) - Area of shaft
= 
= 77.7 cm²
= 0.007774 m²
Work done in overcoming atmospheric pressure:
Wₐ = PAZ
=101.3 kPa * 0.007774 * 0.82
= 0.637 kJ
total work = work done by shaft + work to overcome atmospheric pressure = 0.367 + 1.37
= 1.737 kJ Ans
