Answer: 1766.667 Ω = 1.767kΩ
Explanation:
V=iR
where V is voltage in Volts (V), i is current in Amps (A), and R is resistance in Ohms(Ω).
3mA = 0.003 A
Rearranging the equation, we get
R=V/i
Now we are solving for resistance. Plug in 0.003 A and 5.3 V.
R = 5.3 / 0.003
= 1766.6667 Ω
= 1.7666667 kΩ
The 6s are repeating so round off to whichever value you need for exactness.
If two vehicles arrive at an uncontrolled intersection almost simultaneously, the driver of the car that arrived at the intersection last must give the right of way. The driver on the left has the right of way if you both arrive at the intersection at the same moment.
Which driver should yield at an uncontrolled T-intersection?
The driver of the car turning must give way to all oncoming cross traffic when two cars are approaching an uncontrolled "T" crossroads. You need to stop and give way to vehicles and pedestrians while entering a public road from a driveway or private road.
What is right of way?
When two vehicles or pedestrians are approaching from opposite directions, moving at a speed, and close enough to create a collision, one has the right to move forward legally before the other, unless the other gives way.
Learn more about uncontrolled intersection: brainly.com/question/14312150
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Answer:
Q= 6491.100 kJ/s
Explanation:
Air-Fuel Ratio:
For a combustion reaction the proportion of air that is present in a gaseous substance responsible for the reaction,this proportion is known as air-fuel ratio.The air fuel ratio is calculated using the combustion reaction for the substance.
Considering reaction for the Ethanol as
C₂H₅OH +XO₂(O₂+3.76N₂)→ aCO₂+bH₂O+cN₂
Balancing the equation we get;
a=2,
2b=6
∴ b=3
xO₂=3
The air-fuel ratio
A/F = XO₂+H₂O+xN₂× mass of N₂/mass(fuel)
3.31×31.9+11.28×28.013/46.069
= 8.943
Equivalent ratio = 0.7,
so, heat transfer
Q= m ×Cp×ΔT
= 75×0.7×112.4(1500-400)
Q= 6491.100 kJ/s
1kJ/s=1000w
∴ Q= 6491100 W
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Answer:
The force between the charges when the separation decreases to 0.7 meters equals 126.955 Newtons
Explanation:
We know that for two point charges of magnitude 
the magnitude of force between them is given by
where
is constant
is the separation between the charges
Initially when the charges are separated by 2.4 meters the force can be calculated as
Now when the separation is reduced to 0.7 meters the force is similarly calculated as
Applying value of the constant we get
Thus 
The question is incomplete. The complete question is :
The solid rod shown is fixed to a wall, and a torque T = 85N?m is applied to the end of the rod. The diameter of the rod is 46mm .
When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the strains vary linearly along radial lines. Within the proportional limit, the stress also varies linearly along radial lines. If point A is located 12 mm from the center of the rod, what is the magnitude of the shear stress at that point?
Solution :
Given data :
Diameter of the rod : 46 mm
Torque, T = 85 Nm
The polar moment of inertia of the shaft is given by :
J = 207.6 
So the shear stress at point A is :
Therefore, the magnitude of the shear stress at point A is 4913.29 MPa.
