Answer:
The current in the primary is 0.026 A
Explanation:
Using the formula
I1 = (V1/V2)*I2
we have
I1 = (6.4/120)*0.500
I1 = 0.026 A
Answer:
The first one is: His weight on the Earth before take-off and the weight after take-off back on Earth once he gets back should be recorded as his Independent variable and his dependent variable.
The second one is: If he gained the weight back that he had lost while on the trip then you should disregard them unless that was the weight he was when he weighed himself after he got back.
The Third one is: The mass of an object is the amount of matter it contains, regardless of its volume or any forces acting on it. … Gravity is a force that attracts objects toward the Earth. The weight of the object is defined as the force caused by gravity on a mass.
Explanation:
I took the quiz earlier. Hope this Helps you.
The speed of a car travelling over a hill that has a radius of curvature should not exceed a certain speed other it will topple. This speed is related to the radius of curvature and the gravitational acceleration as shown below:
V^2 = Rg, where V = maximum speed, R = Radius of curvature, g = gravitational acceleration.
Substituting;
V = Sqrt (Rg) = Sqrt (120*9.81) = 34.31 m/s
Answer:
34.3 m/s
Explanation:
Newton's Second Law states that the resultant of the forces acting on the car is equal to the product between the mass of the car, m, and the centripetal acceleration 
(because the car is moving of circular motion). So at the top of the hill the equation of the forces is:
where
(mg) is the weight of the car (downward), with m being the car's mass and g=9.8 m/s^2 is the acceleration due to gravity
R is the normal reaction exerted by the road on the car (upward, so with negative sign)
v is the speed of the car
r = 0.120 km = 120 m is the radius of the curve
The problem is asking for the speed that the car would have when it tires just barely lose contact with the road: this means requiring that the normal reaction is zero, R=0. Substituting into the equation and solving for v, we find:
The magnitude of the force required to stop the weight in 0.333 seconds is 67.6 N.
<h3>
Magnitude of required force to stop the weight</h3>
The magnitude of the force required to stop the weight in 0.333 seconds is calculated by applying Newton's second law of motion as shown below;
F = ma
F = m(v/t)
F = (mv)/t
F = (5 x 4.5)/0.333
F = 67.6 N
Thus, the magnitude of the force required to stop the weight in 0.333 seconds is 67.6 N.
Learn more about force here: brainly.com/question/12970081
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