A 30 N rightward force is exerted on a 7 kg object. There is a 3 N force of friction acting on the object. What is the Weight of
1 answer:
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Explanation:
Below is an attachment containing the solution.
1) 5.5 N
When the ball is at the bottom of the circle, the equation of the forces is the following:
where
T is the tension in the string, which points upward
mg is the weight of the string, which points downward, with
m = 0.158 kg being the mass of the ball
g = 9.8 m/s^2 being the acceleration due to gravity
is the centripetal force, which points upward, with
v = 5.22 m/s being the speed of the ball
R = 1.1 m being the radius of the circular trajectory
Substituting numbers and re-arranging the formula, we find T:
2) 3.9 N
When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:
And by substituting the numerical values, we find
3) 2.3 N
When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is
And substituting the numerical values and re-arranging it, we find
4) 3.3 m/s
The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:
and re-arranging the equation, we find
Answer:
Mass of the steel cube = 7800 kg
Volume of the steel = 1.025 cubic centimetre
Explanation:
Given:
The density of the steel = 7.8
Side of the cube = 12 cm
<u>(1)The mass of steel cube :</u>
We know that,
We are given with density and sides of the cube
then volume of the cube
=
=
= 1000 cubic centimetre
Now
mass = 7800 kg
<u>(2)volume of steel:</u>
Given the mass = 8 kg
Substituting the values
volume = 1.025 cubic centimetre