Calculate the acceleration of a car (in m/s^2) that accelerates from 0 to 30 m/s in 6 s along a straight road.
1 answer:
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Answer:
c.
Explanation:
We are given that
Acceleration due to gravity on the moon=
Acceleration due to gravity on the earth=
Net force due to am on an object on moon=
There is no friction and no drag force and there is no gravity involved
Then, the force acting on an object on earth=
(given)
Hence, option c is true.
Kinetic energy: the energy of motion
Work: the change in kinetic energy
Power: the rate of work done
Explanation:
The kinetic energy of an object is the energy possessed by the object due to its motion. Mathematically, it is given by:
where
m is the mass of the object
v is its speed
The work done an object is the amount of energy transferred; according to the energy-work theorem, it is equal to the change in kinetic energy of an object:
where
is the final kinetic energy
is the initial kinetic energy
Finally, the power is the rate of work done per unit time. Mathematically, ti can be expressed as
where
W is the work done
t is the time elapsed
Learn more about kinetic energy, work and power:
#LearnwithBrainly
Answer:
T1 = 131.4 [N]
T2 = 261 [N]
Explanation:
To solve this problem we must make a sketch of how will be the semicircle, for this reason we conducted an internet search, to find the scheme of the problem. This scheme is attached in the first image.
Then we make a free body diagram, with this free body diagram, we raise the forces that act on the body. Since it is a problem involving static equilibrium, the sum of forces in any direction and moments must be equal to zero.
By performing a sum of forces on the Y axis equal to zero we can find an equation that relates the forces of tension T1 & T2.
The second equation can be determined by summing moments equal to zero, around the point of application of the T1 force. In this way we find the T2 force.
The value of T2, is replaced in the first equation and we can find the value for T1.
Therefore
T1 = 131.4 [N]
T2 = 261 [N]
The free body diagram and the developed equations can be seen in the second attached image.
