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(a) The work done by the force applied by the tractor is 79,968.47 J.
(b) The work done by the frictional force on the tractor is 55,977.93 J.
(c) The total work done by all the forces is 23,990.54 J.
<h3>
Work done by the applied force</h3>
The work done by the force applied by the tractor is calculated as follows;
W = Fd cosθ
W = (5000 x 20) x cos(36.9)
W = 79,968.47 J
<h3>Work done by frictional force</h3>
W = Ffd cosθ
W = (3500 x 20) x cos(36.9)
W = 55,977.93 J
<h3>Net work done by all the forces on the tractor</h3>
W(net) = work done by applied force - work done by friction force
W(net) = 79,968.47 J - 55,977.93 J
W(net) = 23,990.54 J
Learn more about work done here: brainly.com/question/25573309
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Answer:
m = 788.2[kg]
Explanation:
The potential energy of a body is defined as the product of mass by gravitational acceleration by height. And it can be calculated by means of the following equation.
where:
Epot = potential energy = 63405 [J]
m = mass [kg]
g = gravity acceleration = 9.81[m/s²]
h = elevation = 8.2[m]
Now replacing:
![63405=m*9.81*8.2\\m=788.2[kg]](https://tex.z-dn.net/?f=63405%3Dm%2A9.81%2A8.2%5C%5Cm%3D788.2%5Bkg%5D)
a) uniform velocity
b) zero or no acceleration
c) (see picture)
EXPLANATION:
(see picture)
Answer: -0.84 rad/sec (clockwise)
Explanation:
Assuming no external torques act on the system (man + turntable), total angular momentum must be conserved:
L1 = L2
L1 = It ω + mm. v . r = 81.0 kg . m2 .21 rad/s – 56.0 kg. 3.1m/s . 3.1 m
L1 = -521.15 kg.m2/sec (1)
(Considering to the man as a particle that is moving opposite to the rotation of the turntable, so the sign is negative).
Once at rest, the runner is only a point mass with a given rotational inertia respect from the axis of rotation, that can be expressed as follows:
Im = m. r2 = 56.0 kg. (3.1m)2 = 538.16 kg.m2
The total angular momentum, once the runner has come to an stop, can be written as follows:
L2= (It + Im) ωf = -521.15 kg.m2/sec
L2= (81.0 kg.m2 + 538.16 kg.m2) ωf = -521.15 kg.m2/sec
Solving for ωf, we get:
ωf = -0.84 rad/sec (clockwise)
