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oee [108]
2 years ago
5

A truck of mass 10000kg moving through a sloppy road of length 20km with the help of 5000N force. If the truck climbs the vertic

al height of 200m and it's efficiency is 80% calculate the additional mass that can be carried by the truck​ ?​
Physics
1 answer:
gizmo_the_mogwai [7]2 years ago
4 0

Answer:

30000

Explanation:

Statement of the given problem,

A truck of mass 10000 kg is moving through a sloppy road of length 20 km with the help of 5000 N force. If the truck climb the vertical height of 200 m and its efficiency is 80% calculate the additional mass that can be carried by the truck?

Let M denotes the additional mass (in kg) that can be carried by the truck.

From above data we determine that,

P.E. gained by total mass at the vertical height of 200 m = 80% of work done by the applied force [road elevation = tan^-1 (200/20000) = 0.57° (assumed negligible)]

Hence we get following relation,

(10000 + M)*g*200 = 5000*20*1000*80/100 [g = gravitational acceleration]

or 10000 + M = 50000*8/g = 50000*8/10 [g = 10 m/s^2 (assumed)]

or M = 40000 - 10000 = 30000 (kg) [Ans]

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1. A television set falls from rest from a bridge 100m above the river surface.
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3 0
2 years ago
a system of four particles moves along one dimension. the center of mass of the system is at rest, and the particles do not inte
Zinaida [17]

The velocity of the particle 4 at time, t = 2.83 s, is -14.1 m/s.

The given parameters;

m1 = 1.45 kg, v1(t) = (6.09m/s) + (0.299m/s^2) × t

m2 = 2.81 kg, v2(t) = (7.83m/s) + (0.357m/s^2) × t

m3 = 3.89 kg, v3(t) = (8.09m/s) + (0.405m/s^2) × t

m4 = 5.03kg

The velocity of the center mass of the particles is calculated as;

McmVcm = m1v1 + m2v2 +m3v3+m4v4

Vcm= m1v1 + m2v2 +m3v3 +m4v4/ Mcm

0 = m1v1 + m2v2 +m3v3 +m4v4/ Mcm

m1v1 + m2v2 +m3v3+m4v4 = 0

m4v4 = -(m1v1 + m2v2 +m3v3)

v4 =-(m1v1 + m2v2 +m3v3)/ m4

The velocity of particle 1 at time, t = 2.83 s;

vi = 6.09 + 0.299× 2.83

v1 = 6.94 m/s

The velocity of particle 3 at time, t = 2.83 s;

v2 = 7.83 + 0.357 × 2.83

v2 = 8.84 m/s

The velocity of particle 3 at time, t = 2.83 s;

v3 = 8.09 + 0.405 × 2.83

v3 = 9.24 m/s

The velocity of particle 3 at time, t = 2.83 s;

v4 = - (m1v1 + m2v2 + m3v3)/m4

v4 = -(1.45×6.94 + 2.81×8.84 + 3.89×9.24)/5.03

v4 = -14.4 m/s

Thus, the velocity of the particle 4 at time, t = 2.83 s, is -14.1 m/s.

Learn more about Velocity here:

brainly.com/question/18084516

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5 0
1 year ago
A railroad car with a mass of 30,000 kg is moving at 2.0 m/s when it runs into an at-rest freight car with an equal mass. The ca
11Alexandr11 [23.1K]
The total momentum of the system is preserved through the collision.

Note that momentum is
P = m*v
where m = mass
v = velocity.

Initial momentum:
P1 = (30000 kg)*(2 m/s) = 60000 (kg-m)/s for the moving car
P2 = 0 for the starionary car.

Final momentum:
P3 = (30000 + 30000)*v = 60000v (kg-m)/s

Because momentum is preserved,
P3 = P1 + P2
60000v = 60000
v = 1 m/s
The final velocity is 1 m/s.

Answer: 1.0 m/s
4 0
3 years ago
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