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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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Why would we expect protogalactic clouds with relatively high density to form an elliptical galaxy rather than a spiral galaxy?
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7 0
2 years ago
Help pls i need this right now
pantera1 [17]

Answer:

The x-component of F_{3} is 56.148 newtons.

Explanation:

From 1st and 2nd Newton's Law we know that a system is at rest when net acceleration is zero. Then, the vectorial sum of the three forces must be equal to zero. That is:

\vec F_{1} + \vec F_{2} + \vec F_{3} = \vec O (1)

Where:

\vec F_{1}, \vec F_{2}, \vec F_{3} - External forces exerted on the ring, measured in newtons.

\vec O - Vector zero, measured in newtons.

If we know that \vec F_{1} = (70.711,70.711)\,[N], \vec F_{2} = (-126.859, 46.173)\,[N], F_{3} = (F_{3,x},F_{3,y}) and \vec O = (0,0)\,[N], then we construct the following system of linear equations:

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\Sigma F_{y} = 70.711\,N + 46.173\,N+F_{3,y} = 0\,N (3)

The solution of this system is:

F_{3,x} = 56.148\,N, F_{3,y} = -116.884\,N

The x-component of F_{3} is 56.148 newtons.

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