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Helga [31]
3 years ago
15

How much work is required to lift an object with a mass of 5.0 kilograms to a height of 3.5 meters?

Physics
2 answers:
KATRIN_1 [288]3 years ago
7 0
Work is Force multiplied by distance: work = Fd

work = mgh

work = (5.0 kg)(9.8 m/s)(3.5 m)

work = 171.5 joules or for two sigfigs, 170 joules
Veronika [31]3 years ago
6 0
I believe it would be 172 J (or 1.72 · 10² J)
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in a closed system three objects have the following momentum: 11 kg* m/s, -65 kg*m/s and -100 kg m/s. the objects collide and mo
guajiro [1.7K]

Explanation:

The momentum of the three objects are as follow :

11 kg-m/s, -65 kg-m/s and -100 kg-m/s

Before collision, the momentum of the system is :

P_i=11+(-65)+(-100)\\\\P_i=-154\ kg-m/s

After collison, they move together. It means it is a case of inelastic collision. In this type of collision, the momentum of the system remains conserved.

It would mean that, after collision, momentum of the system is equal to the initial momentum.

Hence, final momentum = -154 kg-m/s.

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3 years ago
A small fan has blades that have a radius of 0.0600 m. When the fan is turned on, the tips of the blades have a tangential accel
zhuklara [117]

Answer:

α = 395 rad/s²

Explanation:

Main features of uniformly accelerated circular motion

A body performs a uniformly accelerated circular motion   when its trajectory is a circle and its angular acceleration is constant  (α = cte). In it the velocity vector is tangent at each point to the trajectory and, in addition, its magnitude varies uniformly.

There is tangential acceleration (at) and is constant.

at = α*R     Formula (1)

where

α  is the angular acceleration

R is the radius of the circular path

There is normal or centripetal acceleration that determines the change in direction of the velocity vector.

Data

R = 0.0600 m   :blade radius

at = 23.7 m/s² : tangential acceleration of the blades

Angular acceleration of the blades (α)

We replace data in the formula (1)

at = α*R  

23.7 = α*(0.06)

α = (23.7) / (0.06)

α = 395 rad/s²

7 0
2 years ago
a disk of a radius 50 cm rotates at a constant rate of 100 rpm. what distance in meters will a point on the outside rim travel d
vlabodo [156]

Answer: 50π m ≈ 157 m

Explanation:

100 rev/min (2π rad/rev) / (60 sec/min) = 3⅓π rad/s

d = ωrt = 3⅓π(0.50)(30) = 50π m ≈ 157 m

8 0
2 years ago
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