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Sindrei [870]
3 years ago
8

When a forklift raises an object, is it doing positive or negative work on the object? Which type of work is the

Physics
1 answer:
saul85 [17]3 years ago
6 0
The answer:

-> When the forklift raises an object, it is doing positive work. This is because electrical energy is converted in to kinetic energy that is required for the forklift to raise the object.

-> However the resultant force of the object reacts on to the forklift, causing force of drag. This slightly weighs the forklift downwards due to how much mass the object weighs.
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Please help on this one?
DedPeter [7]

The answer is C. Hope this helps.

3 0
3 years ago
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This painting represents the milky way galaxy as it might appear from a distance. Identify the indicated features.
aksik [14]

The indicated features of how the painting represents the milky way galaxy include the following:

  • Spiral arm
  • Bulge
  • Disk

<h3>What is Milky way?</h3>

This is the galaxy which comprises of our solar system and has a milky appearance when it's viewed from the earth.

The painting represented it as a result of the spiral arm and other features which are listed above.

Read more about Milky way here brainly.com/question/13956361

5 0
2 years ago
What is the answer?​
scoundrel [369]

"Choice-C" is the answer.

7 0
3 years ago
Two objects made of the same material are heated to 60 oC and 90 oC. According to
nlexa [21]
Assuming that there is in a vacuum, the two object will cool at the same rate, because the objects are made of the same material they will have the same cooling rate, assuming the surrounding temperature is the same.
4 0
3 years ago
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You need to design a banked curve at the new circular Super 100 Raceway. The radius of the track is 800 m and cars typically tra
uysha [10]

Answer:

33.1^{\circ}

Explanation:

Let's start by writing the equations of the forces along the two directions:

- Vertical:

N cos \theta = mg

where

N is the normal reaction

\theta is the angle between the road and the horizontal

(mg) is the weight of the car, with m being its mass and g the acceleration of gravity

- Horizontal:

N sin\theta = m \frac{v^2}{r}

where

v is the speed of the car

r is the radius of the turn

Dividing the 2nd equation by the 1st one, we get:

tan \theta = \frac{v^2}{rg}

In this problem:

r = 800 m (radius of the turn)

v=160 mi/h = 71.5 m/s is the speed

g=9.8 m/s^2

Substituting, we find:

\theta= tan^{-1} (\frac{v^2}{rg})=tan^{-1}(\frac{(71.5)^2}{(800)(9.8)})=33.1^{\circ}

8 0
4 years ago
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