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crimeas [40]
3 years ago
15

A forklift lifts 5 boxes from the ground to a height of 2 meters (m). The boxes push down with a force of 1000 newtons (N). How

much work was done by the forklift to lift the boxes?
Physics
1 answer:
Nata [24]3 years ago
8 0
Hello!

Answer:

2000 J

Explanation

Work equation is expressed as:

W=F.d.Cos \alpha

Where:

F: Applied force
d: traveled distance
α: Angle between the direction of the force and the direction of the movement. (in this case, both of the direction are the same, so the angle is 0°)

By substituting:

F=1000N.2m.Cos(0)=2000N.m=2000 J

Have a nice day!
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What is the work-energy theorem equation?
grin007 [14]

Answer:

W = Fd = KE =1/2mv²

Explanation:

not sure if that's what your looking for but i'm pretty sure this is it.

7 0
3 years ago
Read 2 more answers
50 points please help ASAP! !
Contact [7]

Answer:

The sum of the initial and final velocity is divided by 2 to find the average. The average velocity calculator uses the formula that shows the average velocity (v) equals the sum of the final velocity (v) and the initial velocity (u), divided by 2.

4 0
3 years ago
An 80-kgkg quarterback jumps straight up in the air right before throwing a 0.43-kgkg football horizontally at 15 m/sm/s . How f
vladimir1956 [14]

Answer:

V = 0.0806 m/s

Explanation:

given data

mass quarterback = 80 kg

mass football = 0.43 kg

velocity = 15 m/s

solution

we consider here momentum conservation is in horizontal direction.

so that here no initial momentum of the quarterback

so that final momentum of the system will be 0

so we can say

M(quarterback) ×  V = m(football) × v (football)   ........................1

put here value we get

80 ×  V  = 0.43  × 15

V = 0.0806 m/s

5 0
3 years ago
Ruff, the 50 cm tall Labrador Retriever stands 3m from a plane mirror and looks at his image. What is Ruffs image position and h
GaryK [48]
Ruff's image is 50m behind the mirror surface and the image is also 3m tall.

This is because it is a plane mirror.
5 0
3 years ago
Two stationary positive point charges, charge 1 of magnitude 3.90 nC and charge 2 of magnitude 1.80 nC, are separated by a dista
soldi70 [24.7K]

Answer:

v = 7793150 m/s

Explanation:

First, we are going to calculate the electrical potential in the point middle between the two charges

Remember that the electrical potential can be calculated as:

v = \frac{kQ}{r}

                 Where     k = 8.9874 x 10^{9} \frac{Nm^{2} }{C^{2} }

and it is satisfy the superposition principle, thus

v = \frac{8.9874x10^{9}(3.90x10^{-9} ) }{0.23} +  \frac{8.9874x10^{9}(1.80x10^{-9} ) }{0.23}

v = 222.73v

The electrical potential at 10 cm from charge 1 is:

v = \frac{8.9874x10^{9}(3.90x10^{-9} ) }{0.1} +  \frac{8.9874x10^{9}(1.80x10^{-9} ) }{0.36}

v = 395.44 v

Since the work - energy theorem, we have:

q\Delta v = \frac{mv^{2} }{2}

                     where q is the electron's charge and m is the electron's mass

Therefore:

v = \sqrt{\frac{2q\Delta v}{m} }

v = 7793150 m/s

6 0
3 years ago
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