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Alinara [238K]
2 years ago
10

A child carries a 3N book at a constant velocity of 4 meters across a horizontal floor. What is the net work done?

Physics
1 answer:
s2008m [1.1K]2 years ago
6 0
This problem is getting you used to the definition of 'work' = (force) x (distance).

It's hard to really imagine, but technically, when an object moves horizontally,
no work is done, because there's no lifting against the force of gravity.

Another way to look at it:  The potential energy of the object never changes,
because its height never changes, so no energy ever goes into it ot gets
taken out of it.

Also, if the object moves at a constant speed, then it never accelerated,
and that means that no force ever acted on it.

So technically, when an object moves horizontally at a constant speed,
the work done on it is zero.
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What do understand by the efficiency of a machine? By using a block and tackel a man can raise a load of 720 N by an effort of 1
motikmotik

Answer:

Efficiency of a machine is how well the machine works and what the machine is capable of doing.

Mechanical advantage=Load/Effort.

720/180=4

6 0
3 years ago
A devout halloweener not only dressed as an astronaut, but travelled to the moon for the full experience. The astronaut jumps on
nikdorinn [45]

Answer:

2.78 m

Explanation:

At the peak, the velocity is 0.

Given:

a = -1.6 m/s²

v₀ = 2.98 m/s

v = 0 m/s

x₀ = 0 m

Find:

x

v² = v₀² + 2a(x - x₀)

(0 m/s)² = (2.98 m/s)² + 2(-1.6 m/s²) (x - 0 m)

x = 2.775 m

Rounded to 3 sig-figs, the astronaut halloweener reaches a maximum height of 2.78 meters.

5 0
3 years ago
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!
Solnce55 [7]

I think the correct answer is D: Potential Energy.

3 0
3 years ago
Read 2 more answers
Muscular Strength exercises focus on high______and low___
Advocard [28]

Answer:

d d d d dd d d d d d dd d d d d dd d

4 0
3 years ago
Why is it important to know the direction of the force applied to a moving object and the direction in which the object is movin
ELEN [110]
C is correct.  The work-force relation is given by W=F·d, where F is force vector, and d is the displacement vector.  The dot is the dot product, which is a measure of how parallel the two vectors are.  It can be restated as the product of two vector magnitudes times the cosine of the angle between them.  Therefore work is a scalar, not a vector, since the dot product returns a scalar.  
W=Fdcos(\theta)
3 0
3 years ago
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