All categories

2 years ago

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

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.

You might be interested in

Deadpool is doing a superhero landing from a 21 meter tall building what would his velocity be right before he hits the ground?

Morgarella [4.7K]

Answer:

I think he would be dead poggers

Explanation:

4 0

2 years ago

A child's top is held in place upright on a frictionless surface. The axle has a radius of ????=3.21 mm . Two strings are wrappe

tekilochka [14]

Answer:

Angular momentum, $L=6.47\times 10^{-3}\ m$

Explanation:

It is given that,

Radius of the axle, $r=3.21\ mm=3.21\times 10^{-3}\ m$

Tension acting on the top, T = 3.15 N

Time taken by the string to unwind, t = 0.32 s

We know that the rate of change of angular momentum is equal to the torque acting on the torque. The relation is given by :

$\tau=\dfrac{dL}{dt}$

Torque acting on the top is given by :

$\tau=F\times r$

Here, F is the tension acting on it. Torque acting on the top is given by :

$\tau=2F\times r$

$2T\times r=\dfrac{L}{t}$

$L=2T\times r \times t$

$L=2\times 3.15\times 3.21\times 10^{-3}\times 0.32$

$L=6.47\times 10^{-3}\ m$

So, the angular momentum acquired by the top is $6.47\times 10^{-3}\ m$ . Hence, this is the required solution.

7 0

3 years ago

A car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of 1.85 m/s2

Tasya [4]

Answer:

Coefficient of friction = 0.836

Explanation:

If v be the speed after one quarter of the circular path

v² = 2as = 2 x 1.85 x 2πr/4 ; v²/r = 1.85 x 3.14 = 5.8

tangential acceleration = 5.8 m/s²

radial acceleration = v² /r = 5.8

total acceleration = √2 x 5.8

m x√2 x 5.8 = m x g xμ

μ = √2 x 5.8 / 9.8 = 0.836

7 0

3 years ago

An air-track glider attached to a spring oscillates between the 10.0 cm mark and the 57.0 cm mark on the track. The glider compl

pickupchik [31]

Answer:

a. 2.1 s

b.0.48 Hz

c. A=24cm

d. 72cm/s

Explanation:

An air-track glider attached to a spring oscillates between the 10.0 cm mark and the 57.0 cm mark on the track. The glider completes 15.0 oscillations in 31.0 s.What are the (a) period, (b) frequency, (c) amplitude, and (d) maximum speed of the glider?

What are the period,

period is the time taken for a wave particle to make one complete oscillation

a) 31 / 15 = 2.066 seconds

= 2.1 s

(b) frequency : this the number of oscillation made in one seconds.

it is also the inverse of the period.

= oscillations / time

= 15/31= 0.48 Hz

amplitude = 1/2 of the difference of oscillation marks

= 1/2(57-10) = 47/2cm

23.5cm

A=24cm

(d) maximum speed of the glider?

V=ωA

angular frequency *Amplitude

V=a*pi*f*amplitude

2π x frequency x amplitude = maximum speed

= 2π x .48 x 24

=72.38 cm/s

72cm/s

5 0

3 years ago

A train takes 2h to reach station B from station A and 3h to return back to A.The distance between the station is 200km, then it

serg [7]

Answer:

80km/hr

Explanation:

Total distance: 400km

Total time 5 hours

Average speed= distance/time

400/5= 80km/hr

8 0

3 years ago