Ask question

Ask question

All categories

3 years ago

14

A forward horizontal force of 3 lb is used to pull a 60 lb sled at constant velocity on a frozen pond. the coefficent of (kineti

c) friction is:
1. 0.2
2. 0.05
3. 0.5
4. 20
5. 2

1 answer:

puteri [66]3 years ago

7 0

It’s either 0.05 or 20. Assuming that the coefficient friction is a damping factor, I feel like 0.05 would be correct m

You might be interested in

If an object is moving to the right and there is a net force acting on it to the left, what will happen to the object?

Ksju [112]

Hdjnenhdhdbbdbddbbdjjdjdjjejekrkrk

7 0

3 years ago

An ant travels 2.78 cm (West) and then turns and travels 6.25 cm (South 40 degrees East). What is the ant's total displacement?

andrey2020 [161]

$From\ cosine\ theorem:\\\\
c^2=a^2+b^2-2abcos(\angle between\ a\ and\ b)\\\\
a=2,78\\b=6,25\\
\angle between\ a\ and\ b=90+50=140^{\circ}\\\\
c^2=2,78^2+6,25^2-2*2,78*6,25cos(140^{\circ})\\\\
c^2=7,7284+39,0625-34,75*(-0,77)\\\\c^2=46,7909+26,7575\\\\c^2=735484\\\\c=8,58cm\\\\Total\ displacement\ is\ equal\ to\ 8,58cm.$

5 0

3 years ago

Your body is receiving backround radiation all the tim. Explain

adoni [48]

The uniform microwave radiation remaining from the Big Bang.

So, your body is always having background radiation and that means space!

3 0

3 years ago

Match these items.

horrorfan [7]

Iron oxide = small region within a magnet
drop or hammer =man-made magnet
strength decreases rapidly with distance lines of force

4 0

3 years ago

Tarzan swings on a 26.2 m long vine initially inclined at an angle of 28° from the vertical. (a) What is his speed at the bottom

Marianna [84]

Answer

given,

length of the swing = 26.2 m

inclined at an angle = 28°

let, the initial height of the Tarzan be h

h = L (1 - cos θ)

a) initial velocity v₁ = 0 m/s

final velocity of Tarzan = v_f

law of conservation of energy

PE_i + KE_i = PE_f + KE_f

$mgh_i + \dfrac{1}{2}mv_i^2= mgh_f + \dfrac{1}{2}mv_f^2$

$mgh_i + 0 = 0 + \dfrac{1}{2}mv_f^2$

$mgh_i = \dfrac{1}{2}mv_f^2$

$v_f = \sqrt{2gh_i}$

$= \sqrt{2gL(1- cos\theta)}$

$= \sqrt{2\times 9.8 \times 26.2(1- cos 28^0)}$

= 7.75 m/s

the speed tarzan at the bottom of the swing

v_f = 7.75 m/s

b)initial speed of the = 3 m/s

$mgh_i + \dfrac{1}{2}mv_i^2= mgh_f + \dfrac{1}{2}mv_f^2$

$mgh_i + 0 = 0 + \dfrac{1}{2}mv_f^2$

$mgh_i+ \dfrac{1}{2}mv_i^2 = \dfrac{1}{2}mv_f^2$

$gh_i+ \dfrac{1}{2}v_i^2 = \dfrac{1}{2}v_f^2$

$v_f = \sqrt{v_1^2+2gh_i}$

$v_f = \sqrt{3^2+2\times 9.8 \times (1- cos 28^0)}$

v_f= 11.29 m/s

3 0

3 years ago