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Ainat [17]
3 years ago
9

A 6450kg gorilla slides down the side of a 20m tall mountain at 47degrees. If the coefficient of kinetic friction is 0.23, what

is the net work done on the gorilla?
80kg Keoni Mabini is pulled 33m up makakilo hill at 45 degrees by Brayden Pfister with aforce of 200 N. If the coefficient of kinetic friction between the Keoni’s okole and the hillis 0.11, what is the change in her kinetic energy?

Anilov rolls Brayden’s 3.5kg ball 10m. If she exerts force of 12N and the coefficient offriction between the ball and the ground is 0.31, what is the ball’s final speed?

PLEASE HELP ASAP
Physics
1 answer:
blondinia [14]3 years ago
5 0

Answer:

I have no idea but next time don't put so many points.

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An elephant's legs have a reasonably uniform cross section from top to bottom, and they are quite long, pivoting high on the ani
xz_007 [3.2K]

Answer:

is this the full question?

Explanation:

8 0
3 years ago
PLEASE HELP ASAP!!!!!!!!!!!!
lianna [129]

The first choice on the list is the correct one.

7 0
3 years ago
A block of mass m=1.2 kg is held at rest against a spring with a force constant k=730N/m. Initially the spring is compressed a d
igor_vitrenko [27]

Answer:

Explanation:

potential energy of compressed spring

= 1/2 k d²

= 1/2 x 730 d²

= 365 d²

This energy will be given to block of mass of 1.2 kg in the form of kinetic energy .

Kinetic energy after crossing the rough patch

= 1/2 x 1.2 x 2.3²

= 3.174 J

Loss of energy

= 365 d² - 3.174  

This loss is due to negative work done by frictional force

work done by friction = friction force x width of patch

= μmg d ,   μ = coefficient of friction , m is mass of block , d is width of patch

= .44 x 1.2 x 9.8 x .05

= .2587 J

365 d² - 3.174   = .2587

365 d² = 3.4327

d² = 3.4327 / 365

= .0094

d = .097 m

= 9.7 cm

If friction increases , loss of energy increases . so to achieve same kinetic energy , d will have to be increased so that initial energy increases so compensate increased loss .

5 0
3 years ago
Metals in group 1 on the periodic table most commonly form which type of ion
givi [52]

Answer:

Group 1 - the alkali metals. The Group 1 elements in the periodic table are known as the alkali metals.

Explanation:

7 0
3 years ago
A large fraction of the ultraviolet (UV) radiation coming from the sun is absorbed by the atmosphere. The main UV absorber in ou
irakobra [83]

Answer:

λ = 3.2 x 10⁻⁷ m = 320 nm

Explanation:

The relationship between the velocity of electromagnetic waves (UV rays) and the their frequency is:

v = fλ

where,

v = c = speed of the electromagnetic waves (UV rays) = speed of light

c = 3 x 10⁸ m/s

f = frequency of the electromagnetic waves (UV rays) = 9.38 x 10¹⁴ Hz

λ = wavelength of the electromagnetic waves (UV rays) = ?

Therefore, substituting the values in the relation, we get:

3 x 10⁸ m/s = (9.38 x 10¹⁴ Hz)(λ)

λ = (3 x 10⁸ m/s)/(9.38 x 10¹⁴ Hz)

<u>λ = 3.2 x 10⁻⁷ m = 320 nm</u>

So, the radiation of <u>320 nm</u> wavelength is absorbed by Ozone.

3 0
3 years ago
Read 2 more answers
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