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snow_tiger [21]
3 years ago
11

Help me please ()- :))

Physics
2 answers:
ratelena [41]3 years ago
8 0

Answer:

A pardon

Explanation:

Hope this helps

Darya [45]3 years ago
3 0

Answer:

that would probably be a A.Pardon

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FLUKE
Thepotemich [5.8K]

Answer:

cant understand anything

Explanation:

3 0
3 years ago
A cooling fan is turned off when it is running at 850 rev/min. It turns 1500 revolutions before it comes to a stop. (a) What was
8_murik_8 [283]

Answer

given,

cooling fan revolution = 850 rev/min

fan turns before revolution = 1500 revolutions

\omega = 850 \dfrac{2\pi}{60}

\omega = 89\ rad/s

θ = 1500 revolution

θ = 1500 x 2 x π

θ = 9424.78 rad

a) using equation of rotation

ω² = ω₀² + 2 α θ

ω = 0 because body comes to rest

0 = 89² + 2 x α x 9424.78

α = -0.42 rad/s²

b) time take for the fan to stop

ω = ω₀ + α t

0 = 89 - 0.42 t

t = \dfrac{89}{0.42}

t = 211.9 s

5 0
4 years ago
A hydraulic machine can be used to lift extremely heavy objects. Why is the fluid in the hydraulic machine a liquid rather than
Sauron [17]
Hi,

The force that acts on hydraulic machine is heavy therefore the content must be something that cannot be compressed by that kind of force, the gas can easily be compressed while a liquid is nearly impossible to.
4 0
3 years ago
Read 2 more answers
Question 2 of 10
Arlecino [84]

Answer:

B: Energy that is transformed is neither created or destroyed

Explanation:

4 0
2 years ago
Given a second class lever with a distance of 5.00 feet from the fulcrum to the effort and a distance of 33.0 inches from the re
leva [86]

Answer:

The correct answer is C. 45.5 lbs.

Explanation:

In a second class lever, the load is located between the point in which the force is exerted and the fulcrum.

The formula for any problem involving a lever is:

F_ed_e=F_ld_l

Where F_e is the effort force, d_e is the total length of the lever, F_l is the load that can be lifted and d_l is the distance between the point of the effort and the fulcrum.

The parameter of the formula that you need is F_l:

F_l=\frac{F_ed_e}{d_l}

The conversion from feet to inches is 1 ft is equal to 12 inches. In this case, 5 ft are equal to 60 inches.

F_l=\frac{25*60}{33}

F_l=45.5 lbs

7 0
4 years ago
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