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olga_2 [115]
3 years ago
9

A cheetah starts from rest and accelerated at 8.7 m/s^2 for 3s. How far did the cheetah go in that time

Physics
1 answer:
nika2105 [10]3 years ago
4 0

Answer:

cheetah goes 52.2 m in that time.

Explanation:

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Which will result in positive buoyancy and cause the object to float?
Wewaii [24]

the answer should be:

When the buoyant force is equal to the force of gravity

4 0
3 years ago
Read 2 more answers
E)
guapka [62]

Answer:

Pressure, P = 32666.66 Pa

Explanation:

It is given that,

Surface area of foot of Bimaba is 150 cm² or 0.015 m².

Her weight is 50 kg

We need to find the pressure does she exert on the ground, as she stands on  her one foot. The force acting per unit area is called pressure. It can be given by :

P=\dfrac{F}{A}\\\\P=\dfrac{mg}{A}\\\\P=\dfrac{50\times 9.8}{0.015}\\\\P=32666.66\ Pa

So, the pressure is 32666.66 Pa.

7 0
3 years ago
Please do number 25! Explain how you got your answer with detail to get Brainliest! Thank you!
Ad libitum [116K]
John weighs 200 pounds.
In order to lift himself up to a higher place, he has to exert force of 200 lbs.

The stairs to the balcony are 20-ft high.
In order to lift himself to the balcony, John has to do
(20 ft) x (200 pounds)  =  4,000 foot-pounds of work.

If he does it in 6.2 seconds, his RATE of doing work is
(4,000 foot-pounds) / (6.2 seconds)  =  645.2 foot-pounds per second.

The rate of doing work is called "power".

(If we were working in the metric system (with SI units),
the force would be in "newtons", the distance would be in "meters",
1 newton-meter of work would be 1 "joule" of work, and
1 joule of work per second would be 1 "watt".
Too bad we're not working with metric units.)

So back to our problem.

John has to do 4,000 foot-pounds of work to lift himself up to the balcony,
and he's able to do it at the rate of 645.2 foot-pounds per second.

Well, 550 foot-pounds per second is called 1 "horsepower".

So as John runs up the steps to the balcony, he's doing the work
at the rate of

           (645.2 foot-pounds/second) / (550 ft-lbs/sec per HP)

=  1.173 Horsepower.  GO JOHN !

(I'll betcha he needs a shower after he does THAT 3 times.)
_______________________________________________

Oh my gosh !  Look at #26 !  There are the metric units I was talking about.

Do you need #26 ?

I'll give you the answers, but I won't go through the explanation,
because I'm doing all this for only 5 points.

a).  5
b).  750 Joules
c).  800 Joules
d).  93.75%

You're welcome.

And #27 is 0.667 m/s .
7 0
3 years ago
Sally takes two bar magnets and randomly tapes one end of each bar magnet. she labels the magnets A and B. She brings the taped
vlabodo [156]
Since the two taped poles of the magnets labeled A and B attracts one to each other, we know that the two taped poles are oppsosite.

So, you can predict with total certainty that when she brings the taped end of the third magnet (magnet C) near each of the first two magntes, in one case they will attract each other and in the other case they will repele mutually.

You are certain of that because, since the taped poles of the first two magnets are opposite, the pole of the third magnet has to be equal to one of the two first taped poles and opposite to the other of the two firest taped poles.
8 0
3 years ago
A circular loop of wire with radius 2.00 cm and resistance 0.600 Ω is in a region of a spatially uniform magnetic field B⃗ that
Anika [276]

Answer:

Incomplete questions

Let assume we are asked to find

Calculate the induced emf in the coil at any time, let say t=2

And induced current

Explanation:

Flux is given as

Φ=NAB

Where

N is number of turn, N=1

A=area=πr²

Since r=2cm=0.02

A=π(0.02)²=0.001257m²

B=magnetic field

B(t)=Bo•e−t/τ,

Where Bo=3T

τ=0.5s

B(t)=3e(−t/0.5)

B(t)=3exp(-2t)

Therefore

Φ=NAB

Φ=0.001257×3•exp(-2t)

Φ=0.00377exp(-2t)

Now,

Induce emf is given as

E= - dΦ/dt

E= - 0.00377×-2 exp(-2t)

E=0.00754exp(-2t)

At t=2

E=0.00754exp(-4)

E=0.000138V

E=0.138mV

b. Induce current

From ohms laws

V=iR

Given that R=0.6Ω

i=V/R

i=0.000138/0.6

i=0.00023A

i=0.23mA

5 0
3 years ago
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