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GREYUIT [131]
2 years ago
9

A boy and a girl are on a spinning merry-go-round. The boy is at a radial distance of 1.2 m from the central axis; the girl is a

t a radial distance of 1.8 m from the central axis. Which is true?A- Boy and girl have zero tangential and angular accelerations.B- The girl has a larger angular acceleration than the boy.C- The boy has a larger tangential acceleration than the girl.D- The boy has a larger angular acceleration than the girl.E- The girl has a larger tangential acceleration than the boy.
Physics
1 answer:
Licemer1 [7]2 years ago
8 0

Answer:

E) True. The girl has a larger tangential acceleration than the boy.

Explanation:

In this exercise they do not ask us to say which statement is correct, for this we propose the solution to the problem.

Angular and linear quantities are related

          v = w r

          a = α r

the boy's radius is r₁ = 1.2m the girl's radius is r₂ = 1.8m

as the merry-go-round rotates at a constant angular velocity this is the same for both, but the tangential velocity is different

          v₁ = w 1,2 (boy)

          v₂ = w 1.8 (girl)

whereby

          v₂> v₁

reviewing the claims we have

          a₁ = α 1,2

          a₂ = α 1.8

          a₂> a₁

A) False. Tangential velocity is different from zero

B) False angular acceleration is the same for both

C) False. It is the opposite, according to the previous analysis

D) False. Angular acceleration is equal

E) True. You agree with the analysis above,

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Answer:

Its heat capacity is higher than that of any other liquid or solid, its specific heat being 1 cal / g, this means that to raise the temperature of 1 g of water by 1 ° C it is necessary to provide an amount of heat equal to a calorie . Therefore, the heat capacity of 1 g of water is equal to 1 cal / K.

Explanation:

The water has a very high heat capacity, a large amount of heat is necessary to raise its temperature 1.0 ° K. For biological systems this is very important because the cellular temperature is modified very little in response to metabolism. In the same way, aquatic organisms, if water did not possess that quality, would be very affected or would not exist.

This means that a body of water can absorb or release large amounts of heat, with little temperature change, which has a great influence on the weather (large bodies of water in the oceans take longer to heat and cool than the ground land). Its latent heats of vaporization and fusion (540 and 80 cal / g, respectively) are also exceptionally high.

6 0
3 years ago
One of the largest barometers ever built was an oil-filled barometer constructed in Leicester, England in 1991. The oil had a he
ZanzabumX [31]

Answer:

ρ = 830.32 kg/m³

Explanation:

Given that

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Pressure P = 1.013 x 10⁵ Pa

Lets take density of the liquid =ρ

The pressure due to liquid P given as

P = ρ g h

Now by putting the all values in the above equation

1.013 x 10⁵ Pa = ρ x 10 x 12.2                 ( take g =10 m/s²)

ρ = 830.32 kg/m³

Therefore the density of oil is 830.32  kg/m³

5 0
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During a vaporization a substance changes from what to what
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A substance changes from liquid to gas
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Ja-Anna, has a mass of 45 kg, climbed 3m up a ladder to rescue her cat from a tree. how much work against gravity did she do?
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3 years ago
An electron is moving east in a uniform electric field of 1.55 N/C directed to the west. At point A, the velocity of the electro
valkas [14]

Answer:

Final velocity of electron, v=6.45\times 10^5\ m/s    

Explanation:

It is given that,

Electric field, E = 1.55 N/C

Initial velocity at point A, u=4.52\times 10^5\ m/s

We need to find the speed of the electron when it reaches point B which is a distance of 0.395 m east of point A. It can be calculated using third equation of motion as :

v^2=u^2+2as........(1)

a is the acceleration, a=\dfrac{F}{m}

We know that electric force, F = qE

a=\dfrac{qE}{m}

Use above equation in equation (1) as:

v^2=u^2+\dfrac{2qEs}{m}

v^2=(4.52\times 10^5\ m/s)^2+2\times \dfrac{1.6\times 10^{-19}\ C\times 1.55\ N/C}{9.1\times 10^{-31}\ kg}\times 0.395\ m

v = 647302.09 m/s

or

v=6.45\times 10^5\ m/s

So, the final velocity of the electron when it reaches point B is 6.45\times 10^5\ m/s. Hence, this is the required solution.

3 0
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