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2 years ago

8

What is the final speed of an object that starts from rest and accelerates uniformly at 3.0 meters per second2 over a distance o

f 6.0 meters? (1) 8.0 m/s (2) 6 m/s (3) 16 m/s (4) 64 m/s

1 answer:

jasenka [17]2 years ago

7 0

Answer:

16 m/s sir

Explanation:

because i am in collage

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How many miles away is the sun to the earth?

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<span>92.96 million mi..........</span>

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A spinning tossed baseball veers off course in the direction ofA) reduced air pressure on the ball.B) increased air pressure on

katovenus [111]

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A) reduced air pressure on the ball.

Explanation:

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2 years ago

When Simon grows (let’s say he doubles his mass) what happens to his GPE when he is at the balcony?

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His gravitational potential energy will increase as well.

Explanation:

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gpe = mass × gravitational field strength × height

From the formula above, we can conclude that as the mass of a body increases, it's gpe increases too.

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2 years ago

Calculate the pressure exerted on the floor by the boy standing on both feet if the weight of the boy is 40kg. Assume that the a

Jlenok [28]

Answer:

P = 1333.33 N

Explanation:

The pressure exerted by the boy on the floor can be calculated by the following equation:

$P = \frac{F}{A}$

where,

P = Pressure exerted by the boy = ?

F = Force Applied = Weight of Boy = 40 kg = 40 N (since 1 kg = 1N)

A = Area of application of force = 2(Area of one show) = 2(6 cm x 25 cm)

A = 2(0.06 m x 0.25 m) = 0.03 m²

Therefore,

$P = \frac{40\ N}{0.03\ m^2}\\\\$

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2 years ago

A string of mass 60.0 g and length 2.0 m is fixed at both ends and with 500 N in tension. a. If a wave is sent along this string

Darya [45]

Answer:

a

The speed of wave is $v_1 = 129.1 \ m/s$

b

The new speed of the two waves is $v = 129.1 \ m/s$

Explanation:

From the question we are told that

The mass of the string is $m = 60 \ g = 60 *10^{-3} \ kg$

The length is $l = 2.0 \ m$

The tension is $T = 500 \ N$

Now the velocity of the first wave is mathematically represented as

$v_1 = \sqrt{ \frac{T}{\mu} }$

Where $\mu$ is the linear density which is mathematically represented as

$\mu = \frac{m}{l}$

substituting values

$\mu = \frac{ 60 *10^{-3}}{2.0 }$

$\mu = 0.03\ kg/m$

So

$v_1 = \sqrt{ \frac{500}{0.03} }$

$v_1 = 129.1 \ m/s$

Now given that the Tension, mass and length are constant the velocity of the second wave will same as that of first wave (reference PHYS 1100 )

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3 years ago