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Morgarella [4.7K]
3 years ago
15

A ball of mass of 0.5kg is dropped from a height of 2m.

Physics
1 answer:
erastovalidia [21]3 years ago
3 0

Answer:

6.3m/s

Explanation:

Given parameters:

Mass of the ball  = 0.5kg

Height  = 2m

Unknown:

Velocity when the ball hits the ground  = ?

Solution:

Since the potential  energy is transformed into kinetic energy;

            P.E  = K.E

              mgh  = \frac{1}{2} m v²

     cancelling m;

               gh  =  \frac{1}{2} v²

                v² = 2gh

                v  = √2gh

Insert the parameters and solve;

               v  = √2gh  = √ 2 x 9.8 x 2  = 6.3m/s

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Answer:

Explanation:

atomic number

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3 years ago
A mass is tied to a string and swung in a horizontal circle with a constant angular speed. show answer No Attempt If this speed
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Answer:

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Explanation:

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m is the mass, v is the velocity and r is the radius.

It follows that F \propto v^2, provided m and r are constant.

When v is doubled, the new force, F_1, is

F_1 = \dfrac{m(2v)^2}{r} = \dfrac{4mv^2}{r} = 4\dfrac{mv^2}{r} = 4F

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8 0
4 years ago
You are trying to hear your friend give directions to new store in town. But from your distance (1 point) of 15 m you only hear
Marat540 [252]

Answer:

option D

Explanation:

given,

Intensity of sound = 20 dB

distance = 15 m

intensity of sound is increased to = 50 dB

distance between the sound level = ?

Using relation

L_2 = L_1 - |20(log \dfrac{r_2}{r_1})|

L₁ = 20 dB        L₂ = 50 dB         r₁ = 15 m      r₂ = ?

log (\dfrac{r_2}{r_1}) = \dfrac{L_1 -L_2}{20}

\dfrac{r_2}{r_1}= 10^{\dfrac{|L_1 -L_2|}{20}}

r_2 =r_1 10^{\dfrac{|L_1 -L_2|}{20}}

r_2 =15 \times 10^{\dfrac{|20-50|}{20}}

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7 0
3 years ago
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Andre45 [30]

Answer:

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6 0
3 years ago
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The local high school is installing new bleachers at the stadium and must also add handrails to meet code. The students know the
Advocard [28]

Answer:

The handrails must be approximately 10.63 meters long

Explanation:

The given parameters are;

The height of the bleachers, h = 8 m

The depth of the bleachers, d = 7 m

The length of the hand rails to go along the bleachers from bottom to top is given by Pythagoras' Theorem as follows;

The length of the hand rail = √(d² + h²)

∴ The length of the hand rail = √(7² + 8²) = √113 ≈ 10.63

In order for the handrails to go along the bleachers from top to bottom, they must be approximately 10.63 meters long.

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