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kaheart [24]
4 years ago
12

A soccer ball is kicked horizontally off a bridge with a heigh of 36m. The ball travels 25m horizontally before it hits the pave

ment below. What was the soccer ball's speed when it was first kicked?
Physics
1 answer:
kow [346]4 years ago
5 0

Answer:

The initial velocity was U=22.14m/s

Explanation:

Step one :

Applying the third equation of motion

v² = u²+ 2as

Where v= Final velocity

U =initial velocity

a= acceleration due to gravity

S= distance or displacement

Step two :

V= 0

a= 9.81m/s²

S=25m

U=?

Step three :

Substituting into the equation we have

0²=U²+2*9.81*25

0=U²+490.5

U²=-490.5

U=√490.5

U=22.14m/s

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ki77a [65]
The answer is C. in sort of a way. You can't technically see black matter. However, it is holding the galaxies together.
4 0
4 years ago
Is the wavelength comparable to the size of atoms?
Helen [10]

It totally depends on what kind of wave you're talking about.

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-- the shortest wave that would be called an "X-ray" is 0.01 nanometer long.     You'd have to line up 2.5 billion of <u>those</u> babies to cover an inch.  Hold on to these for a second ... there's one more kind of wave to mention.

-- This brings us to "gamma rays" ... our name for the shortest of all electromagnetic waves.  To be a gamma ray, it has to be shorter than 0.01 nanometer.

Talking very very very very roughly, atoms range in size from about 0.025 nanometers to about 0.26 nanometers.

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5 0
3 years ago
A
malfutka [58]

Answer:

<em>The penny will hit the ground at 6.39 seconds</em>

Explanation:

<u>Free Fall</u>

The penny is dropped from a height of y=200 m. The equation of the height on a free-fall motion is given by:

\displaystyle y=\frac{gt^2}{2}

Where g=9.8\ m/s^2, and t is the time.

Solving for t:

\displaystyle t=\sqrt{\frac{2y}{g}}

Using the value y=200:

\displaystyle t=\sqrt{\frac{2*200}{9.8}}

t=6.39 sec

The penny will hit the ground at 6.39 seconds

4 0
4 years ago
An aluminum wire is held between two clamps under zero tension at room temperature. Reducing the temperature, which results in a
Blababa [14]

Answer:

\frac{\Delta L}{L} =5.37\times 10^{-4}

Explanation:

Given:

  • cross sectional area of the wire, A=5.75\times 10^{-6}\ m^2
  • density of aluminium wire, \rho=2.7\times 10^3\ kg.m^{-3}
  • young's modulus of the material, E=7\times 10^{10}\ N.m^{-2}
  • wave speed, v=118\ m.s^{-1}

<u>We have mathematical expression for strain as:</u>

\frac{\Delta L}{L} =\frac{\sigma}{E} ...............................(1)

and since, \sigma =\frac{T}{A}

where, T = tension force in the wire

equation (1) becomes:

\frac{\Delta L}{L} =\frac{T}{A.E} ............................(2)

<u>Also velocity ofwave in tensed wire:</u>

v=\sqrt{\frac{T}{\mu} } ...................................(3)

where: \mu= linear mass density of the wire

\therefore \mu=\rho \times A

Now, equation (3) becomes

v=\sqrt{\frac{T}{\rho \times A} }

T=v^2.\rho \times A ............................(4)

Using eq. (2) & (4) for tension T

v^2.\rho \times A=A.E\times \frac{\Delta L}{L}

\frac{\Delta L}{L} =\frac{v^2.\rho}{E}

putting the respective values

\frac{\Delta L}{L} =\frac{118^2\times 2.7\times 10^3}{7\times 10^{10}}

\frac{\Delta L}{L} =5.37\times 10^{-4}

6 0
3 years ago
A block with mass m1m1m_1 is placed on an inclined plane with slope angle ααalpha and is connected to a second hanging block tha
vredina [299]

Answer:

Explanation:

m₂ is hanging vertically and m₁ is placed on inclined plane . Both are in limiting equilibrium so on m₁ , limiting friction will act in upward direction as it will tend to slip in downward direct . Tension in cord connecting the masses be T .

For equilibrium of m₁

m₁ g sinα= T + f where f is force of friction

m₁ g sinα= T + μsx m₁ g cosα

m₁ g sinα -  μs x m₁ g cosα = T

For equilibrium of m₂

T = m₂g

Putting this value in equation above

m₁ g sinα -  μs x m₁ g cosα = m₂g

m₂ = m₁ sinα -  μs x m₁ cosα

6 0
3 years ago
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