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enyata [817]
3 years ago
7

You are on your balcony and notice some bad squirrels digging in your garden directly below. You start tossing your pistachios a

t them to get them out of your yard. If the height of your guard rail is 43 inches above the deck, which is at a standard height of 10 feet above the ground, and you toss the pistachios straight down with a speed of 7.4 m/s, then how long does it take for the pistachio to reach the ground?
Physics
1 answer:
Rashid [163]3 years ago
3 0

Answer:

0.43 s

Explanation:

We have the following parameters:

Initial velocity, u = 7.4 m/s

Acceleration of gravity, g = 9.8 m/s^2

Distance, s = 43 in + 10 ft = 1.092 m + 3.048 m = 4.14 m

Time, t = ?

Using the equation of motion s=ut +\frac{1}{2}gt^2, we have

4.14 = 7.4t + 0.5\times9.8t^2

4.9t^2 + 7.4t - 4.14 =0

Using the quadratic formula \dfrac{-b\pm\sqrt{b^2-4ac}}{2a} where a = 4.9, b = 7.4 and c = - 4.14, and solving for the positive value of t only, we have

t = 0.43 s

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A 100-kg tackler moving at a speed of 2.6 m/s meets head-on (and holds on to) an 92-kg halfback moving at a speed of 5.0 m/s. Pa
DIA [1.3K]

Given that,

Mass of trackler, m₁ = 100 kg

Speed of trackler, u₁ = 2.6 m/s

Mass of halfback, m₂ = 92 kg

Speed of halfback, u₂ = -5 m/s (direction is opposite)

To find,

Mutual speed immediately after the collision.

Solution,

The momentum of the system remains conserved in this case. Let v is the mutual speed after the collision. Using conservation of momentum as :

m_1u_1+m_2u_2=(m_1+m_2)V\\\\V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\V=\dfrac{100\times 2.6+92\times (-5)}{(100+92)}\\\\V=-1.04\ m/s

So, the mutual speed immediately after the collision is 1.04 m/s but in opposite direction.

3 0
3 years ago
Sunburn is caused by
Snowcat [4.5K]

Answer:

too much exposure to the sun's rays

5 0
3 years ago
Read 2 more answers
If the wave represents a sound wave, explain how increasing amplitude will affect the loudness of the sound? If we decrease the
Viktor [21]

Answer:

Explanation:

Think of a sound wave like a wave on the ocean, or lake... It's not really water moving, as much as it's energy moving through the water. Ever see something floating on the water, and notice that it doesn't come in with the wave, but rides over the top and back down into the trough between them? Sound waves are very similar to that. If you looked at a subwoofer speaker being driven at say... 50 cycles a second, you'd actually be able to see the speaker cone moving back and forth. The more power you feed into the speaker, the more it moves back and forth, not more quickly, as that would be a higher frequency, but further in and further out, still at 50 cycles per second. Every time it pushed out, it's compressing the air in front of it... the compressed air moves away from the speaker's cone, but not as a breeze or wind, but as a wave through the air, similar to a wave on the ocean

More power, more amplitude, bigger "wave", louder ( to the human ear) sound.

If you had a big speaker ( subwoofer ) and ran a low frequency signal with enough power in it, you could hold a piece of paper in front of it, and see the piece of paper move in and out at exactly the same frequency as the speaker cone. The farther away from the speaker you got, the less it'd move as the energy of the sound wave dispersed through the room.

Sound is a wave

We hear because our eardrums resonates with this wave I.e. our ear drums will vibrate with the same frequency and amplitude. which is converted to an electrical signal and processed by our brain.

By increasing the amplitude our eardrums also vibrate with a higher amplitude which we experience as a louder sound.

Of course when this amplitude is too high the resulting resonance tears our eardrums so that they can't resonate with the sound wave I.e. we become deaf

6 0
2 years ago
A star is moving away from an observer at 1% of the speed of light. At what wavelength would the observer find an emission line
Ivan

Answer:

  λ = 5940 Angstroms

Explanation:

This is an exercise of the relativistic Doppler effect

        f’= f  √((1- v / c) / (1 + v / c))

Where the speed in between the strr and the observer is positive if they move away

Let's use the relationship

         c = λ f

         f = c /λ

We replace

              c /λ’ = c /λ  √ ((1- v / c) / (1 + v / c))

              λ = λ’ √ ((1- v / c) / (1 + v / c))

Let's calculate

             v = 0.01 c

             v = 0.01 3 10⁸

             v=  3 10⁶ m / s

             λ = 6000 √ [(1- 3 10⁶/3 10⁸) / (1+ 3 10⁶/3 10⁸)]

             λ = 6000 √ [0.99 / 1.01]

             λ = 5940 Angstroms

6 0
3 years ago
The maximum number of orbits in an atom is _____
svp [43]

Correct answer is:

<h2>The maximum number of orbits in an atom is <u>Seven.</u></h2><h3>Explanation:</h3>

Every energy level has a limited one orbital including two electrons. The orbits are settled in the sub-levels and there can be further than 1 sub-level as the number of energy levels rises. On energy level 1, there is 1 sub-level and 1 orbital. Energy level 2 can possess 2 sub-levels and 2 orbitals. These remain to develop as you progress from the nucleus of the atom, closing up with an infinite potential number of levels and orbits.

8 0
3 years ago
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