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strojnjashka [21]
3 years ago
15

A tennis ball is dropped from 1.62 m above the

Physics
1 answer:
natita [175]3 years ago
8 0

Answer:

-5.63 m/s

Explanation:

Given:

y₀ = 1.62 m

y = 0 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: v

v² = v₀² + 2a (y − y₀)

v² = (0 m/s)² + 2(-9.8 m/s²) (0 m − 1.62 m)

v = -5.63 m/s

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1)Calculate the mass of a girl that has a weight of 113 Newtons.
Blababa [14]

Answer:

Explanation:We should know that weight = mass * gravity.

That is weight equals mass times gravity.

Gravity is a force of attraction between any two bodies in the universe. It is directly proportional to product of their masses and inversely proportional to the square of the distance between them.

Gravity is generally measured in terms of acceleration due to gravity, denoted as g. For Earth it is, 9.8 m/s². And for moon, it is about 1.62 m/s².

On Earth, your weight is 70 kg = W

W = mass x 9.8

70 = mass x 9.8

Your mass is 70/ 9.8

i.e approximately 7.14

Weight at the Moon, W' = 7.14 x 1.62

Hence, your weight on the surface of the moon is just 11.56 kg.

Congratulations, you've lost about 58.14 kilograms without any hard exercise. And you're as light as a Sweedish Vallhund! Cheers!

3 0
3 years ago
1. A listener stands 20.0 m from a speaker that pumps out music with a power output of 100.0 W.
marta [7]

(1.a) The surface area being vibrated by the time the sound reaches the listener is 5,026.55 m².

(1.b) The intensity of the sound wave as it reaches the person listening is 0.02 W/m².

(1.c) The relative intensity of the sound as heard by the listener is 103 dB.

(2.a) The speed of sound if the air temperature is 15⁰C is 340.3 m/s.

(2.b) The frequency of the sound heard by the suspect is 614.3 Hz.

<h3>Surface area being vibrated</h3>

The surface area being vibrated by the time the sound reaches the listener is calculated as follows;

A = 4πr²

A = 4π x (20)²

A = 5,026.55 m²

<h3>Intensity of the sound</h3>

The intensity of the sound is calculated as follows;

I = P/A

I = (100) / (5,026.55)

I = 0.02 W/m²

<h3>Relative intensity of the sound</h3>

B = 10log(\frac{I}{I_0} )\\\\B = 10 \times log(\frac{0.02}{10^{-12}} )\\\\B = 103 \ dB

<h3>Speed of sound at the given temperature</h3>

v= 331.3\sqrt{1 + \frac{T}{273} } \\\\v = 331.3\sqrt{1 + \frac{15}{273} } \\\\v = 340.3 \ m/s

<h3>Frequency of the sound</h3>

The frequency of the sound heard is determined by applying Doppler effect.

f_o = f_s(\frac{v \pm v_0}{v \pm v_s} )

where;

  • -v₀ is velocity of the observer moving away from the source
  • -vs is the velocity of the source moving towards the observer
  • fs is the source frequency
  • fo is the observed frequency
  • v is speed of sound

f_0 = f_s(\frac{v-v_0}{v- v_s} )

f_0 = 512(\frac{340.3 - 10}{340.3 - 65} )\\\\f_0 = 614.3 \ Hz

Learn more about intensity of sound here: brainly.com/question/17062836

3 0
1 year ago
Explain briefly how solar energy is used to generate electricity
valkas [14]

Answer:

Solar radiation may be converted directly into electricity by solar cells (photovoltaic cells). In such cells, a small electric voltage is generated when light strikes the junction between a metal and a semiconductor (such as silicon) or the junction between two different semiconductors.

Explanation:

Pls mark me as brainliest

7 0
2 years ago
A bobsled zips down an ice track, starting from rest at the top of a hill with a vertical height of 150m. Disregarding friction,
olchik [2.2K]
<span>The velocity would be 54.2 m/s We would use the equation 1/2mv^2top+mghtop = 1/2mv^2bottom+mghbottom where m is the mass of the bobsled(which can be ignored), vtop/bottom is the velocity of the bobsled at the top or bottom, g is gravity, and htop/bottom is the height of the bobsled at the top or bottom of the hill. Since the velocity of the bobsled at the top of the hill and height at the bottom of the hill are zero, 1/2mv^2top and mghbottom will equal zero. The equation will be mghtop=1/2mv^2bottom. Thus we would solve for v.</span>
5 0
3 years ago
Read 2 more answers
A spring stretches by 0.0208 m when a 3.39-kg object is suspended from its end. How much mass should be attached to this spring
sashaice [31]

Answer:

COMPLETE QUESTION

A spring stretches by 0.018 m when a 2.8-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 3.0 Hz?

Explanation:

Given that,

Extension of spring

x = 0.0208m

Mass attached m = 3.39kg

Additional mass to have a frequency f

Let the additional mass be m

Using Hooke's law

F= kx

Where F = W = mg = 3.39 ×9.81

F = 33.26N

Then,

F = kx

k = F/x

k = 33.26/0.0208

k = 1598.84 N/m

The frequency is given as

f = ½π√k/m

Make m subject of formula

f² = ¼π² •(k/m

4π²f² = k/m

Then, m4π²f² = k

So, m = k/(4π²f²)

So, this is the general formula,

Then let use the frequency above

f = 3Hz

m = 1598.84/(4×π²×3²)

m = 4.5 kg

4 0
3 years ago
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