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pav-90 [236]
3 years ago
7

When you speak or sing, you force air through your ?

Physics
1 answer:
zaharov [31]3 years ago
6 0
You force air from your lungs up through your larynx which is commonly called the voicebox
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Katrina is a teen from Australia who participates in curling, a sport in which players slide stones across the ice toward a targ
goldenfox [79]
Personally i would say advertising but I can also see where it could be globalization because the Olympics is about different countries coming together and having friendly competition. 
6 0
3 years ago
Read 2 more answers
You pull with a force of 295 N on a rope that is attached to a block of mass 22 kg, and the block slides across the floor at a c
Sergeeva-Olga [200]

Answer:

Fnet = 0

Explanation:

  • Since the block slides across the floor at constant speed, this means that it's not accelerated.
  • According Newton's 2nd Law, if the acceleration is zero, the net force on the sliding mass must be zero.
  • This means that there must be a friction force opposing to the horizontal component of the applied force, equal in magnitude to it:

       F_{appx} = F_{app} * cos \theta = 295 N * cos 35 = 242 N  (1)

  • In the vertical direction, the block is not accelerated either, so the sum of the normal force and the vertical component of the applied force, must be equal in magnitude to the force of gravity on the block:

      F_{appy} = F_{app} * cos \theta = 295 N * sin 35 = 169 N  (2)

⇒    169 N + Fn = Fg = 216 N  (3)

  • This means that there must be a normal force equal to the difference between Fappy and Fg, as follows:
  • Fn = 216 N - 169 N = 47  N (4)

6 0
2 years ago
Compare the atmospheric pressure of two places 500m and 2 km respectively above sea-level. Give reason for your answer.
Lera25 [3.4K]

Answer:

Explanation:

The average pressure at mean sea-level (MSL) in the International Standard Atmosphere (ISA) is 1013.25 hPa, or 1 atmosphere (atm), or 29.92 inches of mercury. Pressure (p), mass (m), and the acceleration due to gravity (g), are related by P = F/A = (m*g)/A, where A is surface area.

3 0
3 years ago
Whqt do you mean by geo-stationery satellite? Explain.​
Ratling [72]

Answer:

Description: When a geosynchronous satellite is placed directly above the Equator with a circular orbit and angular velocity identical to that of the Earth, the satellite is known as a geostationary satellite

Explanation:

These satellites appear to be stationary above a particular point which is due to the synchronization. This type of satellite provides a distinct benefit of fixing the receiving antennas at one place, hence making them more economical than regular tracking antennas. Over the years, these satellites have helped in revolutionizing the global communications, weather forecasting and television broadcasting. When the orbit of a geosynchronous satellite is not aligned directly over the equator, the orbit is known as an inclined orbit.

8 0
2 years ago
A block of mass 0.1 kg is attached to a spring of spring constant 21 N/m on a frictionless track. The block moves in simple harm
bogdanovich [222]

Answer:

A) 2.75 m/s  B) 0.1911 m    C) 0.109 s

Explanation:

mass of block = M =0.1 kg

spring constant = k = 21 N/m

amplitude = A = 0.19 m

mass of bullet = m = 1.45 g = 0.00145 kg

velocity of bullet = vᵇ = 68 m/s

as we know:

Angular frequency of S.H.M = ω₀ = \sqrt\frac{k}{M}

                                                       = \sqrt\frac{21}{0.1}

                                                       = 14.49 rad/sec

<h3>A) Speed of the block immediately before the collision:</h3>

displacement of Simple Harmonic  Motion is given as:

                                x = A sin (\omega t + \phi)\\

Differentiating this to find speed of the block immediately before the collision:

                    v=\frac{dx}{dt}= A\omega_{o} cos (\omega_{o}t =\phi}\\

As bullet strikes at equilibrium position so,

                                  φ = 0

                                   t= 2nπ

                             ⇒ cos (ω₀t + φ) = 1

                             ⇒ v= A\omega_{o}

                                       v=(.19)(14.49)\\v= 2.75 ms^{-1}

<h3>B) If the simple harmonic motion after the collision is described by x = B sin(ωt + φ), new amplitude B:</h3>

S.H.M after collision is given as :

                              x= Bsin(\omega t + \phi)

To find B, consider law of conservation of energy

K.E = P.E\\K.E= \frac{1}{2}(m+M)v^{2}  \\P.E = \frac{1}{2} kB^{2}

\frac{m+M}{k} v^{2} = B^{2} \\B =\sqrt\frac{m+M}{k} v\\B = \sqrt\frac{.00145+0.1}{21} (2.75)\\B = .1911m

<h3>C) Time taken by the block to reach maximum amplitude after the collision:</h3>

Time period S.H.M is given as:

T=2\pi \sqrt\frac{m}{k}\\ for given case\\m= m=M\\then\\T=2\pi \sqrt\frac{m+M}{k}

Collision occurred at equilibrium position so time taken by block to reach maximum amplitude is equal to one fourth of total time period

T=\frac{\pi }{2}\sqrt\frac{m+M}{k} \\T=0.109 sec

5 0
3 years ago
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