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marshall27 [118]
3 years ago
11

Which best describes the motion of air particles when a transverse wave passes through them?

Physics
1 answer:
Minchanka [31]3 years ago
3 0
C.
The particles move perpendicular to the direction of the wave.
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A farmer lifts his hay bales into the top loft of his barn by walking his horse forward with a constant velocity of 1 ft/s. Dete
Lesechka [4]

Answer:

The velocity of the hay bale is - 0.5 ft/s and the acceleration is 6.25\times 10^{- 3} ft/s^{2}

Solution:

As per the question:

Constant velocity of the horse in the horizontal, v_{x} = 1 ft/s

Distance of the horse on the horizontal axis, x = 10 ft

Vertical distance, y = 20 ft

Now,

Apply Pythagoras theorem to find the length:

20^{2} + 10^{2} = l^{2}

l^{2}= 500

Now,

x^{2} + y^{2} = 500                            (1)

Differentiating equation (1) w.r.t 't':

2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0

x\frac{dx}{dt} = - y\frac{dy}{dt}

where

\frac{dx}{dt} = Rate of change of displacement along the horizontal

\frac{dy}{dt} = Rate of change of displacement along the vertical

v_{x} = velocity along the x-axis.

v_{y} = velocity along the y-axis

xv_{x} = -yv_{y}

v_{y} = - 10\times \frac{1}{20} = - 0.5 ft/s

|v_{y}| = 0.5\ ft/s

Acceleration of the hay bale is given by the kinematic equation:

v_{y}^{2} = u_{y} + 2ay

(-0.5)^{2} =0 + 2ay

0.25 = 2ay

\frac{0.25}{2y} = a

a = \frac{0.25}{2\times 20} = 6.25\times 10^{- 3} ft/s^{2}

7 0
3 years ago
En un experimento de calorimetría, 0.50 kg de un metal a 100°C se añaden a 0.50 kg de agua a 20°C en un vaso de calorímetro de a
Maru [420]

Answer:

c=0.14J/gC

Explanation:

A.

2) The specific heat will be the same because it is a property of the substance and does not depend on the medium.

B.

We can use the expression for heat transmission

Q=mc(T_2-T_1)

In this case the heat given by the metal (which is at a higher temperature) is equal to that gained by the water, that is to say

Q_1=-Q_2

for water we have to

c = 4.18J / g ° C

replacing we have

c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}

I hope this is useful for you

A.

2) El calor específico será igual porque es una propiedad de la sustancia y no depende del medio.

B.

Podemos usar la expresión para la transmisión de calor

Q=mc(T_2-T_1)

En este caso el calor cedido por el metal (que está a mayor temperatura) es igual al ganado por el agua, es decir

Q_1=-Q_2

para el agua tenemos que

c=4.18J/g°C

reemplazando tenemos

c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}

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The muscular system brings what to the body?
kow [346]
The muscular system brings strength and endurance to the body. It helps us perform everyday activities. As well as soaks up water to keep us hydrated longer.
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The velocity selector in in a mass spectrometer consists of a uniform magnetic field oriented at 90 degrees to a uniform electri
NeTakaya

Answer:

50k/h is the answer to iy

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How many genders are there?
Dvinal [7]

Answer:2

Explanation:

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