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Maru [420]
3 years ago
9

You visit the Grand Canyon on a particularly hot day, and like all good tourists, you yell "Hello!". Would you expect to hear th

e echo sooner or later than on a cooler day? Why?
Physics
1 answer:
RUDIKE [14]3 years ago
7 0

Answer:

Hear echo sooner on a hot day than on a cooler day.

Explanation:

Temperature is one of the factors that determine the speed of sound. Speed of sound, which in this case is echo is calculated considering distance and time. Speed of sound is directly proportional to the temperature , therefore, when the temperature is hot, the speed of sound is faster than when the temperature is cold and vice versa. In conclusion, you would expect to hear the echo “hello” sooner on a hot day compared to a cold day provided all the other factors are kept constant.

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Which lists imaging techniques that use wave behaviors in order of resolution from the best resolution to the worst?
ElenaW [278]
<span>C.CT scan, X-ray imaging, MRI 
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3 0
3 years ago
Question 7 (2 points)
DiKsa [7]
This has to do with independent and dependent variables. The independent variable is what affects the dependant variable, so the question is, does a students mark on a test affect their hours of sleep or does their hours of sleep affect their test results? Which one makes more sense? I would say the latter.
6 0
2 years ago
A yoyo with a mass of m = 150 g is released from rest as shown in the figure.
avanturin [10]

(1) The linear acceleration of the yoyo is 3.21 m/s².

(2) The angular acceleration of the yoyo is 80.25 rad/s²

(3) The  weight of the yoyo is 1.47 N

(4) The tension in the rope is 1.47 N.

(5) The angular speed of the yoyo is 71.385 rad/s.

<h3> Linear acceleration of the yoyo</h3>

The linear acceleration of the yoyo is calculated by applying the principle of conservation of angular momentum.

∑τ = Iα

rT - Rf = Iα

where;

  • I is moment of inertia
  • α is angular acceleration
  • T is tension in the rope
  • r is inner radius
  • R is outer radius
  • f is frictional force

rT - Rf = Iα  ----- (1)

T - f = Ma  -------- (2)

a = Rα

where;

  • a is the linear acceleration of the yoyo

Torque equation for frictional force;

f = (\frac{r}{R} T) - (\frac{I}{R^2} )a

solve (1) and (2)

a = \frac{TR(R - r)}{I + MR^2}

since the yoyo is pulled in vertical direction, T = mg a = \frac{mgR(R - r)}{I + MR^2} \\\\a = \frac{(0.15\times 9.8 \times 0.04)(0.04 - 0.0214)}{1.01 \times 10^{-4} \ + \ (0.15 \times 0.04^2)} \\\\a = 3.21 \ m/s^2

<h3>Angular acceleration of the yoyo</h3>

α = a/R

α = 3.21/0.04

α = 80.25 rad/s²

<h3>Weight of the yoyo</h3>

W = mg

W = 0.15 x 9.8 = 1.47 N

<h3>Tension in the rope </h3>

T = mg = 1.47 N

<h3>Angular speed of the yoyo </h3>

v² = u² + 2as

v² = 0 + 2(3.21)(1.27)

v² = 8.1534

v = √8.1534

v = 2.855 m/s

ω = v/R

ω = 2.855/0.04

ω = 71.385 rad/s

Learn more about angular speed here: brainly.com/question/6860269

#SPJ1

3 0
2 years ago
Alan starts from his home and walks 1.3 km east to the library. He walks an additional 0.68 km east to a music store. From there
zepelin [54]

Answer: final Displacement = 0 km, total distance covered =7 km

Explanation:

Given the following :

From home to library = 1.3 km East

Library to music store = 0.68km East

Music store to friend's house = 1.1km North

Friend's house to grocery store = 0.42 km North

Displacement is the net change in distance traveled.

Eastward distance :

To = (1.3 + 0.68)km = 1.98km East

Fro = (0.68 + 1.3)km = 1.98 km East

Δ distance = (1.98 - 1.98) = 0

Northward direction:

To = (1.1 + 0.42)km = 1.52km north

Fro = (0.42 + 1.1)km = 1.52km North

Δ distance = (1.98 - 1.98) = 0

Hence final Displacement = 0

Total distance covered = 2 × (1.3 + 0.68 + 1.1 + 0.42) = 2 × 3.5

= 7km

3 0
3 years ago
Un neumático sin cámara, soporta una presión de 1.5 atm cuando la temperatura ambiente es de 300°K. ¿Qué presión llegará a sopor
arlik [135]

Answer:

El neumático soportará una presión de 1.7 atm.

Explanation:

Podemos encontrar la presión final del neumático usando la ecuación del gas ideal:

PV = nRT

En donde:

P: es la presión

V: es el volumen

n: es el número de moles del gas

R: es la constante de gases ideales

T: es la temperatura

Cuando el neumático soporta la presión inicial tenemos:

P₁ = 1.5 atm

T₁ = 300 K

V_{1} = \frac{nRT_{1}}{P_{1}}  (1)  

La presión cuando T = 67 °C es:

P_{2} = \frac{nRT_{2}}{V_{2}}   (2)

Dado que V₁ = V₂  (el volumen del neumático no cambia), al introducir la ecuación (1) en la ecuación (2) podemos encontrar la presión final:

P_{2} = \frac{nRT_{2}}{V_{2}} = \frac{nRT_{2}}{\frac{nRT_{1}}{P_{1}}} = \frac{P_{1}T_{2}}{T_{1}} = \frac{1.5 atm*(67 + 273)K}{300 K} = 1.7 atm  

Por lo tanto, si en el transcurso de un viaje las ruedas alcanzan una temperatura de 67 ºC, el neumático soportará una presión de 1.7 atm.

Espero que te sea de utilidad!

4 0
3 years ago
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