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WINSTONCH [101]
3 years ago
10

A sinusoidal wave is traveling along a rope. The oscillator that generates the wave completes 43.0 vibrations in 29.0 s. A given

crest of the wave travels 430 cm along the rope in 14.0 s. What is the wavelength of the wave?
Physics
1 answer:
Aliun [14]3 years ago
8 0

Answer:

wavelength is 20.71 cm

Explanation:

given data

wave = 43.0 vibrations

time = 29 s

wave travels = 430 cm

time t2 = 14 s

to find out

wavelength

solution

we know here wavelength formula that is

wavelength = velocity / frequency  .........................1

here frequency = 43 / 29 = 1.4827 Hz

and velocity = 430 / 14 = 30.71 cm/s

so here from equation 1 put all value

wavelength = velocity / frequency

wavelength = 30.71 / 1.4827

wavelength = 20.71 cm

so wavelength is 20.71 cm

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Part A What will be the equilibrium temperature when a 227 g block of copper at 283 °C is placed in a 155 g aluminum calorimeter
stellarik [79]

Answer:

T = 20.84°C

Explanation:

From the law of conservation of energy:

Heat Lost by Copper Block = Heat Gained by Aluminum Calorimeter + Heat Gained by Water

m_cC_c\Delta T_c = m_wC_w\Delta T_w + m_aC_a\Delta T_a

where,

m_c = mass of copper = 227 g

m_w = mass of water = 844 g

m_a = mass of aluminum = 155 g

C_c = specific heat capacity of calorimeter = 385 J/kg.°C

C_w = specific heat capacity of water = 4200 J/kg.°C

C_a = specific heat capacity of aluminum = 890 J/kg.°C

\Delta T_c = change in temperature of copper = 283°C - T

\Delta T_w = change in temperature of water = T - 14.6°C

\Delta T_a = change in temperature of aluminum = T - 14.6°C

T = equilibrium temperature = ?

Therefore,

(227\ g)(385\ J/kg.^oC)(283^oC-T)=(844\ g)(4200\ J/kg.^oC)(T-14.6^oC)+(155\ g)(890\ J/kg.^oC)(T-14.6^oC)\\\\24732785\ J - (87395\ J/^oC) T = (3544800\ J/^oC) T - 51754080\ J+ (137950\ J/^oC) T-2014070\ J\\\\24732785\ J +51754080\ J+2014070\ J = (3544800\ J/^oC) T+(137950\ J/^oC+(87395\ J/^oC) T\\\\78560935\ J = (3770145\ J/^oC) T\\\\T = \frac{78560935\ J}{3770145\ J/^oC}

<u>T = 20.84°C</u>

8 0
2 years ago
A 570 kg elevator accelerates downwards at 1.5 m/s2 for the first 13 m of its motion.
jeka94
  • Mass of the elevator (m) = 570 Kg
  • Acceleration = 1.5 m/s^2
  • Distance (s) = 13 m
  • Let the force be F.
  • We know, F = ma,
  • Therefore, F = (570 × 1.5) N = 855 N
  • Angle between distance and force (θ) = 0°
  • We know, work done = F s Cos θ
  • Therefore, work done by the cable during this part
  • = (855 × 13 × Cos 0°) J
  • = (855 × 13 × 1) J
  • = 11115 J

<u>Answer</u><u>:</u>

<u>1</u><u>1</u><u>1</u><u>1</u><u>5</u><u> </u><u>J</u>

Hope you could get an idea from here.

Doubt clarification - use comment section.

6 0
2 years ago
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