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vlabodo [156]
3 years ago
5

How much thermal energy does it take to raise the temperature of 3.5 kg of water from 25°C to 46°C? The specific heat of water i

s 4.18 kJ/(kg.C)
Physics
2 answers:
Alex_Xolod [135]3 years ago
6 0

Q = mcΔT

Q = (3.5)(4.18)(46-25)

Q = 307.23 kJ

astra-53 [7]3 years ago
4 0

Answer:310 kJ

Explanation:

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Explanation:

a = -g = -9.80 m/s squared

d o  = 0

v o =  0

t = 1.8s

<h3>unkown:</h3>

d = ?

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A police officer is using her radar to check speeds. She is moving south at 50 mph. You are moving
Luda [366]

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the relative speed should be 40mph

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Calcula el valor de la velocidad de las ondas sonoras en el agua sabiendo que su
dybincka [34]
  1. La velocidad de las ondas sonoras es aproximadamente 1469,694 metros por segundo.
  2. La longitud de onda de las ondas sonoras es 1,470 metros.

1) Inicialmente, debemos determinar la velocidad de las ondas sonoras a través del agua (v), en metros por segundo:

v = \sqrt{\frac{K}{\rho} } (1)

Donde:

  • K - Módulo de compresibilidad, en newtons por metro cuadrado.
  • \rho - Densidad del agua, en kilogramos por metro cúbico.

Si sabemos que \rho = 1\times 10^{3}\,\frac{kg}{m^{3}} y K = 2,16\times 10^{9}\,\frac{N}{m^{2}}, entonces la velocidad de las ondas sonoras es:

v = \sqrt{\frac{2,16\times 10^{9}\,\frac{N}{m^{2}}}{1\times 10^{3}\,\frac{kg}{m^{3}} } }

v\approx 1469,694\,\frac{m}{s}

La velocidad de las ondas sonoras es aproximadamente 1469,694 metros por segundo.

2) Luego, determinamos la longitud de onda (\lambda), en metros, mediante la siguiente fórmula:

\lambda = \frac{v}{f} (2)

Donde f es la frecuencia de las ondas sonoras, en hertz.

Si sabemos que v\approx 1469,694\,\frac{m}{s} y f = 1000\,hz, entonces la longitud de onda de las ondas sonoras es:

\lambda = \frac{1469,694\,\frac{m}{s} }{1000\,hz}

\lambda = 1,470\,m

La longitud de onda de las ondas sonoras es 1,470 metros.

Para aprender más sobre las ondas sonoras, invitamos a ver esta pregunta verificada: brainly.com/question/1070238

6 0
2 years ago
Find the range of a projectile launched at an angle of 30° with an initial velocity of 20m/s.​
Tems11 [23]

Answer:

<em>The range is 35.35 m</em>

Explanation:

<u>Projectile Motion</u>

It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.

Being vo the initial speed of the object, θ the initial launch angle, and g=9.8m/s^2 the acceleration of gravity, then the maximum horizontal distance traveled by the object (also called Range) is:

\displaystyle d={\frac  {v_o^{2}\sin(2\theta )}{g}}

The projectile was launched at an angle of θ=30° with an initial speed vo=20 m/s. Calculating the range:

\displaystyle d={\frac  {20^{2}\sin(2\cdot 30^\circ )}{9.8}}

\displaystyle d={\frac  {400\sin(60^\circ )}{9.8}}

d=35.35\ m

The range is 35.35 m

7 0
3 years ago
A series circuit has a 12-volt power source and two resistors of 1 ohm and 2 ohms respectively. How many amps will flow when
Andrej [43]

Answer:

ayoooooooo

Explanation:

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2 years ago
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