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Reil [10]
3 years ago
11

A 2100 kg car starts from rest and accelerates at a rate of 2.6 m/s2 for 4.0 s. Assume that the force acting to accelerate the c

ar is acting in the same direction as its motion. How much work has the car done?
Physics
1 answer:
Reika [66]3 years ago
8 0
When the system is experiencing a uniformly accelerated motion, there are a set of equations to work from. In this case, work is energy which consist solely of kinetic energy. That is, 1/2*m*v2. First, let's find the final velocity.

a = (vf - v0)/t
2.6 = (vf - 0)/4
vf = 10.4 m/s

Then W = 1/2*(2100 kg)*(10.4 m/s)2
W = 113568 J = 113.57 kJ
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if the pressure on a gas increases, the volume of the gas also increases is that statement true or false?
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Suppose a plot of inverse wavelength vs frequency has slope equal to 0.119, what is the speed of sound traveling in the tube to
strojnjashka [21]

Answer:

8.40 m/s

Explanation:

Slope of the plot is 0.119

Slope of a plot is given by the change in y direction divided by the change in x direction

Here, the y axis represents inverse wavelength and the x axis represents frequency.

f = Frequency (Hz, assumed)

v = Phase velocity (m/s, assumed)

λ = Wavelength (m, assumed)

So, slope

m=\frac{\frac{1}{\lambda}}{f}

Now,

\lambda=\frac{v}{f}\\\Rightarrow \lambda^{-1}=\frac{f}{v}

\\\Rightarrow m=\frac{\frac{f}{v}}{f}\\\Rightarrow 0.119=\frac{1}{v}\\\Rightarrow v=\frac{1}{0.119}\\\Rightarrow v=8.40\ m/s

The speed of sound travelling in the tube is 8.40 m/s

5 0
3 years ago
Cuanto cambia la entropía de 0.50 kg de vapor de mercurio [Lv: 2.7 x 10⁵ j/kg ] al calentarse en su punto de ebullición de 357°
lord [1]

Answer:

La entropía del vapor de mercurio cambia en 214.235 joules por Kelvin.

Explanation:

Por definición de entropía (S), medida en joules por Kelvin, tenemos la siguiente expresión:

dS = \frac{\delta Q}{T} (1)

Donde:

Q - Ganancia de calor, en joules.

T - Temperatura del sistema, en Kelvin.

Ampliamos (1) por la definición de calor latente:

dS = \frac{L_{v}}{T}\cdot dm (1b)

Donde:

m - Masa del sistema, en kilogramos.

L_{v} - Calor latente de vaporización, en joules

Puesto que no existe cambio en la temperatura durante el proceso de vaporización, transformamos la expresión diferencial en expresión de diferencia, es decir:

\Delta S = \frac{\Delta m \cdot L_{v}}{T}

Como vemos, el cambio de la entropía asociada al cambio de fase del mercurio es directamente proporcional a la masa del sistema. Si tenemos que m = 0.50\,kg,L_{v} = 2.7\times 10^{5}\,\frac{J}{kg} and T = 630.15\,K, entonces el cambio de entropía es:

\Delta S = \frac{(0.50\,kg)\cdot \left(2.7\times 10^{5}\,\frac{J}{kg} \right)}{630.15\,K}

\Delta S = 214.235 \,\frac{J}{K}

La entropía del vapor de mercurio cambia en 214.235 joules por Kelvin.

3 0
3 years ago
Identify the vibrating media in three different<br> types of musical instruments.
mart [117]

Explanation:

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String. String. ...

Sound. hole. ...

Bow. Bow.

3 0
3 years ago
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