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3 years ago

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The latent heat of fusion of alcohol is 50 kcal/kg and its melting point is -54oC. It has a specific heat of 0.60 in its liquid

state. Calculate the energy needed to melt 2.0 kg of solid alcohol at -54o and raise it to a temperature of -4°C.
40. kcal
100 kcal
160 kcal
280 kcal

1 answer:

sashaice [31]3 years ago

5 0

Answer:

C

Explanation:

To melt the alcohol

Heat needed = M . L = 2 . 25 = 50 kcal

To warm up the alcohol

Heat needed = M . sp. ht. . ∆t = 2 . 0.6 . 100 = 120 kcal

Total heat needed = 170 kcal

Assuming that 0.6 kcal/ kg / ˚C is the specific heat and that the answer is wanted in kcal ( a rather odd unit to be in use here.)

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X- rays. It can damage living tissue

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Answer:

Bouyancy

Explanation:

Bouyancy occurs when the upthrust exerted on an object is equal to the weight of object displaced. It is mostly applicable to low density objects for example balloon. When balloon is displaced in water, it floats. This is due to the effect of the upthrust acting on the balloon which allows the balloon to float and which is opposite the weight.

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A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t1 = 11.9 seconds to get up to it

Degger [83]

Answer:

a) $\omega_1=8.168\,rad.s^{-1}$

b) $n_1=7.735 \,rev$

c) $\alpha_1 =0.6864\,rad.s^{-2}$

d) $\alpha_2=4.1454\,rad.s^{-2}$

e) $t_2=1.061\,s$

Explanation:

Given that:

initial speed of turntable, $N_0=0\,rpm\Rightarrow \omega_0=0\,rad.s^{-1}$

full speed of rotation, $N_1=78 \,rpm\Rightarrow \omega_1=\frac{78\times 2\pi}{60}=8.168\,rad.s^{-1}$

time taken to reach full speed from rest, $t_1=11.9\,s$

final speed after the change, $N_2=120\,rpm\Rightarrow \omega_2=\frac{120\times 2\pi}{60}=12.5664\,rad.s^{-1}$

no. of revolutions made to reach the new final speed, $n_2=11\,rev$

(a)

∵ 1 rev = 2π radians

∴ angular speed ω:

$\omega=\frac{2\pi.N}{60}\, rad.s^{-1}$

where N = angular speed in rpm.

putting the respective values from case 1 we've

$\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}$

$\omega_1=8.168\,rad.s^{-1}$

(c)

using the equation of motion:

$\omega_1=\omega_0+\alpha . t_1$

here α is the angular acceleration

$78=0+\alpha_1\times 11.9$

$\alpha_1 = \frac{8.168 }{11.9}$

$\alpha_1 =0.6864\,rad.s^{-2}$

(b)

using the equation of motion:

$\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1$

$8.168^2=0^2+2\times 0.6864\times n_1$

$n_1=48.6003\,rad$

$n_1=\frac{48.6003}{2\pi}$

$n_1=7.735\, rev$

(d)

using equation of motion:

$\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2$

$12.5664^2=8.168^2+2\alpha_2\times 11$

$\alpha_2=4.1454\,rad.s^{-2}$

(e)

using the equation of motion:

$\omega_2=\omega_1+\alpha_2 . t_2$

$12.5664=8.168+4.1454\times t_2$

$t_2=1.061\,s$

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sergeinik [125]

$v^2-{v_0}^2=2a(x-x_0)$

dónde $v$ es la velocidad de la esfera, $v_0$ es suya velocidad inicial, $a=-g$ la aceleración debida a la gravedad, $x$ la posición, y $x_0$ la posición inicial. Tomamos $x_0=0\,\mathrm m$ a referirse a la posición de la esfera en el momento que la esfera fue lanzada.

Entonces

$v^2-\left(30\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-10\,\dfrac{\mathrm m}{\mathrm s^2}\right)(40\,\mathrm m)$

$\implies v^2=100\,\dfrac{\mathrm m^2}{\mathrm s^2}\implies v=\pm10\,\dfrac{\mathrm m}{\mathrm s}$

Esto nos dice que la esfera alcanza una altura de 40 m en dos momentos - una vez hacia arriba y una vez hacia abajo. Sin embargo, independientemente del signo de la velocidad, sabemos que suya magnitud es 10 m/s, y así tenemos una rapidez de 10 m/s también en ambos momentos.

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A drag racer starts her car from rest and accelerates at 5.5 m/s2 for the entire distance of 523 m. How long did it take the car

ANEK [815]

Answer:

t = 13.7 s or t = 14 s with proper significant figures

Explanation:

The initial speed is 0 m/s since the car starts from rest, acceleration is 5.5 m/s2 and distance is 523 m.

Since we have initial speed, acceleration and distance we can use the following formula to find the time. We can now use algebra to work out our answer.

d = vt + $\frac{1}{2}$ at²

523 = (0)t + ( $\frac{1}{2}$ )(5.5)t²

523 = 2.8t²

186.8 = t²

13.7 s = t

(t = 14 s with proper significant figures)

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