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3 years ago

How much heat is absorbed by a 74g iron skillet when its temperature rises from 7oC to 28oC?

1 answer:

5 0

The specific heat of iron (assuming the skillet is pure iron) is 0.45 J/g-C. Then we use the equation q = mc(dT) to determine the total heat absorbed. This gives:
q = (74 g)(0.45 J/g-C)(28 - 7 C) = (74)(0.45)(21) = 699.3 J

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A 0.1 kg tennis ball is brought from a speed of 15m/s to a speed of 45 m/s when hit by a tennis racket. If this done over 0.05 s

Hitman42 [59]

45 lolololololololol

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3 years ago

Refrigerant-134a enters an adiabatic compressor as saturated vapor at -24C and leaves at 0.8 MPa and 60C. When the mass flow r

sergey [27]

Answer:

= 287kW

Explanation:

Knowing the enthalpy data, we have to

$h1=239.16kJ/kg\\h2=296.81kJ/kg$

So,

$E_i_n_p_u_t=mh_1$

Here,

m=mass flow rate

h= Enthalpy of refrigerant at the compressor

Replacing

= 1.2 × 239.16

= 287kW

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3 years ago

What is the S.I Unit of electric flux?

Arada [10]

Answer:

Volt metre (Vm)

or Newton/Coulomb x (Metre)^2 (N/C x m^2)

8 0

3 years ago

Two swimmers A and B, of weight 190 lb and 125 lb, respectively, are at diagonally opposite corners of a floating raft when they

Harrizon [31]

Answer:

a) 0.618 ft/s

b) 3.04 ft/s

Explanation:

Givens:

Weight of swimmer A $W_{A}$ = 190 Ib.

Weight of swimmer B $W_{B}$ = 125 Ib.

Weight of the raft $W_{R}$ = 300 Ib.

Swimmer A walks toward swimmer B relative to the raft with a speed

$V_{A/R}$ = 2 ft/s

a) Conservation of linear momentum

$m_{A} v_{A} +m_{B} v_{B} +m_{R} v_{R} =0..........(1)\\v_{A/R}=v_{A} -v_{R}\\\v_{A}=v_{A/R}+v_{R}.................(2)$

Since swimmer B does not move

$v_{B} =v_{R}...............(3)$

Substitute from (2) and (3) into (1)

$m_{A} (v_{A/R} +v_{R} )+m_{B} v_{R} +m_{R} v_{R}=0\\(m_{A}+m_{B}+m_{R})v_{R} =-m_{A}v_{A/R}\\v_{R}=\frac{-m_{A}v_{A/R}}{m_{A}+m_{B}+m_{R}} \\v_{R}=0.618ft/s$

b) if the raft not to move $v_{R}=0$

from (2)

$v_{A} =v_{A/R}$

substitute in (1)

$m_{A} v_{A/R} +m_{B} v_{B}+m_{R} (0)=0\\v_{B}=\frac{W_{A}v_{A/R}}{W_{B}} \\v_{B}=3.04ft/s$

3 0

3 years ago

An air tank of volume 1.5 m3 is initially at 800 kPa and 208C. At t 5 0, it begins exhausting through a converging nozzle to sea

Serggg [28]

Answer:

(a) m = 0.141 kg/s

(b) t = 47.343 s

Explanation:

Given that:

The volume of air in the tank V = 1.5 m³

The initial temperature in the tank is supposed to be 20° C and not 208 C;

So $T_o = 20^0 C = ( 20 +273) K = 293K$

The initial pressure in the tank $P_o= 800 \ kPa$

The throat area $A_t$ = 0.75 cm²

To find the initial mass flow in kg/s.

Lets first recall that:

Provided that $\dfrac{P_{amb}}{P_{tank}}< 0.528$ , then the flow is choked.

Then;

$\dfrac{P_{amb}}{P_{tank}}= \dfrac{101.35}{800}$

$\dfrac{P_{amb}}{P_{tank}}= 0.1266$

From what we see above, it is obvious that the ratio is lesser than 0.528, therefore, the flow is choked.

Now, for a choked nozzle, the initial mass flow rate is determined by using the formula:

$m = \rho \times A \times V$

where;

$\rho = \rho_o \bigg ( \dfrac{2}{k+1} \bigg) ^{\dfrac{1}{k-1}}$

$\rho =\dfrac{P_o}{RT_o} \bigg ( \dfrac{2}{k+1} \bigg) ^{\dfrac{1}{k-1}}$

$\rho =\dfrac{800 \times 10^3}{287 \times 293} \bigg ( \dfrac{2}{1.4+1} \bigg) ^{\dfrac{1}{1.4-1}}$

$\rho =9.51350( 0.8333 ) ^{2.5}$

$\rho =6.03 \ kg/m^3$

$T = T_o \bigg ( \dfrac{2}{k+1}\bigg)$

where;

$T_o = 293 \ K$

$T = 293 \bigg ( \dfrac{2}{1.4+1}\bigg)$

$T = 293 \bigg ( \dfrac{2}{2.4}\bigg)$

$T = 293 ( 0.8333)$

T = 244.16 K

During a critical condition when Mach No. is equal to one;

$V = a = \sqrt{kRT}$

$V = \sqrt{1.4 \times 287 \times 244.16}$

$V = \sqrt{98103.488}$

V = 313.214 m/s

Thus, the initial mass flow rate $m = \rho \times A \times V$

m = 6.03 × 0.75 × 10⁻⁴ × 313.214

m = 0.141 kg/s

(b)

The mass balance formula for the control volume surrounding the tank can be expressed as:

$\dfrac{d}{dt}(\rho_o V) = \dfrac{d}{dt} \bigg ( \dfrac{P_o}{RT_o} V\bigg)$

$= \dfrac{V}{RT_o}\dfrac{dP_o}{dt}= -m$

When the air mass flow rate is:

$m = 0.6847 \dfrac{P_oA}{\sqrt{RT_o}}$

Thus; replacing $m = 0.6847 \dfrac{P_oA}{\sqrt{RT_o}}$ in the previous equation; we have:

$\dfrac{V}{RT_o}\dfrac{dP_o}{dt}= - 0.6847 \dfrac{P_oA}{\sqrt{RT_o}}$

$\dfrac{dP_o}{P_o}= -0.6847 \dfrac{A\sqrt{RT_o}}{V} \ dt$

Taking the differential of both sides from 0 → t

$In(P_o)^t_o = -0.6847 \dfrac{A\sqrt{RT}}{V} \times t$

$In \bigg ( \dfrac{P(t)}{P(0)} \bigg) = -0.6847 \dfrac{A\sqrt{RT_o}}{V}\times t$

$\dfrac{P(t)}{P(0)} =exp \bigg ( -0.6847 \dfrac{A\sqrt{RT_o}}{V}\times t \bigg )$

So, when the pressure P = 500 kPa, the time required is:

$\dfrac{500}{800} =exp \bigg ( -0.6847 \dfrac{0.75 \times 10^{-4}\sqrt{287 \times 293}}{1.5}\times t \bigg )$

t = 47.343 s

(c)

Let us recall that:

The choking on the nozzle occurred when $\dfrac{P_{amb}}{P_{tank}} = 0.528$

$\dfrac{P_{amb}}{0.528} = P_{tank}$

$\dfrac{101.35}{0.528} = P_{tank}$

$P_{tank}= 191.95 \ kPa \\ \\ P_{tank} \simeq 192 \ kPa$

From $\dfrac{P(t)}{P(0)} =exp \bigg ( -0.6847 \dfrac{A\sqrt{RT_o}}{V}\times t \bigg )$ ; the time required for $P_{tank} \simeq 192 \ kPa$ is:

$\dfrac{192}{800} =exp \bigg ( -0.6847 \dfrac{0.75 \times 10^{-4}\sqrt{287 \times 293}}{1.5}\times t \bigg )$

By solving:

t = 143.745 s

7 0

3 years ago