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2 years ago

7

a 10kg box car rolls at 6m/s and crashes into a 20kg box car that is at rest. After the crash, both cars are stuck together. wha

t is their velocity?

1 answer:

Goshia [24]2 years ago

5 0

Answer:

Their velocity is 0m/s because the first box was only 10 kg and the second box was double the weight

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he "escape velocity from Earth (the speed required to escape Earth's gravity) is 2.5 x 10 miles per hour. What is this speed in

Natasha_Volkova [10]

Answer: $11.17\ \text{ m/s}$

Explanation:

Given : The escape velocity : $v=2.5\times10\text{ miles per hour}$

We know that 1 mile = 1609 meters (approx)

and 1 hour= 3600 seconds

To convert escape velocity 2.5 x 10 miles per hour into m/s , we need to multiply it by 1609.34 and divide it by 3600.

Thus, the escape velocity in m/s is given by :-

$v=2.5\times10\times\dfrac{1609}{3600}\\\\=11.1736111111\approx11.17\text{ m/s}$

Hence, the speed in m/s = 11.17

4 0

3 years ago

A 30-cm-diameter, 4-m-high cylindrical column of a house made of concrete ( k = 0.79 W/m⋅K, α = 5.94 × 10 −7 m2/s, rho = 1600kg

PilotLPTM [1.2K]

Answer:

a) Time it will taken for the column surface temperature to rise to 27°C is

17.1 hours

b) Amount of heat transfer is 5320 kJ

c) Amount of heat transfer until the surface temperature reaches 27°C is 4660 kJ

Explanation:

Given that;

Diameter D = 30 cm

Height H = 4m

heat transfer coeff h = 14 W/m².°C

thermal conductivity k = 0.79 W/m.°C

thermal diffusivity α = 5.94 × 10⁻⁷ m²/s

Density p = 1600 kh/m³

specific heat Cp = 0.84 Kj/kg.°C

a)

the Biot number is

Bi = hr₀ / k

we substitute

Bi = (14 W/m².°C × 0.15m) / 0.79 W/m.°C

Bi = 2.658

From the coefficient for one term approximate of transient one dimensional heat conduction The constants λ₁ and A₁ corresponding to this Biot number are,

λ₁ = 1.7240

A₁ = 1.3915

Once the constant J₀ = 0.3841 is determined from corresponding to the constant λ₁

the Fourier number is determined to be

[ T(r₀, t) -T∞ ] / [ Ti - T∞] = A₁e^(-λ₁²t') J₀ (λ₁r₀ / r₀)

(27 - 28) / (14 - 28) = (1.3915)e^-(17240)²t (0.3841)

t' = 0.6771

Which is above the value of 0.2. Therefore, the one-term approximate solution (or the transient temperature charts) can be used. Then the time it will take for the column surface temperature to rise to 27°C becomes

t = t'r₀² / ₐ

= (0.6771 × 0.15 m)² / (5.94 x 10⁻⁷ m²/s)

= 23,650 s

= 7.1 hours

Time it will taken for the column surface temperature to rise to 27°C is

17.1 hours

b)

The heat transfer to the column will stop when the center temperature of column reaches to the ambient temperature, which is 28°C.

Maximum heat transfer between the ambient air and the column is

m = pV

= pπr₀²L

= (1600 kg/m³ × π × (0.15 m)² × (4 m)

= 452.389 kg

Qin = mCp [T∞ - Ti ]

= (452.389 kg) (0.84 kJ/kg.°C) (28 - 14)°C

= 5320 kJ

Amount of heat transfer is 5320 kJ

(c)

the amount of heat transfer until the surface temperature reaches to 27°C is

(T(0,t) - T∞) / Ti - T∞ = A₁e^(-λ₁²t')

= (1.3915)e^-(1.7240)² (0.6771)

= 0.1860

Once the constant J₁ = 0.5787 is determined from Table corresponding to the constant λ₁, the amount of heat transfer becomes

(Q/Qmax)cyl = 1 - 2((T₀ - T∞) / ( Ti - T∞)) ((J₁(λ₁)) / λ₁)

= 1 - 2 × 0.1860 × (0.5787 / 1.7240)

= 0.875

Q = 0.875Qmax

Q = 0.875(5320 kJ)

Q = 4660 kJ

Amount of heat transfer until the surface temperature reaches 27°C is 4660 kJ

6 0

2 years ago

10) Um viajante, ao desembarcar no aeroporto de Londres, observou que o valor da temperatura do ambiente na escala Fahrenheit é

joja [24]

Answer:

The observed temperature was 10º Celsius or 50º Fahrenheit.

Explanation:

The traveler observed that the temperature in Fahrenheit is five times the value of the temperature in Celsius, therefore:

$F = 5*C$

A Fahrenheit temperature relates to a Celsius one by the following expression:

$F = \frac{9}{5}*C + 32$

Using the second expression on the first, we can solve for the temperature in Celsius, this is done below:

$\frac{9}{5}*C + 32 = 5*C\\\frac{9}{5}*C - 5*C = -32\\\frac{9*C - 25*C}{5} = -32\\\frac{-16*C}{5} = -32\\-16*C = -32*5\\-16*C = -160\\C = \frac{-160}{-16} = 10\º\text{C}\\F = 5*C = 5*10 = 50\º\text{F}$

The observed temperature was 10º Celsius or 50º Fahrenheit.

7 0

3 years ago

The terminals of a 0.70 Vwatch battery are connected by a 80.0-m-long gold wire with a diameter of 0.200 mm What is the current

Komok [63]

Answer:

$I=0.047A$

Explanation:

Let's use Ohm's law:

$V=IR$

or

$I=\frac{V}{R}$ (1)

Where:

$V=Voltage\\I=Current\\R=Electrical\hspace{2 mm}Resistance$

We know the value of the voltage V, so we need to find the value of R in order to find I. Fortunately there is a relation between the resistivity of a conductor and its electrical resistance given by:

$R=\rho*\frac{l}{A}$ (2)

Where:

$R=Electrical\hspace{2 mm}Resistance\\l=Length\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}conductor=80m\\A=Cross\hspace{2 mm}sectional\hspace{2 mm}area\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}conductor=1.256637061*10^{-7} \\\rho=Electrical\hspace{2 mm}resistivity\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}material=2.35*10^{-8}$

Keep in mind that the electrical resistivity of the gold is a known constant which is $\rho_g_o_l_d=2.35*10^{-8}$ and the cross sectional area of the conductor is calculated as:

$A=\pi *(r^{2})=\pi *(0.0002m)^{2} =1.256637061*10^{-7} m^{2}$

Because we have a wire in this case, so we assume a cylindrical geometry.

Now replacing our data in (2)

$R=(2.35*10^{-8})*\frac{80}{1.256637061*10^{-7} } =14.96056465\Omega$

Finally, we know R and V, so replacing these values in (1) we will be able to find the current:

$I=\frac{0.7}{14.96056465}\approx0.047A$

7 0

3 years ago

# 5 what will most likely happen when an air mas of low temperature exists above a water body at a higher temperature?

laila [671]

Answer: Heat will transfer from the water to the air. When a mass of air moves on a warmer surface it is heated by its base. Then thermal instability develops in the lower layers and then extends upwards. If the air initially contained inversions, these are destroyed and a strong gradient is established uniformly in the lower troposphere temperature.

3 0

2 years ago

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