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dimaraw [331]
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?​
Physics
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

You might be interested in
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
Read 2 more answers
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