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3 years ago

Compare the properties of Titan’s atmosphere with those of Earth’s atmosphere.

Physics

1 answer:

Aloiza [94]3 years ago

5 0

Answer and Explanation:

Comparison between the Titan's atmosphere and earth atmosphere

Titan atmosphere is more denser than the earth atmosphere
The quantity of nitrogen is more in titan atmosphere than earth atmosphere titan atmosphere have about 98 % of nitrogen in the other hand earth atmosphere has only 78 %
There are no oxygen present in titan atmosphere while in earth atmosphere it is present
Its air is not suitable for breath but in earth atmosphere we can breath

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A car moves with the speed of 120m/s for 4 minutes ,calculate the distance covered by the car

Vsevolod [243]

Answer:

960 m

Explanation:

Given that,

Speed = 120 m/s
Time taken = 4 minutes

We have to find the distance covered.

Firstly, let's convert time in seconds.

→ 1 minute = 60 seconds

→ 4 minutes = (4 × 60) seconds

→ 4 minutes = 240 seconds

Now, we know that,

→ Distance = Speed × Time

→ Distance = (4 × 240) m

→ Distance = 960 m

Therefore, distance covered is 960 m.

7 0

3 years ago

An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate

OverLord2011 [107]

Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 × $10^{-4}$ m

dynamic viscosity = 1.75 × $10^{-5}$ Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ $\frac{du}{dy}$

= µ $\frac{v}{h}$ ............1

put here value

= 1.75× $10^{-5}$ × $\frac{v}{10^{-4}}$

= 0.175 v

and

area between air and puck is given by

Area = $\frac{\pi }{4} d^{2}$

area = $\frac{\pi }{4} 0.1^{2}$

area = 7.85 × $\frac{v}{10^{-3}}$ m²

force on puck is express as

Force = × area

force = 0.175 v × 7.85 × $10^{-3}$

force = 1.374 × $10^{-3}$ v

and now apply newton second law

force = mass × acceleration

- force = $mass \frac{dv}{dt}$

- 1.374 × $10^{-3}$ v = $0.03 \frac{0.9v - v }{t}$

t = $\frac{0.1 v * 0.03}{1.37*10^{-3} v}$

time = 2.18

so time required after impact for a puck is 2.18 seconds

3 0

3 years ago

A playground merry-go-round has a mass of 50 kg and a diameter of 4.0 m. There are 4 children who want to ride on it. They have

mixer [17]

Answer:

B) I1 = 1680 kg.m^2 I2 = 1120 kg.m^2

C) V = 0.84m/s T = 29.92s

D) ω2 = 0.315 rad/s

Explanation:

The moment of inertia when they are standing on the edge:

$I1 = 1/2*M*R^2 + (m1+m2+m3+m4)*R^2$ where M is the mass of the merry-go-round.

I1 = 1680 kg.m^2

The moment of inertia when they are standing half way to the center:

$I2 = 1/2*M*R^2 + (m1+m2+m3+m4)*(R/2)^2$

I2 = 1120 kg.m^2

The tangencial velocity is given by:

V = ω1*R = 0.84m/s

Period of rotation:

T = 2π / ω1 = 29.92s

Assuming that there is no friction and their parents are not pushing anymore, we can use conservation of the angular momentum to calculate the new angular velocity:

I1*ω1 = I2*ω2 Solving for ω2:

ω2 = I1*ω1 / I2 = 0.315 rad/s

5 0

3 years ago

Chứng minh V=kq/r từ mối liên hệ giữ E và V

Marat540 [252]

Answer:V=Aq=KQr

Explanation:

Không biết V bạn kí hiệu ở đây là gì nhỉ? Có phải là điện thế?

Điện thế tại 1 điểm trong điện trường được định nghĩa là công làm vật dịch chuyển từ vị trí đó đến vô cùng. V = A/q

Chứng minh thì được, nhưng chỉ e bạn không có hiểu biết về nguyên hàm, tích phân nên không hiểu.

- Xét tại vị trí cách điện tích Q một đoạn x, khi đó điện tích q sẽ chịu 1 lực: dF=KQ.qx2

Điện tích q dịch chuyển 1 khoảng dx rất nhỏ. Khi đó công do lực điện trường gây ra là:

dA=dF.dx=KQ.qx2dx

Công để dịch chuyển điện tích q từ vị trí r đến vô cùng là:

A=∫∞rdA=∫∞rKQ.qx2dx=KQ.qr−KQ.q∞=KQ.qr

Theo đúng định nghĩa: V=Aq=KQr

4 0

3 years ago

A coil with an inductance of 2.8 H and a resistance of 12 Ω is suddenly connected to an ideal battery with ε = 89 V. At 0.086 s

Thepotemich [5.8K]

Answer:

The stored energy is 140.7 watt.

The thermal energy is 62.7 watt.

The delivered energy is 203.4 watt.

Explanation:

Given that,

Inductance = 2.8 H

Resistance = 12 Ω

Potential $\epsilon_{0}=89\ V$

Time = 0.086 s

(a). We need to calculate the energy stored in the magnetic field

Using formula of current

$i=i_{max}(1-e^(\frac{-t}{\tau}))$

Using formula of energy

$U=\dfrac{1}{2}Li^2$

On differentiating

$\dfrac{dU}{dt}=Li\frac{di}{dt}$

$\dfrac{dU}{dt}=L\dfrac{d}{dt}(i_{max}(1-e^(\frac{-t}{\tau}))$

Again differentiating

$\dfrac{dU}{dt}=\dfrac{\epsilon^2}{R}(1-e^{\frac{-t}{\tau}})e^{\frac{-t}{\tau}}$

$\dfrac{dU}{dt}=\dfrac{\epsilon^2}{R}(1-e^{\frac{-\t\times R}{L}})e^{\frac{-t\times R}{L}}$

Put the value into the formula

$\dfrac{dU}{dt}=\dfrac{(89)^2}{12}(1-e^{\dfrac{-0.086\times12}{2.8}})e^{\dfrac{-0.086\times12}{2.8}}$

$\dfrac{dU}{dt}=140.7\ watt$

(b). We need to calculate the thermal energy

Using formula of thermal energy

$P=i^2R$

$P=\dfrac{\epsilon^2}{R}(1-e^{\frac{-t}{\tau}})^2$

Put the value into the formula

$P=\dfrac{89^2}{12}(1-e^{\dfrac{-0.086\times12}{2.8}})^2$

$P=62.7\ Watt$

(c). We need to calculate the delivered energy by the battery

Using formula of energy

$P'=P+\dfrac{dU}{dt}$

$P'=62.7+140.7$

$P'=203.4\ watt$

Hence, The stored energy is 140.7 watt.

The thermal energy is 62.7 watt.

The delivered energy is 203.4 watt.

5 0

3 years ago