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ozzi
3 years ago
15

Define the Earth's atmosphere

Physics
1 answer:
Alex Ar [27]3 years ago
7 0
<span> The blanket of gas on the surface of a planet or satellite. Note: The </span>atmosphere<span> of the </span>Earth<span> is roughly eighty percent nitrogen and twenty percent oxygen, with traces of other gases. (See ionosphere, stratosphere, and troposphere.)

</span>
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Standard twist-on wire connectors are not suitable for terminating or splicing aluminum or copper-clad aluminum conductors.
Rainbow [258]

Answer:

True.

Explanation:

Twist-on wire connectors are a sort of electrical conduit used only to fasten two or more small tension (or extra-low tension) electrical conductors.They are also called wire nuts, wire connectors or cone connectors( because of their conical shape). They are not suitable for for terminating or splicing aluminum or copper-clad aluminum conductors.

Hence the above statement is true.

4 0
3 years ago
A fire hose has an inside diameter of 6.40 cm. Suppose such a hose carries a flow of 40.0 L/s starting at a gauge pressure of 1.
inessss [21]

Answer:

The Reynolds numbers for flow in the fire hose.

Explanation:

Given that,

Diameter = 6.40 cm

Rate of flow = 40.0 L/s

Pressure P=1.62\times10^{6}\ N/m^2

We need to calculate the Reynolds numbers for flow in the fire hose

Using formula of rate of flow

Q=Av

v=\dfrac{Q}{A}

Where, Q = rate of flow

A = area of cross section

Put the value into the formula

v=\dfrac{40.0\times10^{-3}}{3.14\times(3.2\times10^{-2})^2}

v=12.44\ m/s

We need to calculate the Reynolds number

Using formula of the Reynolds number

n_{R}=\dfrac{2\rho\times v\times r}{\eta}

Where, \eta =viscosity of fluid

\rho =density of fluid

Put the value into the formula

n_{R}=\dfrac{2\times100\times12.44\times3.2\times10^{-2}}{1.002\times10^{-3}}

n_{R}=7.945\times10^{5}

Hence, The Reynolds numbers for flow in the fire hose.

3 0
3 years ago
Two planets P1 and P2 orbit around a star S in circular orbits with speeds v1 = 40.2 km/s, and v2 = 56.0 km/s respectively. If t
Readme [11.4K]

Answer: 3.66(10)^{33}kg

Explanation:

We are told both planets describe a circular orbit around the star S. So, let's approach this problem begining with the angular velocity \omega of the planet P1 with a period T=750years=2.36(10)^{10}s:

\omega=\frac{2\pi}{T}=\frac{V_{1}}{R} (1)

Where:

V_{1}=40.2km/s=40200m/s is the velocity of planet P1

R is the radius of the orbit of planet P1

Finding R:

R=\frac{V_{1}}{2\pi}T (2)

R=\frac{40200m/s}{2\pi}2.36(10)^{10}s (3)

R=1.5132(10)^{14}m (4)

On the other hand, we know the gravitational force F between the star S with mass M and the planet P1 with mass m is:

F=G\frac{Mm}{R^{2}} (5)

Where G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

In addition, the centripetal force F_{c} exerted on the planet is:

F_{c}=\frac{m{V_{1}}^{2}}{R^{2}} (6)

Assuming this system is in equilibrium:

F=F_{c} (7)

Substituting (5) and (6) in (7):

G\frac{Mm}{R^{2}}=\frac{m{V_{1}}^{2}}{R^{2}} (8)

Finding M:

M=\frac{V^{2}R}{G} (9)

M=\frac{(40200m/s)^{2}(1.5132(10)^{14}m)}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}} (10)

Finally:

M=3.66(10)^{33}kg (11) This is the mass of the star S

4 0
4 years ago
Using the law of conservation of energy, what is the kinetic energy at B? K.E. = 3, 6, or 4.8
musickatia [10]

Answer:

3

Explanation:

7 0
3 years ago
If a 15 N box is lifted a distance of 3 m, how much work is done?
Naily [24]

Answer:

W=45J

Explanation:

W=Fd

W=15(3)=45

W=45J

7 0
3 years ago
Read 2 more answers
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