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3 years ago

Which of the following statements are true?Check all that apply.

Physics

1 answer:

MaRussiya [10]3 years ago

4 0

Answer:

1.The Sun is located at one of the foci of the planets' elliptical orbits.

2.The path of the planets around the Sun is elliptical in shape.

Explanation:

As per Kepler's law of planet motion we know that all planets revolve around the sun in elliptical path in such a way that position of Sun must be at one of the focii of the path

So all planets are in elliptical path always

Position of sun is always at one of the focus

so correct answer will be

1.The Sun is located at one of the foci of the planets' elliptical orbits.

2.The path of the planets around the Sun is elliptical in shape.

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A long copper rod of diameter 2.0 cm is initially at a uniform temperature of 100°C. It is now exposed to an air stream at 20°C

Lisa [10]

Answer:

t = 4.0 min

Explanation:

given data:

diameter of rod = 2 cm

T_1 = 100 degree celcius

Air stream temperature = 20 degree celcius

heat transfer coefficient = 200 W/m2. K

WE KNOW THAT

copper thermal conductivity = k = 401 W/m °C

copper specific heat Cp = 385 J/kg.°C

density of copper = 8933 kg/m3

charateristic length is given as Lc

$Lc = \frac{V}{A_s}$

$Lc = \frac{\frac{\pi D^2}{4} L}{\pi DL}$

$Lc = \frac{D}{4}$

$Lc = \frac{0.02}{6} = 0.005 m$

Biot number is given as $Bi = \frac{hLc}{k}$

$Bi = \frac{200*0.005}{401}$

Bi = 0.0025

As Bi is greater than 0.1 therefore lumped system analysis is applicable

so we have

$\frac{T(t) - T_∞}{Ti - T_∞} = e^{-bt}$ ............1

where b is given as

$b = \frac{ hA}{\rho Cp V}$

$b = \frac{ h}{\rho Cp Lc}$

$b = \frac{200}{8933*385*0.005}$

b = 0.01163 s^{-1}

putting value in equation 1

$\frac{25-20}{100-20} = e^{-0.01163t}$

solving for t we get

t = 4.0 min

6 0

3 years ago

One ring of radius a is uniformly charged with charge +Q and is placed so its axis is the x-axis. A second ring with charge –Q i

kati45 [8]

Answer:

The force exerted on an electron is $7.2\times10^{-18}\ N$

Explanation:

Given that,

Charge = 3 μC

Radius a=1 m

Distance = 5 m

We need to calculate the electric field at any point on the axis of a charged ring

Using formula of electric field

$E=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

$E_{1}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

Put the value into the formula

$E_{1}=\dfrac{9\times10^{9}\times3\times10^{-6}\times5}{(1^2+5^2)^{\frac{3}{2}}}$

$E_{1}=1.0183\times10^{3}\ N/C$

Using formula of electric field again

$E_{2}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times(-3\times10^{-6})\times5}{((0.5)^2+5^2)^{\frac{3}{2}}}$

$E_{2}=-1.064\times10^{3}\ N/C$

We need to calculate the resultant electric field

Using formula of electric field

$E=E_{1}+E_{2}$

Put the value into the formula

$E=1.0183\times10^{3}-1.064\times10^{3}$

$E=-0.045\times10^{3}\ N/C$

We need to calculate the force exerted on an electron

Using formula of electric field

$E = \dfrac{F}{q}$

$F=E\times q$

Put the value into the formula

$F=-0.045\times10^{3}\times(-1.6\times10^{-19})$

$F=7.2\times10^{-18}\ N$

Hence, The force exerted on an electron is $7.2\times10^{-18}\ N$

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