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3 years ago

Total energy of an orbiting satellite is negative of its Kinetic or potential energy

Physics

1 answer:

Over [174]3 years ago

5 0

Total energy of an orbiting satellite is negative of Its [Kinetic energy].

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7. A force stretches a wire by 1 mm. a. A second wire of the same material has the same cross section and twice the length. How

Lera25 [3.4K]

Answer:

(a) The second wire will be stretched by 2 mm

(b) The third wire will be stretched by 0.25 mm

Explanation:

Tensile stress on every engineering material is given as the ratio of applied force to unit area of the material.

σ = F / A

Tensile strain on every engineering material is given as the ratio of extension of the material to the original length

δ = e / L

The ratio of tensile stress to tensile strain is known as Young's modulus of the material.

$Y = \frac{FL}{Ae}$

Part A

cross sectional area and applied force are the same as the original but the length is doubled

$\frac{FL_1}{A_1e_1} = \frac{FL_o}{A_oe_o} \\\\\frac{L_1}{e_1} = \frac{L_o}{e_o}\\\\e_1 = \frac{L_1e_o}{L_o} \\\\But, L_1 =2L_o\\\\e_1 = \frac{2L_oe_o}{L_o}\\e_1 = 2e_o$

The second wire will be stretched by 2 mm

Part B

a third wire with the same length but twice the diameter of the first

$\frac{FL}{A_1e_1} = \frac{FL}{A_oe_o} \\\\\frac{1}{A_1e_1} = \frac{1}{A_oe_o}\\\\\frac{4}{\pi d_1^2e_1} = \frac{4}{\pi d^2_oe_o}\\\\\frac{1}{d_1^2e_1} = \frac{1}{d^2_oe_o}\\\\d_1^2e_1 = d^2_oe_o\\\\e_1 = \frac{d^2_oe_o}{d_1^2} \\\\e_1 =(\frac{d_o}{d_1})^2e_o\\\\But, d_1 = 2d_o\\\\e_1 =(\frac{d_o}{2d_o})^2e_o\\\\e_1 =(\frac{1}{2})^2e_o\\\\e_1 =(\frac{1}{4})e_o$

e₁ = ¹/₄ x 1 mm = 0.25 mm

The third wire will be stretched by 0.25 mm

3 0

4 years ago

What is the force that opposes motion and works against the downward pull? A) friction B) gravity C) weight D) acceleration

charle [14.2K]

I'm not entirely sure, but I believe it is A Friction. because gravity pulls down, weight isn't a force, and acceleration doesn't oppose motion

6 0

4 years ago

Consider an electron with charge −e and mass m orbiting in a circle around a hydrogen nucleus (a single proton) with charge +e.

alexandr1967 [171]

Answer:

$v=\sqrt{k\frac{e^2}{m_e r}}$ , $2.18\cdot 10^6 m/s$

Explanation:

The magnitude of the electromagnetic force between the electron and the proton in the nucleus is equal to the centripetal force:

$k\frac{(e)(e)}{r^2}=m_e \frac{v^2}{r}$

where

k is the Coulomb constant

e is the magnitude of the charge of the electron

e is the magnitude of the charge of the proton in the nucleus

r is the distance between the electron and the nucleus

v is the speed of the electron

$m_e$ is the mass of the electron

Solving for v, we find

$v=\sqrt{k\frac{e^2}{m_e r}}$

Inside an atom of hydrogen, the distance between the electron and the nucleus is approximately

$r=5.3\cdot 10^{-11}m$

while the electron mass is

$m_e = 9.11\cdot 10^{-31}kg$

and the charge is

$e=1.6\cdot 10^{-19} C$

Substituting into the formula, we find

$v=\sqrt{(9\cdot 10^9 m/s) \frac{(1.6\cdot 10^{-19} C)^2}{(9.11\cdot 10^{-31} kg)(5.3\cdot 10^{-11} m)}}=2.18\cdot 10^6 m/s$

7 0

4 years ago

16. Two capacitors have an equivalent

Gennadij [26K]

Answer:

C1 + C2 = 30 parallel connection

C1 * C2 / (C1 + C2) = 7.2 series connection

C1 * C2 = 7.2 * (C1 + C2) = 216

C2 + 216 / C2 = 30 using first equation

C2^2 + 216 = 30 C2

C2^2 - 30 C2 + 216 = 0

C2 = 12 or 18 solving the quadratic

Then C1 = 18 or 12

5 0

3 years ago

A gas cylinder contains argon atoms (m=40.0 u). The temperature is increased from 286 K (13°C) to 362 K (89°C) (a) What is the c

Rama09 [41]

Answer:

(a). The change in the average kinetic energy per atom is $1.57\times10^{-21}\ eV$ .

(b). The change in vertical position is 2413 m.

Explanation:

Given that,

Mass = 40.0 u

The increased temperature from 286 K to 362 K.

(a). We need to calculate the change in the average kinetic energy per atom

Using formula of kinetic energy

$\Delta K.E=\dfrac{3}{2}k\Delta t$

Put the value into the formula

$\Delta K.E=\dfrac{3}{2}\times1.38\times10^{-23}\times(362-286)$

$\Delta K.E=1.57\times10^{-21}\ eV$

(b). The change in potential energy of the container due to change in the vertical position

We need to calculate the change in vertical position

Using formula of potential energy

$\Delta U=mg\Delta h$

$\Delta h =\dfrac{\Delta U}{mg}$

$\Delta h=\dfrac{1.57\times10^{-21}}{40.0\times1.66\times10^{-27}\times9.8}$

$\Delta h=2412.7=2413\ m$

Hence, (a). The change in the average kinetic energy per atom is $1.57\times10^{-21}\ eV$ .

(b). The change in vertical position is 2413 m.

4 0

4 years ago

Total energy of an orbiting satellite is negative of its Kinetic or potential energy​

Total energy of an orbiting satellite is negative of its Kinetic or potential energy