The ionosphere lies between the mesosphere and exosphere is it true or false

Engineers and science fiction writers have proposed designing space stations in the shape of a rotating wheel or ring, which wou

Gennadij [26K]

Answer:

a. The station is rotating at $1.496 \frac{rev}{min}$

b. the rotation needed is $2.8502 \frac{rev}{min}$

Explanation:

We know that the centripetal acceleration is

$a_{c}= \omega ^2 r$

where $\omega$ is the rotational speed and r is the radius. As the centripetal acceleration is feel like an centrifugal acceleration in the rotating frame of reference (be careful, as the rotating frame of reference is <u>NOT INERTIAL,</u> the centrifugal force is a fictitious force, the real force is the centripetal).

The rotational speed is :

$2.7 \frac{m}{s^2} = \omega ^2 * 110 \ m$

$\omega ^2 = \frac{2.7 \frac{m}{s^2}} {110 \ m}$

$\omega = \sqrt{ 0.02454 \frac{rad^2}{s^2} }$

$\omega = 0.1567 \frac{rad}{s}$

Knowing that there are $2\pi \ rad$ in a revolution and 60 seconds in a minute.

$\omega = 0.1567 \frac{rad}{s} \frac{1 \ rev}{2\pi \ rad} \frac{60 \ s}{1 \ min}$

$\omega = 1.496 \frac{rev}{min}$

The rotational speed needed is :

$9.8 \frac{m}{s^2} = \omega ^2 * 110 \ m$

$\omega ^2 = \frac{9.8 \frac{m}{s^2}} {110 \ m}$

$\omega = \sqrt{ 0.08909 \frac{rad^2}{s^2} }$

$\omega = 0.2985 \frac{rad}{s}$

Knowing that there are $2\pi \ rad$ in a revolution and 60 seconds in a minute.

$\omega = 0.2985 \frac{rev}{min} \frac{1 \ rev}{2\pi \ rad} \frac{60 \ s}{1 \ min}$

$\omega = 2.8502 \frac{rev}{min}$

3 0

3 years ago

Read 2 more answers

8. Fig. 4.1 shows a heavy ball B of weight W suspended from a fixed beam by two ropes P and Q.

mart [117]

Answer:

The resultant tension of the two ropes is approximately 42.4 N

The length of the line representing the resultant tension is approximately 8.48 cm

Please find included with the answer the scale drawing created with Microsoft Word

Explanation:

The given parameters are;

The tension in rope P, $T_P$ = 30 N

The tension in rope Q, $T_Q$ = 30 N

The angle the rope, 'P', makes with the horizontal = 45°

The angle the rope, 'Q', makes with the horizontal = 45°

The scale factor of the scale diagram, S.F. = 5.0 N/cm

By the resolution of forces at equilibrium, we have;

The sum of the vertical forces, $\Sigma F_y$ = $T_P_y$ + $T_Q_y$ + W = 0

∴ W = -( $T_P_y$ + $T_Q_y$ )

W = -(30 × sin(45°) + 30 × sin(45°)) = -42.4264068712

The weight of the heavy ball, W ≈ 42.4 N acting downwards

The sum of the horizontal forces, $\Sigma F_x$ = $T_P_x$ + $T_Q_x$ = 0

The length of the resultant force, W = W/(S.F.) ≈ 42.4 N/(5.0 N/cm) = 8.48 cm

The drawing of the vectors using the scale factor of 5.0 N/cm is created using Microsoft Word is included

3 0

3 years ago

a 70 kg skydiver opens her parachute. The force due to air resistance is now 1200 N. what is the acceleration of the skydiver

Natasha_Volkova [10]

Around 50 I think so

7 0

3 years ago

Suppose we measure the energy stored in some inductor to be E when there is a current I running through it. If I double the curr

slavikrds [6]

Answer:

If I double the current in the inductor, the new total energy will become 4E (option f).

Explanation:

The coil or inductor is a passive component made of an insulated wire that stores energy in the form of a magnetic field due to its form of coiled turns of wire, through a phenomenon called self-induction. In other words, inductors store energy in the form of a magnetic field. The energy stored in the space where there is a magnetic field in the inductor is:

$E=\frac{1}{2} *L*I^{2}$

where E is Energy [J], L is Inductance [H] and I is Current [A].

If you double the current in the inductor, then the new value of the current is I'= 2*I. So replacing the new total energy is:

$E'=\frac{1}{2} *L*I'^{2}=\frac{1}{2} *L*(2*I)^{2}=\frac{1}{2} *L*4*I^{2}=4*\frac{1}{2} *L*I^{2}$

Then:

$E'=4*E$

<em><u>If I double the current in the inductor, the new total energy will become 4E (option f).</u></em>

3 0

3 years ago

Transmission Lines and Health. Currents in dc transmission lines can be 100 A or higher. Some people are concerned that the elec

Blababa [14]

Calculate the magnetic field strength at the ground. Treat the transmission line as infinitely long. The magnetic field strength is then given by:

B = μ₀I/(2πr)

B = magnetic field strength, μ₀ = magnetic constant, I = current, r = distance from line

Given values:

μ₀ = 4π×10⁻⁷H/m, I = 170A, r = 8.0m

Plug in and solve for B:

B = 4π×10⁻⁷(170)/(2π(8.0))

B = 4.25×10⁻⁶T

The earth's magnetic field strength is 0.50G or 5.0×10⁻⁵T. Calculate the ratio of the line's magnetic field strength to earth's magnetic field strength:

4.25×10⁻⁶/(5.0×10⁻⁵)

= 0.085

= 8.5%

The transmission line's magnetic field strength is 8.5% of that of earth's natural magnetic field. This is no cause for worry.

8 0

3 years ago