Ask question

Ask question

All categories

3 years ago

13

Radio transmissions show observers in Houston that the International Space Station is 323 miles away, the Chi- nese space statio

n Tiangong is 462 miles away, and the angle International Space Station–Houston–Tiangong is 109 . How far apart are the two space stations?

1 answer:

Komok [63]3 years ago

7 0

Answer:

The two space stations are 644.1653 miles far apart.

Explanation:

The distance of Huston and the International Space Station is 323 miles.

The distance of Huston and the Chinese space station Tiangong is 462 miles.

Angle between the two stations = 109⁰

So, applying the law of cosines as;

c² = a² + b² - 2abcosC

From the picture shown below, a = 323 miles and b = 462 miles

So,

c² = 323² + 462² + 2(323)(462)cos109 = 104329 + 213444 - 298452cos109

cos109 = -0.3256

So,

c² = 414948.9712

<u>c = 644.1653 miles</u>

You might be interested in

Consider a wod raft with length = 5.0m, width 3.0 m and height = 1.0m, with density =600kg/m^3. There is an object of a mass of

diamong [38]

Answer:

0.63333m

Explanation:

If the wood raft partially sinks, it means the water exerts a force upwards, equal to the weight of the displaced water by the part sank that is in magnitude equal to the weight of the raft and the objects on it but the opposite direction. There fore, we need to calculate the mass of the raft and add it to the object mass for the total.

$m_r=V_r\rho_r$

Where $V_r=5\times3\times1=15m^3$ is the volume of the raft and $\rho=600kg/m^3$ is the density.

$m_r=(600)(15)=9000Kg$

The total mass is 9500Kg (The raft plus the object), and this is equal to the mass of the water displaced. Now find the volume of water that has this mass:

$V=m/\rho$

$V_w=m_w/\rho_w=\frac{9500}{1000} =9.5m^3$

With this volume you can calculate at what height of the raft the water will be. The base of the raft will be the same, just calculate the new height.

$V=h\times B\\h=V/B=\frac{9.5}{5\times3}=\frac{9.5}{15}=0.63333m$

7 0

3 years ago

What is the basic form of matter which cannot be broken down any further?

Law Incorporation [45]

Answer:

An element

Explanation:

An element is a pure substance that cannot be broken down any smaller.

4 0

2 years ago

Read 2 more answers

zlopas [31]

Bronchial Tubes
hope this helps :)

7 0

3 years ago

Read 2 more answers

Please Help!!!! I WILL GIVE BRAINLIEST!!!!!!!!!!!!!

Bas_tet [7]

Given info

d = 0.000250 meters = distance between slits

L = 302 cm = 0.302 meters = distance from slits to screen

$\theta_8 = 1.12^{\circ}$ = angle to 8th max (note how m = 8 since we're comparing this to the form $\theta_m$ )

$x_n = x_5 = 3.33 \text{ cm} = 0.0333 \text{ meters}$ (n = 5 as we're dealing with the 5th minimum )

---------------

Method 1

$d\sin(\theta_m) = m\lambda\\\\0.000250\sin(\theta_8) = 8\lambda\\\\8\lambda = 0.000250\sin(1.12^{\circ})\\\\\lambda = \frac{0.000250\sin(1.12^{\circ})}{8}\\\\\lambda \approx 0.000 000 61082633\\\\\lambda \approx 6.1082633 \times 10^{-7} \text{meters}\\\\ \lambda \approx 6.11 \times 10^{-7} \text{ meters}\\\\ \lambda \approx 611 \text{ nm}$

Make sure your calculator is in degree mode.

-----------------

Method 2

$\Delta x = \frac{\lambda*L*m}{d}\\\\L*\tan(\theta_m) = \frac{\lambda*L*m}{d}\\\\\tan(\theta_m) = \frac{\lambda*m}{d}\\\\\tan(\theta_8) = \frac{\lambda*8}{0.000250}\\\\\tan(1.12^{\circ}) = \frac{\lambda*8}{0.000250}\\\\\lambda = \frac{1}{8}*0.000250*\tan(1.12^{\circ})\\\\\lambda \approx 0.00000061094306 \text{ meters}\\\\\lambda \approx 6.1094306 \times 10^{-7} \text{ meters}\\\\\lambda \approx 611 \text{ nm}\\\\$

-----------------

Method 3

$\frac{d*x_n}{L} = \left(n-\frac{1}{2}\right)\lambda\\\\\frac{0.000250*3.33}{302.0} = \left(5-\frac{1}{2}\right)\lambda\\\\0.00000275662251 \approx \frac{9}{2}\lambda\\\\\frac{9}{2}\lambda \approx 0.00000275662251\\\\\lambda \approx \frac{2}{9}*0.00000275662251\\\\\lambda \approx 0.00000061258279 \text{ meters}\\\\\lambda \approx 6.1258279 \times 10^{-7} \text{ meters}\\\\\lambda \approx 6.13 \times 10^{-7} \text{ meters}\\\\\lambda \approx 613 \text{ nm}\\\\$

There is a slight discrepancy (the first two results were 611 nm while this is roughly 613 nm) which could be a result of rounding error, but I'm not entirely sure.

7 0

3 years ago

A ring with an 18mm diameter falls off a scientist's finger into the solenoid in the lab. The solenoid is 25 cm long, 5.0 cm in

Troyanec [42]

Answer:

The value is $\epsilon = 3.84 *10^{-5} \ V$

Explanation:

From the question we are told that

The diameter of the ring is $d = 18 \ mm = 0.018 \ m$

The length of the solenoid is $l = 25 \ cm = 0.25 \ m$

The diameter of the solenoid is $D = 5.0 \ cm = 0.05 \ m$

The number of turns is N = 1500

The change in current in the solenoid is $\Delta I = 20 \ A$

The time taken is $\Delta t = 1 \ s$

Generally the radius of the ring is

$r = \frac{d}{2}$

=> $r = \frac{0.018 }{2}$

=> $r = 0.009 \ m$

Generally the area of the ring is mathematically represented as

$A = \pi r^2$

=> $A = 3.142 * 0.009^2$

=> $A = 2.545 *10^{-4}\ m^2$

Generally the induced emf is mathematically represented as

$\epsilon = A * \frac{dB}{dt}$

Here

$\frac{dB }{dt} = \mu_o * \frac{N}{l} *\frac{ \Delta I }{\Delta t}$

Here $\mu_o$ is the permeability of free space with value

$\mu_o = 4\pi *10^{-7} \ N/A^2$

So

$\frac{dB }{dt} = 4\pi * 10^{-7} * \frac{1500}{0.25} *\frac{20 }{1}$

=> $\frac{dB }{dt} = 0.150816\ T/s$

So

$\epsilon = 0.150816 * 2.545 *10^{-4}$

=> $\epsilon = 3.84 *10^{-5} \ V$

3 0

2 years ago