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vagabundo [1.1K]

3 years ago

12

You are in a hot air balloon, 350 m above the flat Texas plains. You look out toward the horizon. How far out can you see--that

is, how far is your horizon? The Earth's radius is about 6400 km.

1 answer:

pshichka [43]3 years ago

3 0

Refer to the diagram shown below.

x = the distance (horizon) that you can see from the balloon,

Because the radius of the earth is 6400 km, and the balloon is 350 n above the earth's surface, therefore
h = 6400 + 0.350 = 6400.35 km.

From the Pythagorean theorem, obtain
x = √(6400.35² - 6400²) = 66.934 km

Answer: Your horizon is 66.934 km (or 66,934 m)

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What causes a sealed bag of vegetables to expand when it is heated in a microwave oven?

Fynjy0 [20]

Vegetables has moisture and during rise in the temperature the gas sealed and generated from moisture expands. Gas when heated or experience rise in temperature expands.

3 0

3 years ago

A father places his daughter in a swing that is 0.60\,\text{m}0.60m0, point, 60, start text, m, end text above ground. Then he r

schepotkina [342]

This question involves the concepts of the law of conservation of energy and kinetic energy.

The girl's fastest speed is "3.7 m/s".

According to the law of conservation of energy, the girl will have the fastest speed at mean position, which will be calculated as follows:

Loss in Potential Energy = Gain in Kinetic Energy

$mg\Delta h=\frac{1}{2}mv^2\\\\v=\sqrt{2g\Delta h}$

where,

v = maximum speed = ?

g = acceleration due to gravity = 9.81 m/s²

Δh = change in height = 1.3 m - 0.6 m = 0.7 m

Therefore,

$v=\sqrt{2(9.81\ m/s^2)(0.7\ m)}$

<u>v = 3.7 m/s</u>

<u></u>

Learn more about the Law of Conservation of Energy here:

brainly.com/question/381281?referrer=searchResults

5 0

2 years ago

A solid cylindrical bar conducts heat at a rate of 25 W from a hot to a cold reservoir under steady state conditions. If both th

expeople1 [14]

Answer:

Using the new cylinder the heat rate between the reservoirs would be 50 W

Explanation:

Conduction could be described by the Law of Fourierin the form: $Q=kA\frac{T_1-T_2}{L}$ where $Q$ is the rate of heat transferred by conduction, $k$ is the thermal conductivity of the material, $T_1$ and $T_2$ are the temperatures of each heat deposit, $A$ is the cross area to the flow of heat, and ${L}$ is the distance that the flow of heat has to go.
For the original cylinder the Fourier's law would be: $kA_1\frac{T_1-T_2}{L_1}=25W$ , and if $A_1=\frac{\pi D_{1}^{2}}{4}$ , then the expression would be: $k\frac{\pi D_1^{2}}{4} \frac{T_1-T_2}{L_1}=25W$ where $D_1$ is the diameter of the original cylinder, and ${L_1}$ is the length of the original cylinder.
For the new cylinder, in the same fashion that for the first, Fourier's Law would be: $Q_2=k\frac{\pi D_2^2}{4}\frac{T_1-T_2}{L_2}$ ,where $Q_2$ is the heat rate in the second case, $D_2$ and ${L_2$ are the new diameter and length.
But, $D_2=2D_1$ and $L_2=2L_1$ , substituting in the expression for $Q_2$ : $Q_2=k\frac{\pi (2D_1)^2}{4}\frac{T_1-T_2}{2L_1}$ .
Rearranging: $Q_2=\frac{2^2}{2}(k\frac{\pi D_1^2}{4}\frac{T_1-T_2}{L_1})$ .
In the last declaration of $Q_2$ , it could be noted that the expressión inside the parenthesis is actually $Q_1$ , then: $Q_2=\frac{2^2}{2}(25W)=50W$ .
<u>It should be noted, that the temperatures in the hot and cold reservoirs never change.</u>

7 0

4 years ago

A remote controlled car travels at a uniform speed of 3.5 m/s and reaches its destination in 5 minutes. How far did the controll

vovangra [49]

Converting minutes to seconds
1 minute = 60 seconds
5 minutes = 5 x 60 = 300 seconds.

distance
Time = - - - - - - - -
speed

Distance = time x speed = 300 x 3.5 = 1050 meters.
The car traveled 1050 meters.

7 0

3 years ago

Una carga de -10Mc está situada a 20cm delante de otra carga de 5 Mc. Calcular la fuerza electrostática en Newton ejercida por u

tino4ka555 [31]

Answer:

a) force between them is attraction, b) F = 1.125 10⁻² N

Explanation:

In this case the electric force is given by Coulomb's law

F = $k \frac{q_1q_2}{r^2}$

In the exercise they give us the values of the loads

q1 = - 10 mC = -10 10⁻³ C

q2 = 5 mC = 5 10⁻³ C

d = 20 cm = 0.20 m

let's calculate

F = 9 10⁹ $\frac{10 \ 10^{-3} \ 5 \ 10^{-3}}{0.20^2}$

F = 1.125 10⁻² N

To find the direction of the force we use that charges of the same sign repel each other, as in this case there is a positive and a negative charge, the force between them is attraction

7 0

3 years ago