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2 years ago

This type of soil is found in warm, humid climates. Question 4 options: pedocal laterite pedalfer humus

Physics

1 answer:

Oksi-84 [34.3K]2 years ago

7 0

Answer:

Laterite is your correct answer

Explanation:

Well i took the quiz and made a %100 meaning i am correct.

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Radar is used to determine distances to various objects by measuring the round-trip time for an echo from the object. (a) How fa

zloy xaker [14]

Answer:

150000000000 m

0.0000005 seconds

33.33 ns

Explanation:

Speed of electromagnetic waves through vacuum = $3\times 10^8\ m/s$

Echo time = 1000 seconds

Echo time is the time taken to reach the object and come back to the observer

Distance = Speed×Time

$Distance=3\times 10^8\times \frac{1000}{2}=150000000000\ m$

Venus is 150000000000 m away from Earth

Time = Distance / Speed

$Time=\frac{75}{3\times 10^8}=0.00000025\ seconds$

Echo time will be twice the time

$Echo\ time=2\times 0.00000025=0.0000005\ seconds$

The echo time will be 0.0000005 seconds

Difference in time = Difference in distance / Speed

$\Delta t=\frac{\Delta d}{v}\\\Rightarrow \Delta t=\frac{10}{3\times 10^8}\\\Rightarrow \Delta=33.33\ ns$

The accuracy by which I will be able to measure the echo time is 33.33 ns

5 0

3 years ago

What is the frequency of this wave?

Anastasy [175]

In telecommunication systems, Carrier frequency is a technical term used to indicate: ... The frequency of the unmodulated electromagnetic wave at the output of a conventional amplitude-modulated (AM-unsupressed carrier), or frequency-modulated (FM), or phase-modulated (PM) radio transmitter.

6 0

2 years ago

Read 2 more answers

How do I calculate the tension in the horizontal string?

matrenka [14]

ANSWER

T₂ = 10.19N

EXPLANATION

Given:

• The mass of the ball, m = 1.8kg

First, we draw the forces acting on the ball, adding the vertical and horizontal components of each one,

In this position, the ball is at rest, so, by Newton's second law of motion, for each direction we have,

$\begin{gathered} T_{1y}-F_g=0_{}_{}_{} \\ T_2-T_{1x}=0 \end{gathered}$

The components of the tension of the first string can be found considering that they form a right triangle, where the vector of the tension is the hypotenuse,

$\begin{gathered} T_{1y}=T_1\cdot\cos 30\degree \\ T_{1x}=T_1\cdot\sin 30\degree \end{gathered}$

We have to find the tension in the horizontal string, T₂, but first, we have to find the tension 1 using the first equation,

$T_1\cos 30\degree-m\cdot g=0$

Solve for T₁,

$T_1=\frac{m\cdot g}{\cos30\degree}=\frac{1.8kg\cdot9.8m/s^2}{\cos 30\degree}\approx20.37N$

Now, we use the second equation to find the tension in the horizontal string,

$T_2-T_1\sin 30\degree=0$

Solve for T₂,

$T_2=T_1\sin 30\degree=20.37N\cdot\sin 30\degree\approx10.19N$

Hence, the tension in the horizontal string is 10.19N, rounded to the nearest hundredth.

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4 months ago

Why do we see the color black?

Inessa [10]

We see black colour in absence of light. Black colour absoorbs all the light, ( just opposite of white) and hence we do not see any colour. Black colour is also related to darkness.

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2 years ago

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MissTica

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