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3 years ago

Unscramble this caybnetruopergs * starts with "T" ends with "S" Please help me

1 answer:

8 0

Typebars
Turbocars
Teacups
Teapoys
Tourneys
Toecaps
Trepangs
Tabourers
Trouncers
Troupers
Terraces
Trances
Turbans
Toupees
Toucans
Sorry if this is not what you want me to do and I just threw a bunch of words at you.

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A woman takes her dog rover for a walk on a leash. She pulls on the leash with a force of 30.0 N at an angle of 29° above the ho

IrinaK [193]

19.8 N force is tending to lift Rover vertically off the ground.

<h3>What is horizontal and vertical component?</h3>

The horizontal velocity component ( $v_{x}$ ) describes the influence of the velocity in displacing the projectile horizontally. The vertical velocity component ( $v_{y}$ ) describes the influence of the velocity in displacing the projectile vertically.

According to the question,

The women pulls the dog with a force of 30 N at an angle of 29° from the horizontal.

Horizontal component= 30cos(29°) = 22.2 N

Vertical component = 30sin(29°) = 19.8 N

Therefore,

The horizontal component would tend to make the dog move forward and the vertical component would tend lift it off the ground.

Hence,

19.8 N force is tending to lift Rover vertically off the ground.

Learn more about horizontal and vertical component here:

brainly.com/question/11776718

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4 0

2 years ago

What is 7.832 rounded to the nearest 10th?

Galina-37 [17]

When rounding you look at the number right after the one you want to round. The tenth place is the one right after the decimal. If the number after it is more than five it goes up by one, it it's less than five it stays the same. So since the number after the 8 is a 3 it stays the same.

The answer is 7.8

3 0

3 years ago

Radiation from the Sun The intensity of the radiation from the Sun measured on Earth is 1360 W/m2 and frequency is f = 60 MHz. T

Zina [86]

a) Total power output: $3.845\cdot 10^{26} W$

b) The relative percentage change of power output is 1.67%

c) The intensity of the radiation on Mars is $540 W/m^2$

Explanation:

The intensity of electromagnetic radiation is given by

$I=\frac{P}{A}$

where

P is the power output

A is the surface area considered

In this problem, we have

$I=1360 W/m^2$ is the intensity of the solar radiation at the Earth

The area to be considered is area of a sphere of radius

$r=1.5\cdot 10^{11} m$ (distance Earth-Sun)

Therefore

$A=4\pi r^2 = 4 \pi (1.5\cdot 10^{11})^2=2.8\cdot 10^{23}m^2$

And now, using the first equation, we can find the total power output of the Sun:

$P=IA=(1360)(2.8\cdot 10^{23})=3.845\cdot 10^{26} W$

The energy of the solar radiation is directly proportional to its frequency, given the relationship

$E=hf$

where E is the energy, h is the Planck's constant, f is the frequency.

Also, the power output of the Sun is directly proportional to the energy,

$P=\frac{E}{t}$

where t is the time.

This means that the power output is proportional to the frequency:

$P\propto f$

Here the frequency increases by 1 MHz: the original frequency was

$f_0 = 60 MHz$

so the relative percentage change in frequency is

$\frac{\Delta f}{f_0}\cdot 100 = \frac{1}{60}\cdot 100 =1.67\%$

And therefore, the power also increases by 1.67 %.

In this second case, we have to calculate the new power output of the Sun:

$P' = P + \frac{1.67}{100}P =1.167P=1.0167(3.845\cdot 10^{26})=3.910\cdot 10^{26} W$

Now we want to calculate the intensity of the radiation measured on Mars. Mars is 60% farther from the Sun than the Earth, so its distance from the Sun is

$r'=(1+0.60)r=1.60r=1.60(1.5\cdot 10^{11})=2.4\cdot 10^{11}m$