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gtnhenbr [62]
3 years ago
6

What’s gravitational pull of the earth

Physics
2 answers:
DENIUS [597]3 years ago
8 0

Explanation:

9.807 m/s squared is the gravitational pull of planet Earth.

OLga [1]3 years ago
3 0

Answer:

9.8 m/s^2

Explanation:

9.8 meters per second squared. A common misconception is that heavier objects fall faster, but gravity affects all things equally.

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Which is true about a surface wave? ANSWER FAST
Anna [14]

Answer:

D

Explanation:

6 0
3 years ago
The unit used to measure electric current is the ampere (A). Now, assume that the current delivered at a wall socket reaches the
pav-90 [236]

Answer:

T = 0.017s

Explanation:

period is the time it takes a particle to make one oscillation

An electric current is periodic in nature

The current reaches 3.8A  ten times.  

So there must have been 10 cycles (10 periods) in 0.17s.   let 'T'  be the period:

T=\frac{t}{n}

t is the total time interval

n is the number of oscillations

T=\frac{0.17}{10}

10T = 0.17

T = 0.17/10 = 0.017s

8 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:

a)

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

b)

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 %.

c)

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

Now we can find the radiation intensity with the equation

I=\frac{P}{A}

Where the area is

A=4\pi r'^2 = 4\pi(2.4\cdot 10^{11})^2=7.24\cdot 10^{23} m^2

And substituting,

I=\frac{3.910\cdot 10^{26}}{7.24\cdot 10^{23}}=540 W/m^2

Learn more about electromagnetic radiation:

brainly.com/question/9184100

brainly.com/question/12450147

#LearnwithBrainly

4 0
3 years ago
What is the economic system in the united states?
AnnyKZ [126]

Answer:

A. Capitalist economy

Explanation:

The United States is a capitalist economy. Hope this helps, thank you !!

3 0
3 years ago
In a football game, the running back takes a handoff and begins running toward midfield at 3.21 yards/s . As he moves through hi
Inessa [10]

Given Information:

Initial speed = u = 3.21 yards/s

Acceleration = α  = 1.71 yards/s²

Final speed = v = 7.54 yards/s

Required Information:

Distance = s = ?

Answer:

Distance = s = 13.61

Explanation:

We are given the speeds and acceleration of the runner and we want to find out how much distance he covered before being tackled.

We know from the equations of motion,

v² = u² + 2αs

Where u is the initial speed of the runner, v is the final speed of the runner, α is the acceleration of the runner and s is the distance traveled by the runner.

Re-arranging the above equation for distance yields,

2αs = v² - u²

s = (v² - u²)/2α

s = (7.54² - 3.21²)/2×1.71

s = 46.55/3.42

s = 13.61 yards

Therefore, the runner traveled a distance of 13.61 yards before being tackled.

8 0
3 years ago
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