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3 years ago

How much force do you push down on the Earth with? (Hint: 1kg is about 2 pounds).

Physics

2 answers:

e-lub [12.9K]3 years ago

6 0

The force with which I push down on the Earth is PRECISELY equal to the force with which the Earth pushes UP on me.

I weigh EXACTLY the same amount on the Earth as the Earth weighs on me. Find THAT amount, and you'll have the answer to the question.

(Hint: 1 kg is a mass, and 2 pounds is a force, so they can never be equal. 1kg weighs about 2.2 pounds IF it happens to be on Earth. But if the same 1kg is on the Moon, then it weighs about 5.8 ounces. And if it happens to be on Mars, then it weighs about 13.3 ounces. The same kilogram weighs different weights in different places.)

On a good day, I push down on the Earth with about 890 Newtons of force. No matter how hard I try, I can never push more or less than that.

Gennadij [26K]3 years ago

3 0

Answer: 100lbs

Explanation: The Earth pushes you down at 100lbs, so you push down the earth by 100lbs which is enough to keep you firmly attached to the ground and not allow you to jump more than a couple of feet.

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3 years ago

An automobile fuel tank is filled to the brim with 45 L of gasoline (12 gal) at 10°C. Immediately afterward, the vehicle is park

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Answer:

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Explanation:

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3 years ago

Many physical quantities are connected by inverse square laws, that is, by power functions of the form f(x)=kx^(-2). In particul

goldenfox [79]

Answer:

4 times

Explanation:

Since the equation for the illumination of an object, i.e. the brightness of the light, is <em>inversely proportional to the square of the distance from the light source</em>, the form of the function is:

f(x) = k.x⁻²

Where x is the distance between the object and the light force, k is the constant of proportionality, and f(x) is the brightness.

Then, if you move halfway to the lamp the new distance is x/2 and the new brightness (call if F) is :

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Then, you have found that the light is 4 times as bright as it originally was.

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2 years ago

12.51 A parallel RLC circuit, which is driven by a variable frequency 2-A current source, has the following values: R = 1 kΩ, L

Anastaziya [24]

Answer:

BW = 100 rad/s

wlow = 452.49 rad/s

whigh = 552.49 rad/s

V(jwlow) =1414.21 < 45°V

V(jwhigh) =1414.21 <-45°V

Explanation:

To calculate bandwidth we have formula

BW = 1/RC

BW = 1/ 1000x10x10^¯6

BW = 100 rad/s

We will first calculate resonant frequency and quality factor for half power frequencies.

For resonant frequency

wo = 1/(SQRT LC)

wo = 1/SQRT 400×10¯³ × 10×10^¯6

wo = 500 rad/s

For Quality

Q = wo / BW

Q = 500/100

Q = 5

wlow = wo [-1/2Q+ SQRT (1/2Q)² + 1]

wlow = 500 [-1/2×5 + SQRT (1/2×5)² + 1]

wlow = 452.49 rad/s

whigh = wo [1/2Q+ SQRT (1/2Q)² + 1]

whigh = 500 [1/2×5 + SQRT (1/2×5)² + 1]

whigh = 552.49 rad/s

We will start with admittance at lower half power frequency

Y(jwlow) = (1/R) + (1/jwlow L) + (jwlow C)

Y(jwlow) = (1/1000) + (1/j×452.49×400×10¯³) + (j×452.49×10×10^¯6)

Y(jwlow) = 0.001 - j5.525×10¯³ + j4.525×10¯³

Y(jwlow) = (1-j).10¯³ S

Voltage across the network is calculated by ohm's law

V(jwlow) = I/Y(jwlow)

V(jwlow) = 2/(1-j).10¯³

V(jwlow) = 1414.2 < 45°V

Now we will calculate the admittance at higher half power frequency

Y(jwhigh) = (1/R) + (1/jwhigh L) + (jwhigh C)

Y(jwhigh) = (1/1000) + (1/j×552.49×400×10¯³) + (j×552.49×10×10^¯6)

Y(jwhigh) = 0.001 - j4.525×10¯³ + j5.525×10¯³

Y(jwhigh) = (1+j).10¯³ S

Voltage across network will be calculated by ohm's law

V(jwhigh) = I/Y(jwhigh)

V(jwhigh) = 2/(1+j).10¯³

V(jwhigh) = 1414.2 < - 45°V

6 0

3 years ago

nolan ryan has the record for having the speediest fastball in baseball. he could pitch one at 148 i/sec. what is that speed in

Klio2033 [76]

The speed of the ball is 101miles/hr.

A mile is a unit of length that is exactly 1,609.344 metres long. Similarly, 5,280 feet or 1,760 yards make up one mile. The mile is an imperial and common US measurement of distance.

We just have to deal with unit conversions.

One mile is 5280 feet, or 1 ft = 0.000189

The speed of the ball in miles per hour is

$\frac{148ft}{1sec} . \frac{1mile}{5280ft} .\frac{60s}{1min} .\frac{60min}{1hr}$

So, the speed of the ball in miles per hour is 101miles/hr.

Learn more about miles here;

brainly.com/question/23245414

#SPJ4

5 0

1 year ago

How much force do you push down on the Earth with? (Hint: 1kg is about 2 pounds).​

How much force do you push down on the Earth with? (Hint: 1kg is about 2 pounds).