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matrenka [14]
3 years ago
5

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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A single slit of width 0.50 mm is illuminated with light of wavelength 500 nm, and a screen is placed 120 cm in front of the sli
Vaselesa [24]

Answer:

a) 2.4 mm

b) 1.2 mm

c) 1.2 mm

Explanation:

To find the widths of the maxima you use the diffraction condition for destructive interference, given by the following formula:

m\lambda=asin\theta

a: width of the slit

λ: wavelength

m: order of the minimum

for little angles you have:

y=\frac{m\lambda D}{a}

y: height of the mth minimum

a) the width of the central maximum is 2*y for m=1:

w=2y_1=2\frac{1(500*10^{-9}m)(1.2m)}{0.50*10^{-3}m}=2.4*10^{-3}m=2.4mm

b) the width of first maximum is y2-y1:

w=y_2-y_1=\frac{(500*10^{-9}m)(1.2m)}{0.50*10^{-3}m}[2-1]=1.2mm

c) and for the second maximum:

w=y_3-y_2=\frac{(500*10^{-9}m)(1.2m)}{0.50*10^{-3}m}[3-2]=1.2mm

4 0
2 years ago
Traveling waves propagate with a fixed speed usually denoted as v (but sometimes c). The waves are called __________ if their wa
Snezhnost [94]

Answer:

Periodic.

Explanation:

Electromagnetic waves is a propagating medium used in all communications device to transmit data (messages) from the device of the sender to the device of the receiver.

Generally, the most commonly used electromagnetic wave technology in telecommunications is radio waves.

Radio waves can be defined as an electromagnetic wave that has its frequency ranging from 30 GHz to 300 GHz and its wavelength between 1mm and 3000m. Therefore, radio waves are a series of repetitive valleys and peaks that are typically characterized of having the longest wavelength in the electromagnetic spectrum.

Basically, as a result of radio waves having long wavelengths, they are mainly used in long-distance communications such as the carriage and transmission of data.

Generally, a fixed speed is used for the propagation of traveling waves and this speed is usually denoted with the variable "v" or sometimes "c."

Furthermore, if the waveform of a traveling wave is repeated every time at specific intervals T, it is referred to as periodic wave.

Mathematically, the period of a traveling wave is given by the formula;

Period = \frac {1}{T}

Where;

T is the time measured in seconds.

5 0
2 years ago
Ball B is suspended from a cord of length l attached to cart A, which can roll freely on a frictionless, horizontal track. The b
Kazeer [188]

Answer:

a.Velocity of A and Velocity of B will be opposite in direction.

b.Both velocities have the same direction.

c. The two objects will move in a direction of some angle α.  

Explanation:

When the angle gets maximum the velocities of the two objects pull each other and move in opposite direction. But when the two objects make an angle zero degree with each other their motion is in the same direction.  

7 0
3 years ago
Because the time was the same for each segment you know the speed was the same for each segment
netineya [11]

Answer:

Market segmentation helps you send the right message, every time, by efficiently targeting specific groups of consumers.

Explanation:

3 0
2 years ago
A 4.00 m long, massless beam rests horizontally on a support 3.00 m from the left
Rus_ich [418]

If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Given the data in the question;

  • Length of the massless beam;L = 4.00m
  • Distance of support from the left end; x = 3.00m
  • First mass; m1 = 31.3 kg
  • Distance of beam from  the left end( m₁ is attached to ); x_1 = ?
  • Second mass; m_2 = 61.7 kg
  • Distance of beam from  the right of the support( m₂ is attached to ); x_1 = 0.273m

Now, since it is mentioned that the beam is in static equilibrium, the Net Torque on it about the support must be zero.

Hence, m_1g( x-x_1) = m_2gx_2

we divide both sides by g

m_1( x-x_1) = m_2x_2

Next, we make x_1, the subject of the formula

x_1 = x - [ \frac{m_2x_2}{m_1} ]

We substitute in our given values

x_1 = 3.00m - [ \frac{61.7kg\ * \ 0.273m}{31.3kg} ]

x_1 = 3.00m - 0.538m

x_1 = 2.46m

Therefore, If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Learn more; brainly.com/question/3882839

6 0
3 years ago
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