Someone is trying to balance a (110cm) plank with certain forces.



a) Calculate the moment of the 60N force (about O), then name its type.


b) Calculate the moment of the 30N force (about O), then name its type

c) Will the plank balance? If not, which way will it tip?

d) What extra force would be needed at (B) to balance the plank?

Answer:

I think it is A

Explanation:

It is A because if you calculate the weight minus the F, then you have to divide that by the amount of absorbation (the lb) before timsing it by 1000.

To find the horizontal and vertical components of the tension force in para sailing, we can use the basic trigonometric ratios of sine, cosine and tangent.

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The formula for the horizontal component (Fx), which is the angle the force makes with the horizontal, is Ftens•cosine(). Ftens• sine() can be used to compute the upward component (Fy), where is the angle that the force produces with the horizontal.

The portion of a force that works horizontally with regard to a body is known as its horizontal component. Its vertical component is known as its vertical component.

The pulling of the box is with the rope horizontal. When the rope is angled Because Is less than, the horizontal component (= F cos ) is less. As a result, it is the proper response.

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Answer:

1. c

2. d

3. a

4. d

5. c

Explanation:

im pretty sure im right

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Yes, the skater is accelerating.

Acceleration is a change in speed OR direction OR both.  Since she's skating in a circle, her direction is constantly changing, so she is accelerating.

At A, her velocity points down.

At B, her velocity points right.

At C, her velocity points up.

At D, her velocity points left.

Answer:

  B. Digital signals are more secure

Explanation:

You want to know an advantage of transmitting information using digital signals.

Digital signals are characterized by having a few discrete states. In the simplest case, those states are {transmitting, not transmitting}. Modern digital signals are often transmitted using numerous combinations of phases, frequencies, and amplitudes.

Analog signals are characterized by a continuum of the characteristic used to convey the information. When amplitude modulation is used, the information of interest is conveyed by the amplitude of the signal. Similarly, when frequency or phase modulation is used, the information is conveyed by the frequency or phase of the signal, respectively.

That is, analog signals have a continuous range of values.

Any signal that is broadcast will have a signal level that decreases with distance from the source of the broadcast. Any signal conveyed through a medium will be subjected to distortions and losses due to the interaction of the signal and the medium. Hence all signals become weaker as the distance from the source increases.

The receiver of a signal will, in general, receive signals other than the one intended. These are referred to as "noise." A transmission channel may have many sources of noise, including other transmitters, Cosmic Background Radiation, thermal noise in the components of the receiver or transmission medium, and others.

For analog signals, noise can directly affect the amplitude or phase being used to represent the signal of interest. Hence it changes the received value so it is not the same as the transmitted value.

The states of digital signals are designed to be sufficiently far apart that any corruption by noise will not affect the value as interpreted by the receiver. Thus, in terms of corruption of the information, we can say digital signals are more secure.

__

Additional comment

In radio transmission, noise is generally an addition to the signal. The weaker the analog signal, the greater the effect of noise. Broadcast radio began as AM radio: amplitude modulated. As the signal weakens, noise becomes a greater portion of the amplitude, so the receiver hears "static" where they want to hear voice or music.

FM, frequency modulated, radio replaced AM radio in order to avoid this issue with noise. The frequency of a signal is less likely to be corrupted. However, as the signal weakens, it can be difficult to determine its frequency, so that method of analog transmission also succumbs to noise.

Digital transmission in a noisy channel can also be corrupted by noise. Ultimately, the rate of information transmission in a given channel depends on the channel bandwidth and the signal-to-noise ratio. The integrity of a signal can be improved through coding that allows missing or corrupted bits of information to be determined based on the other bits that are received.

"Security" of a transmission can have different meanings. Here, we understand it to refer to the integrity of the information being conveyed. It could also relate to the difficulty of intercepting the transmission, or the difficulty of understanding the meaning of the received signal. Digital signals may or may not be more "secure" in these other meanings of the word.

Explanation:

Fmax = Fc = mv^2/r

[tex]m \frac{ {v}^{2} }{r} = f\frac{}{max} \\ {v}^{2} = \frac{(0.50 \: m)(55 \: n)}{1.0 \: kg} = 27.5 \: \frac{ {m}^{2} }{ {s}^{2} } \\ v = 5.2 \: \frac{m}{s} [/tex]

If a longitudinal force exceeding 3000 N is applied. Young's modulus for that object equals DL = 0.0139m.

The greater the modulus, or the lesser elasticity strain that arises from the application of a given stress, the stronger that medium is. Young's modulus, sometimes referred to as mechanical performance, is a metric used to determine how stiff a material is. In other words, it is readily bent or stretched.

Young's modulus really measures a material's ability to withstand changes in length when compressed or under stress along its length. Divide the primary stresses by the strain to get Young's modulus, commonly known also as young's modulus.

Young's modulus Y= stress/strain

= (F/A) /(DL/L)

F=3000N,

A=0.54 cm²,

L=2.5cm,

Y=0.1 Gpa

0.1 * 10⁹ = (3000/(0.54* 10⁻⁴) (2.5 ⋆ 0.01/DL)

DL = 0.0139m

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Precipitation amount, storm frequency, and severity, storm duration, kind of precipitation, wind during storm, and wind during evaporation are the main precipitation and storm elements that influence interruption loss (Kenneth, 1996).

All procedures that stop rainfall from instantly reaching the soil are referred to as interceptions. The two main ways that vegetation intercepts water are by catching it on leaf surfaces throughout the canopy and by depositing litter on the ground.

Precipitation that is intercepted by leaves, plant branches, and the forest floor prevents it from reaching the soil and instead stays on the surface.

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Answer:

a)    F_net = 6.48 10⁻¹⁸ ( [tex]\frac{1}{x^2} + \frac{1}{(0.300-x)^2}[/tex] ),   b) x = 0.15 m

Explanation:

a) In this problem we use that the electric force is a vector, that charges of different signs attract and charges of the same sign repel.

The electric force is given by Coulomb's law

         F =[tex]k \frac{q_2q_2}{r^2}[/tex]

Since when we have the two negative charges they repel each other and when we fear one negative and the other positive attract each other, the forces point towards the same side, which is why they must be added.

          F_net= ∑ F = F₁ + F₂

let's locate a reference system in the load that is on the left side, the distances are

left side - electron       r₁ = x

right side -electron     r₂ = d-x

let's call the charge of the electron (q) and the fixed charge that has equal magnitude Q

we substitute

          F_net = k q Q  ( [tex]\frac{1}{r_1^2}+ \frac{1}{r_2^2}[/tex])

          F _net = kqQ  ( [tex]\frac{1}{x^2} + \frac{1}{(d-x)^2}[/tex] )

let's substitute the values

          F_net = 9 10⁹  1.6 10⁻¹⁹ 4.50 10⁻⁹ ( [tex]\frac{1}{x^2} + \frac{1}{(0.30-x)^2}[/tex] )

          F_net = 6.48 10⁻¹⁸ ( [tex]\frac{1}{x^2} + \frac{1}{(0.300-x)^2}[/tex] )

now we can substitute the value of x from 0.05 m to 0.25 m, the easiest way to do this is in a spreadsheet, in the table the values ​​of the distance (x) and the net force are given

x (m)        F (N)

0.05        27.0 10-16

0.10          8.10 10-16

0.15          5.76 10-16

0.20         8.10 10-16

0.25        27.0 10-16

b) in the adjoint we can see a graph of the force against the distance, it can be seen that it has the shape of a parabola with a minimum close to x = 0.15 m

Answer: Various forms or states of water are the parts of water cycle

Explanation:

The water cycle initiates with the evaporation of liquid water that can on the surface of earth or a part of water body (hydrosphere). The evaporated water becomes the part of water vapors. The water vapor is the gaseous form of water. The water vapors accumulate in the higher atmosphere. They condense and precipitate in the form of rain or liquid water or solid water in the form of snow that is the solid form of water.

The mass of copper heated from a temperature of 27°C to 95°C is 45.72 g.

Mass is the quantity of matter contained in a body, The units of mass are

To calculate the mass of the copper, we use the formula below.

Formula:

From the question,

Given:

From the question,

Given:

Substitute these values into equation 1

Hence, the mass of copper is 45.72 g.

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Answer:

Explanation:

Following are the three important ways by which a body can get charged. ⇒ By conduction :When an uncharged body is made in contact with a charged body, the charge flow into the non charged body and the body gets charged. ⇒ By rubbing: When a body is rubbed with another body, both of them gets charged.

Answer:

[tex]\boxed {\boxed {\sf C. \ The \ Mitochondria}}[/tex]

Explanation:

Let's examine each organelle and its function.

A. The Nucleus

This organelle is the control center. It regulates a cell's activities and processes. It also contains the DNA or genetic information.

B. The Cytoplasm

The cytoplasm is a jelly like liquid that fills the inside the cell. Organelles and molecules are suspended in it.

C. The Mitochondria

This is also called the powerhouse of the cell, because it produces energy! This is the site of cellular respiration, the process that turns oxygen and glucose into ATP (energy), water, and carbon dioxide.

D. The Cell Membrane

This covers and protects the entire cell. It keeps important substances in and allows them to enter the cell, while preventing harmful ones from entering.

Based on this information, we see that the mitochondria is the organelle producing energy for the cell and choice C is correct.

Answer:

Name and describe at least SIX engineering challenges that the Mars copter, Ingenuity, had to overcome in order to be built/to travel to Mars/to function on Mars. BE SPECIFIC with your answers!!

Answer:

total potential energy is 383.18.

Explanation:

It is because that when we hold something at the influence of gravity than work is done . Similarly we know that

Potential Energy is Equal to the product of m g and h.Here weight gives the product of mass and acceleration due to gravity.so

PE = mgh

= 112.7×3.4

=383.18 joule

Unit of energy is joule

The ranking of the voltages, V₁, V₂, V₃, and V₄ in the order of decreasing voltage value is; V₂ > V₃ > V₁ > V₄

A voltage is the difference in potential or pressure that pushes or causes the flow of electric current in a circuit or conducting loop.

The current flowing through the circuit can be found using the following formula;

[tex]I=\dfrac{V}{R}[/tex]

The formula for conductivity is presented as follows;

[tex]\sigma =\dfrac{l}{R\times A}[/tex]

Where;

σ = The conductivity of the material

l = The length

R = The resistance to electricity

A = The cross-sectional area of the conductor

Therefore;

[tex]R=\dfrac{V}{I}[/tex]

[tex]R=\dfrac{l}{A\times \sigma}[/tex]

[tex]\dfrac{V}{I} =\dfrac{l}{A\times \sigma}[/tex]

[tex]V=\dfrac{l \times I}{A\times \sigma}[/tex]

Based on the dimensions of the cylinders, obtained from a similar question, we get;

[tex]V_1=\dfrac{3\cdot L \times I}{\pi \cdot D^2\times \sigma} = 3\times \dfrac{ L \times I}{\pi \cdot D^2\times \sigma}[/tex]

[tex]V_2= \dfrac{2\cdot L \times I}{\pi \cdot \dfrac{D^2}{4} \times \sigma} = \dfrac{8\cdot L \times I}{\pi \cdot D^2\times \sigma} = 8\times \dfrac{ L \times I}{\pi \cdot D^2\times \sigma}[/tex]

[tex]V_3=\dfrac{L \times I}{\pi \cdot \dfrac{D^2}{4} \times \sigma} =\dfrac{4\cdot L \times I}{\pi \cdot D^2\times \sigma}= 4\times \dfrac{L \times I}{\pi \cdot D^2\times \sigma}[/tex]

[tex]V_4=\dfrac{L \times I}{\pi \cdot D^2\times \sigma} = 1\times \dfrac{ L \times I}{\pi \cdot D^2\times \sigma}[/tex]

Therefore;

V₁ = 3·V₄

V₂ = 8·V₄

V₃ = 4·V₄

The rank of the voltages in the order of decreasing value is therefore;

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Answer:

3m/s^-2

Explanation:

a=(v-u)/t where v is the final velocity at 28 m/s and u is the initial velocity at 10m/s and t is the time taken at 6 seconds

a=(28-10)/6 = 3

By changing the frequency of an EM wave, both the energy and phase of the wave is changing. The correct option is e.

Frequency is the number of times a wave passes through a particular point in one second.

The time interval to 1° of phase change is inversely proportional to the frequency.

There is a change in phase with changing frequency.

Therefore, by changing the frequency of an EM wave, the energy and phase of the wave is changing.

The correct option is e.

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Two forces are said to be balanced when their strengths are equal but their directions of action are opposing.

The forces are balanced if the pullers are exerting equal force but going in the opposite direction on either side of the rope.

Forces that are balanced can cancel one another out. The thing stays in place whenever there is a balanced force.

To lift, turn, move, open, close, push, pull, and many other things requires forces. You apply force to a ball when you throw it in order for it to travel through the air. An object can be under the influence of multiple forces at once.

Therefore, Two forces are said to be balanced when their strengths are equal but their directions of action are opposing.

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a) The magnitude of the force F on the −10 nC charge is  0.00134 Newton.

b) The direction of the force F on the −10 nC charge is 41.99° with horizontal line.

Electric force is the attractive or repulsive interaction between any two charged things.

a) the magnitude of the force F on the −10 nC charge = √{k (q₁Q/r₁²)²+(q₂Q/r₂²)²}

= kQ √( (q₁/r₁²)²+ (q₂/r₂²)²}

= 9×10⁹×10×10⁻⁹ √{ (1×10⁻⁹/(0.01)²)² +(10×10⁻⁹/(0.03)²)² }

= 0.00134 Newton.

b)  the direction of the force F on the −10 nC charge

= tan⁻¹( (1×10⁻⁹/(0.01)²) ÷(10×10⁻⁹/(0.03)²) )

= 41.99°

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work done =force /area

=500/50

power =work done * time taken

Answer:

The opposition offered by the conductor to the current flowing through it is known as resistance.

we know,

Resistance is directly proportional to the length of the conductor.Hence,the resistance offered by the long wire will be more than that of the short wire.

Also, Resistance is inversely proportional to the area of cross section of conductor.So,The Resistance in the thick wire will be less than that of the thin wire.

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The value of x is 0.04m and restoring force is 9 N

(a) From the figure, the spring force balances force due to gravity.

    So, [tex]\frac{1}{2} kX^{2} = mg[/tex]

    Where, k = spring constant = 50N/m

                 X = compressed distance = L+x = 0.1+x

                 m = mass = 50 g = 0.05Kg

                 g = acceleration due to gravity = 10

Putting these values in above equation, we get: x= 0.04m

(b) Using Hooke's law, F = kx

     Where, k = spring constant =50N/m

                  x = distance = 180mm = 0.18 m

    Putting these values in above equation, we get restoring force F = 9N

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Wind turbines, hydroelectric, and geothermal heat have been relied upon for decades to provide a renewable and clean energy source to people around the world. These alternative energy technologies provide optimal clean and renewable energy sources for many communities and countries striving for sustainable development.