pLEasE aNSWer CorRReCtly

Answer:

This question is incomplete

Explanation:

This question is incomplete because there are some parameters that are missing. However, when calculating the average emf and speed/velocity is provided, then

average emf = Brvl

where Br is the vertical component of the earth's magnetic field (in weber per square meter, Wb/m²)

v is velocity (in m/s)

l is the length (in meter) of the wing span

Answer:

a) letting in-flowing and/or out-flowing tide flow through turbines in a dam.

Explanation:

Tidal power makes use of the energy from the tidal force and wave action in order to generate electricity. it is a predictable source of energy.

Tidal Barrages

The system allows tides to enter, seawater flows via the dam and is trapped into the basin when the tides subside and system’s gates close.

When the tides start to move out, the gates in the dam are opened up this consist of turbines and water begins to flow out, hitting the turbines this eventually turn to produce energy.

Power is produced when the tidal range, which is the difference between low and high tide, has to be more than 5 meters.

Answer:

D

Explanation:

When light makes contact with any object it naturally slows down, and the more dense the object the slower it travels.

Answer:

The same as

Explanation:

First of all, they are in a merry go round, which is a circular one. And thus, we would be dealing with circular motions.

The angular velocity of a merry go round, in this instance, is given by

w = Δ/Δt

The formula has nothing whatsoever to do with distance, and as such, both Rasheed and Sofia would have the same angular velocity.

Transferring this further, the angular acceleration is given as

α = Δw/Δt

Remember I said the velocity has nothing to do with distance, well, so does the acceleration as we can see from the formula stated. And therefore, both Rasheed and Sofia would have the same angular acceleration

Answer: Multiply the time by the acceleration due to gravity to find the velocity when the object hits the ground. If it takes 9.9 seconds for the object to hit the ground, its velocity is (1.01 s)*(9.8 m/s^2), or 9.9 m/s.

Explanation:

The measurement of length and time is calculated the average speed with the expression

         [tex]v_{avg} = \frac{\Delta x}{\Delta t}[/tex]

Average velocity is defined as the change in displacement between the interval time

      [tex]v_{avg} = \frac{\Delta x}{\Delta t}[/tex]

where [tex]v_{avg}[/tex] is the average velocity, Δx is the displacement in the interval, and Δt the time.

To carry out this measurement, it must be taken into account that the velocity is a vector, for which we define a reference system, it can be a vertical axis with the upward direction as positive, therefore the downward displacement is negative.

A tape measure is placed and the interval during which the scalar distance measurement is performed is defined, the most as is to use time as an independent variable (controlled by the researcher) that is measured with a stopwatch.

The displacement is measured for each given time interval and these two quantities are divided. The sign of the velocity is negative because it goes down

Using the measurements of length and time, the average velocity is calculated with the expression

         [tex]v_{avg} = \frac{\Delta x}{\Delta t}[/tex]t

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Answer:

Following are the solution to this question

Explanation:

please find the complete question in the attached file.

In point a:

The answer is "bottom".

In point b:

Using formula:

[tex]T= mg + \frac{m V^2}{2}[/tex]

[tex]\to 100= 0.2 \times 9.81 + \frac{0.2 \times V^2}{0.5}[/tex]

[tex]\to 100= 1.962+ \frac{0.2 + V^2}{0.5}\\\\\to 100- 1.962= \frac{0.2 + V^2}{0.5}\\\\\to 98.038= \frac{0.2 + V^2}{0.5}\\\\\to 49.019=0.2+V^2\\\\\to 48.819=V^2\\\\ \to 6.987 \ \frac{m}{s}[/tex]

Answer:

The combined force of the catapult and engines is 44545.455 newtons.

Explanation:

Let suppose that platform of the aircraft carrier is horizontal, such that changes in gravitational potential energy can be neglected. In addition, effects of non-conservative forces are neglected. By using Principle of Energy Conservation and Work-Energy Theorem we find that steam catapult-jet aircraft is represented by:

[tex]K_{1}+W_{1\rightarrow 2} = K_{2}[/tex] (1)

Where:

[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Initial and final translational kinetic energies of the jet fighter, measured in joules.

[tex]W_{1\rightarrow 2}[/tex] - Work done on the jet fighter due to the combined force of the catapult and engines, measured in joules.

By applying definitions of translational kinetic energy and work, we expand and simplify the equation above as follows:

[tex]F\cdot \Delta s = \frac{1}{2}\cdot m\cdot (v_{2}^{2}-v_{1}^{2})[/tex]

[tex]F = \frac{m\cdot (v_{2}^{2}-v_{1}^{2})}{2\cdot \Delta s}[/tex] (2)

Where:

[tex]F[/tex] - Combined force of the catapult and engines, measured in newtons.

[tex]m[/tex] - Mass of the jet fighter, measured in kilograms.

[tex]\Delta s[/tex] - Length of the catapult, measured in meters.

[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final speeds of the jet aircraft, measured in meters per second.

If we know that [tex]m = 4000\,kg[/tex], [tex]v_{1} = 0\,\frac{m}{s}[/tex], [tex]v_{2} = 70\,\frac{m}{s}[/tex] and [tex]\Delta s = 220\,m[/tex], then the combined force of the catapult and engines is:

[tex]F = \frac{(4000\,kg)\cdot \left[\left(70\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]}{2\cdot (220\,m)}[/tex]

[tex]F = 44545.455\,N[/tex]

The combined force of the catapult and engines is 44545.455 newtons.

The combined force of the catapult and engines is 44,545.5 N.

The given parameters;

The combined force of the catapult and engines is determined by applying work energy theorem as shown below;

[tex]W = \Delta E\\\\Fd = \frac{1}{2}m(v^2 -u^2) \\\\220F = \frac{1}{2}(4000)(70^2 - 0^2) \ \\\\220 F = 9,800,000\\\\F = \frac{9,800,000}{220} \\\\F = 44,545.5 \ N[/tex]

Thus, the combined force of the catapult and engines is 44,545.5 N.

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Answer:

Vertical velocity = 8.28 m/s

Horizontal velocity = 28.3 m/s

__________________________________________________________

We are Given:

Initial velocity of the ball = 40 m/s

Angle of elevation of the ball = 45°

Time interval = 2 seconds

__________________________________________________________

Vertical and Horizontal Components:

Vertical Component:

The vertical component of the ball can also be written as:

40*Sin(45°)

40*(1/√2)                                                          [since Sin(45°) = 1/√2]

40/√2 m/s

Horizontal Component:

The horizontal component of the ball can also be written as:

40*Cos(45°)

40*(1/√2)                                               [Since Cos(45°) = 1/√2]

40/√2 m/s

__________________________________________________________

Speed of the Ball after 2 seconds:

Horizontal Velocity:

Since the horizontal velocity is not affected by any force during the flight, the horizontal velocity after 2 seconds will also be 40/√2 m/s

Since √2 = 1.41:

Horizontal velocity = 28.3 m/s

Vertical Velocity:

v = u + at                            [First equation of motion]

v = 40/√2 + (-10)(2)          

[g is negative because it is acting against the direction of motion]

v = 40/√2 - 20

v = [tex]\frac{40-20\sqrt{2} }{\sqrt{2} }[/tex]                          [taking the LCM]

v = 8.28 m/s                    [√2 = 1.41]

Answer: C

Explanation:

The characteristics and problems of telescopes we can find the correct answer to the question that the Hubble telescope helps to answer is:

           D) Are there more galaxies beyond the milky way

Telescopes are instruments that can perceive very small amounts of light and have very large magnifications, in general to be able to see objects increasingly weak in intensity, telescopes become larger and the observation times of the object increase.

When we increase the observation times, the problem is that atmospheric movement changes the observation conditions and makes it difficult to obtain good images. One way to solve the problem was with telescopes in near space, the environmental disturbance is eliminated.

When the Hubble telescope was placed in space, there were already many answers but they continue to wonder what the limit of the universe is, this telescope allowed us to observe a body much farther away than ground-based telescopes, which is why it answered the question how many and to where there are galaxies out of the milky way.

Let's review the statements given to see the correct ones;

A) False. Copernicus clarified that the Sun is the center of the solar system

B) False. The center of the universe is not known, but the sun is a star in one of the arms of the milky way that is not the center of the universe either.

C) False. It has been known for a century what the movement of the planets in the solar system

D) True. The telescope in space allowed us to expand the limit of vision of the universe and respond to where there are galaxies outside the Milky Way.

In conclusion with the characteristics and problems of telescopes we can find the correct answer to the question that the Hubble telescope helped to answer is:

      D) Are there more galaxies beyond the milky way

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Answer:

f= ma

Explanation:

v-u/ t

30-0/10

3m/s

maas 1500

1500*3

4500

Answer:

.....

Explanation:

Answer:

The answer is "[tex]5.06 \times 10^{-6} \ kg \ m^2[/tex]"

Explanation:

[tex]\to E_1=0..............(i)\\\\\to E_2= \frac{mV^2}{2} +\frac{Iw^2}{2} - mgh.............(ii)\\\\ \Delta E=0\\\\\to mgh= \frac{mV^2}{2} +\frac{Iw^2}{2} \\\\ \to 2 \ mgh= mV^2 +Iw^2\\\\ \to 2 \ mgh- mV^2 =Iw^2\\\\ \to m(2gh- V^2) =Iw^2\\\\ \to I= \frac{m(2gh- V^2)}{w^2}[/tex]

       [tex]= 5.06 \times 10^{-6} \ kg \ m^2[/tex]

Answer:

Non-zero digits

Explanation:

umm (ꏿ﹏ꏿ;)

Answer:

There's three rules on determining how many important quantities are in a number: Non-zero digits are always predominant. Any zeros between two significant digits are always major. The final zero or dangling zeros in the decimal segment are really only important.

I hope this answered your question, have a nice day!

Answer:

The answer is D. In the direction of the wave

Explanation:

I took the test, hope this helps! :)

Answer:

A

Explanation:

The image formed by the concave mirror is always virtual. An enlarged image is caused when the concave mirror is positioned too near to the object.

When a hollow spherical is divided into pieces and the exterior surface of each cut portion is painted, it forms a mirror, with the inner surface reflecting the light.

A concave mirror is a name for this sort of mirror. An enlarged and fake image is caused when the concave mirror is positioned too near to the object.

When the distance between the item and the mirror is increased, however, the size of the image decreases, and a true picture is generated.

The picture created by the concave mirror may be manipulated. The image formed by the concave mirror is always virtual.

Hence, option A is correct.

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Answer:

Shape and position

Explanation:

Hope this helps! :)

Answer:

i didnt know ther was differnt types

Explanation:

Answer:

a. Please see the attached graph and drawing combined created with Microsoft Excel

b. Approximately 3.2217 m/s

c. Approximately 14.0 m/s  

d. Approximately 11.05 meters

Explanation:

a. The given parameters are;

The height of the wall, h = 10.0 m

The distance from the base of the wall the car lands = 4.60 m

The time, t, it takes the car to land is given by the equation for free fall as follows;

h = 1/2·g·t²

Where;

g = The acceleration due to gravity = 9.81 m/s²

From the equation for free fall, we have;

h/(1/2·g) = t²

∴ t = √(h/(1/2·g) ) = √(10/(1/2·9.81) ) ≈ 1.4278

The time it takes the car to land, t ≈ 1.4278 seconds

b. The horizontal speed of the car = Horizontal distance/(Time) = 4.6/1.4278 ≈ 3.2217 m/s

The horizontal speed of the car before it drives off the wall  ≈ 3.2217 m/s

c. The car's vertical speed just before impact is given by the following equation;

v = u + gt

Where;

u = The initial vertical speed = 0

t = The time it takes before impact ≈ 1.4278 seconds

∴ v = 9.81 × 1.4278 ≈ 14.0 m/s  

The car's vertical speed just before impact, v ≈ 14.0 m/s  

d. Whereby the car is subject to the gravitational field of the moon, we have;

Gravitational force per kilogram = 1.7 N/kg

∴ Gravitational acceleration = 1.7 m/s².

The time it takes the car to land whilst subject to the gravitational field of the moon is therefore;

[tex]t_{Moon}[/tex] = √(h/(1/2·g)) = √(10/(1/2 × 1.7)) ≈ 3.43

[tex]t_{Moon}[/tex] ≈ 3.43 seconds

The horizontal distance covered, at the car's horizontal speed in the time of free fall ≈ 3.2217 m/s ×  3.43 seconds ≈ 11.05 meters

The horizontal distance covered, at the car's horizontal speed in the time of free fall  = The distance the car will land from the base of the wall ≈ 11.05 meters

The distance the car will land from the base of the wall ≈ 11.05 meters.

Answer:

density of the sediment, size of the sediment, and velocity of the agent of erosion.

Explanation:

Have a nice day!

Answer: b., c., e.

Explanation:

Answer:

T = 12732 [N]

Explanation:

To solve this problem we must use newton's second law which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

By means of the attached free body diagram, we can better understand the solution to this problem.

As we see in the diagram the tension Force T lifts the car upwards so the movement is upwards. Therefore we have two forces one upward (positive) of the force T and the other negative downward due to the weight of the vehicle.

As the movement is up the acceleration is also up (with a positive sign).

ΣF = m*a

where:

F = forces [N] (units of Newtons)

m = mass = 1200 [kg]

a = acceleration = 0.8 [m/s²]

T - 1200*9.81 = 1200*(0.8)

T = 12732 [N]

Answer:

12,720

Explanation:

From the graph, it is clear that, the velocity is at a time of 1 s is highest. The velocity at 1 second corresponds to 1250 km/hr. Then it decreases with time.

The velocity - time graph shows the change in velocity with respect to time. The velocity is placed in y -axis and time is given in x - axis. The slope of the curve in velocity - time graph gives the acceleration of the object.

Similarly, the position of the object in meter after a t seconds can be determined from the velocity - time graph. It is the rate of change in velocity of the object.

From the graph, it is clear that, the curve has its peak at 1 second. After that the peak descends down. Hence, the maximum velocity of the car is at a time of 1 second at which the velocity is 1250 km/hr.

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Answer:

2.51 m/s

Explanation:

Given that,

Angular speed of a fan, [tex]\omega=0.5\ rev/s = 3.14\ rad/s[/tex]

Length of the fan arm, r = 0.8 m

We need to find the speed of a point on the end of the fan arm at this time. Let v is the speed. It is given in terms of angular speed is given by :

[tex]v=r\omega\\\\v=0.8\times 3.14\\\\v=2.51\ m/s[/tex]

So, the required speed is 2.51 m/s.

Answer:

The answer is the 4th one

Explanation:

Good luck man

Answer:

B. soft tissue injuries

Explanation:

i took the test on edge and its really the only one that would make sense if you payed attention to the lesson  :)

Answer:

32.67km

Explanation:

~= round off

11.2km/h~3.11m/s

175minutes=10500s

10500*3.11~32, 666.67m

=32.66667km

~ 32.67km

AnswerAmontons's law. If the temperature is increased, the average speed and kinetic energy of the gas molecules increase. ... If the gas volume is decreased, the container wall area decreases and the molecule-wall collision frequency increases, both of which increase the pressure exerted by the gas (Figure 1).:

Explanation:

Answer:

The value is  [tex]H = 0.036 \ kg/m^3[/tex]

Explanation:

From the question we are told that

   The saturation density of air is [tex]\rho = 0.04 \ kg/m^3[/tex]

   The temperature for this saturation density to occur  [tex]T_s = 96^o F[/tex]

    The relative humidity at this temperature is  [tex]R_h = 90\% = 0.90[/tex]

Generally the relative humidity is mathematically represented as

      [tex]R_H = \frac{H}{\rho}[/tex]

Here H represents humidity

=> [tex]0.90 = \frac{H}{0.04}[/tex]

=>  [tex]H = 0.90 * 0.04[/tex]

=>  [tex]H = 0.036 \ kg/m^3[/tex]