list 2 application of convertional current​

Answer:

F = 60 N

Explanation:

By the Newton's second law, the force is equal to mass times acceleration:

F = m•a

F = 5•12

F = 60 N

Answer:

22.1meter/min

Explanation:

Change 488s to min.

Speed= Distance÷Time

so 180m ÷ 122/15

= 22.1 (3s.f.) meter / min.

Answer:

2.5 Hz

Explanation:

3.0m/s / 1.2 m=2.5 1/s=2/5 Hz

The frequency of a wave with speed of 3.0 m/s and has a wavelength of 1.2m is 2.5 Hz.

The frequency of a wave is the number of cycles per second.

Given the wavelength is 1.2 m and speed v =3.0m/s, then the frequency of the wave is

v =fλ

f = 3 / 1.2

f = 2.5 Hz

Thus, the frequency of the wave is 2.5 Hz.

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Answer:

The momentum of an object can be calculated using the formula: momentum = mass x velocity. So, in this case, the momentum of the truck would be:

momentum = 2,000 kg x 35 m/s = 70,000 kg·m/s

So the momentum of the truck is 70,000 kg·m/s if it is moving with a velocity of 35 m/s.

Answer:   1470 N

Explanation:   mass = 15

                   weight is the product of mass and force due to gravity

                     AS WE KNOW THAT

                        W = mG

                        W = 15 X 9.8

                         W =  1470 N

Explanation:

Formula:

F*255.92778=K

83 F *255.92778=301.48k

therefore the answer is:

301.48K

The speed of the mass is approximately 0.942 m/s.

The numerical value of the centripetal acceleration is approximately 0.591 m/s².

To find the speed of the mass, we need to know the circumference of the circular path it follows.

The circumference of a circle can be calculated using the formula:

Circumference = 2 * π * radius

Given that the mass is located 1.5 meters from the axis of rotation, the radius (r) is 1.5 meters.

Circumference = 2 * π * 1.5

Circumference ≈ 9.42 meters

Since the mass makes one revolution in 10 seconds, we can calculate its speed by dividing the circumference by the time taken:

Speed = Circumference / Time

Speed = 9.42 meters / 10 seconds

Speed ≈ 0.942 meters per second (m/s)

Therefore, the speed of the mass is approximately 0.942 m/s.

Now, let's calculate the centripetal acceleration.

The centripetal acceleration (ac) can be calculated using the formula:

ac = (v²) / r

Given:

m (mass) = 0.1 kg

r (radius) = 1.5 meters

v (velocity) = 0.942 m/s

Substituting the values into the formula, we can calculate the centripetal acceleration:

ac = (v²) / r

ac = (0.942 m/s)² / 1.5 meters

ac ≈ 0.591 m/s²

Therefore, the numerical value of the centripetal acceleration is approximately 0.591 m/s².

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True, due to the protective nature of the ozone layer and the absorption of UV radiation, conducting ultraviolet astronomy from space is necessary to gather accurate and detailed data about celestial objects and phenomena emitting UV light.

The ozone layer plays a crucial role in protecting the Earth's surface from harmful ultraviolet (UV) radiation. It absorbs a significant portion of the Sun's UV rays, preventing them from reaching the surface and potentially causing damage to living organisms.

As a result, to observe and study ultraviolet astronomy, which involves the detection and analysis of celestial objects and phenomena emitting UV radiation, observations must be conducted from space. By observing from space, astronomers can bypass the Earth's atmosphere, including the ozone layer, and directly capture the UV light from celestial sources.

UV astronomy satellites and telescopes, such as the Hubble Space Telescope and the International Ultraviolet Explorer (IUE), have been specifically designed and deployed to conduct observations in the ultraviolet spectrum. These instruments are situated above the Earth's atmosphere, enabling them to capture and analyze UV radiation without interference from atmospheric absorption.

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Answer:

the formula v = f×lambda.

v= 4.25× 5.6

therefore speed is

23.8 meters per second

The top reaches an angular velocity of 28.28 rad/s when the string is fully unwound.

We can use the kinematic equation of rotational motion to solve this problem. The third kinematic equation relates the final angular velocity of a rotating object to its initial angular velocity, angular acceleration, and the angle through which it rotates:

\(\omega_f^2 = \omega_i^2 + 2\alpha\theta\)

where θ is the angle through which the object rotates.

The number of turns the body turns to unwind the string is calculated as,

\(n=L/(2\pi r)\)

\(n=80/(2\pi \times 2)\)

\(n=6.366\)

Therefore the angle turned by the body in unwinding the string is calculated as,

\(\theta = 2\pi \times 6.366 \ rad\)

\(\theta = 39.99\  radians\)

The initial angular velocity of the top is zero, the angular acceleration is \(10 \ rad/s^2\), and we need to find the final angular velocity when the string is completely unwound.

Substituting the values given in the problem into the equation, we get:

\(\omega_f^2 = 0 + 2\times 10 \times 39.99\)

Simplifying, we get:

\(\omega _f = 28.28 \ rad/s\)

Therefore, the final angular velocity of the top when the string is completely unwound is 28.28 rad/s.

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Answer:

Im sorry but i really need these points. FORGIVE ME

Explanation:

Answer:

Valence Electrons

Explanation:

The electrons of an atom that can participate in the formation of chemical bonds with other atoms. they are the furthest electrons from nucleus

Answer:

No it can not

Explanation: Kinetic energy is the energy of motion so it can not be negative the kinetic energy can only be at a point of "0" which is when its not moving. (I hope this helped) :))

Answer:

emotions

Explanation:

emotions are how you feel and can happen any time

Hope it helps <333

Answer:

emotions affect your body and how you feel

Answer:

C

Explanation:

Gravity from earth compress buried sediment which form rock......

Answer:

because mars is far

Explanation:

so if he shouts louder than he will hear it quicker

Answer:

d = 100.56 m

Explanation:

It is given that,

Force acting on the dog is 22.4 N

Angle at which the force is applied is 115 º

Work done is - 42.3 J

We need to find the distance covered by the dog. The work done by an object is given by :

\(W=Fd\cos\theta\)

d = distance covered by the dog

\(d=\dfrac{W}{\cos\theta}\\\\d=\dfrac{-42.5}{\cos(115)}\\\\d=100.56\ m\)

So, the dog moves a distance of 100.56 m.

Answer:

4.47 M.

Explanation:

I do Acellus and just got it correct!

The magnitude of the momentum of a 5 kg object moving with a velocity of v = (-6 m/s) + (8 m/s)ý is 50 kg m/s. To find the magnitude of the momentum, we need to use the formula p = m*v, where p is the momentum, m is the mass, and v is the velocity.
In this case, we have m = 5 kg and v = (-6 m/s) + (8 m/s)ý.
First, we need to find the magnitude of the velocity vector. This can be done using the Pythagorean theorem:
|v| = √((-6 m/s)² + (8 m/s)²) = √(36 + 64) = √100 = 10 m/s
Now, we can plug in the values into the formula:
p = m*v = (5 kg)*(10 m/s) = 50 kg m/s
Therefore, the magnitude of the momentum is 50 kg m/s, which corresponds to option c.

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The magnitude of the momentum of the object is 50 kg m/s. Here option C is the correct answer.

The momentum of an object is defined as the product of its mass and velocity. Mathematically, it can be represented as:

p = mv

where p is the momentum, m is the mass, and v is the velocity.

In this problem, the object has a mass of 5 kg and is moving with a velocity of v = (-6 m/s) + (8 m/s)ý. We can calculate the magnitude of the momentum by finding the magnitude of the velocity vector and multiplying it by the mass of the object.

The magnitude of the velocity vector:

|v| = sqrt((-6 m/s)^2 + (8 m/s)^2) = sqrt(100) = 10 m/s

Therefore, the magnitude of the velocity vector is 10 m/s.

Momentum:

p = mv = (5 kg) x (10 m/s) = 50 kg m/s

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a) Using Kepler's third law, the orbital period of planet B located at 1.0 AU from the sun can be calculated. b) Given an orbital period of 4 years for planet C, we can determine its average distance from the sun.

Kepler's third law states that the square of the orbital period (P) of a planet is proportional to the cube of its average distance (a) from the sun. Mathematically, it can be expressed as:

\(\[P^2 = a^3\]\)

Given that planet B is located at an average distance of 1.0 AU from the sun, we can substitute this value into the equation to solve for P:

\(\[P^2 = (1.0 \, \text{AU})^3\]\)

Taking the square root of both sides, we find:

\(\[P = \sqrt{(1.0 \, \text{AU})^3}\]\)

Evaluating the expression, we get:

\(\[P \approx 1.0 \, \text{year}\]\)

Therefore, the orbital period of planet B is approximately 1.0 year.

Similarly, using Kepler's third law, we can solve for the average distance (a) of planet C from the sun. We have the equation:

\(\[P^2 = a^3\]\)

Given an orbital period (P) of 4 years, we can substitute this value into the equation to solve for a:

\(\[(4 \, \text{years})^2 = a^3\]\)

Simplifying, we get:

\(\[16 \, \text{years}^2 = a^3\]\)

Taking the cube root of both sides, we find:

\(\[a = \sqrt[3]{16 \, \text{years}^2}\]\)

Evaluating the expression, we get:

\(\[a \approx 2.52 \, \text{AU}\]\)

Therefore, if planet C has an orbital period of 4 years, its average distance from the sun is approximately 2.52 AU.

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Answer:

Explanation:idek

The net force acting on the body will be 250 N since there is significant friction also. Thus, the acceleration of the 50 Kg body is 5 m/s².

Friction is a kind of force acting on a body to resist it from motion. Thus, frictional force will always be negative. The frictional force acting on a body is the the product of the coefficient of friction and the normal force.

Here the normal force = 1250 N

coefficient of friction = 1.2

Frictional force = 1250 N × 1.2 = 1500 N

Net force = 1500 N - 1250 N = 250 N

Acceleration = net force/ mass

                     = 250 N/50 Kg

                     = 5 m/s²

Therefore, the acceleration of the body from rest is 5 m/s².

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Answer:

no i did not observe anything

Explanation:

When you are standing in front of a flat mirror and focus a camera on your image, the camera will focus at a distance equal to the distance between the mirror and the object from the camera. Here, the camera focuses on the image of the person standing in front of the mirror.

When you are standing in front of a flat mirror and focus a camera on your image, the camera will focus at a distance equal to the distance between the mirror and the object from the camera. Here, the camera focuses on the image of the person standing in front of the mirror. Therefore, if the camera is in focus when set for a distance of 5.56 m, the object is also 5.56 m away from the mirror, as the image and the object are at the same distance from the mirror.

Image:
The image shows the object and its image reflected by a flat mirror. The distance between the mirror and the object, which is equal to the distance between the mirror and its image, is represented by "d." To find the distance between the person and the mirror, we have to take the distance between the person and the image into account. Since the image is the same distance behind the mirror as the person is in front of the mirror, the total distance between the person and the mirror is twice the distance between the person and the image. The total distance "D" is given by:

D = 2dwhere d = 5.56 m

Thus, the person is standing at a distance of D = 2 x 5.56 = 11.12 m from the mirror. Therefore, the distance between the person and the mirror is 11.12 m. So, the answer is 11.12 m.  In conclusion, the distance between the person and the mirror is 11.12 m.

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When viewed above the North Pole, the Earth rotates counterclockwise, from west to east. This is also called a prograde rotation.

The Earth rotates around the Sun-also counterclockwise viewed from above the North Pole, clockwise observe from the South Pole-to give us our years. The plane of the Earth's orbit is called the sphere where aureole happens. Because of this direction of rotation, we see the sun rising every day in the east and position in the west.

As we viewed above the earth revolves empty around the sun each year when viewed from above the north pole. We know that the Earth is viewed once in twenty-four hours.

So we can conclude that Earth when seen from the top of the north pole, will be seen revolve in anti-clockwise.

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Answer:

The time required to come to a stop will increase by a factor of four. This is because the acceleration is constant, and the stopping distance is proportional to the square of the speed, so doubling the speed quadruples the stopping distance.

The correct nozzle to use with medium-expansion foam is a **medium expansion foam nozzle**.

Medium-expansion foam is a type of fire-suppressing foam that expands to a moderate volume, typically 20 to 200 times its original liquid volume. It is commonly used in firefighting scenarios where a balance between suppression effectiveness and foam coverage is desired.

To properly apply medium-expansion foam, a dedicated medium-expansion foam nozzle is used. This specialized nozzle is designed to deliver the foam solution at the correct flow rate and generate the desired expansion ratio. It is typically equipped with adjustable settings to control the foam application, such as flow rate and expansion ratio.

The medium-expansion foam nozzle is different from other nozzles, such as low-expansion foam nozzles or high-expansion foam generators, which are used for different types of foam applications. Using the correct nozzle ensures that the foam is produced and deployed effectively, providing optimal fire suppression capabilities and coverage.

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Given:

• Acceleration of rocket = 28.0 m/s²

• Mass of astronaut = 95 kg

Let's find the apparent weight of the astranaut.

In a direct rocket motion, as we are coming towards the Earth the weight reduces while as the rocket is going up, you feel more weighted.

Now, to find the apparent weight of the astranaut, apply the formula:

Where:

Aw is the apparent weight

m is the mass = 95 kg

g is acceleration due to gravity = 9.8 m/s²

a is the acceleration of the rocket = 28.0 m/s²

Thus, we have:

Therefore, the apparent weight of the astranaut is 3.591 kN

ANSWER:

3.591 kN