A mass of 10. kg is placed on the end of a 0.50-meter pendulum. What is the period of the pendulum?

Answer:

C) Momentum

Explanation:

Refers to an objects mass in motion.

Answer:

C . Magnetic dipoles is the correct

Answer:

b). movement of charged particles.

Explanation:

These charges create the nagnetic dipoles.

The image distance for an object distance of 19.8 cm is -10.5 cm.

The image distance for an object distance of 6.8 cm is -5.8 cm.

A diverging lens is a lens that is thinner in the middle than at the edges. When light passes through a diverging lens, it spreads out and diverges, making objects appear smaller. The focal length of a lens is the distance between the center of the lens and the point where light rays converge or diverge. In the case of a diverging lens, the focal length is negative. Part 1: To find the image distance for an object distance of 19.8 cm, we can use the formula: 1/f = 1/do + 1/di where f is the focal length, do is the object distance, and di is the image distance. Substituting the given values, we get: 1/(-16.6) = 1/19.8 + 1/di Solving for di, we get: di = -10.5 cm Therefore, the image distance for an object distance of 19.8 cm is -10.5 cm. Part 2: To find the image distance for an object distance of 6.8 cm, we can use the same formula as before: 1/(-16.6) = 1/6.8 + 1/di Substituting the given values, we get: di = -5.8 cm Therefore, the image distance for an object distance of 6.8 cm is -5.8 cm. It is important to note that the negative sign in the image distance indicates that the image is virtual, upright, and smaller than the object.

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Answer:

Final velocity = 10 m/s

Time taken to travel 100 meter = 8.16 second (Approx.)

Explanation:

Given:

Acceleration = 2 m/s²

Initial velocity = 0 m/s

Find:

Velocity after 5 seconds

Time taken to travel 100 meter

Computation:

Using first equation of motion

v = u + at

v = 0 + (2)(5)

Final velocity = 10 m/s

Using Second equation of motion

s = ut + (1/2)(a)(t²)

100 = (0)(t) + (1/2)(3)(t²)

100 = (1/2)(2)(t²)

100  = (t²)

t = 10

Time taken to travel 100 meter = 8.16 second (Approx.)

The models that predict galaxies should be getting farther apart now are; accelerating model, critical model,- and coasting model.

Observations show that the expansion of the universe is accelerating, such that the velocity at which a distant galaxy recedes from the observer is continuously increasing with time.

The accelerating model of the galaxies predicts that galaxies should be getting farther apart now.

From the model presented in the graph critical model of the galaxies also predict that galaxies should be getting farther apart now.

Another model that predicts that galaxies should be getting farther apart now is coasting model.

These models can be seen in the graph as they increase proportional with time. That is the distance between galaxies increase with increase in time measured in years.

Thus, we can conclude that there  models ( accelerating, critical and coasting model) predicts the relative position of the galaxies from us.

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Answer:

Greater.

Explanation:

This pressure difference will be greater if the scuba diver were in seawater and went to the same depth because the seawater have salts which increases the density of water as compared to freshwater. Salt in water increases the density which automatically increases the pressure on the diver so that's why we can say that the pressure will be increases for the scuba diver in seawater as compared to freshwater.

a) To find the unit vector r^ that points from flan 2 to flan 1, we need to subtract the position vector of flan 2 from that of flan 1 and then divide by the magnitude of the resulting vector:

r→ = r→ 1 - r→ 2 = 〈2.0∗105 - 9.0∗105, 4.0∗105 - 4.0∗105, 0 - 0〉 = 〈-7.0∗105, 0, 0〉

| r→ | = sqrt((-7.0∗10^5)^2 + 0^2 + 0^2) = 7.0∗10^5 m

r^ = r→ / | r→ | = 〈-1, 0, 0〉

b) The magnitude of the gravitational force on flan 2 due to flan 1 is given by the formula:

F = G * m1 * m2 / r^2

where G is the gravitational constant (6.6743 × 10^-11 Nm^2/kg^2), m1 and m2 are the masses of flan 1 and flan 2 respectively, and r^ is the unit vector pointing from flan 2 to flan 1. Substituting the given values, we get:

F = 6.6743 × 10^-11 * 5.0×10^21 * 2.0×10^21 / (7.0×10^5)^2 ≈ 2.23 × 10^18 N

c) To find the gravitational force vector on flan 2 due to flan 1, we multiply the magnitude F by the unit vector r^:

F→ = F * r^ = (2.23 × 10^18) * 〈-1, 0, 0〉 = 〈-2.23 × 10^18, 0, 0〉 N

d) To predict the velocity of flan 2 in one time step Δt=14400 seconds later, we can use the formula for the gravitational force between two objects:

F = m * a

where F is the gravitational force, m is the mass of the object, and a is its acceleration. We can solve for the acceleration of flan 2:

a = F / m2 = (-2.23 × 10^18) / (2.0×10^21) = -1.115 m/s^2

The negative sign indicates that the acceleration is in the opposite direction of the unit vector r^. Using the formula for constant acceleration:

v = v0 + a * Δt

where v0 is the initial velocity (which is zero in this case), we can find the velocity of flan 2 after one time step:

v = 0 + (-1.115) * 14400 = -16056 m/s

Again, the negative sign indicates that the velocity is in the opposite direction of the unit vector r^.

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Explanation:

you move 10 -³ up

so: 100* 10³ /5.5*10

100000/55

which is approximately 18182

Answer:

200 meters per minute

Explanation:

You run the 1600 meters in a total of 8 minutes, so the average speed if 200 meters per minute.

Answer:

lamp oil.

Explanation: the ping pong ball doesnt sink through the lamp oil.

Answer:

Explanation:

Using the equation of motion v² = u²+2aS

v is the final velocity = 120km/hr

120km/hr = 120 * 1000/1 * 3600 = 33.3m/s

u is the initial velocity = 0m/s

a is the acceleration

S is the distance covered = 240m

On substituting the given parameters

33.3² = 0²+2a(240)

33.3² = 480a

1110 = 480a

a = 1110/480

a = 2.3125m/s²

Hence the minimum constant acceleration that the aircraft require to be airborne after a takeoff run of 240 m is 2.3125m/s²

Answer:

C) The book's inertia carried it forward.

When the bus stops suddenly, the book tends to remain in motion due to its inertia. The book was at rest on the seat of the bus, and when the bus stopped suddenly, the book continued moving forward with the same speed and direction it had before the bus stopped. As a result, the book slid off the seat and onto the floor.

At room temperature,300 K, will a miscibility limit emerge. At the temperature less than, the miscibility will fall below 1%.

Temperature directs the hotness or coldness of a body. In clear terms, it is the method of finding the kinetic energy of particles within an entity. Faster the motion of particles, more the temperature.

Hence, at room temperature,300 K, will a miscibility limit emerge. At the temperature less than, the miscibility will fall below 1%.

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Answer:

i think its 400 miles driven

Explanation:

attempted to do it...

All these people perform similar duties due to their similar profession.

All these People are the astronomers so they used various instruments that are present at their time in order to study the heavenly bodies such as stars and planets present in the sky. They also analyze their findings with the help of researches and experiments. They also develop theories that are based on personal observations and tested the theories of other astronomers in order to verify their theories.

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Answer: Internal expansive pore fluid pressure within a host material with a high specific surface is expected to be higher than that for a host material with a lower specific surface.

This is because a material with a high specific surface area has more surface area for the fluid to interact with and absorb into, leading to higher internal pore fluid pressure. In contrast, a material with a lower specific surface area has less surface area for the fluid to interact with, leading to lower internal pore fluid pressure.

Therefore, assuming a constant porous host material type and non-wetting absorbing fluid, internal expansive pore fluid pressure within a host material with a high specific surface is expected to be higher than that for a host material with a lower specific surface.

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With the use of the formula SinФ = nλ / d, there are 16 spectral orders which can be seen when it is illuminated by white light.

Given that a grating has 2000 slits/cm. That is,

d = 0.01 / 2000

d = 5 x \(10^{-6}\) m

The wavelength λ = (700 - 400) nm

λ = 300 x \(10^{-9}\) m

To calculate how many full spectral orders that can be seen (400 to 700 nm) when it is illuminated by white light, we will use the below formula

SinФ = nλ / d

Φ = \(Sin^{-1}\) (nλ / d)

When n = 1

Φ = \(Sin^{-1}\)(300 x \(10^{-9}\)/ 5 x \(10^{-6}\) )

Φ = 3.4 degrees

when n = 2

Φ = \(Sin^{-1}\)(2 x 300 x \(10^{-9}\)/ 5 x \(10^{-6}\) )

Ф = 6.9 degrees

When n = 3

Ф = \(Sin^{-1}\)(3 x 300 x \(10^{-9}\)/ 5 x \(10^{-6}\) )

When n = 16

Ф = \(Sin^{-1}\)(16 x 300 x \(10^{-9}\)/ 5 x \(10^{-6}\) )

Ф = \(Sin^{-1}\)(0.96)

Ф = 73.7 degrees

when n = 17

Ф = \(Sin^{-1}\)(17 x 300 x \(10^{-9}\)/ 5 x \(10^{-6}\) )

Ф = \(Sin^{-1}\)(1.05)

Ф = Error ( that is, it does not exist)

Therefore, there are 16 spectral orders which can be seen when it is illuminated by white light.

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Answer:

projectile motion

total time of flight= 2V° Sintheta / g

answer is 3second

Hope This would help you!

The characteristic that should appear at point 4 is "jaws and vertebral column."

Jaws and a vertebral column are important adaptations that allowed for increased mobility and feeding capabilities in early vertebrates, eventually leading to the evolution of more complex organisms.

A vertebral column provides support and protection for the spinal cord, while also allowing for movement and flexibility. It allows an organism to move in various directions, as well as providing structural support for the rest of the body.

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32.04cm  the circumference of a circle with a radius of 5.1cm

C=2πr=2·π·5.1≈32.04425cm

The length of a circle's edge in straight lines is known as its circumference. Although the term perimeter is exclusively applied to polygons, it is identical to the perimeter of a geometric form.

We all know that a circle's radius is the distance a point on the circle's perimeter travels from its centre. The circle's diameter is a straight line that goes through its centre and has its endpoints on the circle. Radius is therefore equal to half of diameter.

The distance around a circle is its circumference. The diameter is the circumference of a circle measured from centre to centre.

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If a surface at 27 °C emits radiation at a rate of 100 W.  the wavelength of the maximum amount of radiation emitted by the second surface is approximately 8.86 micrometers.

a. To calculate the rate at which an identical surface at 54 °C emits radiation, we can use the Stefan-Boltzmann law:

P = σAε(T^4)

where P is the rate of radiation emitted, σ is the Stefan-Boltzmann constant (5.67 x 10^-8 W/m^2K^4), A is the surface area, ε is the emissivity (assumed to be 1 for a blackbody surface), and T is the absolute temperature in Kelvin.

Let's assume that the surface area and emissivity are the same for both surfaces. Then we can write:

P1/P2 = (T1/T2)^4

where P1 is the rate of radiation emitted by the first surface at 27 °C, P2 is the rate of radiation emitted by the second surface at 54 °C, T1 is the absolute temperature of the first surface, and T2 is the absolute temperature of the second surface.

Converting the temperatures to Kelvin:

T1 = 27 + 273 = 300 K

T2 = 54 + 273 = 327 K

Plugging in the values:

P1/P2 = (300/327)^4

P2 = P1/(300/327)^4

P2 = 100/(300/327)^4

P2 ≈ 402 W

Therefore, an identical surface at 54 °C emits radiation at a rate of approximately 402 W.

b. To determine the wavelength of the maximum amount of radiation emitted by each surface, we can use Wien's displacement law:

λ_max = b/T

where λ_max is the wavelength of maximum radiation, b is Wien's displacement constant (2.898 x 10^-3 mK), and T is the absolute temperature in Kelvin.

For the first surface at 27 °C:

T = 27 + 273 = 300 K

λ_max = 2.898 x 10^-3 mK / 300 K

λ_max ≈ 9.66 x 10^-6 m

Therefore, the wavelength of the maximum amount of radiation emitted by the first surface is approximately 9.66 micrometers.

For the second surface at 54 °C:

T = 54 + 273 = 327 K

λ_max = 2.898 x 10^-3 mK / 327 K

λ_max ≈ 8.86 x 10^-6 m

Therefore, the wavelength of the maximum amount of radiation emitted by the second surface is approximately 8.86 micrometers.

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With a 7ms1 horizontal velocity, an object R departs from a platform XY and lands at Q. The length PQ is 0.45m if the identical object falls freely from Xy to P in 0.3 seconds.

There is no possible for the items to have the same start y-component velocity unless they're shot with distinct initial speeds given that each is launched at a various perspectives. The objects must have been shot at different speeds if they had the same maximum height.

We can observe that over the longest-range the signal reaches its maximum output value, 1, at an output data angle of 90 degrees Punts 2 are at a 90-degree angle, so is at a 45-degree angle. In other words, a projectile moves and furthest when it is fired at a 45-degree angle.

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Answer:130

Explanation:just answered it

The negative outcome indicates a flaw in the information provided or how the issue was put up because the flow rate cannot be fictional or negative.

It can be expressed as:

P1 + (1/2)ρv1^2 + ρgh1 = P2 + (1/2)ρv2^2 + ρgh2

the terms involving h cancel out in the equation, and we can solve for v1:

P1 + (1/2)ρv1^2 = P2 + (1/2)ρv2^2

v1 = √(P2 - P1))/ρ)

Substituting the given values for P1, P2, and ρ into the equation,

\(v1 = \sqrt(1.22 x 10^5 - 1.65 x 10^5))/1000) = \sqrt(-0.43)\)

B) The same formula (v2) may be used to get the flow rate in the top part.

v2 = sqrt((2(P1 - P2))/ρ)

\(v2 = \sqrt{2} (1.65 x 10^5 - 1.22 x 10^5))/1000) =\sqrt(0.86) ≈ 0.93 m/s\)

C) The following formula may be used to get the volume flow rate (Q) via the pipe:

Q = Av

A1 = πr1\(1^{2}\); at point 2, the cross-sectional area A2 = πr\(2^2\).

we get:

Q = A1v1 = πr1^2v1

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The statement that a teacher sitting in her rolling chair pushes against the wall and moves in the opposite direction exemplifies the third law of motion by Newton.

Newton's third law of motion states that for every action, there is an equal and opposite reaction.

This law describes the nature of a force as the result of a mutual and simultaneous interaction between an object 1 and a second object 2.

According to Newton's third law of motion, whenever two objects A and B interact with each other, they both exert equal but opposite forces upon each other.

Therefore, the statement that a teacher sitting in her rolling chair pushes against the wall and moves in the opposite direction exemplifies the third law of motion by Newton.

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