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In old Polaroid cameras, the image was projected on film that developed inside the camera. The developed image was then ejected from the camera as a printed photograph. The standard film size for one popular camera was 79 mm square. The film was 116 mm behind the lens.
If you wanted a picture of your 1.8-m-tall friend to fill half the frame, how far away from you did she need to stand?

100 points + Brainliest to correct answer

The lab experiment found that light travels faster in Mystery A compared to Mystery B, with average speeds of 3.0 and 2.8, respectively. The increase in the index of refraction for the second medium led to a higher angle of refraction, resulting in light bending more. These findings have practical implications for optics and communications.

1. Light would travel faster in Mystery A since the average speed of light in Mystery A (3.0) is higher than Mystery B (2.8).

2. Increasing the index of refraction for the second medium leads to an increase in the angle of refraction. When light comes in at a lower angle, it bends more.

3. In conclusion, this lab experiment showed that the speed of light in a material is influenced by the material's index of refraction. Mystery A had a higher average speed of light compared to Mystery B, indicating that light travels faster in Mystery A. Additionally, the angle of refraction increased as the index of refraction for the second medium was increased. These findings have practical applications in the field of optics and communications.

Hence,The laboratory experiment discovered that, with average speeds of 3.0 and 2.8, respectively, light moves more quickly in Mystery A than Mystery B. Light bent more as a result of the second medium's increased index of refraction due to a higher angle of refraction. For optics and communications, these findings have real-world applications.

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Answer:

(A) Total Displacement = 12 km

(B) t = 38.33 min

(C) Average Velocity = 18.8 km/h

Explanation:

(A)

The displacement of the trip is the sum of both distances:

Total Displacement = 10 km + 2 km

Total Displacement  = 12 km

(B)

First we calculate the time taken by the car:

s = vt

where,

s = distance covered by car = 10 km

v = speed of car = 72 km/h

t = time taken = ?

Therefore,

t = (10 km)/(72 km/h)

t = (0.139 h)(60 min/1 h)

t = 8.33 min

now, we add the 40 min of walking in this:

t = 8.33 min + 30 min

t = 38.33 min

(C)

The average velocity is given as:

Average Velocity = Total Displacement/t = (12 km/38.33 min)(60 min/1 h)

Average Velocity = 18.8 km/h

Answer:

Explanation:

Answer:

(A) Total Displacement = 12 km

(B) t = 38.33 min

(C) Average Velocity = 18.8 km/h

Explanation:

(A)

The displacement of the trip is the sum of both distances:

Total Displacement = 10 km + 2 km

Total Displacement  = 12 km

(B)

First we calculate the time taken by the car:

s = vt

where,

s = distance covered by car = 10 km

v = speed of car = 72 km/h

t = time taken = ?

Therefore,

t = (10 km)/(72 km/h)

t = (0.139 h)(60 min/1 h)

t = 8.33 min

now, we add the 40 min of walking in this:

t = 8.33 min + 30 min

t = 38.33 min

(C)

The average velocity is given as:

Average Velocity = Total Displacement/t = (12 km/38.33 min)(60 min/1 h)

Average Velocity = 18.8 km/h

Answer:

F = Bev

Explanation:

B is magnetic field density

e is the electron charge

v is the electron velocity

Answer:

\(force = mass \times acceleration \\ 100 = 50 \times a \\ a = 2 \: {ms}^{ - 2} \\ from : \: \: v = u + at \\ 150 = 100 + (2 \times t) \\ 50 = 2t \\ time = 25 \: seconds\)

Answer:

C

Explanation:

The intermolecular forces between the water molecule is less binding than that of the copper molecule. Hence the water would take a shorter time to be converted to vapour where the temperature of boiling is constant however the temperature of that of the copper molecule keeps increasing.

Answer:hinwjdheiohuddwrr

87u7t

Stochastic impacts of radiation allude to those that happen arbitrarily and are not reliant upon the portion got. These impacts are related to the likelihood of events and incorporate disease and hereditary changes. Non-stochastic impacts, then again, have a limit, and their seriousness increments with expanding portions.

Models incorporate radiation consumption and intense radiation conditions. Understanding the qualification among stochastic and non-stochastic impacts of radiation is significant in fields like radiation security, atomic medication, and radiobiology.

It assists in setting radiation with dosing limits, creating well-being rules, and carrying out suitable radiation safeguarding measures. By separating these impacts, experts can evaluate and deal with the dangers related to openness to ionizing radiation all the more successfully.

This information guides choices in regard to radiation wellbeing conventions, word-related openness limits, and the improvement of radiation therapy systems in medication.

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What is the acceleration of the cart at t=8 seconds?

Hence the answer us letter a) 0 m/s^2.

That's all I know, Hope it help :)

A wave with a wavelength of 4.0 m moves with a speed of 2.0 m/s will have frequency of 0.5 Hz.

Given,

The wavelength of the wave, λ = 4.0 m

Speed of the wave, v = 2.0 m/s

Frequency of the wave can be calculated as;

ν =v/λ

= 2/4

=1/2 Hz

Or, 0.5 Hz

Hence, the frequency of the wave will be 0.5 Hz.

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The density of the plastic block of cork is 600 kg/m³.

The density of the oil is 800 kg/m³.

The specific gravity of the cork in water is the fraction of the cork submerged in the water.

S.G = 1 - ²/₅

S.G = ³/₅ = 0.6

The density of the cork is calculated as follows;

\(S.G = \frac{\rho_c}{\rho _w} \\\\\rho_c = S.G \times \rho_w \\\\\rho_c = 0.6 \times 1000\ kg/m^3\\\\\rho_c = 600 \ kg/m^3\)

The density of the oil is calculated as follows;

S.G = ⁶/₈ = 0.75

\(S.G = \frac{\rho_c }{\rho_o} \\\\\rho_o = \frac{\rho_c }{S.G} \\\\\rho_o =\frac{600 \ kg/m^3}{0.75} \\\\\rho_o = 800 \ kg/m^3\)

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Answer:

3.87 s

Explanation:

When the y component = 0, the ball is at its apex and 1/2 of the way through its flight.

 vy = 19 + at      where a = - 9.81 m/s

  0 = 19 - 9.81t shows the one-half-flight 't' = 1.94 s

      double this result to get total 't' = 3.87 s

We can use the kinematic equation for constant acceleration to solve this problem .The mass of the bin is 40 kg.

Kinematics is the study of how points, objects, and groups of objects move without considering the factors that generate those motions. The part of classical mechanics known as kinematics addresses the motion of points, objects, and systems made up of collections of items.

\(x = 1/2 * a * t^2 + v0 * t\)

\(a = 2(x - v0 * t) / t^2\)

In this case, x = 100 m, t = 10 s, v0 = 0, and the net force acting on the bin is:

F_net = F_applied - F_friction

F_net = 0

Therefore, we can solve for the mass of the bin:

F_applied - F_friction = ma

100 N - 20 N = m * a

Simplifying, we get:

80 N = m * a

\(a = 2(x - v0 * t) / t^2\)

\(= 2(100 m - 0) / (10 s)^2\)

\(= 2 m/s^2\)

Substituting this value into the previous equation, we get:

\(80 N = m * 2 m/s^2\)

Simplifying, we get:

m = 40 kg

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Answer:

I think A is correct.

Explanation:

If it's not I'm sorry probably my fault but I hope this helps:)

The complete expression for the magnitude of the electric field generated by the minute plastic segment is: |dE| = 1/(4πε₀) * ρ * dr / \(r^2\)

The magnitude of the electric field generated by a point charge q at a distance r from the charge is given by Coulomb's law:

E = k*q/\(r^2\)

where k is the Coulomb constant.

For a small segment of charge dq, we can consider it as a point charge located at its center, which is at a distance r from the point where we want to calculate the electric field. The length of the segment is dr, and its charge density is ρ = dq/dr. Therefore, the charge q of the segment is given by:

q = ρ * dr

Substituting this expression for q into Coulomb's law, we get:

dE = k*(ρ*dr)/\(r^2\)

Simplifying this expression, we obtain the magnitude of the electric field generated by the minute plastic segment:

|dE| = kρdr/\(r^2\)

where |dE| denotes the magnitude of the electric field, ρ is the charge density of the segment, dr is the length of the segment, r is the distance from the segment to the point where we want to calculate the electric field, and k is the Coulomb constant:

k = 1/(4πε₀)

where ε₀ is the electric constant, also known as the permittivity of free space.  |dE| = 1/(4πε₀) * ρ * dr / \(r^2\)

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RESPUESTA

a) C = 3.54 x 10⁻⁹ F = 3.54 nF

b) Q = 35.4 x 10⁻⁶ C = 35.4 μC

c) E = 2,000,000 V/m

EXPLICACIÓN

a) La capacitancia de un capacitor de placas paralelas se define como la razón entre la magnitud de la cantidad de carga en cada placa y la diferencia de potencial aplicado. A su vez, ésta depende de la geometría del capacitor. La ecuación que la define es:

donde C es la capacitancia, A es el área de las placas y d es la distancia que separa las placas. Además ε0 es la permitividad del vacío. En general la fórmula también incluye la permitividad relativa del material dielétrico (el material entre las placas del capacitor) pero en este problema no nos interesa, dado que el material entre las placas de este capacitor es el vacío.

De esta manera, si reemplazamos los datos de este problema:

• ε0 = 8.85 x 10⁻¹² F/m

• A = 2 m²

• d = 5 mm

Nota que tanto la permitividad como el área de las placas tienen unidades de metros, pero la distancia de separación entre placas está en milímetros, por lo tanto debemos convertirla a metros:

Ahora sí, reemplazamos en la ecuación de arriba y obtenemos la capacitancia:

Cuando escribimos una cantidad en notación científica, y el exponente del 10 es -9 podemos no escribir esta parte y usar el prefijo nano. De esta manera, decimos que la capacitancia es de 3.54 nF.

b) La carga de las placas podemos obtenerla de la otra ecuación para definir la capacitancia:

C es la capacitancia que encontramos en el punto a), V es la diferencia de potencial aplicada y Q lo que estamos buscando, la carga de las placas. Del enunciado, tenemos que la diferencia de potencial es V = 10,000V. Reemplazando y resolviendo para Q:

Si movemos el punto decimal un lugar hacia la derecha tenemos: 35.4 x 10⁻⁶ C, que puede llevar el prefijo de micro. Para el prefijo micro usamos la letra griega mu (μ). Entonces la carga de las placas es 35.4 μC

c) Finalmente, para encontrar la magnitud del campo eléctrico, utilizaremos la relación entre la diferencia de potencial y la distancia entre las placas:

Reemplazando con V = 10,000V y d = 0.005m, el campo eléctrico es:

Answer:

My youngins heartless, so they ain't playin' no games

We really want 'em dead, he got hit up close range

He fuc__ked up in the head, he wanna see some more brains

On that corner, I couldn't stay up out that do_0pe game

My cousin got indicted dealin' coc__aine

She an Insta>:gram addict, she want more fame

I used to starve, now I'm blowing up like pro__pane

Told my inner self, "I promise you I won't change"

Explanation:

Doubling the velocity or the mass will Increase the Kinetic Energy because the Kinetic energy has a direct proportional relationship with the mass and the velocity.

Given Data

The expression for kinetic energy is given as

Kinetic Energy = 1/2*mv^2

From the formula as seen above, we can see that the mass and the velocity has a direct impact on the Kinetic energy, hence. increasing any of the values will mean a direct and equivalent increase in the Kinetic Energy.

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Answer:

false

Explanation:

it cant defined the messy and clean states

The description that tells two processes that scientists think move Earth's lithospheric plates is convection in the mantle and pressure of magma on the edge of the plate.

The Earth's lithosphere is the rocky outer part of Earth which is made up of the brittle crust and the top part of the upper mantle.

The Earth's lithosphere deflects the convections and as the convections churn clockwise of anticlockwise, they drag the  lithosphere with it via friction an this is what is stipulated to cause tectonic plate movements.  

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Answer: convection in the mantle and pressure of magma on the edge of the plate

Explanation: I took the unit test

B. The object moves forward at 1 cm/s, stops, and then increases velocity to 2 cm/s.

Velocity = change in position / change in time

Velocity = (3 cm - 0 cm)/(3 s - 0 s)

Velocity = 1 cm/s

The displacement of the object between 3 seconds to 6 seconds is zero (0).

Velocity = (11 cm - 3 cm)/(10 s - 6s)

Velocity = 2 cm/s

Thus, the object moves forward at 1 cm/s, stops, and then increases velocity to 2 cm/s.

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Answer:

B

Explanation:

Answer:

\(1.694*10^3KW/hr\)

Explanation:

Formula:

\(KW/hr=E_{(change)}\)

\(P=\frac{E}{t}\)

\((KW)(s)=1000J\)

\((3600)(KW)(s)=3600*1000J\)

\(KW=3.6*10^6\)

\(\frac{6.1*10^9}{3.6*10^6}=\frac{x}{1KW}\)

\(1.694*10^3KW/hr\)

Hope it helps

As the temperature of the balloon is lowered, its volume will decrease as well.

The kinetic theory of matter  says that all matter consists of many, very small particles which are constantly moving or in a continual state of motion.

According to Charles law, the volume of a fixed mass of a gas is directly proportional to its absolute temperature.

Based on Charles law, as the temperature of the balloon is lowered, its volume will decrease as well because the average kinetic energy of the gas molecules has reduced, the rate of gas collision decreased, which will cause a drop in the volume of the balloon.

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The direction of the resultant force on the ball = 28.3°

Sue's foot exerts a force of 28.85 N East

Jenny's foot exerts a force of 15.53 N North

This can be represented diagrammatically as:

The magnitude of the resultant force, R, is calculated using the Pythagora's theorem

The resultant force on the ball is 32.76N

The direction of the resultant force is given as:

The direction of the resultant force on the ball = 28.3°

Answer:

This problem involves the concept of a random walk, which is a mathematical model of a path consisting of a succession of random steps.

The question asks for the distance, R(n), from the center of a star after n steps of a photon, assuming a 2D random walk.

The random walk in two dimensions has a step length of A_i and the direction of the steps is uniformly distributed in [0, 2π). The change in position after each step can be written in Cartesian coordinates (Δx, Δy), where Δx = A_i cos(θ_i) and Δy = A_i sin(θ_i).

The displacement from the center after n steps is given by the vector sum of all the individual steps. This vector sum can be written in terms of its Cartesian coordinates, (X, Y), where X = Σ Δx and Y = Σ Δy. This sum over n random vectors is itself a random variable. The net displacement R(n) from the center of the star after n steps is given by the magnitude of the net displacement vector:

R(n) = √(X² + Y²)

Because each step is independent and has a random direction, the expected value of the cosine and sine for any step is zero. This means that the expected values of X and Y are both zero.

However, the mean square displacement is not zero. Because the steps are independent, the mean square displacement in each direction is additive. For a 2D random walk:

<X²> = Σ <(Δx)²> = n <(A cos θ)²> = n A²/2

<Y²> = Σ <(Δy)²> = n <(A sin θ)²> = n A²/2

Because <X²> = <Y²>, we can write:

<R²> = <X²> + <Y²> = n A²

So, the root mean square distance (the square root of the mean square displacement) after n steps is:

R(n) = √(<R²>) = √(n) * A

Therefore, the distance R(n) that the photon is expected to be from the center of the star after n steps grows as the square root of the number of steps, with each step having a length A. Please note that this result holds for a 2D random walk. A real photon in a star would be performing a 3D random walk, which would have slightly different characteristics.

Answer:

673km

Explanation:

Average acceleration is how much the car increases (on average) per second.  So, since it increases by 30 m/s in 8 s, then dividing 30 by 8 will give you an average acceleration of 3.75 m/s^2 (remember that acceleration is ever increasing, so the unit is s^2, not just s)

Answer:

The standard (metric) units that would be discussed at primary school would include: grams and kilograms, centimetres, metres and kilometres, millilitres and litres (though children also learn about imperial units in Year 5 maths).

Answer:

Standard units are commonly used units of measurement. For an example; grams and kilograms, centimetres.

Answer:

u = v-at

Explanation:

this is my answer, on how I have come up with this formula

The mass of Car B is -6000 kg.

To solve this problem, we can apply the principle of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision.

Therefore, we can write the equation for the conservation of momentum as:

(mass of Car A * velocity of Car A) + (mass of Car B * velocity of Car B) = (mass of Car A + mass of Car B) * velocity after collision

Let's substitute the given values into the equation:

(2000 kg * 10 m/s) + (mass of Car B * 0 m/s) = (2000 kg + mass of Car B) * (-5 m/s)

Simplifying the equation:

20000 kg*m/s = -5 m/s * (2000 kg + mass of Car B)

Dividing both sides by -5 m/s:

-4000 kg = 2000 kg + mass of Car B

Subtracting 2000 kg from both sides:

mass of Car B = -4000 kg - 2000 kg

mass of Car B = -6000 kg

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