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Which statement about thin lenses is correct? In each case, we are considering only a single lens. A. A diverging lens always produces a virtual inverted image. B. A converging lens always produces a real inverted image. C. A converging lens sometimes produces a real erect image. D. A diverging lens produces a virtual erect image only if the object is located within the focal point of the lens. E. A diverging lens always produces a virtual erect image.

Answers

Answer 1

Answer:

A diverging lens always produces a virtual erect image.

The general lens formula is given as;

$(1)/(F) = (1)/(U) + (1)/(V)$

Where;

U = object distance
V = image distance
F = focal length of the lens

A lens can be converging or diverging.

A converging lens produces a virtual image when the object is placed in front of the focal point. The image can also be real when the object is placed beyond focal point.

The image produced by a diverging lens is always virtual and upright.

Thus, we can conclude that a diverging lens always produces a virtual erect image.

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Answer 2

Answer:

Answer:

E) true. The image is always virtual and erect

Explanation:

In this exercise we are asked to find the correct statements,

for this we can use the constructor equation

1 / f = 1 / p + 1 / q

where f is the focal length, p the distance to the object and q the distance to the image

In diverging lenses, the focal length is negative and the image is virtual and erect

In convergent lenses, the positive focal length, if the object is farther than the focal length, the image is real and inverted, and if the object is at a shorter distance than the focal length, the image is virtual and straight.

With this analysis let's review each statement

A) False. The image is right

B) False. The type of image depends on where the object is with respect to the focal length

C) False. The real image is always inverted

D) False. The image is always virtual

E) true. The image is always virtual and erect

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HELP ASAP PLEASE!!!In which direction(s) does the ground shake during an earthquake?
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B. up and down
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Answers

Answer: D i am pretty sure

Explanation:

Answer:

all

Explanation:

John, who has a mass of 65kg stands at rest on the ice. He catches a 10kg ball that is thrown to him at 5m/s.

Answers

The momentum of John after catching the ball is 50 kg.m/s.

"Your question is not complete, it seems to be missing the following information";

find John's momentum

The given parameters;

mass of John, m = 65 kg
mass of the ball caught by John, m' = 10 kg
initial velocity of John, u = 0
initial velocity of the ball, v = 5 m/s

Apply the principles of conservation of linear momentum to determine the momentum of John.

The momentum of John is calculated as follows;

P = mu + mv

P = (65 x 0) + (10 x 5)

P = 0 + 50

P = 50 kg.m/s

Thus, the momentum of John after catching the ball is 50 kg.m/s.

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Two workers are sliding 300 kg crate across the floor. One worker pushes forward on the crate with a force of 400 N while the other pulls in the same direction with a force of 290 N using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor

Answers

Answer:

The kinetic coefficient of friction of the crate is 0.235.

Explanation:

As a first step, we need to construct a free body diagram for the crate, which is included below as attachment. Let supposed that forces exerted on the crate by both workers are in the positive direction. According to the Newton's First Law, a body is unable to change its state of motion when it is at rest or moves uniformly (at constant velocity). In consequence, magnitud of friction force must be equal to the sum of the two external forces. The equations of equilibrium of the crate are:

$\Sigma F_(x) = P+T-\mu_(k)\cdot N = 0$ (Ec. 1)

$\Sigma F_(y) = N - W = 0$ (Ec. 2)

Where:

$P$ - Pushing force, measured in newtons.

$T$ - Tension, measured in newtons.

$\mu_(k)$ - Coefficient of kinetic friction, dimensionless.

$N$ - Normal force, measured in newtons.

$W$ - Weight of the crate, measured in newtons.

The system of equations is now reduced by algebraic means:

$P+T -\mu_(k)\cdot W = 0$

And we finally clear the coefficient of kinetic friction and apply the definition of weight:

$\mu_(k) =(P+T)/(m\cdot g)$

If we know that $P = 400\,N$ , $T = 290\,N$ , $m = 300\,kg$ and $g = 9.807\,(m)/(s^(2))$ , then:

$\mu_(k) = (400\,N+290\,N)/((300\,kg)\cdot \left(9.807\,(m)/(s^(2)) \right))$

$\mu_(k) = 0.235$

The kinetic coefficient of friction of the crate is 0.235.

Final answer:

The calculation of the coefficient of kinetic friction involves setting the total force exerted by the workers equal to the force of friction, as the crate moves at a constant speed. The coefficient of kinetic friction is then calculated by dividing the force of friction by the normal force, which is the weight of the crate. The coefficient of kinetic friction for the crate on the floor is approximately 0.235.

Explanation:

To calculate the coefficient of kinetic friction, we first must understand that the crate moves at a constant velocity, indicating that the net force acting on it is zero. Thus, the total force exerted by the workers (400 N + 290 N = 690 N) is equal to the force of friction acting in the opposite direction.

Since the frictional force (F) equals the normal force (N) times the coefficient of kinetic friction (μk), we can write the equation as F = μkN. Here, the normal force is the weight of the crate, determined by multiplying the mass (m) of the crate by gravity (g), i.e., N = mg = 300 kg * 9.8 m/s² = 2940 N.

Next, we rearrange the equation to solve for the coefficient of kinetic friction: μk = F / N. Substituting the known values (F=690 N, N=2940 N), we find: μk = 690 N / 2940 N = 0.2347. Thus, the coefficient of kinetic friction for the crate on the floor is approximately 0.235.

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A uniform rod of length L rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end. The rod is initially at rest. A bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its center and becomes embedded in it. The mass of the bullet is one-fourth the mass of the rod. What is thefinal angular speed of the rod?

Answers

The value of final angular speed of the uniform rod which rests on the frictionless horizontal surface is,

$\omega=(6v)/(19L)$

What is angular speed of a body?

The angular speed of a body is the rate by which the body changed its angle with respect to the time. It can be given as,

$\omega= (\Delta \theta)/(\Delta t)$

A uniform rod of length L rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end.

The rod is initially at rest. A bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its center and becomes embedded in it.

The mass of the bullet is one-fourth the mass of the rod. The diagram for the above condition is attached below.

In the attached image the angular momentum about the point A is constant just before and after the collision. Thus,

$L_i=L_f\nL_i=(m)/(4)*(vL)/(2)=I\omega\n(m)/(4)*(vL)/(2)=I\omega$

Put the value of inertia as,

$(m)/(4)*(vL)/(2)=\left((mL^2)/(3)+(mL^2)/(4*4)\right)\n(mvL)/(8)*(vL)/(2)=\left((19mL^2)/(48)\right)$

Solving it further we get,

$(v)/(8)=(19)/(48)L\omega\n\omega=(6v)/(19L)$

Hence, the value of final angular speed of the uniform rodwhich rests on the frictionless horizontal surface is,

$\omega=(6v)/(19L)$

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Answer: a) 0.315 (V/L)

Explanation:

From Conservation of angular momentum, we know that

L1 = L2 ,

Therefore MV L/2 = ( Irod + Ib) x W

M/4 x V x L/2 = (M (L/2)^2 + 1/3xMxL^2) x W

M/8 X VL = (ML^2/16 + ML^2 /3 )

After elimination we have,

V/8 = 19/48 x L x W

W = 48/8 x V/19L = 6/19 x V/L

Therefore W = (0.136)X V/L

A chair of weight 100 N lies atop a horizontal floor; the floor is not frictionless. You push on the chair with a force of F = 43.0 N directed at an angle of 35.0 degrees below the horizontal and the chair slides along the floor. Using Newton's laws, calculate n, the magnitude of the normal force that the floor exerts on the chair. Express your answer in newtons.

Answers

Answer:

The normal force will be "122.8 N".

Explanation:

The given values are:

Weight,

W = 100 N

Force,

F = 40 N

Angle,

θ = 35.0°

As we know,

⇒ $N=W+FSin \theta$

On substituting the given values, we get

⇒ $= 100N+40N \ Sin \theta$

⇒ $=100N+22.8$

⇒ $=122.8 \ N$

Final answer:

To calculate the magnitude of the normal force, analyze the forces acting on the chair and find the sum of the vertical component of the applied force and the weight of the chair. The normal force is equal to the sum of the vertical component of the applied force and the weight of the chair.

Explanation:

To calculate the magnitude of the normal force that the floor exerts on the chair, we need to analyze the forces acting on the chair. The weight of the chair is acting vertically downwards with a magnitude of 100 N. The force you apply to the chair is acting at an angle of 35.0 degrees below the horizontal. Since the chair is sliding, there must be a friction force opposing its motion. Using Newton's laws, we can find the normal force.

Resolve the applied force into horizontal and vertical components. The horizontal component is given by F_cos(35°) and the vertical component is given by F_sin(35°).
The horizontal component of the applied force is balanced by the friction force. Therefore, the friction force is equal to the horizontal component of the applied force.
The vertical component of the applied force and the weight of the chair contribute to the normal force. So, the normal force is equal to the sum of the vertical component of the applied force and the weight of the chair.

Therefore, n = F_sin(35°) + weight = 43.0 N * sin(35°) + 100 N.

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What happens to a black body radiator as it increases in temperature? A. it gives off a range of electromagnetic radiation of shorter wavelengths.
B. It gives off only one wavelength of electromagnetic radiation
C. It releases only ultraviolet waves of electromagnetic radiation
D. It becomes hotter but gives off less electromagnetic radiation

Answers

The black body radiator as it increases in temperature gives off a range of electromagnetic radiation of shorter wavelengths so, the option A is correct.

What is radiation?

Radiation is the movement of atomic and subatomic particles as well as waves, such as those that define X-rays, heat rays, and light rays. Radiation of both types, from cosmic and earthly sources, is constantly being thrown at all matter.

The characteristics and behavior of radiation, as well as the matter it interacts with, are outlined in this article, which also explains how energy is transferred from radiation to its surroundings.

The effects of such an energy transfer to living matter, including the typical effects on numerous biological processes, are given a great deal of attention (e.g., photosynthesis in plants and vision in animals).

Thus, the black body radiator gives off a range of electromagnetic radiation of shorter wavelengths.

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Answer: A

Explanation:

Answer is a hope this helps guys!

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