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Y_Kistochka [10]
2 years ago
6

A student obtained a sharp image of the grill of a window on a screen, using a convex lens. For getting better results, the teac

her suggested focusing on a distant tree instead of the grill. In which the direction should the lens be moved for this purpose?​
Physics
1 answer:
viktelen [127]2 years ago
4 0

Answer:

Towards the screen

Explanation:

When the image distance increases, object distance decreases. Thus, distance between the mirror and screen will decrease. So, the mirror should be moved towards the screen. When the image distance increases, object distance decreases.

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A pitcher throws a curveball that reaches the catcher in 0.61 s. The ball curves because it is spinning at an average angular ve
sashaice [31]

Answer:

22.36 rad

Explanation:

Applying,

ω = θ/t.............. Equation 1

Where ω  = angular velocity, θ = angular displacement of the baseball, t = time

make θ the subject of the equation

θ = ωt............... Equation 2

From the question,

Given: ω = 350 rev/min = 350(0.10472) = 36.652 rad/s, t = 0.61 s

Substitute these values into equation 1

θ = 0.61(36.652)

θ = 22.36 rad

Hence the angular displacement of the baseball is  22.36 rad

4 0
3 years ago
Energy slowly leaks outward through the diffusion of photons that repeatedly bounce off ions and electrons
stepan [7]

Energy slowly leaks outward through the radiative diffusion of photons that repeatedly bounce off ions and electrons.

<h3>What is radiative diffusion?</h3>

A radiation zone is a layer of a star's core where energy is mostly carried toward the outside by radiative diffusion and thermal conduction rather than convection.

As photons, energy passes through the radiation zone as electromagnetic radiation.

The radiative diffusion of photons that repeatedly bounce off ions and electrons progressively drains energy outward.

Hence,radiative diffusion is correct answer.

To learn more about radiative diffusion refer:

brainly.com/question/3598352

#SPJ4

7 0
3 years ago
What is the benefit of developing the atomic mass unit as a standard unit of mass?
Alexeev081 [22]

Answer:

The atomic mass unit is 1/12 of an atom of carbon 12, and is a very small amount to represent in kilograms:

1m_{u}=1.66054x10^{-27}kg

m_{u} is atomic mass unit.

This is why the benefits of the atomic mass unit is that it makes the representation of atomic masses easier in terms of the simplicity of the numbers that are used to represent the masses. Also using the atomic mass unit it is easier to compare the masses of different atoms, These numbers would be very small and would require negative powers of 10 to represent them, so it is more convenient to use the atomic mass unit.

3 0
3 years ago
When will you say a body is in a) uniform acceleration (b) non uniform acceleration
s2008m [1.1K]

Answer:

See below ~

Explanation:

Part (a) :

We can say a body is in uniform acceleration if the acceleration of the object remains constant with respect to time throughout its motion.

Part (b) :

We can say a body is non-uniform acceleration if the acceleration of the body varies with respect to time throughout its motion.

7 0
2 years ago
A 1.50-m string of weight 0.0125 N is tied to the ceil- ing at its upper end, and the lower end supports a weight W. Ignore the
Elena L [17]

The wave equation is missing and it is y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)

Answer:

A) 0.0534 seconds

B) 0.67N

C) 41

D) (8.50 mm)cos(172 rad/m x + 4830 rad/s t)

Explanation:

we are given weight of string = 0.0125N

Thus, since weight = mg

Then, mass of string = 0.0125/9.8

Mass of string = 1.275 x 10⁻³ kg

Length of string; L= 1.5 m .

mass per unit length; μ = (1.275 x 10⁻³)/1.5

μ = 0.85 x 10⁻³ kg/m

We are given the wave equation: y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)

Now if we compare it to the general equation of motion of standing wave on a string which is:

y(x,t) = Acos(Kx − ω t)

We can deduce that

angular velocity;ω = 4830 rad/s

Wave number;k = 172 rad/m

A) Velocity is given by the formula;

V = ω/k

Thus, V = 4830/172 m/s

V = 28.08 m /s

Thus time taken to go up the string = 1.5/28.08 = 0.0534 seconds

B) We know that in strings,

V² = F/μ

Where μ is mass per unit length and V is velocity.

Thus, F = V²*μ =28.08² x 0.85 x 10⁻³

F = 0.67N

C) Formula for wave length is given as; wave length;λ = 2π /k

λ = 2 x π/ 172

λ = 0.0365 m

Thus, number of wave lengths over whole length of string

= 1.5/0.0365 = 41

D) The equation for waves traveling down the string

= (8.50 mm)cos(172 rad/m x + 4830 rad/s t)

8 0
4 years ago
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