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3 years ago

Image caughton Screen is called

Physics

2 answers:

natka813 [3]3 years ago

5 0

Answer:

Real Image

Explanation:

Images which are formed on the screen by the actual intersection of light rays are called real images.

bixtya [17]3 years ago

4 0

Answer:

<h2>Virtual image</h2>

Explanation:

<h3>Virtual image can be caught on a screen</h3>

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How long does for one rotation of earth

stellarik [79]

Answer: 24 hours

Explanation: One day is one rotation of the Earth, but it takes the earth 365 days to do one revolution or one year

8 0

3 years ago

The two measurements necessary for calculating average speed are

alisha [4.7K]

The correct answer is Option (C) distance and time

Explanation:

Average speed of any object is defined as the total distance that object travels over the time it takes to travel that distance. In other words, average speed is the total distance divided by the elapsed time.

$Average \thinspace Speed = \frac{Total \thinspace Distance}{Elapsed \thinspace Time}$

Therefore, as you can see in the above equation, the two measurements that are essential for the calculation of the average speed are the (total) distance and the (elapsed) time.

Hence, the correct option is C.

5 0

3 years ago

Ch 27 HW Exercise 27.12 10 of 20 Constants A horizontal rectangular surface has dimensions 2.80 cm by 3.15 cm and is in a unifor

deff fn [24]

Answer:

Magnetic field, B = 0.88 T

Explanation:

It is given that,

The dimension of rectangular surface is 2.80 cm by 3.15 cm. The area of rectangular surface is, $A=8.82\ cm^2=0.000882\ m^2$

Angle between the uniform magnetic field and the horizontal, $\theta=31$

Magnetic flux, $\phi=4\times 10^{-4}\ Wb$

Let B is the magnitude of magnetic field in which the rectangular surface is placed. It is given by :

$\phi=BA\ cos\theta$

$\theta$ is the angle between magnetic field and the area

Here, $\theta=90-31=59^{\circ}$

$B=\dfrac{\phi}{A\ cos\theta}$

$B=\dfrac{4\times 10^{-4}}{0.000882\times cos(59)}$

B = 0.88 T

So, the magnitude of magnetic field is 0.88 T. Hence, this is the required solution.

7 0

3 years ago

Water, with a density of 1000 kg/m3, flows out of a spigot, through a hose, and out a nozzle into the air. The hose has an inner

stepan [7]

Answer:

P₁ = 2.3506 10⁵ Pa

Explanation:

For this exercise we use Bernoulli's equation and continuity, where point 1 is in the hose and point 2 in the nozzle

P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

A₁ v₁ = A₂ v₂

Let's look for the areas

r₁ = d₁ / 2 = 2.25 / 2 = 1,125 cm

r₂ = d₂ / 2 = 0.2 / 2 = 0.100 cm

A₁ = π r₁²

A₁ = π 1.125²

A₁ = 3,976 cm²

A₂ = π r₂²

A₂ = π 0.1²

A₂ = 0.0452 cm²

Now with the continuity equation we can look for the speed of water inside the hose

v₁ = v₂ A₂ / A₁

v₁ = 11.2 0.0452 / 3.976

v₁ = 0.1273 m / s

Now we can use Bernoulli's equation, pa pressure at the nozzle is the air pressure (P₂ = Patm) the hose must be on the floor so the height is zero (y₁ = 0)

P₁ + ½ ρ v₁² = Patm + ½ ρ v₂² + ρ g y₂

P₁ = Patm + ½ ρ (v₂² - v₁²) + ρ g y₂

Let's calculate

P₁ = 1.013 10⁵ + ½ 1000 (11.2² - 0.1273²) + 1000 9.8 7.25

P₁ = 1.013 10⁵ + 6.271 10⁴ + 7.105 10⁴

P₁ = 2.3506 10⁵ Pa

7 0

3 years ago

The period of a wave is the reciprocal of its ______________.

Hunter-Best [27]

Frequency

hope it helps !

6 0

3 years ago

Image caughton Screen is called​

Image caughton Screen is called