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3 years ago

A(n) ____________ image cannot be projected and forms where light rays appear to originate.

Physics

2 answers:

IrinaVladis [17]3 years ago

6 0

A virtual image is the answer.

Sedbober [7]3 years ago

6 0

<h3><u>Answer;</u></h3>

Virtual image

A<em><u> virtual image</u></em> cannot be projected and forms where light rays appear to originate.

<h3><u>Explanation</u>;</h3>

<em><u>A virtual image is an image that can not be formed on a screen. This type of an image is formed when the rays of light appear to meet at a point after reflection or refraction. Virtual images are always erect or upright.</u></em>
<em><u>A real image on the other hand, is an image that can be formed on a screen. Real image is formed when the rays of light meet at some point after reflection or refraction. Real images are always inverted.</u></em>

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Valentin [98]

Answer:

v = 8.09 m/s

Explanation:

For this exercise we use that the work done by the friction force plus the potential energy equals the change in the body's energy.

Let's calculate the energy

starting point. Higher

Em₀ = U = m gh

final point. To go down the slope

Em_f = K = ½ m v²

The work of the friction force is

W = fr L cos 180

to find the friction force let's use Newton's second law

Axis y

N - W_y = 0

N = W_y

X axis

Wₓ - fr = ma

let's use trigonometry

sin θ = y / L

sin θ = 11/110 = 0.1

θ = sin⁻¹ 0.1

θ = 5.74º

sin 5.74 = Wₓ / W

cos 5.74 = W_y / W

Wₓ = W sin 5.74

W_y = W cos 5.74

the formula for the friction force is

fr = μ N

fr = μ W cos θ

Work is friction force is

W_fr = - μ W L cos θ

Let's use the relationship of work with energy

W + ΔU = ΔK

-μ mg L cos 5.74 + (mgh - 0) = 0 - ½ m v²

v² = - 2 μ g L cos 5.74 +2 (gh)

v² = 2gh - 2 μ gL cos 5.74

let's calculate

v² = 2 9.8 11 - 2 0.07 9.8 110 cos 5.74

v² = 215.6 -150.16

v = √65.44

v = 8.09 m/s

6 0

3 years ago

A lab cart is loaded with different masses and moved at various velocities

Sedaia [141]

A lab cart is loaded with different masses and moved at various constant velocities? the anser should be

1.0m/s → 4kg

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3 years ago

Consider a 2.54-cm-diameter power line for which the potential difference from the ground, 19.6 m below, to the power line is 11

tiny-mole [99]

Answer:

The line charge density is $1.59\times10^{-4}\ C/m$

Explanation:

Given that,

Diameter = 2.54 cm

Distance = 19.6 m

Potential difference = 115 kV

We need to calculate the line charge density

Using formula of potential difference

$V=EA$

$V=\dfrac{\lambda}{2\pi\epsilon_{0}r}\times\pi r^2$

$\lambda=\dfrac{V\times2\epsilon_{0}}{r}$

Where, r = radius

V = potential difference

Put the value into the formula

$\lambda=\dfrac{115\times10^{3}\times2\times8.8\times10^{-12}}{1.27\times10^{-2}}$

$\lambda=1.59\times10^{-4}\ C/m$

Hence, The line charge density is $1.59\times10^{-4}\ C/m$

4 0

3 years ago

If the mass of a moving object were quartered, it’s inertia would be

AVprozaik [17]

Answer:

Quartered

Explanation:

Because you're a liberal.

8 0

3 years ago

when a metal ball is heated through 30°c,it volume becomes 1.0018cm^3 if the linear expansivity of the material of the ball is 2

Vlad1618 [11]

Answer:

The original volume of the metal sphere is approximately 1 cm³

Explanation:

The given parameters are;

The temperature change of the metal ball, ΔT = 30°C = 30 K

The new volume of the metal ball, V₂ = 1.0018 cm³

The linear expansivity of the material ball, α = 2.0 × 10⁻⁵ K⁻¹

We have;

d₂ = d₁·(1 + α·ΔT)

Where;

d₁ = The original diameter of the metal ball

d₂ = The new diameter of the ball

From the volume of the ball, V₂, we have;

V₂ = 1.0018 cm³ = (4/3)×π×r₂³

Where;

r₂ = The new radius = d₂/2

∴ V₂ = 1.0018 cm³ = (4/3)×π×(d₂/2)³

∴ d₂ = ∛(2³ × 1.0018 cm³/((4/3) × π)) ≈ 1.241445 cm

d₁ = d₂/(1 + α·ΔT)

∴ d₁ ≈ 1.241445 cm/(1 + 2.0 × 10⁻⁵·K × 30 K) ≈ 1.24070058 cm

The original volume of the metal ball, V₁ = (4/3)×π×(d₁/2)³

∴ V₁ = (4/3)×π×(1.24070058/2)³ ≈ 0.99999902845 cm³ ≈ 1 cm³

The original volume of the metal sphere, V₁ ≈ 1 cm³.

8 0

3 years ago