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iVinArrow [24]
3 years ago
11

How does a real image differ from a virtual image?

Physics
2 answers:
Snezhnost [94]3 years ago
6 0

Answer:

one contains datta that can take up hard drive or disc space weras a picture takes up space on walls or photo albums

Explanation:

goldfiish [28.3K]3 years ago
3 0
Real image is formed when the rays of light after reflection or refraction actually meet at some point whereas a virtual image is formed when the rays of light after reflection or refraction appear to meet at a point
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What will change the velocity of a periodic wave?
Georgia [21]
If it's a mechanical wave, then its speed depends on the physical characteristics of the medium.

If it's an electromagnetic wave, then its speed depends on the
electrical characteristics of the medium.

Either way, the properties of the medium determine the wave speed.
You want to change the speed ?  You have to change the properties
of the medium.
7 0
3 years ago
Select the correct statement to describe when a sample of liquid water vaporizers into water vapor
AnnyKZ [126]
From what I know; When a sample of liquid water vaporizes into water vapor, the electrons in the water sped up due to heat. 
4 0
3 years ago
A 1.05 kg block slides with a speed of 0.865 m/s on a frictionless horizontal surface until it encounters a spring with a force
djyliett [7]

Answer:

a) U = 0 J    

k = 0.393 J

E = 0.393 J

b) U = 0.0229J

k = 0.370 J

E = 0.393 J

c) U = 0.0914 J

k = 0.302 J

E = 0.393 J

d) U = 0.206 J

k = 0.187 J

E = 0.393 J

e) U = 0.366 J

k = 0.027 J

E = 0.393 J

Explanation:

Hi there!

The equations of kinetic energy and elastic potential energy are as follows:

k = 1/2 · m · v²

U = 1/2 · ks · x²

Where:

m = mass of the block.

v = velocity.

ks = spring constant.

x = displacement of the string.

a) When the spring is not compressed, the spring potential energy will be zero:

U = 1/2 · ks · x²

U = 1/2 · 457 N/m · (0 cm)²

U = 0 J

The kinetic energy of the block will be:

k = 1/2 · m · v²

k = 1/2 · 1.05 kg · (0.865 m/s)²

k = 0.393 J

The mechanical energy will be:

E = k + U = 0.393 J + 0 J = 0.393 J

This energy will be conserved, i.e., it will remain constant because there is no work done by friction nor by any other dissipative force (like air resistance). This means that the kinetic energy will be converted only into spring potential energy (there is no thermal energy due to friction, for example).

b) The spring potential energy will be:

U = 1/2 · 457 N/m · (0.01 m)²

U = 0.0229 J

Since the mechanical energy has to remain constant, we can use the equation of mechanical energy to obtain the kinetic energy:

E = k + U

0.393 J = k + 0.0229 J

0.393 J - 0.0229 J = k

k = 0.370 J

c) The procedure is now the same. Let´s calculate the spring potential energy with x = 0.02 m.

U = 1/2 · 457 N/m · (0.02 m)²

U = 0.0914 J

Using the equation of mechanical energy:

E = k + U

0.393 J = k + 0.0914 J

k = 0.393 J - 0.0914 J = 0.302 J

d) U = 1/2 · 457 N/m · (0.03 m)²

U = 0.206 J

E = 0.393 J

k = E - U = 0.393 J - 0.206 J

k = 0.187 J

e) U = 1/2 · 457 N/m · (0.04 m)²

U = 0.366 J

E = 0.393 J

k = E - U = 0.393 J - 0.366 J = 0.027 J.

4 0
3 years ago
An air conditioner running with R-134a on a cycle executed under the saturationdome between the pressure limits of 0.8 MPa and 0
ohaa [14]

Answer:

The COP of the system is = 4.6

Explanation:

Given data

Higher pressure  = 1.8 M pa

Lower pressure = 0.12 M pa

Now we have to find out high & ow temperatures at these pressure limits.

Higher temperature corresponding to pressure 1.8 M pa

T_{H} = 62.9 °c = 335.9 K

Lower temperature corresponding to pressure 0.2 M pa

T_{L} = - 10.1 °c = 262.9 K

COP of the system is given by

COP = \frac{T_{L} }{T_{H} -T_{L}   }

COP = \frac{335.9}{335.9 -262.9}

COP = 4.6

Therefore the COP of the system is = 4.6

8 0
3 years ago
A horizontal force of magnitude 30.2 N pushes a block of mass 3.50 kg across a floor where the coefficient of kinetic friction i
marshall27 [118]

Answer:

A) 89.39 J

B) 30.39J

C) 23.8 J

Explanation:

We are given;

F = 30.2N

m = 3.5 kg

μ_k = 0.646

d = 2.96m

ΔEth (Block) = 35.2J

A) Work done by the applied force on the block-floor system is given as;

W = F•d

Thus, W = 30.2 x 2.96 = 89.39 J

B) Total thermal energy dissipated by the whole system which includes the floor and the block is given as;

ΔEth = μ_k•mgd

Thus, ΔEth = 0.646 x 3.5 x 9.8 x 2.96 = 65.59J

Now, we are given the thermal energy of the block which is ΔEth (Block) = 35.2J.

Thus,

ΔEth = ΔEth (Block) + ΔEth (floor)

Thus,

ΔEth (floor) = ΔEth - ΔEth (Block)

ΔEth (floor) = 65.59J - 35.2J = 30.39J

C) The total work done is considered as the sum of the thermal energy dissipated as heat and the kinetic energy of the block. Thus;

W = K + ΔEth

Therefore;

K = W - ΔEth

K = 89.39 - 65.59 = 23.8J

3 0
3 years ago
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