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3 years ago

15

What's the optical illusion also known as?

1 answer:

shepuryov [24]3 years ago

8 0

Answer:

An optical illusion (also called a visual illusion) is an illusion caused by the visual system and characterized by a visual percept that arguably appears to differ from reality. ... Cognitive visual illusions are the result of unconscious inferences and are perhaps those most widely known.

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what is the specific heat capacity of silver metal if 55.00g of the absored 47.3 calories of heat and temperature rise 15.°c

igor_vitrenko [27]

Answer:

$0.240 J/(g^{\circ}C)$

Explanation:

When an amount of energy Q is supplied to a sample of substance of mass m, the temperature of the substance increases by $\Delta T$ according to the equation

$Q=mC_s \Delta T$

where

$C_s$ is the specific heat capacity of the substance

In this problem, we have:

m = 55.0 g is the mass of the sample of silver

$Q = 47.3 \cdot 4.184 = 197.9 J$ is the amount of energy supplied to the sample

$\Delta T = 15^{\circ}C$ is the change in temperature of the sample

Solving the equation for $C_s$ , we find the specific heat capacity of silver:

$C_s = \frac{Q}{m \Delta T}=\frac{197.9}{(55.0)(15)}=0.240 J/(g^{\circ}C)$

3 0

4 years ago

What is the efficient cause if acceleration

vladimir1956 [14]

Answer:

acceleration= velocity ÷ time

Explanation:

the question is outrageous

4 0

3 years ago

Read 2 more answers

Find the heat given off if 1 kg of 125 degree C steam is cooled to 50 degree C water.

Gnom [1K]

T_i=125°C=398K
T_f=50°C=323K
c=4184J/kg°K
m=1kg
∆T=T_f-T_i=398-323=75K

$\\ \sf{:}\Rrightarrow Q=mc\Delta T$

$\\ \sf{:}\Rrightarrow Q=1(4184)(75)$

$\\ \sf{:}\Rrightarrow Q=313800J$

$\\ \sf{:}\Rrightarrow Q=313.8KJ$

4 0

3 years ago

What is the net force on a race car with a mass of 1200 kg if its acceleration is 32.0m/s2 West?

frosja888 [35]

Answer:

38400N

Explanation:

F=mass×acceleration

=1200×32

=38400

3 0

3 years ago

On Earth, the number flux of solar neutrinos from the p-p chain is:

Nataly [62]

Answer:

Explanation:

The volume of the tank = 50 kton

50 kton = 5 × 10⁷ kg

Since 18 grams of water will contain: 10 electrons × 6.023 × 10²³

Then;

5× 10⁷ kg will contain $( \dfrac{5 \times 10^7 \times 10^3}{18}) \times 10 \times 6.023 \times 10^{23}$

= 1.67 × 10³⁴ electrons

(b)

Suppose:

$f_{neutrino} = \dfrac{2f_o}{26.2 MeV} = 6.7\times 10^{10} \ s^{-1} cm^{-2}$

Then;

10⁻⁶ of $f_{neutrino} = 6.7 \times 10^{10} \times 10^{-6} \ s^{-1} cm^{-2}$

$=6.7 \times 10^{4}\ s^{-1} cm^{-2}$

Thus, the number of high energy neutrinos which will interact with water is:

= $6.7 \times 10^4 \times \sigma$

= $6.7 \times 10^4 \times 10^{-43}$

= $6.7 \times 10^{-39} s^{-1}$

For 1.67 × 10³⁴ electrons, the detection rate is:

$6.7 \times 10^{-39} \times 1.67 \times 10^{34}$

$= 11.19 \times 10^{-5} \ s^{-1}$

= 9.668 per day

3 0

3 years ago