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2 years ago

Which statement best describes the term absolute threshold?

1 answer:

7 0

I believe the correct answer from the choices listed above is option B. Absolute threshold refers to the lowest intensity at which a person can detect stimuli. It <span>is the smallest level of energy required by an external stimulus to be detectable by the human senses. Hope this answers the question.</span>

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A hydrogen discharge lamp emits light with two prominent wavelengths: 656 nm (red) and 486 nm (blue). The light enters a flint-g

mezya [45]

Answer:

The angle between the red and blue light is 1.7°.

Explanation:

Given that,

Wavelength of red = 656 nm

Wavelength of blue = 486 nm

Angle = 37°

Suppose we need to find the angle between the red and blue light as it leaves the prism

$n_{r}=1.572$

$n_{b}=1.587$

We need to calculate the angle for red wavelength

Using Snell's law,

$n_{r}\sin\theta_{i}=n_{a}\sin\theta_{r}$

Put the value into the formula

$1.572\sin37=1\times\sin\theta_{r}$

$\theta_{r}=\sin^{-1}(\dfrac{1.572\sin37}{1})$

$\theta_{r}=71.0^{\circ}$

We need to calculate the angle for blue wavelength

Using Snell's law,

$n_{b}\sin\theta_{i}=n_{a}\sin\theta_{b}$

Put the value into the formula

$1.587\sin37=1\times\sin\theta_{b}$

$\theta_{b}=\sin^{-1}(\dfrac{1.587\sin37}{1})$

$\theta_{b}=72.7^{\circ}$

We need to calculate the angle between the red and blue light

Using formula of angle

$\Delta \theta=\theta_{b}-\theta_{r}$

Put the value into the formula

$\Delta \theta=72.7-71.0$

$\Delta \theta=1.7^{\circ}$

Hence, The angle between the red and blue light is 1.7°.

8 0

2 years ago

Which biome contains mostly coniferous trees and receives 35 to 100 cm of rain per year?

a_sh-v [17]

The question is incomplete as it does not have the options which are:

deciduous forest

taiga (boreal forest)

temperate rainforest

tropical rainforest

Answer:

Taiga (boreal forest)

Explanation:

A Biome refers to the habitat which is occupied by flora and fauna living in similar conditions. These biomes are distinguished based on many features like precipitation, temperature and many other physical factors.

In the given question, the biome which receives an annual rainfall of 35 to 100 cm annually and is mostly covered by the coniferous trees is known as "Taiga biome" which is also known as Boreal forest.

The Taiga biome is one of the largest terrestrial biomes which is present in Eurasia and North America. The biome is characterised by the conifers trees and therefore is also known as the Coniferous trees.

Thus, Taiga (boreal forest) is the correct answer.

4 0

2 years ago

One of the Lady Spartans was falling to the ground after

dem82 [27]

Answer:

Explanation:

Given that,

A lady falling has a final velocity of 4m/s

v = 4m/s

Mass of the lady is 60kg.

m = 60kg

Using conservation of energy, the potential energy of the body from the point where the lady is dropping is converted to the final kinetic energy of the lady.

Therefore,

P.E = K.E(final) = ½mv²

P.E = ½ × 60 × 4²

P.E = 480 J.

0 0

2 years ago

You catch a volleyball (mass 0.270 kg) that is moving downward at 7.50 m/s. In stopping the ball, your hands and the volleyball

sesenic [268]

Explanation:

The work done equals the change in energy.

W = ΔKE

W = 0 − ½mv²

W = -½ (0.270 kg) (-7.50 m/s)²

W = -7.59 J

Work is force times displacement.

W = Fd

-7.59 J = F (-0.150 m)

F = 50.6 N

3 0

2 years ago

Consider steady-state conditions for one-dimensional conduction in a plane wall having a thermal conductivity k = 50 W/m · K and

tatuchka [14]

Answer:

solution:

dT/dx =T2-T1/L

q_x = -k*(dT/dx)

dT/dx= (-20-50)/0.35==> -280 K/m

q_x =-50*(-280)*10^3==>14 kW

Case (2)

dT/dx= (-10+30)/0.35==> 80 K/m

q_x =-50*(80)*10^3==>-4 kW

Case (2)

dT/dx= (-10+30)/0.35==> 80 K/m

q_x =-50*(80)*10^3==>-4 kW

Case (3)

q_x =-50*(160)*10^3==>-8 kW

T2=T1+dT/dx*L=70+160*0.25==> 110° C

Case (4)

q_x =-50*(-80)*10^3==>4 kW

T1=T2-dT/dx*L=40+80*0.25==> 60° C

Case (5)

q_x =-50*(200)*10^3==>-10 kW

T1=T2-dT/dx*L=30-200*0.25==> -20° C

note:

all graph are attached

6 0

2 years ago