How do I solve using the formula: Vf^2=vo^2+2gh ^2 means square

Estimate the number of photons emitted per second from 1.0 cm2 of a person's skin if a typical emitted photon has a wavelength o

Diano4ka-milaya [45]

Answer:

2.63 x 10^18

Explanation:

A = 1 cm^2 = 1 x 10^-4 m^2

λ = 10,000 nm = 10,000 x 10^-9 m = 10^-5 m

T = 37 degree C = 37 + 273 = 310 k

Energy of each photon = h c / λ

where, h is the Plank's constant and c be the velocity of light

Energy of each photon = (6.63 x 10^-34 x 3 x 10^8) / 10^-5 = 1.989 x 10^-20 J

Energy radiated per unit time = σ A T^4

Where, σ is Stefan's constant

Energy radiated per unit time = 5.67 x 10^-8 x 10^-4 x 310^4 = 0.05236 J

Number of photons per second = Energy radiated per unit time / Energy of

each photon

Number of photons per second = 0.05236 / (1.989 x 10^-20) = 2.63 x 10^18

4 0

3 years ago

A sled is moving down a steep hill. The mass of the sled is 50 kg and the net force acting on it is 20 N. What must be done to f

amid [387]

You need to first measure the angle of descent, i.e. the angle the hill makes with the ground. Then identify the forces acting on the sled, split them up into horizontal and vertical components, or into components that are parallel and perpendicular to the hill, and use Newton's second law to determine the components of the sled's acceleration vector.

There are at least 2 forces acting on the sled:

• its weight, pointing downward with magnitude W = m g

• the normal force, pointing perpendicular to the hill and away from the ground with mag. N

The question doesn't specify, but there might also be friction to consider, indicated in the attachment by the vector F pointing parallel to the slope of the hill and opposing the direction of the sled's motion with mag. F.

Splitting up the forces into parallel/perpendicular components is less work. By Newton's second law, the net force (denoted with ∑ or "sigma" here) in a particular direction is equal to the mass of the sled times its acceleration in that direction:

∑ (//) = W (//) = m a (//)

∑ (⟂) = W (⟂) + N = m a (⟂)

where, for instance, W (//) denotes the component of the sled's weight in the direction parallel to the hill, while a (⟂) denotes the component of the sled's acceleration perpendicular to the hill. If there is friction, you need to add -F to the first equation.

If the hill makes an angle of θ with flat ground, then W makes the same angle with the hill so that

W (//) = -m g sin(θ)

W (⟂) = -m g cos(θ)

So we have

-m g sin(θ) = m a (//) → a (//) = -g sin(θ)

-m g cos(θ) + N = m a (⟂) → a (⟂) = 0

where the last equality follows from the fact that the normal force exactly opposes the perpendicular component of the weight. This is because the sled is moving along the slope of the hill, and not into the air or into the ground.

Then the acceleration vector is

a = a (//)

with magnitude

||a|| = a = g sin(θ).

6 0

2 years ago

Marco is conducting an experiment. He knows the wave that he is working with has a wavelength of 32.4 cm. If he measures the fre

liq [111]

Answer: D. the wave has traveled 97.2 cm in 1 second.

Explanation:

Got it right on edge, hope this helps :)

7 0

3 years ago

• What value is used to identify each element in the periodic table?

bearhunter [10]

Answer:

protons

Explanation:

5 0

2 years ago

Colonist who belived in complete

velikii [3]

Answer:

Natural right

Explanation:

That is all what i know

5 0

3 years ago