work done by gravitational force = mass × g × height
= mgh
= 5 × 10 × 8 N
<h3>= 400 N</h3>
The mitochondria is the powerhouse of the cell because it takes nutrients and breaks it down to provide energy for the cell.
Answer:
- Here we use the conservation of momentum theorem.
- m stands for mass, and v stands for velocity. The numbers refer to the respective objects.
- m1v1 + m2v2 = m1vf1 + m2vf2
- Since the equation is perfectly inelastic, the final velocity of both masses is the same. Let’s account for this in our formula.
- m1v1 + m2v2 = vf(m1 + m2)
<u>Let’s substitute in our givens.</u>
(0.002 kg)(700 m/s) + (5 kg)(0 m/s) = vf(0.002 kg + 5 kg)
I assume you are proficient in algebra I, so I will not include the steps to simplify this equation.
Note that I have considered the bullet’s velocity to be in the positive direction,
The answer is vf = 0.280 m/s
Answer:
Power = 21[W]
Explanation:
Initial data:
F = 35[N]
d = 18[m]
In order to solve this problem we must remember the definition of work, which tells us that it is equal to the product of a force for a distance.
Therefore:
Work = W = F*d = 35*18 = 630 [J]
And power is defined as the amount of work performed in a time interval.
Power = Work / time
Time = t = 30[s]
Power = 630/30
Power = 21 [W]
Answer:
correct answer is d
The pinhole has a size comparable with the laser wavelength, so the pinhole diffracts the passing laser beam.
Explanation:
The researcher does not have a refraction problem since the medium on both sides of the pinhole is the same with the same refractive index, the problem he is having is with the diffraction of the laser beam through the pinhole, let's analyze the diffraction process that is described by the expression
a sin θ = m λ
where a is the pinlole size, λ the wavelength of the laser and m an integer.
The laser extends from the maximum of diffraction to the first zero (m = 1) of diffraction
sin θ = λ/a
when analyzing this expression we have some interesting cases
* when λ « a. the sine approaches zero therefore we are in the case of optical geometry, in this case the laser passes through the hole without being diffracted
* when λ ≈ a. the sine function has values between 0 and 1, for which a diffraction of the laser beam occurs, which increases the diameter of the same
* when λ> a. The laser does not pass through the gap since the sine cannot have values greater than 1
After this analysis, we review the answers to the exercise to find that the correct answer is d