Answer:
The work done by Joel is greater than the work done by Jerry.
Explanation:
Let suppose that forces are parallel or antiparallel to the direction of motion. Given that Joel and Jerry exert constant forces on the object, the definition of work can be simplified as:
[tex]W = F\cdot \Delta s[/tex]
Where:
[tex]W[/tex] - Work, measured in joules.
[tex]F[/tex] - Force exerted on the object, measured in newtons.
[tex]\Delta s[/tex] - Travelled distance by the object, measured in meters.
During the first 10 minutes, the net work exerted on the object is zero. That is:
[tex]W_{net} = W_{Joel} - W_{Jerry}[/tex]
[tex]W_{net} = F\cdot \Delta s - F\cdot \Delta s[/tex]
[tex]W_{net} = (F-F)\cdot \Delta s[/tex]
[tex]W_{net} = 0\cdot \Delta s[/tex]
[tex]W_{net} = 0\,J[/tex]
In exchange, the net work in the next 5 minutes is the work done by Joel on the object:
[tex]W_{net} = W_{Joel}[/tex]
[tex]W_{net} = F\cdot \Delta s[/tex]
Hence, the work done by Joel is greater than the work done by Jerry.
How does increasing frequency affect the crests of a wave?
They get higher.
They get closer together.
They get lower.
They get farther apart.
Answer:
they get closer together
Explanation:
A ball is thrown vertically upwards from the roof of a building with an initial velocity of 30 m / s. If it stops in the air 220 m above the ground, what is the height of the building?
Answer:
175 m
Explanation:
Given:
y = 220 m
v₀ = 30 m/s
v = 0 m/s
a = -10 m/s²
Find: y₀
v² = v₀² + 2a (y − y₀)
(0 m/s)² = (30 m/s)² + 2 (-10 m/s²) (220 m − y₀)
y₀ = 175 m
a skydiver jumped out of a plane and fell 9 miles to the ground. his average speed while falling was 174 miles/hour. how much time did the dive last?
Answer:
t = 0.051 seconds
Explanation:
Given that,
A skydiver jumped out of a plane and fell 9 miles to the ground. It means 9 miles is the distance covered.
The average speed of the skydiver is 174 miles/hour
We need to find the time for which he dive last. It means t is the time taken. Total distance covered divided by time taken is called average speed. So,
[tex]v=\dfrac{d}{t}\\\\t=\dfrac{d}{v}\\\\t=\dfrac{9}{174}\\\\t=0.051\ s[/tex]
So, a skydiver dive for 0.051 seconds.
A very thin film of soap, of thickness 170 nm, in between air seems dark. On the other hand, when placed on top of glass some visible light is seen to shine from the film. How can this happen and what is the smallest visible light that creates constructive interference when we place the film on top of glass
Answer:
λ₀ = 2 d n
Explanation:
A soap film is a layer where the lus is reflected on the surface and on the inside of the film, these two reflected rays can interfere with each other either constructively or destructively.
Let's analyze the general conditions of this interference,
* When the ray of light reaches the surface of the film it is reflected, as the index of refraction of the air is less than the index of the film, the reflected ray has a phase change of 180º
* When the ray penetrates the film, its wavelength changes due to the refractive index of the film.
λ = λ₀ / n
where lick is the wavelength in the vacuum or air and n index of refraction of the film, in general this interference is observed perpendicular to the film, so the sine veils 1. the expression for constructive interference taking in what previous remains
2d = (m + ½) λ
the expression for destructive interference remains
2d = m λ
2d = m λ₀ / n
When the film is placed on a glass plate whose index of refraction is greater than the index of refraction of the film, in the reflection in the lower part of the film another phase difference of 180º is created, for which we have a difference of total phase of 180 +180 = 360º, which is equivalent to no phase difference, therefore the two previous equations are interchanged.
Therefore where we had destructive interference now a cosntructive interference happens we can see the reflected light.
Find us the wavelength that this constructive interference creates
2d n = m λ₀
λ₀ = 2 d n / m
To find the minimum wavelength, suppose we observe the first interference pattern m = 1
λ₀ = 2 d n
where d is the thickness of the film and n the index of refraction of the same