Answer:
13,750 N
Yes
Explanation:
Given:
v₀ = 90 km/h = 25 m/s
v = 0 m/s
t = 4 s
Find: a and Δx
a = Δv / Δt
a = (0 m/s − 25 m/s) / (4 s)
a = -6.25 m/s²
F = ma
F = (2200 kg) (-6.25 m/s²)
F = -13,750 N
Δx = ½ (v + v₀) t
Δx = ½ (0 m/s + 25 m/s) (4 s)
Δx = 50 m
Answer:
(b) The electrons, because they have the smaller momentum and, hence, the larger de Broglie wavelength
Explanation:
de Broglie wavelength λ = h / m v
Since both electrons and protons have same velocity , momentum mv will be less for electrons because mass of electron is less .
for electron , momentum is less so . Therefore de Broglie wavelength λ will be more for electrons .
Amount of diffraction that is angle of diffraction is proportional to λ
Therefore electrons having greater de Broglie wavelength will show greater diffraction.
Hi! Decomposers (mainly soil bacterium, fungus, or invertebrate)<span /> are categorized as consumers due to the fact that they consume dead organic matter such as plants and animals. They differ from producers (green plants and some bacteria) because they do not produce their own food using photosynthesis or chemosythensis.
Hope this helped!
Answer:
1.84 hours or 110.4 minutes
Explanation:
8 miles 1.61 km 1 hr
------------ x ----------------- x -------------- = 1.84 hours or 110.4 minutes
1 mile 7 km
Answer:
h'=0.25m/s
Explanation:
In order to solve this problem, we need to start by drawing a diagram of the given situation. (See attached image).
So, the problem talks about an inverted circular cone with a given height and radius. The problem also tells us that water is being pumped into the tank at a rate of
. As you may see, the problem is talking about a rate of volume over time. So we need to relate the volume, with the height of the cone with its radius. This relation is found on the volume of a cone formula:

notie the volume formula has two unknowns or variables, so we need to relate the radius with the height with an equation we can use to rewrite our volume formula in terms of either the radius or the height. Since in this case the problem wants us to find the rate of change over time of the height of the gasoline tank, we will need to rewrite our formula in terms of the height h.
If we take a look at a cross section of the cone, we can see that we can use similar triangles to find the equation we are looking for. When using similar triangles we get:

When solving for r, we get:

so we can substitute this into our volume of a cone formula:

which simplifies to:


So now we can proceed and find the partial derivative over time of each of the sides of the equation, so we get:

Which simplifies to:

So now I can solve the equation for dh/dt (the rate of height over time, the velocity at which height is increasing)
So we get:

Now we can substitute the provided values into our equation. So we get:

so:
