2 years ago

What kind of map might have diagrams of air pressure?

Physics

1 answer:

Katyanochek1 [597]2 years ago

4 0

Synoptic maps can have diagrams of air pressure.

You might be interested in

What is the electric field at the point x=6.5 m? positive e-fields point to the right.?

Mademuasel [1]

6,5 6,4 6,3 6,2 6
if you know,1

5 0

2 years ago

25 POINTS FOR CORRECT ANSWER

castortr0y [4]

No. I do not agree with Stefan. Quite the contrary. I disagree
with his description of "<span>angle of incidence" as the angle between
the surface of the mirror and the incoming ray.

The correct description of "angle of incidence" is </span><span>the angle between
the NORMAL TO the surface of the mirror and the incoming ray.

Thus, the true angle of incidence is the complement of the angle that
Stefan calculates or measures.</span>

5 0

3 years ago

Read 2 more answers

Find the uniform acceleration that causes a car's velocity to change from 20.0 m/s to 105 m/s in and 12.0 s

Sauron [17]

Answer:

7.08 m/s²

Explanation:

Given:

v₀ = 20.0 m/s

v = 105 m/s

t = 12.0 s

Find: a

v = at + v₀

105 m/s = a (12.0 s) + 20.0 m/s

a = 7.08 m/s²

5 0

3 years ago

How did you manage to overcome the difficulties that arose?

Flauer [41]

Answer:

I just toughed it out and talked with friends

Explanation:

4 0

2 years ago

a uniform rod is hung at onen end and is partially submerged in water. If the density of the rod is 5/9 than of wter, find the f

34kurt

Answer:

$\frac{h_{liquid} }{ h_{body} }$ = 5/9

Explanation:

This is an exercise that we can solve using Archimedes' principle which states that the thrust is equal to the weight of the desalted liquid.

B = ρ_liquid g V_liquid

let's write the translational equilibrium condition

B - W = 0

let's use the definition of density

ρ_body = m / V_body

m = ρ_body V_body

W = ρ_body V_body g

we substitute

ρ_liquid g V_liquid = ρ_body g V_body

$\frac{\rho_{body} }{\rho_{liquid} } } = \frac{V_{liquid} }{V_{body} } }$

In the problem they indicate that the ratio of densities is 5/9, we write the volume of the bar

V = A h_bogy

Thus

$\frac{V_{liquid} }{V_{1body} } = \frac{ h_{liquid} }{h_{body} }$

we substitute

5/9 = $\frac{h_{liquid} }{ h_{body} }$

8 0

2 years ago