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Sveta_85 [38]
3 years ago
7

What is the length of the rod as measured by this ruler?

Physics
1 answer:
Simora [160]3 years ago
7 0

2.2 CM is your answer

jfkanfnakcndbcnd

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the ocean floor is, on average, 4267 m below sea level. What is the pressure in the atmosphere at this depth?
hichkok12 [17]
The pressure at the depth h in the ocean is given by (Stevin's law)
p= p_0 + \rho g h
where
p_0 = 1.0 \cdot 10^5 Pa is the atmospheric pressure
and \rho g h is the pressure exerted by the column of water of height h=4267 m, with \rho = 1000 kg/m^3 being the water density and g=9.81 m/s^2.
Substituting, we find
p=1.0 \cdot 10^5 Pa + (1000 kg/m^3)(9.81 m/s^2)(4267 m)=4.20 \cdot 10^7 Pa
We want to convert this into atmospheres: we know that 1 atm corresponds to the atmospheric pressure at sea level, so 1 atm=1.0\cdot 10^5 Pa, therefore we just need to divide by this number:
p= \frac{4.20 \cdot 10^7 Pa}{1.0 \cdot 10^5 Pa/atm} =420 atm
7 0
3 years ago
A 600-kg car traveling at 30.0 m/s is going around a curve having a radius of 120 m that is banked at an angle of 25.0°. The coe
ikadub [295]

Answer:

f_{fr}=1590.85 N

Explanation:

Here the total force at the horizontal components will be equal to the centripetal force on the car. So we will have:

f_{fr}cos(25)+Nsin(25)=m\frac{v^{2}}{r} (1)

  • f(fr) is the friction force
  • N is the normal force

Now, the sum of forces at the vertical direction is equal to 0.

Ncos(25)-mg-f_{fr}sin(25)=0 (2)          

Let's combine (1) and (2) to find f(fr)

f_{fr}=\frac{(mv^{2}/r)-mgtan(25)}{cos(25)+tan(25)sin(25)}

f_{fr}=\frac{(600*30^{2}/120)-600*9.81*tan(25)}{cos(25)+tan(25)sin(25)}  

f_{fr}=1590.85 N

I hope it helps you!

5 0
3 years ago
Three polarizing filters are shown in the figure below. The alignment of each filterâs transmission axis is shown, as measured f
lidiya [134]

Answer: B.

Explanation:

Hope help :p

5 0
3 years ago
State the law of universal gravitation, and use examples to explain how changes in mass and changes in distance affect gravitati
juin [17]
F(g)= Gm1m2/ r^2 If mass is increased, so will the force of gravity because it is in direct relationship with the gravitational force, but if distance is increased, the force of gravity will decrease because it is indirectly related ( since it is on the bottom of the equation)
6 0
3 years ago
If the wavelength of a wave increases does its frequency also increase?
irinina [24]
Frequency decreases whilst wavelength increases and the opposite also occurs
4 0
3 years ago
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