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Tresset [83]
3 years ago
10

A car accellerates from 12.0 m/s to 18 m/s in 4 seconds. What distance did it travel in this time?

Physics
1 answer:
Alona [7]3 years ago
6 0

Answer:

Just go broom broom on ya teacher

Explanation:

cause why not

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Explanation:

Muons in the atmosphere, a component of cosmic rays. Atmospheric muons are an essential component of cosmic ray showers. When a high energy primary particle coming from space collides with a nucleus of the upper atmosphere, it generates a spray of particles which later interact in their turn.

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Four wires meet at a junction. In two of the wires, currents and enter the junction. In one of the wires, current leaves the jun
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Explanation:

Kirchoff's current law: It states that the total current entering at a junction is equal to the total current leaving at the junction. It isbased on conservation of charge.  

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4 0
3 years ago
¿Cuál es la masa aproximada del aire en una habitación de 5.6 m * 3.8 m * 2.8 m?
Nady [450]
It would be 3.15 in meters
8 0
3 years ago
PLEASE HELP ASAP
morpeh [17]
A is the correct answer
7 0
3 years ago
A storage tank containing oil (SG=0.92) is 10.0 meters high and 16.0 meters in diameter. The tank is closed, but the amount of o
blagie [28]

Answer:

a-1 Graph is attached. The relation is linear.

a-2 The corresponding height for 68 kPa Pressure is 7.54 m

a-3 The corresponding weight for 68 kPa Pressure is 1394726kg

b The original height of the column is 5.98 m

Explanation:

Part a

a-1

The graph is attached with the solution. The relation is linear as indicated by the line.

a-2

By the equation

P=\rho \times g \times h

Here

  • P is the pressure which is given as 68 kPa.
  • ρ is the density of the oil whose SG is 0.92. It is calculated as

                                       \rho=S.G \times \rho_{water}\\\rho=0.92 \times 1000 kg/m^3\\\rho=920 kg/m^3\\

  • g is the gravitational constant whose value is 9.8 m/s^2
  • h is the height which is to be calculated

                                        P=\rho \times g \times h\\h=\frac{P}{\rho \times g}\\h=\frac{68 \times 10^3}{920 \times 9.8}\\h=7.54m

So the height of column is 7.54m

a-3

By the relation of volume and density

M=\rho \times V

Here

  • ρ is the density of the oil which is 920 kg/m^3
  • V is the volume of cylinder with diameter 16m calculated as follows

                             V=\pi r^2h\\V=3.14\times (8)^2 \times 7.54\\V=1515.23 m^3

Mass is given as

                             M=\rho \times V\\M=920 \times 1515.23\\M=1394726kg

So the mass of oil leading to 68kPa is 1394726kg

Part b

Pressure variation is given as

                            \Delta P=P_{obs}-P_{atm}\\\Delta P=115-101 kPa\\\Delta P=14 kPa\\

Now corrected pressure is as

P_c=P_g-\Delta P\\P_c=68-14 kPa\\P_c=54 kPa

Finding the value of height for this corrected pressure as

P_c=\rho \times g \times h\\h=\frac{P_c}{\rho \times g}\\h=\frac{54 \times 10^3}{920 \times 9.8}\\h=5.98m

The original height of column is 5.98m

4 0
3 years ago
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