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3 years ago

When a train is traveling along a

Physics

2 answers:

baherus [9]3 years ago

8 0

Railroad should be your answer

Deffense [45]3 years ago

6 0

It's should be a railroad mag dude

You might be interested in

How do i balance WO3 + H2 = W+ H₂O<br>

Levart [38]

Answer:

Hello, I think it'll be:

WO3 + 3H2 = W + 3H₂O

5 0

3 years ago

Read 2 more answers

A bowling ball (mass = 7.2 kg, radius = 0.11 m) and a billiard ball (mass = 0.38 kg, radius = 0.028 m) may each be treated as un

Semenov [28]

Answer:

Explanation:

Given that

Mass of bowling ball M1=7.2kg

The radius of bowling ball r1=0.11m

Mass of billiard ball M2=0.38kg

The radius of the Billiard ball r2=0.028m

Gravitational constant

G=6.67×10^-11Nm²/kg²

The magnitude of their distance apart is given as

r=r1+r2

r=0.028+0.11

r=0.138m

Then, gravitational force is given as

F=GM1M2/r²

F=6.67×10^-11×7.2×0.38/0.138²

F=9.58×10^-9N

The force of attraction between the two balls is

F=9.58×10^-9N

3 0

3 years ago

A normal blood pressure reading is less than 120/80 where both numbers are gauge pressures measured in millimeters of mercury (m

KIM [24]

The pressure value is given by the equation,

$P= \rho gh$

Where,

$\rho$ represents the density of the liquid

g= gravity

h= Heigth

A) For the measurement of the guage pressure we have the data data,

$\rho_{mercury} = 13.6*10^3kg/m^3$

$h=0.0930m$

$g=9.8m/s^2$

Replacing we get,

$P_g=\rho_{mercury}gh$

$[tex]P_g = (13.6*10^3)(9.8)(0.0930)$ P_g = 12395Pa[/tex]

In order to find the Absolute pressure, we perform a sum between the atmospheric pressure and that of the Gauge,

B) The atmospheric pressure at sea level is 101325Pa, assuming ideal conditions, we will take this pressure for our calculation, so

$P_T = P_a+P_g\\P_T = 101325Pa+12395Pa\\P_T = 113720Pa\\P_T = 113.7kPa$

6 0

3 years ago

A 3.42 kg mass hanging vertically from a spring on the Earth (where g = 9.8 m/s2) undergoes simple oscillatory motion. If the sp

sineoko [7]

Answer:

Time period of oscillation on moon will be equal to 3.347 sec

Explanation:

We have given mass which is attached to the spring m = 3.42 kg

Spring constant K = 12 N/m

We have to find the period of oscillation

Period of oscillation is equal to $T=2\pi \sqrt{\frac{m}{K}}$ , here m is mass and K is spring constant

So period of oscillation $T=2\times 3.14\times \sqrt{\frac{3.42}{12}}$

$T=2\times 3.14\times 0.533=3.347sec$

So time period of oscillation will be equal to 3.347 sec

As it is a spring mass system and from the relation we can see that time period is independent of g

So time period will be same on earth and moon

3 0

3 years ago

Please please help. I don’t understand this.

Kazeer [188]

Which questions do you need help? I could help you

5 0

4 years ago