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Triss [41]
3 years ago
6

One of the largest barometers ever built was an oil-filled barometer constructed in Leicester, England in 1991. The oil had a he

ight of 12.2 m. Assuming a pressure of 1.013 × 105 Pa, what was the density of the oil used in the barometer?
Physics
1 answer:
ZanzabumX [31]3 years ago
5 0

Answer:

ρ = 830.32 kg/m³

Explanation:

Given that

Oil head = 12.2 m

h= 12.2 m

Pressure P = 1.013 x 10⁵ Pa

Lets take density of the liquid =ρ

The pressure due to liquid P given as

P = ρ g h

Now by putting the all values in the above equation

1.013 x 10⁵ Pa = ρ x 10 x 12.2                 ( take g =10 m/s²)

ρ = 830.32 kg/m³

Therefore the density of oil is 830.32  kg/m³

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A car starts from rest, speeds up with constant acceleration, and travels 400 meters in 10 seconds. What was the acceleration of
PIT_PIT [208]

Answer:

Acceleration of the car will be a=8m/sec^2

Explanation:

We have given that car starts from rest so initial velocity of the car u = 0 m/sec

And car traveled 400 m in 10 sec

So distance traveled by car s = 400 m

Time taken to compete this distance t = 10 sec

We have to find the acceleration of the car

From second equation of motion we know that s=ut+\frac{1}{2}at^2

So 400=0\times 10+\frac{1}{2}\times a\times 10^2

a=8m/sec^2

So acceleration of the car will be a=8m/sec^2

7 0
3 years ago
How does the speed of a sound wave change as the medium through which it traveks changes?
uranmaximum [27]
Increase as density increase and vise versa.
<span>The wavelength increases when a sound wave travels from a less dense to a more dense medium, the speed increases, and the frequency stays the same.</span>
8 0
3 years ago
ate around its central axis. A rope wrapped around the drum of radius 1.24 m exerts a force of 4.56 N to the right on the cylind
Mashcka [7]

Answer:

Magnitude the net torque about its axis of rotation is 1.3338 Nm

Explanation:

The radius of the wrapped rope around the drum, r = 1.24 m

Force applied to the right side of the drum, F = 4.56 N

The radius of the rope wrapped around the core, r' = 0.57 m

Force on the cylinder in the downward direction, F' = 7.58 N

Now, the magnitude of the net torque is given by:

\tau_{net} = \tau + \tau'

where

\tau = Torque due to Force, F

\tau' = Torque due to Force, F'

\tau = F\times r\tau' = F'\times r'

Now,

\tau_{net} = - F\times r + F'\times r'\tau_{net} = - 4.56\times 1.24 + 7.58\times 0.57 \\\\= - 1.3338\ Nm

The net torque comes out to be negative, this shows that rotation of cylinder is in the clockwise direction from its stationary position.

Now, the magnitude of the net torque:

|\tau_{net}| = 1.3338\ Nm

8 0
3 years ago
The following charges are located inside a submarine: 2.75 μC, −9.00 μC, 27.0 μC, and −43.5 μC.(a) Calculate the net electric fl
Kaylis [27]

Answer with Explanation:

We know that by Gauss's Law

E\cdot Flux=\frac{q_{inside}}{\epsilon _o}

Where

\epsilon _o is permitivity of free space

q_{inside} is the net charge in the hull

Part a)

Applying the given values we get

E\cdot Flux=\frac{(2.75+27-9-43.5)\times 10^{-6}C}{8.85\times 10^{-12}}\\\\\therefore E\cdot flux=-2.5\times 10^{6}

Part b)

Since the sign of the electric flux is negative we conclude that the electric field lines entering the submarine are more than the electric field lines leaving the submarine. Since negative sign implies that flux by negative charges is dominant.

7 0
3 years ago
A hot air balloon of total mass M (including passengers and luggage) is moving with a downward acceleration of magnitude a. As i
frez [133]

Answer:

When balloon moves in the downward direction two forces acts on it.  

    i) Force exerted by air in the upward direction  

    ii) Weight  

According to newton’s second law of motion:  

      Sum of forces = Ma  

                W – F = Ma  

               Mg – F = Ma     …….. (i)  

when some of the mass m is dropped and balloon is moving in upward direction with acceleration a/2 then,  

               F – W = (M-m)a/2

               F – (M-m)g = (M-m)a/2

               F – Mg + mg = Ma/2 – ma/2 ….. (ii)  

Adding equation (i) and (ii)  

              mg = M(3a/2) – ma/2

              m(g + a/2) = M(3a/2)

              m = M(3a/2)/(g + a/2)

5 0
3 years ago
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