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barxatty [35]
3 years ago
10

The level of mercury falls in a barometerwhile taking it to a mountain.​

Physics
1 answer:
Alexxandr [17]3 years ago
8 0

Answer:

This is due to a relative decrease in atmospheric pressure in high places.

Explanation:

Given that atmospheric pressure decreases at the higher point or ground, this reduced atmospheric pressure, however, will be unable to contain the Mercury in the barometer tube.

Therefore, at the top of the mountain where the air pressure is low, the barometer reading ultimately goes down.

Hence, the level of mercury falls in a barometer while taking it to a mountain "due to a relative decrease in atmospheric pressure in high places."

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What is the acceleration of a 1107 kg car as it comes to rest with a force of 2242 N in 6.9 seconds?
xz_007 [3.2K]

acceleration is 2.025m/s².

4 0
4 years ago
A stone is dropped from a cliff and falls 9.44 meters. What is the speed of the stone when it reaches the ground?
Lunna [17]

Answer:

option A

Explanation:

given,                                              

height of the drop of stone = 9.44 m

speed of the stone = ?                          

As the stone is dropped the energy of the stone will be conserved.

using conservation of energy.            

Potential energy = Kinetic energy    

m g h = \dfrac{1}{2} m v^2  

     v = \sqrt{2gh}                  

     v = \sqrt{2\times 9.8 \times 9.44}

     v = \sqrt{185.024}              

            v = 13.60 m/s                      

Hence, the correct answer is option A

3 0
3 years ago
For a damped simple harmonic oscillator, the block has a mass of 1.2 kg and the spring constant is 9.8 N/m. The damping force is
ArbitrLikvidat [17]

Answer:

a) t=24s

b) number of oscillations= 11

Explanation:

In case of a damped simple harmonic oscillator the equation of motion is

m(d²x/dt²)+b(dx/dt)+kx=0

Therefore on solving the above differential equation we get,

x(t)=A₀e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)

where A(t)=A₀e^{\frac{-bt}{2m}}

 A₀ is the amplitude at t=0 and

w' is the angular frequency of damped SHM, which is given by,

w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }

Now coming to the problem,

Given: m=1.2 kg

           k=9.8 N/m

           b=210 g/s= 0.21 kg/s

           A₀=13 cm

a) A(t)=A₀/8

⇒A₀e^{\frac{-bt}{2m}} =A₀/8

⇒e^{\frac{bt}{2m}}=8

applying logarithm on both sides

⇒\frac{bt}{2m}=ln(8)

⇒t=\frac{2m*ln(8)}{b}

substituting the values

t=\frac{2*1.2*ln(8)}{0.21}=24s(approx)

b) w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }

w'=\sqrt{\frac{9.8}{1.2}-\frac{0.21^{2}}{4*1.2^{2}}}=2.86s^{-1}

T'=\frac{2\pi}{w'}, where T' is time period of damped SHM

⇒T'=\frac{2\pi}{2.86}=2.2s

let n be number of oscillations made

then, nT'=t

⇒n=\frac{24}{2.2}=11(approx)

8 0
3 years ago
Where would a Christmas tree be most likely to grow?
sveticcg [70]
Taiga is the answer.

Hope it helps!
3 0
3 years ago
Who wants to do my missing physics work
kotegsom [21]

Answer:

<h2>WHAT IF I HATE PHYSICS ?</h2>

4 0
3 years ago
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