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Rufina [12.5K]
3 years ago
13

A barometer reads 780 mm Hg. Mercury has a density of 1.36 x 10^4 kg /m^3.

Physics
2 answers:
iris [78.8K]3 years ago
7 0

The pressure of the atmosphere, when a barometer reads 780 mm Hg.    Mercury which a density of 1.36 x 10^4 kg /m^3 is B 1.1 x 10^5 N/m^2

This problem can be solved using the formula below

P = dgh................. Equation 1

Where P = Pressure of the atmosphere, d = density of the mercury, h = height of the mercury, g = acceleration due to gravity.

From the question,

Given: d = 1.36×10⁴ kg/m³, h = 780 mm = 0.78 m,

Constant: g = 10 m/s²

Substitute these values into equation 1

P = (1.36×10⁴)(10)(0.78)

P = 10.608×10⁴ N/m²

P ≈ 1.1×10⁵ N/m²

Hence the right answer is B. 1.1×10⁵ N/m²

Learn more about Pressure here: brainly.com/question/23603188

frutty [35]3 years ago
5 0

Answer:

B

Explanation:

Pressure = density × g × height

p \:  = (1.36 \times  {10}^{4} )(10)(780 \times  {10}^{ - 3}  ) \\ p = 1.1 \times  {10}^{5}

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harina [27]
Hello.

The answer is:

It creates a spark.

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3 years ago
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Answer the following question​
Ray Of Light [21]

Answer:

A) OA, AB, BC

B) 25m/s^2

C) see explanation

D) 25

E) Rest

Explanation:

From the Velocity time graph shown:

The positive slope = OA ; This is positive because, it is the point of uniform acceleration on the graph.

Constant slope = AB, the slope here is constant because, AB on the graph is the point of constant velocity.

-ve slope = BC

B) Acceleration of body in path OA.

Acceleration = change in Velocity / time

Acceleration = (150 - 0) / 6

Acceleration = 150/6 = 25m/s^2

C) Path AB is Parallel to the because it marks the period of constant velocity (that is Velocity does not increase or decrease during the time interval).

D) Length of BC

BC corresponds to the distance moved, that velocity / time

Velocity = 150 ; time = 6

Therefore Distance (BC) = 150/6 = 25

E.) Velocity =0 ; Hence body is at rest

5 0
3 years ago
A box rests on top of a flat bed truck. The box has a mass of m = 16.0 kg. The coefficient of static friction between the box an
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Answer:

1) 1.31 m/s2

2) 20.92 N

3) 8.53 m/s2

4) 1.76 m/s2

5) -8.53 m/s2

Explanation:

1) As the box does not slide, the acceleration of the box (relative to ground) is the same as acceleration of the truck, which goes from 0 to 17m/s in 13 s

a = \frac{\Delta v}{\Delta t} = \frac{17 - 0}{13} = 1.31 m/s2

2)According to Newton 2nd law, the static frictional force that acting on the box (so it goes along with the truck), is the product of its mass and acceleration

F_s = am = 1.31*16 = 20.92 N

3) Let g = 9.81 m/s2. The maximum static friction that can hold the box is the product of its static coefficient and the normal force.

F_{\mu_s} = \mu_sN = mg\mu_s = 16*9.81*0.87 = 136.6N

So the maximum acceleration on the block is

a_{max} = F_{\mu_s} / m = 136.6 / 16 = 8.53 m/s^2

4)As the box slides, it is now subjected to kinetic friction, which is

F_{\mu_s} = mg\mu_k = 16*9.81*0.69 = 108.3 N

So if the acceleration of the truck it at the point where the box starts to slide, the force that acting on it must be at 136.6 N too. So the horizontal net force would be 136.6 - 108.3 = 28.25N. And the acceleration is

28.25 / 16 = 1.76 m/s2

5) Same as number 3), the maximum deceleration the truck can have without the box sliding is -8.53 m/s2

3 0
3 years ago
An empty elevator has a mass of 722 kg. It moves between floors at a maximum speed of 6.00 m/s. The elevator is stopped on the 2
FinnZ [79.3K]

Answer:

Explanation:

The lift is going down with acceleration

Initial speed u = 0

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distance s = 15.25 m

acceleration a = ?

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a = 1.18 m /s²

Elevator is going down with acceleration  .

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T = 6223.64 N .

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Answer:

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Explanation:

I hope this helped you

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