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klemol [59]
3 years ago
15

the pressure difference between an oil and water pipe is measured by double fluid manometer as shown in figure below for the giv

en fluid height and specific gravity calculate the pressure difference​
Physics
1 answer:
Ahat [919]3 years ago
8 0
The answer above me is correct
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If 30 N of force is exerted over an area of 20 m2, how much pressure is being applied?
Ludmilka [50]
Pressure= F/A
Given, F=30 N
Area=20 m^2
Putting values in formula,
Pressure = 30/20
=3/2
=1.5 Pascals
4 0
3 years ago
Suppose we wish to use a 6.0 m iron bar to lift a heavy object by using it as a lever. If we place the pivot point at a distance
Ber [7]

Answer:

The maximum load that this person is able to lift is 34.3 N

Explanation:

Applying the balancing torque, the expression is equal:

F₁L₁ = F₂L₂

mgL_{1} =F_{2} L_{2}

Where

g = 9.8 m/s² = gravity

L₁ = 0.8 m

F₂ = 527 N

L₂ = 6 - 0.8 = 5.2 m

Replacing and clearing the mass m:

m=\frac{F_{2}L_{2}  }{gL_{1} } =\frac{527*5.2}{9.8*0.8} =3.5kg

The maximum load that this person is able to lift is:

F = m * g = 3.5 * 9.8 = 34.3 N

8 0
3 years ago
Two sailboats leave a harbor in the bahamas at the same time. the first sails at 23 mph in a direction 330°. the second sails at
pishuonlain [190]

First, let us calculate the total distance that each have taken after 2 hours.

Let’s say that:

A = sailboat which sails at 23 mph in a direction 330°

B = sailboat which sails at 34 mph in a direction 190°

Calculating for distances:

dA = 23 mph (2 hours) = 46 miles

dB = 34 mph (2 hours) = 68 miles

Imagining a Cartesian coordinate, the angle θ between the two sailboats is simply the difference:

θ = 330° - 190°

θ = 140°

We know that from the law of cosines:

c^2 = a^2 + b^2 – 2 a*b*cos θ

Therefore the distance between the two after 2 hours, C, is:

C^2 = 46^2 + 68^2 – 2 (46) (68) cos(140)

<span>C = 107.39 miles</span>

7 0
3 years ago
At a given moment, a plane passes directly above a radar station at an altitude of 6 km. Let θ be the angle that the line throug
Rom4ik [11]

- 187.237 km/hr fast is θ changing 12 min after the plane passes over the radar station

<u>Explanation:</u>

Let the distance x and angle θ be defined as in the figure below. Then

                  \tan \theta=\frac{6}{x}

Now, differentiate with respect to t, we get

                 \sec ^{2} \theta \frac{d \theta}{d t}=-\left(\frac{6}{x^{2}}\right) \frac{d x}{d t}

Now, calculate the travel distance from radar station to plane after 12min

Distance,  x=800 \times \frac{12}{60}=160

Substituting ‘x’ value, we get

                \tan \theta=\frac{6}{160}=\frac{3}{80}

Find the rate of change of theta after 12 min,

                \frac{d \theta}{d t}=-\frac{1}{\sec ^{2} \theta} \times \frac{6}{x^{2}} \times \frac{d x}{d t}

We know, the formula for,

                \sec ^{2} \theta=1+\tan ^{2} \theta=1+\frac{3^{2}}{80^{2}}

So, then, \frac{d x}{d t}=800 \mathrm{km} / \mathrm{hr}(\text { let assume })

               \frac{d \theta}{d t}=-\frac{1}{\left(1+\frac{3^{2}}{80^{2}}\right)} \times \frac{6}{160^{2}} \times 800

               =-\frac{1}{\left(1+\frac{3^{2}}{80^{2}}\right)} \times \frac{6}{160^{2}} \times 800

               =-\frac{1}{1.0014} \times \frac{6}{25600} \times 800

             =-\frac{4800}{25635.84}=-0.187237 \mathrm{rad} / \mathrm{hr}

When express the value in km/he, we get, the change in theta as

              =-187.237 \mathrm{km} / \mathrm{hr}

5 0
3 years ago
Para proteger un computador de sobrecargas eléctricas, Juan coloca un filamento delgado de cobre llamado fusible en su circuito,
Verdich [7]

Answer:

Los fusibles están diseñados de tal forma que estos se "rompen" o se funden, cuando la demanda eléctrica supera un dado valor (cuando demasiada electricidad pasa a través de el).

Una vez el filamento se rompe, la corriente ya no puede circular por el (podes pensar en esta situación como un cable roto, la electricidad no puede circular por este cable)

Entonces, al romperse el filamento, en caso de una sobrecarga eléctrica, el flujo de electricidad se corta, y de esta forma se protege al computador de posibles sobrecargas.

4 0
3 years ago
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