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3 years ago

What measures air pressure.?

2 answers:

8 0

That depends on what type of pressure you are attempting to measure, to measure Atmospheric pressure, you would use a Barometer. To measure things like tires, you could use a Tire Pressure Gauge. For Industrial processes and boilers, you would use a Manometer. For pressure vessels, you would use a Bordon Gauge. <span />

Contact [7]3 years ago

5 0

Manometers Is your answer for that question.

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A slug has a speed of .00242 miles per hour

Delvig [45]

I like fortnite too:):):)

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3 years ago

Explain the difference between mass and weight (in at least 2 sentences)

Irina-Kira [14]

Mass= is how big something is.
Weight= is how heavy something is.
They are different things because weight is talking about heavy... not how big it is.

5 0

3 years ago

Read 2 more answers

A hot air balloon exerts a force of 1200 N while

ValentinkaMS [17]

Answer:

Explanation:

For the explained scenario in the free body force diagram definitely the two forces 1200 N and 800 N should present as they are the acting forces

So A & D rules out.

Then you must think of B & C.

You also know that the weight of the load is always acting downwards as that force is generated by gravitational field of Earth. So 800 N should be downwards not upwards. That rules out B.

So answer is C

(Free body diagram is shown in the graph)

4 0

4 years ago

A uniform 1500-kg beam, 20.0 m long, supports a 15,000-kg printing press

Ghella [55]

Answer:

$\mathbf{F_1=4.41*10^4\ N}$

$\mathbf{F_2 = 1.176*10^5 \ N}$

Explanation:

The missing image of the figure slide is attached in below.

However, from the model, it is obvious that it is in equilibrium.

As a result, the relation of the force and the torque is said to be zero.

i.e.

$\sum F = 0$ and $\sum \tau = 0$

From the image, expressing the forces through the y-axis, we have:

$F_1+F_2 = W_B + W_P \\ \\ \implies 9.8(1500+15000) \\ \\ \implies \mathtt{1.617\times 10^5 \ N}$

Also, let the force $F_1$ be the pivot and computing the torque to determine $F_2$ :

Then:

$F_1(0)+F_2(20.0) = 10.0W_B + 15.0W_P$

$F_2 = \dfrac{((10*1500)+(15*15000))*9.8}{20.0}$

$F_2 = 117600 \ N$

$\mathbf{F_2 = 1.176*10^5 \ N}$

For the force equation:

$F_1+F_2=1.617*10^5 \ N;$

where:

$F_2 = 1.176*10^5 \ N$

Then:

$F_1+1.176*10^5 \ N=1.617*10^5 \ N$

$F_1=1.617*10^5 \ N-1.176*10^5 \ N$

$F_1=44100\ N$

$\mathbf{F_1=4.41*10^4\ N}$

8 0

3 years ago

A box weights 50kg applies friction force of 125N. You apply a force of 150N going in the right direction. What is the total net

Pachacha [2.7K]

Answer:

Net force = 25N and in the right direction

Explanation:

mass of box, m = 50kg

Force of block = 150N

Friction force = 125N

$Net force = Force Applied - Friction force = F - F_f$

150 - 125 = 25N

the force applied to the block is greater than the friction force. there the force applied will overcome the friction force and move in the right direction of the force applied

4 0

3 years ago