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andreyandreev [35.5K]
3 years ago
6

The weight of air measured in units of force per area is called _____.

Physics
2 answers:
laila [671]3 years ago
8 0
The weight of air measured in units of force per area is called air pressure.
ElenaW [278]3 years ago
8 0

Answer:

B. air pressure

Explanation:

The definition of pressure is that it is a force applied to an area.

P=\frac{F}{A}

Because weight is a force (a force is mass multiplied by acceleration, and weight is the product of the mass of an object multiplied by the acceleration of gravity, thus weight is a force), the measurement of the weight of the air over area is the air pressure.

The correct answe is B.

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3. How is using a model to study cells helpful?
4vir4ik [10]

Answer:

Yes, it is very helpful.

Explanation:

It's helpful since in a cell, plant or animal, there are a lot of different things. It's hard to memorize everything and know what they look like. Using a model can help you memorize everything better and even understand it better. If someone asked me where or what something was in a cell I think I would be able to recognize it better.

I hope this helps!

6 0
3 years ago
The speed of light can be measured by;
Bezzdna [24]

Answer:

it's the distance between objects in space

Explanation:  Light travels super fast; but it still takes a long time to travel between objects in space. This is because distances between objects in space are enormous.

And can i please receive a brainliest  and have a good day

4 0
3 years ago
An object is placed on a surface. A student tries to apply various combinations of forces on the object. Which pair of forces wi
AVprozaik [17]

Answer:

See Explanation

Explanation:

The question is incomplete, as there are no diagrams or options to provide more information to the question.

The general explanation is as follows:

For the object not to move

(1): The forces acting on the object must opposite each other. i.e. if force A acts at the right (or positive direction), force B will act at the left (or negative direction).

(2) The two forces must be equal.

So, for instance:

If the pair of forces are 5N and 5N in opposite directions, the object wil not move.

However, if one of the forces is greater, the object will move towards the direction of the greater force.

7 0
3 years ago
The head of a rattlesnake can accelerate at 49 m/s2 in striking a victim. If a car could do as well, how long would it take to r
Anettt [7]
<h2>Time taken is 0.459 seconds</h2>

Explanation:

We have equation of motion v = u + at

     Initial velocity, u = 0 m/s

     Final velocity, v = 81 km/hr = 22.5 m/s    

     Time, t = ?

     Acceleration, a = 49 m/s²

     Substituting

                      v = u + at  

                      22.5 = 0 + 49 x t

                      t = 0.459 seconds

   Time taken is 0.459 seconds

3 0
3 years ago
A swimming pool is 50 ft wide and 100 ft long and its bottom is an inclined plane, the shallow end having a depth of 4 ft and th
Nina [5.8K]

Explanation:

We define force as the product of mass and acceleration.

F = ma

It means that the object has zero net force when it is in rest state or it when it has no acceleration. However in the case of liquids. just like the above mentioned case, the water is at rest but it is still exerting a pressure on the walls of the swimming pool. That pressure exerted by the liquids in their rest state is known as hydro static force.

Given Data:

Width of the pool = w = 50 ft

length of the pool = l= 100 ft

Depth of the shallow end = h(s) = 4 ft

Depth of the deep end = h(d) = 10 ft.

weight density = ρg = 62.5 lb/ft

Solution:

a) Force on a shallow end:

F = \frac{pgwh}{2} (2x_{1}+h)

F = \frac{(62.5)(50)(4)}{2}(2(0)+4)

F = 25000 lb

b) Force on deep end:

F = \frac{pgwh}{2} (2x_{1}+h)

F = \frac{(62.5)(50)(10)}{2} (2(0)+10)

F = 187500 lb

c) Force on one of the sides:

As it is mentioned in the question that the bottom of the swimming pool is an inclined plane so sum of the forces on the rectangular part and triangular part will give us the force on one of the sides of the pool.

1) Force on the Rectangular part:

F = \frac{pg(l.h)}{2}(2(x_{1} )+ h)

x_{1} = 0\\h_{s} = 4ft

F = \frac{(62.5)(100)(2)}{2}(2(0)+4)

F =25000lb

2) Force on the triangular part:

F = \frac{pg(l.h)}{6} (3x_{1} +2h)

here

h = h(d) - h(s)

h = 10-4

h = 6ft

x_{1} = 4ft\\

F = \frac{62.5 (100)(6)}{6} (3(4)+2(6))

F = 150000 lb

now add both of these forces,

F = 25000lb + 150000lb

F = 175000lb

d) Force on the bottom:

F = \frac{pgw\sqrt{l^{2} + ((h_{d}) - h(s)) } (h_{d}+h_{s})   }{2}

F = \frac{62.5(50)\sqrt{100^{2}(10-4) } (10+4) }{2}

F = 2187937.5 lb

7 0
3 years ago
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