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1 year ago

Answer fast , is si unit newton/meter square or pascal or both

Physics

1 answer:

Readme [11.4K]1 year ago

8 0

Answer:

Both

Explanation:

The S.I. unit of pressure is newton/meter square or pascal as both represent the same dimensional value.

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You hear a sound in the distance. Suddenly the sound gets deeper, decreasing in pitch. Which can you assume about the sound wave

oksian1 [2.3K]

Answer:

A. The wavelengths of the new sound waves are longer

Explanation:

This is the Doppler effect which can be best illustraded for the case of a siren of an ambulance approaching us having a greater frequency and getting lower in frequency and deeper as the ambulance passes us.

Since the wavelength is inversely proportional to the frequency it follows the wavelengths are longer when the frequency decreases lowering its pitch and getting deeper.

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2 years ago

A dog barking at the sound of the postal carrier delivering mail is reacting to what type of cue?

Gwar [14]

Answer:

A. External

Explanation:

External stimulus includes touch/pain, vision, smell, taste, and sound.

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1 year ago

Who can help me?? physic question

IgorC [24]

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2 years ago

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s2008m [1.1K]

The answer should be B

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2 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:

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