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3 years ago

A scientist wants to use a model to help present results of his detailed scientific investigation

1 answer:

4 0

A problem cause damage damage to people and people who have been doing the damage. A scientist call you when I am done with science

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Please someone answer this ASAP❗️❗️‼️‼️

kogti [31]

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3 years ago

What must happen to an atom of magnesium in order to become a magnesium ion

Zielflug [23.3K]

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3 years ago

From a set of graphed data the slope of the best fit line is found to be 1.35 m/s and the slope of the worst fit line is 1.29m/s

Svetradugi [14.3K]

Solution:

Let the slope of the best fit line be represented by ' $m_{best}$ '

and the slope of the worst fit line be represented by ' $m_{worst}$ '

Given that:

$m_{best}$ = 1.35 m/s

$m_{worst}$ = 1.29 m/s

Then the uncertainity in the slope of the line is given by the formula:

$\Delta m = \frac{m_{best}-m_{worst}}{2}$ (1)

Substituting values in eqn (1), we get

$\Delta m = \frac{1.35 - 1.29}{2}$ = 0.03 m/s

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3 years ago

A bowling ball has a mass of 7.2 kg and a weight of 70.6 N. It moves down the bowling alley at 1 m/s and strikes a pin with a fo

Inga [223]

The answer is 15.0N its explained in newtons third law hope this helps:)

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3 years ago

Read 2 more answers

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:

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