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3 years ago

Which of the following statements about Masters programs is not correct?

Physics

1 answer:

uranmaximum [27]3 years ago

7 0

The correct answer is C. The level of competition is not very high in most Masters programs.

Explanation:

In sports, the word "master" is used to define athletes older than 30 and that usually are professional or have trained for many years, although novates are also allowed. This means in most cases in Master programs and teams a high level of competition can be expected due to the experience and extensive training of Master athletes. Indeed, many records in the field of sport belong to Master athletes rather than younger athletes. According to this, the incorrect statement is "The level of competition is not very high in most Masters programs".

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A football is kicked straight up from a height of 5 feet with an initial speed of 55 feet per second. The formula h equals negat

makvit [3.9K]

Answer:

3.53 second

Explanation:

The formula for the height is

$h=-16t^{2}+55t+5$

When it hits the ground, the height is zero.

So, put h = 0 in the above equation

$0=-16t^{2}+55t+5$

$16t^{2}-55t-5=0$

$t=\frac{+55\pm \sqrt{55^{2}+4\times 5\times 16}}{2\times 16}$

$t=\frac{+55\pm 57.84}{2\times 16}$

Take positive sign

t = 3.53 second.

Thus, the time taken to hit the ground is 3.53 second.

8 0

3 years ago

What is the mass of 2.47 moles of<br> carbon dioxide CO2 ?

Anna11 [10]

Answer:

CO2 molecule is made up of Carbon (atomic mass =12) and oxygen (atomic mass=16).

So first finding the mass of 1 molecule of CO2 which is equals to

= mass of 1 carbon atom + masses of 2 oxygen atom, we get

= 12+(16*2)= 12+32= 44 a.m.u.

Now 1 molecule of CO2 has mass 44 amu so mass of 1 mole CO2 will be 44 grams.( 1 a.m.u.=1.6729*10^-33 grams. 1 mole = 6.022*10^23, so 44 a.m.u.=73.6076*10^-33 grams approx. For one mole CO2, 73.6076*10^-33*6.022*10^23 which is approximately equals to 44 grams. )

1 mole CO2= 44grams, so 2.5 moles = 44*2.5= 110 grams

So our answer is 110 grams

6 0

3 years ago

Read 2 more answers

A Hooke's law spring is mounted horizontally over a frictionless surface. The spring is then compressed a distance d and is used

zloy xaker [14]

Answer:

The compression is $\sqrt{2} \ d$ .

Explanation:

A Hooke's law spring compressed has a potential energy

$E_{potential} = \frac{1}{2} k (\Delta x)^2$

where k is the spring constant and $\Delta x$ the distance to the equilibrium position.

A mass m moving at speed v has a kinetic energy

$E_{kinetic} = \frac{1}{2} m v^2$ .

So, in the first part of the problem, the spring is compressed a distance d, and then launch the mass at velocity $v_1$ . Knowing that the energy is constant.