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2 years ago

10

Please help with the following question

1 answer:

leva [86]2 years ago

5 0

The correct answer is conner because he has more than everyone else 4 is more than 3.5, 2.5, and 3

You might be interested in

Which data set has the largest range? A. 55, 57, 59, 60, 61, 49, 48 B. 21, 25, 14, 16, 29, 22, 20 C. 12, 15, 16, 19, 18, 15, 27

Simora [160]

Data D has the largest range.

Data A: 61-48=13
Data B: 29-14=15
Data C:27-12=15
Data D:54-31=23

Therefore, Data D has the largest range.

3 0

3 years ago

Potassium is a crucial element for the healthy operation of the human

Degger [83]

Answer:

1

The mass of the Potassium-40 is $m_{40}} = 2.88*10^{-6} kg$

2

The Dose per year in Sieverts is $Dose_s = 26.4 *10^{-10}$

Explanation:

From the question we are told that

The isotopes of potassium in the body are Potassium-39, Potassium-40, and Potassium- 41

Their abundance is 93.26%, 0.012% and 6.728%

The mass of potassium contained in human body is $m = 3.0 g = \frac{3}{1000} = 0.0003 \ kg$ per kg of the body

The mass of the first body is $m_1 = 80 \ kg$

Now the mass of potassium in this body is mathematically evaluated as

$m_p = m * m_1$

substituting value

$m_p = 80 * 0.0003$

$m_p =0.024 kg$

The amount of Potassium-40 present is mathematically evaluated as

$m_{40}} =$ 0.012% * 0.024

$m_{40}} = \frac{0.012}{100} * 0.024$

$m_{40}} = 2.88*10^{-6} kg$

The dose of energy absorbed per year is mathematically represented as

$Dose = \frac{E}{m_1}$

Where E is the energy absorbed which is given as $E = 1.10 MeV = 1.10 * 10^6 * 1.602*10^{-19}$

Substituting value

$Dose = \frac{ 1.10 * 10^6 * 1.602*10^{-19}}{80}$

$Dose = 22*10^{-10} J/kg$

The Dose in Sieverts is evaluated as

$Dose_s = REB * Dose$

$Dose_s = 1.2 * 22*10^{-10}$

$Dose_s = 26.4 *10^{-10}$

3 0

2 years ago

Starting from rest, a 6.79 kg block slides 2.82 m down a rough 20.7 ◦ incline. The coefficient of kinetic friction between the b

Veronika [31]

Answer:

23.52092 J

Explanation:

m = Mass of block = 6.79 kg

s = Sliding distance = 2.82 m

$\theta$ = Angle of slide = 20.7°

$\mu$ = Coefficient of kinetic friction = 0.425

g = Acceleration due to gravity = 9.8 m/s²

Work done by the force of gravity is given by

$W=mgsin\theta\\\Rightarrow W=6.79\times 9.8\times sin20.7\\\Rightarrow W=23.52092\ J$

The work done by the force of gravity is 23.52092 J

8 0

2 years ago

A large container is pulled 20 meters with 40 newtons of force. The power exerted is 64 watts. How long did it take to move the

Usimov [2.4K]

A.(20*40)/64=12.5 hope i helped

3 0

3 years ago

An object at rest on a flat, horizontal surface explodes into two fragments, one seven times as massive as the other. The heavie

leva [86]

To solve the problem it is necessary to apply conservation of the moment and conservation of energy.

By conservation of the moment we know that

$MV=mv$

Where

M=Heavier mass

V = Velocity of heavier mass

m = lighter mass

v = velocity of lighter mass

That equation in function of the velocity of heavier mass is

$V = \frac{mv}{M}$

Also we have that $m/M = 1/7 times$

On the other hand we have from law of conservation of energy that

$W_f = KE$

Where,

W_f = Work made by friction

KE = Kinetic Force

Applying this equation in heavier object.

$F_f*S = \frac{1}{2}MV^2$

$\mu M*g*S = \frac{1}{2}MV^2$

$\mu g*S = \frac{1}{2}( \frac{mv}{M})^2$

$\mu = \frac{1}{2} (\frac{1}{7}v)^2$

$\mu = \frac{1}{98}v^2$

$\mu = \frac{1}{g(98)(5.1)}v^2$

Here we can apply the law of conservation of energy for light mass, then

$\mu mgs = \frac{1}{2} mv^2$

Replacing the value of $\mu$

$\frac{1}{g(98)(5.1)}v^2 mgs = \frac{1}{2}mv^2$

Deleting constants,

$s= \frac{(98*5.1)}{2}$

$s = 249.9m$

7 0

3 years ago