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3 years ago

6

Which term refers to how often a person works out?

1 answer:

Dafna11 [192]3 years ago

6 0

Answer: Frequency

Explanation: Frequency is term which best describes how often a person exercises.

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A transparent film laminated to a fabric is identified in the study material as

siniylev [52]

The transparent film laminated to a fabric is identified in the study material as PVC. PVC or polyvinyl chloride is a type of synthetic resin. For fabric, PVC on fabric is typically plastic-coated polyester, and is more resistant to wear and tear than others.

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3 years ago

Una pelota de golf se lanza con una rapidez de 50 m/seg y un ángulo de elevación de 40º. Calcular:

miss Akunina [59]

Answer:

good morning you are you still

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3 years ago

A large, cylindrical water tank with diameter 3.60 m is on a platform 2.00 m above the ground. The vertical tank is open to the

zysi [14]

To solve this problem it is necessary to apply the concepts related to the geometry of a cylindrical tank and its respective definition.

The volume of a tank is given by

$V = \frac{\pi d^2}{4}h$

Where

d = Diameter

h = Height

Considering that there are two stages, let's define the initial and final volume as,

$V_0 = \frac{\pi d^2}{4}H$

$V_f = \frac{\pi d^2}{4}h$

We know as well by definition that

$1gal = 3.785*10^{-3}m^3$

Then we have for the statement that

$V_f = V_0 -1gal$

$V_f = V_0 - 3.785*10^{-3}$

Replacing the previous data

$\frac{\pi d^2}{4}h = \frac{\pi d^2}{4}H- 3.785*10^{-3}$

$\frac{\pi (3.6)^2}{4}h = \frac{\pi (3.6)^2}{4}(2)- 3.785*10^{-3}$

Solving to get h,

$h = 1.99963m$

Therefore the change is

$\Delta h = H-h$

$\Delta h = 2- 1.99963$

$\Delta h = 3.7*10^{-4}m=0.37mm$

Therefore te change in the height of the water in the tank is 0.37mm

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3 years ago

You and a friend each hold a lump of wet clay. Each lump has a mass of 30 grams. You each toss your lump of clay into the air, w

Vesna [10]

Answer:

$\ \text{m/s}$

Explanation:

$u_1$ = Velocity of one lump = $3x+3y-3z$

$u_2$ = Velocity of the other lump = $-4x+0y-4z$

m = Mass of each lump = $30\ \text{g}$

The collision is perfectly inelastic as the lumps stick to each other so we have the relation

$mu_1+mu_2=(m+m)v\\\Rightarrow m(u_1+u_2)=2mv\\\Rightarrow v=\dfrac{u_1+u_2}{2}\\\Rightarrow v=\dfrac{3x+3y-3z-4x+0y-4z}{2}\\\Rightarrow v=-0.5x+1.5y-3.5z=\ \text{m/s}$

The velocity of the stuck-together lump just after the collision is $\ \text{m/s}$ .

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3 years ago

An object is dropped and is in free fall. Each second, the position of the object is marked. The distance between each mark is m

navik [9.2K]

C. hope this helps :)

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4 years ago