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2 years ago

I need to calculate a coefficient of linear expansion given the following

1 answer:

4 0

The coefficient of linear expansion is

<em>(change of length, feet) / (original length, feet)</em> per °F.

Since the numerator and denominator of the fraction have the same
unit, the coefficient turns out to be just a number, with no unit.

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A conveyor belt is used to move sand from one place to another in a factory. The conveyor is tilted at an angle of 18° above the

brilliants [131]

Answer:

x = 2.044 m

Explanation:

given data

initial vertical component of velocity = Vy = 2sin18

initial horizontal component of velocity = Vx = 2cos18

distance from the ground yo = 5m

ground distance y = 0

from equation of motion

$y = yo+ V_y t +\frac{1}{2}gt^2$

$0 = 5 + 2sin18+ \frac{1}{2}*9.8t^2$

solving for t

t = 1.075 sec

for horizontal motion

$x = V_x t$

x = 2cos18*1.075

x = 2.044 m

8 0

2 years ago

How often is carbon dioxide cycled through the atmosphere

olga55 [171]

Maybe around 350 years, depending on the carbon cycle and the time taken through steps.

5 0

2 years ago

Vector a cross b whole square<br><br>

skad [1K]

Answer:

Kriss go Brrrrrrrrrrr

Explanation:

4 0

2 years ago

A cannon, positioned on a hill, shoots a cannonball horizontally at 23 m/s. The cannonball hits the stone wall 1.96 m below the

irina [24]

Answer: 14. 49 m

Explanation:

We can solve this problem with the following equations:

$x=V_{o} cos \theta t$ (1)

$y-y_{o}=V_{o} sin \theta t-\frac{1}{2}gt^{2}$ (2)

Where:

$x$ is the horizontal distance between the cannon and the ball

$V_{o}=23 m/s$ is the cannonball initial velocity

$\theta=0\°$ since the cannonball was shoot horizontally

$t$ is the time

$y=0$ is the final height of the cannonball

$y_{o}=1.96 m$ is the initial height of the cannonball

$g=9.8 m/s^{2}$ is the acceleration due gravity

Isolating $t$ from (2):

$t=\sqrt{-\frac{2(y-y_{o})}{g}}$ (3)

$t=\sqrt{-\frac{2(0 m-1.96 m)}{9.8 m/s^{2}}}$ (4)

$t=0.63 s$ (5)

Substituting (5) in (1):

$x=(23 m/s) cos(0\°) 0.63 s$ (6)

Finally:

$x=14.49 m$

5 0

2 years ago

An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. What is the minimum coef

navik [9.2K]

An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. The minimum coefficient of static friction needed for this claim to be possible is 0.7

In an inclined plane, the coefficient of static friction is the angle at which an object slide over another.

As the angle rises, the gravitational force component surpasses the static friction force, as such, the object begins to slide.

Using the Newton second law;

$\sum F_x = \sum F_y = 0$