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1 year ago

9

When you drink liquid through a straw, is it more accurate to say the liquid is pushed up the straw rather than sucked up the st

raw? What exactly does the pushing? Defend your answer.

1 answer:

victus00 [196]1 year ago

7 0

Answer:

the liquid is pushed up the straw

Explanation:

Liquid is pushed up the straw . It is not sucked up . It is pushed up by force created by atmospheric pressure difference created. by sucking we create a pressure difference . This pressure difference creates force that pushes liquid in the mouth.

We can not drink liquid by sucking it through straw on moon , because moon has no atmosphere so pushing force is absent there.

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2 years ago

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A lever has a 36cm input arm and a 6cm output arm. find MA.

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6

Explanation:

The mechanical advantage is a factor that measures how input force increases using a machine.

A lever is a simple machine with the fulcrum at the center.

To calculate the mechanical advantage M.A of levers we use the expression below;

M. A = $\frac{F_{b} }{F_{a} }$ = $\frac{a}{b}$

$F_{a}$ = input force

$F_{b}$ = output force

a is the distance of the input force from the fulcrum

b is the distance of the output force from the fulcrum

Given

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b = 6cm

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learn more:

Torque brainly.com/question/5352966

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1 year ago

a pitcher threw a baseball straight up at 35.8 meters per second. what is the ball's velocity after 2.50 seconds?

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Ok the velocity of an object in free fall is given by the equation :

v=v0-gt, where v0 is the original velocity, g is the gravitational constant (9.8 m/s^2) and t is the time.

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1 year ago

A mass weighing 20 pounds stretches a spring 6 inches. The mass is initially released from rest from a point 6 inches below the

hoa [83]

Answer:

Since the spring mass system will execute simple harmonic motion the position as a function of time can be written as $x(t)=Asin(\omega t+\phi)$

'A' is the amplitude = 6 inches (given)

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At equilibrium we have

$mg=kx\\\\k=\frac{mg}{x}$

Applying values we get

$k=40 lb/ft$

thus natural frequency equals

$\omega =\sqrt{\frac{40}{\frac{20}{32}}}\\\\\omega =8s^{-1}$

Thus the equation of motion becomes

$x(t)=6sin(8t+\phi)$

At time t=0 since mass is at it's maximum position thus we have

$A=Asin(\omega t+\phi)\\\\\therefore sin(\omega\times 0+\phi)=1\\\\\phi=\frac{\pi}{2}\\\\\therefore x(t)=Asin(\omega t+\frac{\pi}{2})$

Thus the position of mass at the given times is as follows

1) at $\frac{\pi}{12}$ $x(t)=5.99inches$

2) at $\frac{\pi}{8}$ $x(t)=5.9909inches$

3) at $\frac{\pi}{6}$ $x(t)=5.98397inches$

4) at $\frac{\pi}{4}$ $x(t)=5.9639inches$

5) at $\frac{9\pi}{32}$ $x(t)=5.954inches$

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1 year ago

Wood block 1 in (Figure 1), which has a mass of 1.0 kg, is at rest on a wood ramp. The angle of the ramp is 20º above horizontal

Sergeeva-Olga [200]

Answer:

Explanation:

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Y axis

N₁ - W_y = 0 ( 2)

let's use trigonometry for the weight components

sin θ = Wₓ / W

cos θ = W_y / W

Wₓ = W sin θ

W_y = W cos θ

We write the diagram for the second body.

Note that in the block the positive direction rd upwards, therefore for block 2 the positive direction must be downwards

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we add the equations is 1 and 3

- W₁ sin θ - μ N₁ + W₂ = 0

from equation 2

N₁ = W₁ cos θ

we substitute

-W₁ sin θ - μ (W₁ cos θ) + W₂ = 0

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This is the smallest value that supports the equilibrium system

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2 years ago