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3 years ago

Why is the mechanical advantage of a bottle opener always greater than 1?

1 answer:

6 0

For a lever, the mechanical advantage is the ratio of the lever arms. Since the lever arm of the opener is always greater than that of the output arm, the mechanical advantage is greater than 1.

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Why are food and water the limiting factors that usually have the greatest effect on population size?

maw [93]

Because if you have a large population, there is a need for more food and water, but if you don't have enough food and water to support a large population, that limits it from happening

5 0

3 years ago

Which of the following instrument measure time most accurately

ioda

Answer:

What are the options???

6 0

3 years ago

Read 2 more answers

In a local diner, a customer slides an empty coffee cup down the counter for a refill. The cup slides off the counter and strike

zysi [14]

a) $t=\sqrt{\frac{2h}{g}}$

b) $v=\frac{d}{\sqrt{\frac{2h}{g}}}$

c) $v=\sqrt{d^2(\frac{g}{2h})+(2gh)}$

d) $\theta=tan^{-1}(\frac{2h}{d})$ (radians)

Explanation:

The motion of the cup sliding off the counter is the motion of a projectile, consisting of two independent motions:

- A uniform motion along the horizontal direction

- A uniformly accelerated motion (free fall) along the vertical direction

The time of flight of the cup is entirely determined by the vertical motion, therefore we can use the suvat equation:

$s=ut+\frac{1}{2}at^2$

where here:

$s=h$ (the vertical displacement is the height of the counter)

$u=0$ (the initial vertical velocity of the cup is zero)

$a=g$ (the vertical acceleration is the acceleration of gravity)

Solving for t, we find the time of flight of the cup:

$h=0+\frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}$

To solve this part, we just analyze the horizontal motion of the cup.

Here we know that the horizontal motion of the cup is uniform: this means that is horizontal speed is constant during the whole motion, and it is actually equal to the speed at which the mug leaves the counter.

For a uniform motion, the speed is given by

$v=\frac{d}{t}$

where

d is the distance covered

t is the time taken

Here, the distance covered is $d$ , the distance from the base of the counter, while the time taken is the time of flight:

$t=\sqrt{\frac{2h}{g}}$

Substituting into the previous equation, we find the speed of the mug as it leaves the counter:

$v=\frac{d}{\sqrt{\frac{2h}{g}}}$

Here we want to find the speed of the cup immediately before it hits the floor.

Here we have to consider that while the mug falls, its vertical speed increases, while the horizontal speed remains constant.

Therefore, the horizontal speed of the cup before it hits the ground is:

$v_x=\frac{d}{\sqrt{\frac{2h}{g}}}=d\sqrt{\frac{g}{2h}}$

The vertical speed instead is given by the suvat equation:

$v_y=u_y + at$

where:

$u_y=0$ is the initial vertical velocity

$a=g$ is the acceleration

$t=\sqrt{\frac{2h}{g}}$ is the time of flight

Substituting,

$v_y = 0 +g(\sqrt{\frac{2h}{g}})=\sqrt{2gh}$

The actual speed of the cup just before it hits the floor is the resultant of the horizontal and vertical speeds, so it is:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{d^2(\frac{g}{2h})+(2gh)}$

Just before hitting the floor, the velocity of the cup has two components:

$v_x=d\sqrt{\frac{g}{2h}}$ is the horizontal component (in the forward direction)

$v_y=\sqrt{2gh}$ is the vertical component (in the downward direction)

Since the two components are perpendicular to each other, the angle of the direction is given by the equation

$tan \theta = \frac{v_y}{v_x}$

where here $\theta$ is measured as below the horizontal direction.

Substituting the expressions for $v_x,v_y$ , we find:

$tan \theta = \frac{\sqrt{2gh}}{d\sqrt{\frac{g}{2h}}}=\frac{2h}{d}$

$\theta=tan^{-1}(\frac{2h}{d})$ (radians)

4 0

4 years ago

A transverse wave on a rope is given by y(x,t)= (0.750cm)cos(π[(0.400cm−1)x+(250s−1)t]). part a part complete find the amplitude

Pani-rosa [81]

The amplitude of a wave corresponds to its maximum oscillation of the wave itself.
In our problem, the equation of the wave is
$y(x,t)= (0.750cm)cos(\pi [(0.400cm-1)x+(250s-1)t])$
We can see that the maximum value of y(x,t) is reached when the cosine is equal to 1. When this condition occurs,
$y(x,t)=0.750 cm$
and therefore this value corresponds to the amplitude of the wave.

4 0

4 years ago

Read 2 more answers

A coin released at rest from the top of a tower hits the ground after falling 1.5 s. What is the speed of the coin as it hits th

Inessa05 [86]

initially coin is at rest and then it drop for total time t = 1.5 s

so here the speed of the coin at which it will hit the floor is to be find

$v_f = v_i + at$