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3 years ago

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A skateboarder, starting from rest, rolls down a 10.9-m ramp. When she arrives at the bottom of the ramp her speed is 6.74 m/s.

(a) Determine the magnitude of her acceleration, assumed to be constant. (b) If the ramp is inclined at 26.6 ° with respect to the ground, what is the component of her acceleration that is parallel to the ground

1 answer:

JulijaS [17]3 years ago

4 0

Answer:b

Explanation:b

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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:

a)

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:

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

b)

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:

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

c)

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:

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

d)

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:

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

So

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

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Does gravity affect your weight???

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Weight is a force caused by gravity. The weight of an object is the gravitational force between the object and the Earth. The more mass the object has the greater its weight will be.

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9. If everything in the universe – including atoms and particles stop moving, does time stop? Or does time continue even if ever

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Time stops everything is made out of atoms so if atoms freeze everything freezes

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True or False—An external force is defined as a force generated outside the system of interest that acts on an object inside the

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Answer:

An external force is a force that acts on an object within the system from outside the system. This type of force is different than an internal force, which acts between two objects that are both within the system. The net external force combines these two definitions; it is the total combined external force

Explanation:

ig the answer is true

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