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AysviL [449]
3 years ago
12

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

s the floor at distance d from the base of the counter.
(a) If the height of the counter is h, find an expression for the time t it takes the cup to fall to the floor in terms of the variables h and g.
t =
(b) With what speed does the mug leave the counter? Answer in terms of the variables d, g, and h.
v1 =
(c) In the same terms, what is the speed of the cup immediately before it hits the floor?
v2 =
(d) In terms of h and d, what is the direction of the cup's velocity immediately before it hits the floor?
? = radians below horizontal
Physics
1 answer:
zysi [14]3 years ago
4 0

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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A motorcycle travels 90.0 km/h. How many seconds will it take the motorcycle to cover 2.10 x 103 m?
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4 0
3 years ago
A metal block suspended from a spring balance is submerged in water. You observe that the block displaces 55 cm3 of water and th
DiKsa [7]

Answer:

8977.7 kg/m^3

Explanation:

Volume of water displaced = 55 cm^3 = 55 x 10^-6 m^3

Reading of balance when block is immersed in water = 4.3 N

According to the Archimedes principle, when a body is immersed n a liquid partly or wholly, then there is a loss in the weight of body which is called upthrust or buoyant force. this buoyant force is equal to the weight of liquid displaced by the body.

Buoyant force = weight of the water displaced by the block

Buoyant force = Volume of water displaced x density of water x g

                        = 55 x 10^-6 x 1000 x .8 = 0.539 N

True weight of the body = Weight of body in water + buoyant force

m g = 4.3 + 0.539 = 4.839

m = 0.4937 kg

Density of block = mass of block / volume of block

= \frac{0.4937}{55\times10^{-6}}

Density of block = 8977.7 kg/m^3

4 0
3 years ago
The distance between two particles is 2 centimeters. If the distance is increased to 4 centimeters, the force will be ?
postnew [5]

Answer:

The new force is 1/4 of the previous force.

Explanation:

Given

Initial\ Distance = 2cm ---- r_1

New\ Distance = 4cm --- r_2

Required

Determine the new force

Let the two particles be q1 and q2.

The initial force F1 is:

F_1 = \frac{kq_1q_2}{r_1^2} --- Coulomb's law

Substitute 2 for r1

F_1 = \frac{kq_1q_2}{2^2}

F_1 = \frac{kq_1q_2}{4}

The new force (F2) is

F_2 = \frac{kq_1q_2}{r_2^2}

Substitute 4 for r2

F_2 = \frac{kq_1q_2}{4^2}

F_2 = \frac{kq_1q_2}{4*4}

F_2 = \frac{1}{4}*\frac{kq_1q_2}{4}

Substitute F_1 = \frac{kq_1q_2}{4}

F_2 = \frac{1}{4}*F_1

F_2 = \frac{F_1}{4}

The new force is 1/4 of the previous force.

3 0
2 years ago
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