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3 years ago

what is the force on a 1,000 kilogram elevator that is falling freely under the acceleration of gravity only

1 answer:

4 0

Force is equal to weight, F= ma acceleration "a", is now due to gravity g = 9.8 m/s²

Given: Mass m = 1,000 Kg

F =ma W = mg = (1,000 Kg)(9.8 m/s²)

F = 9,800 N

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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:

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

What are the three examples that show relation between pressure and area in daily life.<br>

makvit [3.9K]

Answer: The area of the edge of a knife's blade is extremely small.

Syringes are used to take blood for blood tests.

When air is sucked out of a drinking straw, the air pressure inside if decreases and the atmospheric pressure outside forces the liquid to go inside the straw.

Explanation:

4 0

3 years ago

Julia can swim at 3.5 km/h in still water. She attempts to head straight north

raketka [301]

1) 3.7 km/hr N19°W

2) she must aim at N20°E

work sheet #6

relative velocity

3 0

3 years ago

At a given time of day, the ratio of the height of an object to the length of its shadow is the same for all objects. If a 4-f

katrin2010 [14]

Answer:35.2 ft

Explanation:

Given

height of stick =4 ft

shadow length =2.8 ft

Angle of elevation of sun is

$tan\theta =\frac{4}{2.8}$

let the height of tree be h

as $\theta$ will remain same thus

$tan\theta =\frac{h}{24.64}$

$\frac{4}{2.8}=\frac{h}{24.64}$

h=35.2 ft

8 0

4 years ago

A spring has been compressed inside a toy what kind of energy is this

zlopas [31]

Answer:

Elastic Potential Energy

Explanation:

Elastic Potential Energy (“Spring Energy”) is the form of energy an object has when it is stretched, compressed, twisted, bent, or otherwise has its shape changed as long as the object resists and will try to return to its original state.

4 0

3 years ago