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2 years ago

9

Janet jumps off a high diving platform with a horizontal velocity of 2.06 m/s and lands in the water 1.8 s later. How high is th

e platform? The acceleration of gravity is 9.8 m/s 2 .

1 answer:

Archy [21]2 years ago

8 0

Initial height of platform is 15.87m or 16m.

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It is Potential energy it's at rest

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3 years ago

How to read a micrometer on a clark cm-100 vickers hardness tester

allsm [11]

Answer:

Explanation:

Equipment manufactured by LECO(8 Corporation, St. Joseph, Michigan is warranted free from defect in material

and workmanship for a period of six months from the date of purchase. Equipment not manufactured by LECO is

covered to the extent of warranty provided by the original manufacturer and this warranty does not cover any

equipment, new or used, purchased from anyone other than the LECO Corporation. All replacement parts shall

be covered under warranty for a period of thirt days from date of purchase. LECO MAKES NO OTHER

REPRESENTATION OR WARRANTY OF ANY OTHER KIND, EXPRESSED OR IMPLIED, WITH RESPECT TO

THE GOODS SOLD HEREUNDER, WHETHER AS TO MERCHANTABILITY, FITNESS FOR PURPOSE, OR

OTHERWISE.

Expendable items such as crucibles, combustion tubes, chemicals and items of like nature are not covered by

this warranty.

LECO's sole obligation under this warranty shall be to repair or replace any part or parts which, to our

satisfaction, prove to be defective upon return prepaid to LECO Corporation, St. Joseph, Michigan. This

obligation does not include labor to install replacement parts, nor does it cover any failure due to accident, abuse,

neglect, or use in disregard of instructions furnished by LECO. In no event shall damages for defective goods

exceed the purchase price of the goods, and LECO SHALL NOT BE LIABLE FOR INCIDENTAL OR

CONSEQUENTIAL DAMAGES WHATSOEVER.

All claims in regard to the parts or equipment must be made within ten (10) days after Purchaser learns of the

facts upon which the claim is based. Authorization must be obtained from LECO prior to returning any other

parts. This warranty is voided by failure to comply with these notice requirements.

NOTICE

The warranty on LECO equipment remains valid only when genuine LECO replacernent parts are employed.

Since LECO has no control over the quality or purity of consumable products not manufactured by LECO, the

specifications for accuracy of results using LECO instruments are not guaranteed unless genuine LECO

consumables are employed in conjunction with LECO instruments. If purchaser defaults in making payment for

any parts or equipment, this warranty shall be void and shall not apply to such parts and equipment. No late

payment or cure of default in payment shall extend the warranty period provided herein.

LECO Corporation is not responsible for damage to any associated instruments, equipment or apparatus nor wil

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The instrument should be operated only by technically qualified individuals who have fully read and understand

these instructions. The instrument should be operated only in accordance with these instructions.

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2 years ago

A 124-kg balloon carrying a 22-kg basket is descending with a constant downward velocity of 20.0 m/ s. A I.O-kg stone is thrown

nadya68 [22]

(a) 296.6 m

The motion of the stone is the motion of a projectile, thrown with a horizontal speed of

$v_x = 15.0 m/s$

and with an initial vertical velocity of

$v_{y0} = -20.0 m/s$

where we have put a negative sign to indicate that the direction is downward.

The vertical position of the stone at time t is given by

$y(t) = h + v_{0y} t + \frac{1}{2}gt^2$ (1)

where

h is the initial height

g = -9.81 m/s^2 is the acceleration due to gravity

The stone hits the ground after a time t = 6.00 s, so at this time the vertical position is zero:

y(6.00 s) = 0

Substituting into eq.(1), we can solve to find the initial height of the stone, h:

$0 = h + v_{0y} y + \frac{1}{2}gt^2\\h = -v_{0y} y - \frac{1}{2}gt^2=-(-20.0 m/s)(6.00 s) - \frac{1}{2}(9.81 m/s^2)(6.00 s)^2=296.6 m$

(b) 176.6 m

The balloon is moving downward with a constant vertical speed of

$v_y = -20 m/s$

So the vertical position of the balloon after a time t is

$y(t) = h + v_y t$

and substituting t = 6.0 s and h = 296.6 m, we find the height of the balloon when the rock hits the ground:

$y(t) = 296.6 m + (-20.0 m)(6.00 s)=176.6 m$

(c) 198.2 m

In order to find how far is the rock from the balloon when it hits the ground, we need to find the horizontal distance covered by the rock during the time of the fall.

The horizontal speed of the rock is

$v_x = 15.0 m/s$

So the horizontal distance travelled in t = 6.00 s is

$d_x = v_x t = (15.0 m/s)(6.00 s)=90 m$

Considering also that the vertical height of the balloon after t=6.00 s is

$d_y = 176.6 m$

The distance between the balloon and the rock can be found by using Pythagorean theorem:

$d=\sqrt{(90 m)^2+(176.6 m)^2}=198.2 m$

(di) 15.0 m/s, -58.8 m/s

For an observer at rest in the basket, the rock is moving horizontally with a velocity of

$v_x = 15.0 m/s$

Instead, the vertical velocity of the rock for an observer at rest in the basket is

$v_y (t) = gt$

Substituting time t=6.00 s, we find

$v_y = (-9.8 m/s)(6.00 s)=-58.8 m/s$

(dii) 15.0 m/s, -78.8 m/s

For an observer at rest on the ground, the rock is still moving horizontally with a velocity of

$v_x = 15.0 m/s$

Instead, the vertical velocity of the rock for an observer on the ground is now given by

$v_y (t) = v_{0y} + gt$

Substituting time t=6.00 s, we find

$v_y = (-20.0 m/s)+(-9.8 m/s)(6.00 s)=-78.8 m/s$

6 0

2 years ago

What occurs when an object travels in a curved path

babymother [125]

<span>When an object travels in a curved path, there must be a force acting toward the center of the circular trajectory. This force is called "centripetal force", and it cause an acceleration of the object, called "centripetal acceleration". The effect of this acceleration is that the velocity of the object changes in direction: however if the circular motion is uniform, the speed (=the magnitude of the velocity) does not change. In this case, the magnitude of the centripetal force is given by
</span> $F=m \frac{v^2}{r}$ <span>
where m is the mass of the object, v its velocity, and r the radius of the circular path.</span>

7 0

3 years ago

A ball is thrown straight up. What will

Papessa [141]

Answer:

velocity at the top: 0 m/s

acceleration at the top: -9.8 m/s²

Explanation:

Assuming up is positive and down is negative;

The velocity of the ball at the top of its path will be 0 m/s and the acceleration will be negative.

The velocity is 0 m/s because the ball does not move at the top of its path, and it switches from a positive velocity to a negative velocity. It must go through 0 in order to go from positive to negative.

The acceleration, however, is always negative no matter where the ball is in its motion. This negative acceleration causes the ball to slow down as it reaches the top, and speed up as it reaches the bottom.

<u>Think about it:</u> If there wasn't a negative acceleration, and it was instead 0, the ball would never come back down and instead keep going in a straight line.

6 0

2 years ago

Read 2 more answers