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KatRina [158]
3 years ago
11

Define mechanical energy useing the example of the skateboarder on the half-pipe explain how kinetic and potential energy relate

to the mechanical energy at diffrent positions
Physics
1 answer:
sergejj [24]3 years ago
7 0
When some one balance skate board on pipe the potential energy is responsible to make it slide beacause of m mass and g gravity and height thus the skate boarder concieves mgh potential energy and moving on pipe with balancing himself using kinetic enrgy ie;1/2 mv^2 and thus gauning speed ..for illustrating well there is a another example the two wheeeler gets balanced while moving instead of standing idle or stationary ie; kinetic energy is responsible for balancing
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A body falls from the top of the tower and during the last second of its fall it fall through 23mvfind height of tower.
DanielleElmas [232]

Answer:

39.7 m

Explanation:

First, we conside only the last second of fall of the body. We can apply the following suvat equation:

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

where, taking downward as positive direction:

s = 23 m is the displacement of the body

t = 1 s is the time interval considered

a=g=9.8 m/s^2 is the acceleration

u is the velocity of the body at the beginning of that second

Solving for u, we find:

ut=s-\frac{1}{2}at^2\\u=\frac{s}{t}-\frac{1}{2}at=\frac{23}{1}-\frac{1}{2}(9.8)(1)=18.1 m/s

Now we can call this velocity that we found v,

v = 18 m/s

And we can now consider the first part of the fall, where we can apply the following suvat equation:

v^2-u^2 = 2as'

where

v = 18 m/s

u = 0 (the body falls from rest)

s' is the displacement of the body before the last second

Solving for s',

s'=\frac{v^2-u^2}{2a}=\frac{18.1^2-0}{2(9.8)}=16.7 m

Therefore, the total heigth of the building is the sum of s and s':

h = s + s' = 23 m + 16.7 m = 39.7 m

7 0
3 years ago
Determine the vector perpendicular to the plane of A= 31+ 6j - 2k and B=4i-j +3k
Sliva [168]

The vector perpendicular to the plane of A = 3i+ 6j - 2k and B = 4i-j +3k is 16 i - 17 j - 27 k

Let r be the vector perpendicular to A and B,

r = A * B

A = 3i + 6j - 2k

B = 4i - j + 3k

a1 = 3

a2 = 6

a3 = - 2

b1 = 4

b2 = - 1

b3 = 3

a * b = ( a2 b3 - b2 a3 ) i + ( a3 b1 - b3 a1 ) j + ( a1 b2 - b1 a2 ) k

a * b = [ ( 6 * 3 ) - ( - 1 * - 2 ) ] i + [ ( - 2 * 4 ) - ( 3 * 3 ) ] j + [ ( 3 * - 1 ) - ( 4 * 6 ) ] k

a * b = 16 i - 17 j - 27 k

The perpendicular vector, r = 16 i - 17 j - 27 k

Therefore, the vector perpendicular to the plane of A = 3i + 6j - 2k and B = 4i - j + 3k is 16 i - 17 j - 27 k

To know more about perpendicular vectors

brainly.com/question/14384780

#SPJ1

5 0
1 year ago
A rock is projected upward from the surface of the moon, at time t = 0.0 s, w a velocity of 30 m/s. The acceleration due to grav
Vinvika [58]
<h2>Answer: 277.777 m</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told that the rock was<u> projected upward from the surface</u>, we will only use the equations related to the Y axis.

In this sense, the movement equations in the Y axis are:

y-y_{o}=V_{o}.t+\frac{1}{2}g.t^{2}    (1)

V=V_{o}-g.t    (2)

Where:

y  is the rock's final position

y_{o}=0  is the rock's initial position

V_{o}=30\frac{m}{s} is the rock's initial velocity

V is the final velocity

t is the time the parabolic movement lasts

g=1.62\frac{m}{s^{2}}  is the acceleration due to gravity at the surface of the moon

As we know y_{o}=0 , equation (2) is rewritten as:

y=V_{o}.t+\frac{1}{2}g.t^{2}    (3)

On the other hand, the maximum height  is accomplished when V=0:

V=V_{o}-g.t=0    (4)

V_{o}-g.t=0    

V_{o}=g.t    (5)

Finding t:

t=\frac{V_{o}}{g}    (6)

Substituting (6) in (3):

y=V_{o}(\frac{V_{o}}{g})+\frac{1}{2}g(\frac{V_{o}}{g})^{2}    (7)

y_{max}=\frac{{V_{o}}^{2}}{2g}    (8)  Now we can calculate the maximum height of the rock

y_{max}=\frac{{(30m/s)}^{2}}{(2)(1.62m/s^{2})}   (9)

Finally:

y_{max}=277.777m  

4 0
3 years ago
What is an easy way to balance chemical equations?
Lorico [155]

write out what you have on both sides, then just use basic multiplication to try and even out both sides. I can help if you need me to balance some for you!!

8 0
3 years ago
A wave seen at the ocean’s shore is which kind of wave?
Lostsunrise [7]
B. Transverse Wave this is the correct answer
3 0
3 years ago
Read 2 more answers
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