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3 years ago

6

35 POINTS! Say If You Drop A Ball from 100 Centimeters. When The ball Bounces, The Ball Does Not Bounce Back Up To 100 Centimete

rs, But According To The Law Of Conversation Of Energy, Energy Can Not Be destroyed So where did the energy go?

1 answer:

diamong [38]3 years ago

8 0

some ball when you bounce it it comes back up but according to gravity the energy goes away

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Please need help fast

iVinArrow [24]

(a) See graph in attachment

The appropriate graph to draw in this part is a graph of velocity vs time.

In this problem, we have a horse that accelerates from 0 m/s to 15 m/s in 10 s.

Assuming the acceleration of the horse is uniform, it means that the velocity (y-coordinate of the graph) must increase linearly with the time: therefore, the velocity-time graph will appear as a straight line, having the final point at

t = 10 s

v = 15 m/s

(b) $1.5 m/s^2$

The average acceleration of the horse can be calculated as:

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time interval

In this problem,

u = 0

v = 15 m/s

t = 10 s

Substituting,

$a=\frac{15-0}{10}=1.5 m/s^2$

(c) 75 m

For a uniformly accelerated motion, the distance travelled can be calculated by using the suvat equation:

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

where

s is the distance travelled

u is the initial velocity

t is the time interval

a is the acceleration

In this problem,

u = 0

t = 10 s

$a=1.5 m/s^2$

Substituting,

$s=0+\frac{1}{2}(1.5)(10)^2=75 m$

(d) See attached graphs

In a uniformly accelerated motion:

- The distance travelled (x) follows the equation mentioned in part c,

$x=ut+\frac{1}{2}at^2$

So, we see that this has the form of a parabola: therefore, the graph x vs t will represents a parabola.

- The acceleration is constant during the motion, and its value is

$a=1.5 m/s^2$ (calculated in part b)

therefore, the graph acceleration vs time will show a flat line at a constant value of 1.5 m/s^2.

6 0

3 years ago

The liver’s role is to:

Lelu [443]

D. The liver's role is to remove harmful substances from the blood.

7 0

3 years ago

There are competitions in which pilots fly small planes low over the ground and drop weights, trying to hit a target. Part A A p

tamaranim1 [39]

Explanation:

A pilot flying a weight; it takes t = 3.4 s to hit the ground, during which it travels a horizontal distance d = 120 m

h = g/2*t^2

h= 4.903*3.4^2

h=56.701 m

V = d/t

V= 120/3.4

V=35.3 m/sec

Now the pilot does a run at the same height but twice the speed.

How much time (t1) does it take the weight to hit the ground?

t1 = t = 3.4 sec (depending just upon height)

5 0

3 years ago

A scene in a movie has a stuntman falling through a floor onto a bed in the room below. The plan is to have the actor fall on hi

IgorC [24]

Answer:

$m=17.79Kg$

Explanation:

In this process energy must be conserved. On the initial stage, there will be only gravitational potential energy, while on the final stage there will be only elastic potential energy, so they will be equal. We write this as:

$U_g=U_e$

Which is the same as:

$mgh=\frac{k \Delta x^2}{2}$

So we can obtain our mass from there, and for our values:

$m=\frac{k \Delta x^2}{2gh}=\frac{(65144 N/m)(0.1333m)^2}{2(9.8m/s^2)(3.32m)}=17.79Kg$

4 0

3 years ago

An electron moves with a constant horizontal velocity of 3.0 × 106 m/s and no initial vertical velocity as it enters a deflector

Ghella [55]

Answer:

a = 5.05 x 10¹⁴ m/s²

Explanation:

Consider the motion along the horizontal direction

$v_{x}$ = velocity along the horizontal direction = 3.0 x 10⁶ m/s

t = time of travel

X = horizontal distance traveled = 11 cm = 0.11 m

Time of travel can be given as

$t = \frac{X}{v_{x}}$

inserting the values

t = 0.11/(3.0 x 10⁶)

t = 3.67 x 10⁻⁸ sec

Consider the motion along the vertical direction

Y = vertical distance traveled = 34 cm = 0.34 m

a = acceleration = ?

t = time of travel = 3.67 x 10⁻⁸ sec

$v_{y}$ = initial velocity along the vertical direction = 0 m/s

Using the kinematics equation

Y = $v_{y}$ t + (0.5) a t²

0.34 = (0) (3.67 x 10⁻⁸) + (0.5) a (3.67 x 10⁻⁸)²

a = 5.05 x 10¹⁴ m/s²

7 0

3 years ago