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2 years ago

6

A large air-filled 0.221-kg plastic ball is thrown up into the air with an initial speed of 11.9 m/s. At a height of 2.77 m, the

ball's speed is 2.85 m/s. What fraction of its original energy has been lost to air friction?

1 answer:

kap26 [50]2 years ago

5 0

Answer:

Initial energy = (M/2)*Vo^2 = 12.80 J

Total energy at 4.7 m and V = 5 m/s:

= (M/2)*5^2 + M*g*4.7 = 1.25 + 4.61 J

Fraction of ball's energy LOST = (12.80-5.86)/12.80

= 0.542

Explanation:

I know it's not the real answer, but this will help you find the real one

Hope this helps!!!

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A ball is launched at an angle of 10 degrees from a 20 meter tall building with a speed of 4 m/s. How long is the ball in the ai

shutvik [7]

Answer:

d) None of the above

Explanation:

$v_{o}$ = inituial velocity of launch = 4 m/s

θ = angle of launch = 10 deg

Consider the motion along the vertical direction

$v_{oy}$ = initial velocity along vertical direction = 4 Sin10 = 0.695

m/s

$a_{y}$ = acceleration along the vertical direction = - 9.8 m/s²

y = vertical displacement = - 20 m

t = time of travel

using the equation

$y=v_{oy} t+(0.5)a_{y} t^{2}$

- 20 = (0.695) t + (0.5) (- 9.8) t²

t = 2.1 sec

consider the motion along the horizontal direction

x = horizontal displacement

$v_{ox}$ = initial velocity along horizontal direction = 4 Cos10 = 3.94 m/s

$a_{x}$ = acceleration along the horizontal direction = 0 m/s²

t = time of travel = 2.1 s

Using the kinematics equation

$x =v_{ox} t+(0.5)a_{x} t^{2}$

x = (3.94) (2.1) + (0.5) (0) (2.1)²

x = 8.3 m

Consider the motion along the vertical direction

$v_{oy}$ = initial velocity along vertical direction = 4 Sin10 = 0.695

m/s

$a_{y}$ = acceleration along the vertical direction = - 9.8 m/s²

$y_{o}$ =initial vertical position at the time of launch = 20 m

$y$ = vertical position at the maximum height = 20 m

$v_{fy}$ = final velocity along vertical direction at highest point = 0 m/s

using the equation

${v_{fy}}^{2}= {v_{oy}}^{2} + 2 a_{y}(y - y_{o})$

$0^{2}= 0.695^{2} + 2 (- 9.8)(y - 20)$

$y$ = 20.02 m

h = height above the starting height

h = $y$ - $y_{o}$

h = 20.02 - 20

h = 0.02 m

7 0

3 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

3 years ago

I need help understanding this concept . Would appreciate it so much. Thank you

Firlakuza [10]

Answer:

Explanation:

Both these questions are based on the Universal Law of Gravitation, which is given by :

F = Gm1m2 / r²

2) F = 6.67 x 10⁻¹¹ x 8 x 10³ x 1.5 x 10³ / 1.5 x 1.5

F = 6.67 x 10⁻⁵ x 8 / 1.5

F = 35.57 x 10⁻⁵ N

3) F = 6.67 x 10⁻¹¹ x 7.5 x 10⁵ x 9.2 x 10⁷ / 7.29 x 10⁴

F = 6.67 x 10⁻³ x 7.5 x 9.2 / 7.29

F = 63.13 x 10⁻³ N

7 0

3 years ago

Indicate whether the statement is true or false. A force can be simply defined as a push or a pull.

Sidana [21]

This B: False, because the definition is lacking.

Force is when two objects interact with one another causing it to either move or not move. In our daily lives there are a lot of times force is exerted upon us, rather force is everywhere and here are the evidences:

*Pushing a cart
*Pushing a wall
*Hitting a baseball bat
*Apple falling down from a tree.
*Balls hitting one another
*A swinging pendulum
*Throwing a paper with stone above it
*Breaking of glass in the floor
*Falling of leaves on the grass <span> </span>

5 0

3 years ago

An experiment is set up to test the angular resolution of an optical device when red light (wavelength ????r ) shines on an aper

Neko [114]

Explanation:

As per Rayleigh criterion, the angular resolution is given as follows:

$\theta=\frac{1.22 \lambda}{D}$

From this expression larger the size of aperture, smaller will be the value of angular resolution and hence, better will be the device i.e. precision for distinguishing two points at very high angular difference is higher.

4 0

4 years ago