A force is applied to a 2.9 kg mass and pro-

A child is trying to throw a ball over a fence. She gives the ball an initial speed of 8.0 m/s at an angle of 40° above the hori

EastWind [94]

Answer:

the child is 1.581 m far from the fence

Explanation:

The diagrammatic illustration that give a better view of what the question denote can be seen in the image attached below.

From the image attached below, let assume that the release point is the origin, then equation of the motion (x) is as follows:

$x - x_o = u_xt$

$\mathtt{x = u_xt \ \ \ since (x_o = 0)}$ ---- (1)

the equation of the motion y is :

$\mathtt{y - y_o =u_yt - 0.5 gt^2}$

$\mathtt{y = u_yt-4.9t^2 \ \ \ since (y_o =0)}$

$\mathtt{ 1= (u \ sin 40^0)t -4.9 \ t^2 }$

$\mathtt{1 = 8 sin 40^0 t - 4.9 t^2}$

$\mathtt{1 = 5.14t - 4.9t^2}$

$\mathtt{4.9t^2 - 5.14t +1 = 0}$

By using the quadratic formula, we have;

$\mathtt{ \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{2a}} }$

where;

a = 4.9, b = -5.14 c = 1

$= \mathtt{ \dfrac{ -(-5.14) \pm \sqrt{(-5.14)^2 - 4(4.9)(1)}}{2(4.9)}} }$

$= \mathtt{ \dfrac{ 5.14 \pm \sqrt{26.4196 -19.6}}{9.8}} }$

$= \mathtt{ \dfrac{ 5.14 \pm \sqrt{6.8196}}{9.8}} }$

$= \mathtt{ \dfrac{ 5.14+ \sqrt{6.8196}}{9.8} \ \ OR \ \ \dfrac{ 5.14- \sqrt{6.8196}}{9.8}} }$

$= \mathtt{ \dfrac{ 5.14+ 2.6114}{9.8} \ \ OR \ \ \dfrac{ 5.14- 2.6114}{9.8}} }$

$= \mathtt{ \dfrac{ 7.7514}{9.8} \ \ OR \ \ \dfrac{ 2.5286}{9.8}} }$

$= \mathbf{ 0.791 \ \ OR \ \ 0.258} }$

In as much as the ball is traveling upward, then we consider t= 0.258sec.

From equation (1)

$\mathtt{x = u_x(0.258)}$

$\mathtt{x = ucos 40^0 (0.258)}$

$\mathtt{x = 8 \ cos 40^0 (0.258)}$

$\mathbf{x = 1.581 \ m}$

Thus, the child is 1.581 m far from the fence

6 0

2 years ago

Jeremy accidently dropped his toy stuffed animal from the balcony of his apartment on the fourth floor. The toy hit the ground a

uysha [10]

F = (mass)(acceleration) = ma

m = 0.25 kg

Vi = 16 m/s

t = 2 s

Vf = 0 m/s (since it was put to stop)

a=(Vf-Vi)/t

a=(0-16)/2

a = 8 m/s^2 (decelerating)

F = ma = (0.25 kg)(8 m/s^2)

F = 2 N

<span>Hope this answer will be a good h<span>elp for you.</span></span>

3 0

3 years ago

Read 2 more answers

A stone is thrown straight up from the edge of a roof, 925 feet above the ground, at a speed of 20 feet per second. Remembering

Black_prince [1.1K]

<h2>Answer: 469 feet</h2>

Explanation:

This problem is a good example of Vertical motion, where the main equation for this situation is:

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

Where:

$y$ is the height of the stone at 6s (the value we want to find)

$y_{o}=925ft$ is the initial height of the stone

$V_{o}=20ft/s$ is the initial velocity of the stone

$t=6s$ is the time at which we need to find the height

$g=32ft/s^{2}$ is the acceleration due to gravity

Having this clear, let's find $y$ from (1):

$y=925ft+(20ft/s)(6s)-\frac{1}{2}(32ft/s^{2})(6s)^{2}$ (2)

Finally:

$y=469ft$ This is the height of the stone at t=6s

4 0

2 years ago

Explain, using numbered steps, how you could find the volume of a small, irregularly shaped rock.

dmitriy555 [2]

<span>1. Get a graduated cylinder. 2. Fill the graduated cylinder to a known amount of water. Record the amount of water in the cylinder. 3. Place rock into the graduated cylinder 4. Measure the new volume of the graduated cylinder with the rock in it. 5. Take the difference of the new volume and the old volume and that is the volume of the rock.</span>

8 0

2 years ago

A physics major is cooking breakfast when he notices that the frictional force between the steel spatula and the Teflon frying p

Semenov [28]

Answer:

$f_n=3.75N$