This experiment is to see if water flows faster out of a smaller can or a larger can.

Irina-Kira [14]

The outer rigid layer of the earth is divided into a couple of dozen “plates” that move around across earths surface relative to each other.

7 0

2 years ago

UkoKoshka [18]

Answer: Their u go i found it their was about 3 pages i did not no what pages u had to do.

Explanation:

Download pdf

7 0

2 years ago

Rzqust [24]

Answer:

Velocity (magnitude) is 98.37 m/s

Explanation:

We use the vertical component of the initial velocity, which is:

$v_{0y}=v_0*sin(45)=\frac{\sqrt{2} }{2}v_0$

Using kinematics expression of vertical velocity (in y direction) for an accelerated motion (constant acceleration, which is gravity):

$v_{y}=v_{0y}+a*t=\frac{\sqrt{2} }{2}v_0-9.8t$

Now we need to find $v_y$ as a function of $v_0$ . We use the horizontal velocity, which is always the same as follow:

$v_x=v_0cos(45\º)=\frac{\sqrt{2} }{2}v_0=v_{t=3}*cos(30\º) \\$

We know the angle at 3 seconds:

$v_y(t=3)=v_{t=3}*sin(30\º)\\v_{t=3}=\frac{v_y}{sin(30\º)}$

Substitute $v_{t=3}$ in $v_x$ and then solve for $v_y$

$\frac{\sqrt{2} }{2}v_0=\frac{v_y*cos(30\º) }{sin(30\º)} \\v_y=\frac{\sqrt{6} }{6}v_0$

With this expression we go back to the kinematic equation and solve it for initial speed

$\frac{\sqrt{6} }{6} v_0 =\frac{\sqrt{2} }{2}v_0-29.4\\v_0(\frac{\sqrt{6}-3\sqrt{2}}{6} )=-29.4\\v_0=98.37 m/s$

3 0

2 years ago

dedylja [7]

We Know,

K.E. = 1/2 mv²

480 = 1/2 (m)(8)²

m = 960/64

m = 15 Kg

So, the mass of the object is 15 Kg

7 0

2 years ago

vlada-n [284]

I uploaded the answer to a file hosting. Here's link:

tinyurl.com/wpazsebu

4 0

2 years ago