Is anyone interested in a pile of bricks, it's free on craigslist?

AveGali [126]

Answer:it is a

Explanation hope this helps .

6 0

3 years ago

A 980 kg roller coaster cart is traveling along a track at 17 m/s before it rolls down a 30 m tall hill (Point A). What will be

MrRissso [65]

The kinetic energy halfway the hill is $2.86\cdot 10^5 J$

Explanation:

If there are no friction forces acting on the cart, we can apply the law of conservation of energy: the mechanical energy of the cart (which is the sum of potential energy + kinetic energy) must be conserved. So we can write:

$U_A +K_A = U_B + K_B$

where

$U_A=mgh_A$ is the initial potential energy, at point A, with

m = 980 kg (mass of the cart)

$g=9.8 m/s^2$ (acceleration of gravity)

$h_A = 30 m$ (height at point A)

$K_A=\frac{1}{2}mv_A^2$ is the initial kinetic energy, at point A , with

$v_A=17 m/s$ (velocity at point A)

$U_B=mgh_B$ is the final potential energy, at point B, where

$h_B = 15 m$ (height at point B)

$K_B=\frac{1}{2}mv_B^2$ is the final kinetic energy, at point B, where

$v_B$ is the velocity at point B

Here we are interested in finding $K_B$ , so by re-arranging the equation and substituting we find:

$K_B = U_A+K_B-U_B = mg(h_A-h_B)+\frac{1}{2}mv_A^2=(980)(9.8)(30-15)+\frac{1}{2}(980)(17)^2=2.86\cdot 10^5 J$

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

8 0

3 years ago

Water enters a cylindrical tank through two pipes at rates of 250 and 100 gal/min. If the level of the water in the tank remains

tester [92]

Answer:

The total amount of water that enters the tank is:

250 gal/min + 100 gal/min = 350gal/min.

Then, if the level of the water remains constant, this means that the water leaves the tank at a rate of 350gal/min.

We know that the diameter of the pipe is 8 inches, then the area of the pipe is:

A = pi*(d/2)^2 = 3.14*(4in)^2 = 50.24in^2

now, the flow can be calculated as:

Q = v*A = (velocity*area)

if we want to write our velocity in inches per minute, then we need to write the entering flow in cubic inches:

1 gallon = 231 in^3

then:

350gal/min = (350*231) in^3/min = 80,850 in^3/min.

Then the water that leaves the tank must be the same amoun, we have:

Q = 80,850 in^3/min. = v*A = v*50.24in^2

v = (80,850in^3/min)/50.24in^2 = 1609.3 in/min.

The velocity of the flow leaving the tank is 1609.3 in/min.

3 0

3 years ago

The three forces acting on a hot-air balloon that is moving vertically are its weight, the force due to air resistance and the u

irinina [24]

Explanation :

The forces acting on hot- air balloon are:

Weight, (W)

Force due to air resistance, (F)

Upthrust force, (U)

Its weight W is acting in downward direction. The upthrust force U acts in upward direction. When the balloon is moving upward, the air resistance is in downward and vice versa.

In this case, the hot-air balloon descends vertically at constant speed.

so, $a=0$