100 point <br> good luck answering this<br> 2+2-4+8+-5+???

How much time is needed for a 15,000 w engine to do 1,800,000 j of work?

Hatshy [7]

We know that;
power=W/t
15000Watts=1800000J/t
t=1800000J/15000Watts
t=120s

8 0

4 years ago

Read 2 more answers

Jnjewenwjewuhduwjhhhubcadsfadyubwqedjqwehjfgwegdh

Gnesinka [82]

Answer: Your search - Jnjewenwjewuhduwjhhhubcadsfadyubwqedjqwehjfgwegdh - did not match any documents.

Suggestions:

Make sure all words are spelled correctly.

Try different keywords.

Try more general keywords

Explanation:

3 0

3 years ago

Read 2 more answers

A firecracker in a coconut blows the coconut into three pieces. Two pieces of equal mass fly off south and west, perpendicular t

Lisa [10]

Explanation:

Let us assume that piece 1 is $m_{1}$ is facing west side. And, piece 2 is $m_{2}$ facing south side.

Let $m_{1} \text{and} m_{2}$ = m and $m_{3}$ = 2m. Hence,

$2mv_{3y} = m(21)$

$v_{3y} = \frac{21}{2}$

= 10.5 m/s

$2mv_{3x} = m(21)$

$v_{3x} = \frac{21}{2}$

= 10.5 m/s

$v_{3} = \sqrt{v^{2}_{3x} + v^{2}_{3y}}$

= $\sqrt{(10.5)^{2} + (10.5)^{2}}$

= 14.84 m/s

or, = 15 m/s

Hence, the speed of the third piece is 15 m/s.

Now, we will find the angle as follows.

$\theta = tan^{-1} (\frac{v_{3y}}{v_{3x}})$

= $tan^{-1}(\frac{10.5}{10.5})$

= $tan^{-1} (45^{o})$

= $45^{o}$

Therefore, the direction of the third piece (degrees north of east) is north of east.

5 0

4 years ago

Which of the following statements is false? In solids, the particles do not move around or change position; they can only vibrat

Andrew [12]

"<span>The particles of gases are the least capable of moving around"
its the opposite, gases have the most capability to move around
</span>

8 0

3 years ago

Read 2 more answers

Two objects moving with a speed v travel in opposite directions in a straight line. The objects stick together when they collide

shutvik [7]

Answer:

A. K1/K2 = 36

B. M/m = 7/5

Explanation:

Part A. In order to calculate the ratio of the final kinetic energy and the initial kinetic energy of the system, you first express the kinetic energy of the system before and after the collision.

Before the collision:

$K_1=\frac{1}{2}Mv^2+\frac{1}{2}mv^2=\frac{1}{2}(M+m)v^2$ (1)

After the collision:

$K_2=\frac{1}{2}(M+m)(\frac{v}{6})^2=\frac{1}{72}(M+m)v^2$ (2)

The quotient between K1 and K2 is given by:

$\frac{K_1}{K_2}=\frac{1/2(M+m)v^2}{1/72(M+m)v^2}=\frac{1/2}{1/72}=36$

The quotient between the kinetic energy before and after the collision is K1/K2 = 36

Part B. To find the relation between the masses you use the momentum conservation law. The total momentum of the system before the collision and after the collision is given by:

$Mv-mv=(m+M)(\frac{v}{6})$ (3)

Where the minus sign in the first member of the equation means that the direction of the motion of both objects are opposite between them.

From the previous equation you can cancel the speed v and solve for M/m:

$(M-m)v=(m+M)(\frac{v}{6})\\\\M-m=\frac{1}{6}(m+M)\\\\M-\frac{1}{6}M=\frac{1}{6}m+m\\\\\frac{5}{6}M=\frac{7}{6}m\\\\\frac{M}{m}=\frac{7}{5}$

The ratio M/m is equal to 7/5

6 0

3 years ago

100 point good luck answering this 2+2-4+8+-5+???