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Lubov Fominskaja [6]

2 years ago

6

When the boy crashes his bumper car into the girl's bumper car, the momentum from his car is transferred to hers. What evidence

from the text supports this statement?

2 answers:

Darya [45]2 years ago

8 0

Answer:

Where is the text?

Explanation:

If you refer to the short sentence you wrote as text, I believe the answer is probably the word "crashes" because it shows how the momentum was transferred.

Guest1 year ago

0 0

idk a,aj fjianamaybe

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What's the momentum of a 3.5-kg boulder rolling down hill at 5 m/s

ICE Princess25 [194]

P = mv
p = 3.5 × 5
p = 17.5 kg .m/s

Hope this helps!

6 0

2 years ago

Determine a formula for the maximum height h that a rocket will reach if launched vertically from the Earth's surface with speed

olga55 [171]

Initially, the energies are:

$U_{i}=-\frac{G M_{\varepsilon} m}{r_{e}} \\
=K_{i}=\frac{1}{2} m v_{0}^{2}$

At final point, the energies are:

$U_{f}=-\frac{G M_{\varepsilon} m}{r_{e}+h} \\
K_{f}=\frac{1}{2} m(0)^{2}=0$

Using conservation law of energy,

$-\frac{G M_{e} m}{r_{e}}+\frac{1}{2} m v_{0}^{2} &=-\frac{G M_{e} m}{r_{\varepsilon}+h} \\
-\frac{G M_{e}}{r_{e}}+\frac{v_{0}^{2}}{2} &=-\frac{G M_{e}}{r_{e}+h} \\
\frac{-2 G M_{e}+r_{e} v_{0}^{2}}{2 r_{e}} &=-\frac{G M_{e}}{r_{e}+h} \\
\frac{r_{e}+h}{G M_{e}} &=\frac{2 r_{e}}{2 G M_{e}-r_{e} v_{0}^{2}}$

The equation is further simplified as:

$r_{e}+h &=\left(\frac{2 r_{e}}{2 G M_{e}-r_{e} v_{0}^{2}}\right) G M_{e} \\
h &=\frac{2 r_{e} G M_{e}}{2 G M_{e}-r_{e} v_{0}^{2}}-r_{e} \\
&=\frac{2 r_{e} G M_{e}-2 r_{e} G M_{e}+r_{e}^{2} v_{0}^{2}}{2 G M_{e}-r_{e} v_{0}^{2}} \\
& h=\frac{r_{e}^{2} v_{0}^{2}}{2 G M_{e}-r_{e} v_{0}^{2}}$

7 0

1 year ago

A baseball thrown at an angle of 60.0° above the horizontal strikes a building 16.0 m away at a point 8.00 m above the point fro

yanalaym [24]

Answer:

a) $v_{o} =16m/s$

b) $v=9.8m/s$

c) $\beta =-35.46º$

Explanation:

From the exercise we know that the ball strikes the building 16m away and its final height is 8m more than the initial

Being said that, we can calculate the initial velocity of the ball

a) First we analyze its horizontal motion

$x=v_{ox}t$

$x=v_{o}cos(60)t$

$v_{o}=\frac{x}{tcos(60)}=\frac{16m}{tcos(60)}$ (1)

That would be our first equation

Now, we need to analyze its vertical motion

$y=y_{o}+v_{oy}t+\frac{1}{2}gt^2$

$y_{o}+8=y_{o}+v_{o}sin(60)t-\frac{1}{2}(9.8)t^2$

Knowing $v_{o}$ in our first equation (1)

$8=\frac{16}{tcos(60)}sin(60)t-\frac{1}{2}(9.8)t^2$

$\frac{1}{2}(9.8)t^2=16tan(60)-8$

Solving for t

$t=\sqrt{\frac{2(16tan(60)-8)}{9.8} } =2s$

So, the ball takes to seconds to get to the other building. Now we can calculate its <u>initial velocity</u>

$v_{o}=\frac{16m}{(2s)cos(60)}=16m/s$

b) To find the <u>magnitude of the ball just before it strikes the building</u> we need to calculate its x and y components

$v_{x}=v_{ox}+at=16cos(60)=8m/s$

$v_{y}=v_{oy}+gt=16sin(60)-(9.8)(2)=-5.7m/s$

So, the magnitude of the velocity is:

$v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{(8m/s)^2+(-5.7m/s)^2}=9.8m/s$

c) The <u><em>direction of the ball</em></u> is:

$\beta=tan^{-1}(\frac{v_{y} }{v_{x}})=tan^{-1}(\frac{-5.7}{8})=-35.46º$

4 0

2 years ago

Suppose the dim-looking headlight on the right is actually a small light on the front of a bicycle. What can you conclude about

aleksley [76]

Answer:

Explanation:

The dimness of two light source, which have the same intensify depends on their distance from the observer. That is why some stars appear brighter than the others, this is due to the distance of each of them from the earth.

The closer a light source, the brighter they appear

Now, if the right lamp has a dim light actually, it shows that the bicycle is closer than it appears.

4 0

2 years ago

Which radioactive isotope would take the least amount of time to become stable? rubidium-91 iodine-131 cesium-135 uranium-238

Leno4ka [110]

The radioactive isotope that would take the least amount of time to become stable is rubidium-91. This is because this isotope is the most stable compared to the rest. This was determined by subtracting its atomic mass by its atomic number. The isotope with the least number of difference is the most stable, while the one with the greatest difference is the most unstable.

Difference:
Rubidium: 54 (most stable)
Iodine: 78
Cesium: 80
Uranium: 146 (least stable)

8 0

2 years ago

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