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Montano1993 [528]

3 years ago

7

A bowling ball rolled with a force of 20 N accelerates at a rate of 5 m/sec2, a second ball rolled with the same force accelerat

es 10 m/sec2. What are the
combined masses of the two balls?

1 answer:

emmasim [6.3K]3 years ago

5 0

Answer:

sorry I need points

Explanation:

IM SO SORRY

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Mars has a mass of about 6.58 × 1023 kg, and its moon Phobos has a mass of about 9.3 × 1015 kg. If the magnitude of the gravitat

NISA [10]

This looks complicated, but it's actually not too tough.

The formula for the gravitational force between two objects is

Force = G (one mass) (other mass) / (distance²) .

The question GAVE us all of those numbers except the distance.
All we have to do is pluggum in, massage it around, and find
the distance.

Force = 4.18 x 10¹⁵ N
G = 6.673 x 10⁻¹¹ N·m²/kg²
One mass = 6.58 x 10²³ kg
Other mass = 9.3 x 10¹⁵ kg .

The only tricky thing about this is gonna be the arithmetic ...
keeping all the exponents straight.

Take the formula for the gravitational force and plug in
everything we know:

Force = (G) · (one mass) · (other mass) / (distance²)

4.18x10¹⁵N = (6.673x10⁻¹¹N-m²/kg²)·(6.58x10²³kg)·(9.3x10¹⁵kg) / (distance²).

Multiply each side by (distance²):

(distance²)·(4.18x10¹⁵N) = (6.673x10⁻¹¹N-m²/kg²)·(6.58x10²³kg)·(9.3x10¹⁵kg)

Divide each side by (4.18 x 10¹⁵ N) :

(distance²)=(6.673x10⁻¹¹N-m²/kg²)·(6.58x10²³kg)·(9.3x10¹⁵kg) / (4.18x10¹⁵N)

That's the end of the Physics and Algebra. The only thing left is Arithmetic.
We have to simplify that whole ugly thing on the right side of the equation,
and then take the square root of each side.

When I crunch down the right side of that equation, I get

(distance²) = 9.769 x 10¹³ m²

and when I take the square root of each side, I get

distance = 9.884 x 10⁶ meters . **

You should check my Arithmetic. **
(Pause occasionally to let your calculator cool off.)

BY THE WAY ...
That "distance" in the equation for gravitational force is the distance
between the CENTERS of the two objects.
This doesn't make much difference for Phobos, because Phobos isn't
much bigger than a big sweet potato. But it does make a difference for
Mars.
The 'distance' we find with all of this nonsense is NOT the distance
between Phobos and the surface of Mars. It's the distance between
Phobos and the CENTER of Mars, so it includes the planet's radius.

** Consulting online resources between Floogle and Flickerpedia,
I found that the orbital distance of Phobos from Mars varies between
9,234 km and 9,517 km. Add the planet's radius to these, and I'm
beginning to feel confidence in the results of my back-of-the-napkin
calculation. But you should still check my Arithmetic.

5 0

3 years ago

The __________ eventually becomes the muscles, skeleton, circulatory system, and other internal organs.

never [62]

Answer:

Mesoderm.....

Explanation:

Endoderm becomes internal organ

4 0

2 years ago

Which of the following correctly describes the relationship between current and voltage as the voltage of a battery increases. R

Y_Kistochka [10]

Answer:_COC1\/2+_H\/2O>_HC1+CO\/2

Explanation:

Need help asap

3 0

2 years ago

Please help ASAP....

Solnce55 [7]

Explanation:

its easy everone can do it .....let me know

5 0

2 years ago

Tennis balls traveling at greater than 100 mph routinely bounce off tennis rackets. At some sufficiently high speed, however, th

Kipish [7]

Answer:

Probability of tunneling is $10^{- 1.17\times 10^{32}}$

Solution:

As per the question:

Velocity of the tennis ball, v = 120 mph = 54 m/s

Mass of the tennis ball, m = 100 g = 0.1 kg

Thickness of the tennis ball, t = 2.0 mm = $2.0\times 10^{- 3}\ m$

Max velocity of the tennis ball, $v_{m} = 200\ mph$ = 89 m/s

Now,

The maximum kinetic energy of the tennis ball is given by:

$KE = \frac{1}{2}mv_{m}^{2} = \frac{1}{2}\times 0.1\times 89^{2} = 396.05\ J$

Kinetic energy of the tennis ball, KE' = $\frac{1}{2}mv^{2} = 0.5\times 0.1\times 54^{2} = 154.8\ m/s$

Now, the distance the ball can penetrate to is given by:

$\eta = \frac{\bar{h}}{\sqrt{2m(KE - KE')}}$

$\bar{h} = \frac{h}{2\pi} = \frac{6.626\times 10^{- 34}}{2\pi} = 1.0545\times 10^{- 34}\ Js$

Thus

$\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}$

$\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}$

$\eta = 1.52\times 10^{-35}\ m$

Now,

We can calculate the tunneling probability as:

$P(t) = e^{\frac{- 2t}{\eta}}$

$P(t) = e^{\frac{- 2\times 2.0\times 10^{- 3}}{1.52\times 10^{-35}}} = e^{-2.63\times 10^{32}}$

$P(t) = e^{-2.63\times 10^{32}}$

Taking log on both the sides:

$logP(t) = -2.63\times 10^{32} loge$

$P(t) = 10^{- 1.17\times 10^{32}}$

6 0

3 years ago