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solmaris [256]
3 years ago
7

Paul’s 10 kg baby sister Susan sits on a mat. Paul pulls the mat across the floor using a rope that is angled 30° above the floo

r. The tension is a constant 30 N and the coefficient of kinetic friction is 0.20. Use work and energy to find Susan’s speed after being pulled 3.0 m.

Physics
2 answers:
kifflom [539]3 years ago
5 0

Answer: speed= 1.95866m/s ~= 2.0 m/s

Explanation: check attachment for explanation

kiruha [24]3 years ago
4 0

Answer:

The speed of Susan is 2.37 m/s

Explanation:

To visualize better this problem, we need to draw a free body diagram.

the work is defined as:

W=F*d*cos(\theta)

here we have the work done by Paul and the friction force, so:

W_p=F_p*d*cos(0)\\F_p=30N*cos(30^o)=26N\\W_p=26*3*(1)=78J

W_f=F_f*d*cos(180)\\F_f=\µ*(10*9.8-30N*sin(30^o))=16.6N\\W_p=16.6*3*(-1)=50J

Now the change of energy is:

W_p-W_f=\frac{1}{2}m*v^2\\v=\sqrt{\frac{2(78J-50J)}{10kg}}\\v=2.37m/s

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What is the highest degrees above the horizon the moon ever gets during the year in the Yakima Valley ?
Ivahew [28]

The trickiest part of this problem was making sure where the Yakima Valley is.
OK so it's generally around the city of the same name in Washington State.

Just for a place to work with, I picked the Yakima Valley Junior College, at the
corner of W Nob Hill Blvd and S16th Ave in Yakima.  The latitude in the middle
of that intersection is 46.585° North.  <u>That's</u> the number we need.

Here's how I would do it:

-- The altitude of the due-south point on the celestial equator is always
(90° - latitude), no matter what the date or time of day.

-- The highest above the celestial equator that the ecliptic ever gets
is about 23.5°. 

-- The mean inclination of the moon's orbit to the ecliptic is 5.14°, so
that's the highest above the ecliptic that the moon can ever appear
in the sky.

This sets the limit of the highest in the sky that the moon can ever appear.

90° - 46.585° + 23.5° + 5.14° = 72.1° above the horizon .

That doesn't happen regularly.  It would depend on everything coming
together at the same time ... the moon happens to be at the point in its
orbit that's 5.14° above ==> (the point on the ecliptic that's 23.5° above
the celestial equator).

Depending on the time of year, that can be any time of the day or night.

The most striking combination is at midnight, within a day or two of the
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5 0
3 years ago
What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of – 30.0 nC?
Nastasia [14]

Answer:

F=1.26*10^{-3}N

Explanation:

Assuming the pith balls as point charges, we can calculate the repulsive force between them, using Coulomb's law:

F=\frac{kq_1q_2}{d^2}

We observe that the magnitude of the electric force is directly proportional to the product of the magnitude of both signed charges(q_1,q_2) and inversely proportional to the square of the distance(d) that separates them.

Replacing the given values, where k is the Coulomb constant:

F=\frac{8.99*10^{9}\frac{N\cdot m^2}{C^2}(-30*10^{-9}C)(-30*10^{-9}C)}{(8*10^{-2}m)^2}\\F=1.26*10^{-3}N

8 0
3 years ago
How many DAYS occur between a new moon and a first quarter moon?
N76 [4]

Answer:

About 29.5

Explanation:

4 0
2 years ago
Read 2 more answers
If a planet has the same mass as the earth, but has twice the radius, how does the surface gravity, g, compare to g on the surfa
shepuryov [24]

Answer:

The surface gravity g of the planet is 1/4 of the surface gravity on earth.

Explanation:

Surface gravity is given by the following formula:

g=G\frac{m}{r^{2}}

So the gravity of both the earth and the planet is written in terms of their own radius, so we get:

g_{E}=G\frac{m}{r_{E}^{2}}

g_{P}=G\frac{m}{r_{P}^{2}}

The problem tells us the radius of the planet is twice that of the radius on earth, so:

r_{P}=2r_{E}

If we substituted that into the gravity of the planet equation we would end up with the following formula:

g_{P}=G\frac{m}{(2r_{E})^{2}}

Which yields:

g_{P}=G\frac{m}{4r_{E}^{2}}

So we can now compare the two gravities:

\frac{g_{P}}{g_{E}}=\frac{G\frac{m}{4r_{E}^{2}}}{G\frac{m}{r_{E}^{2}}}

When simplifying the ratio we end up with:

\frac{g_{P}}{g_{E}}=\frac{1}{4}

So the gravity acceleration on the surface of the planet is 1/4 of that on the surface of Earth.

3 0
3 years ago
intial velocity is 45km/h, final velocity is 0, time taken is 3 seconds,calculate the retradation of the vehicle and distance tr
V125BC [204]

Answer:

v= u+at

0=45×1000×60×60+3a

_3a=16200045

a=5400015m per sec

4 0
3 years ago
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