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Natali [406]
3 years ago
14

Please help solve this inelastic collision question please use GUESS to solve it if you can’t it’s fine

Physics
1 answer:
uranmaximum [27]3 years ago
4 0

Answer:

srfscfsd<vfrz<gfgrfdfsf<fesrf<sdfsre<f

Explanation:

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alexdok [17]

Answer:

10.30928 m/s

Explanation:

6 0
3 years ago
How are the fiducial points of the Celsius and Fahrenheit scales similar?
taurus [48]
The fiducial points of the Celsius<span> and the </span>Fahrenheit<span> temperature </span>scales<span> are the boiling and freezing </span>points<span> of pure water at 1 atm of pressure.

In short, Your Answer would be Option D

Hope this helps!</span>
7 0
3 years ago
You drop a ball from a height of 2.0 m, and it bounces back to a height of 1.5 m. (a) what fraction of its initial energy is los
Scilla [17]

a) At a position of 2.0m, the Initial energy is all made up of the potential energy=m*g*hi<span>
and meanwhile at 1.5 all its energy is also potential energy=m*g*hf 

The percentage of energy remaining is E=m*g*hi/m*g*hf x 100 

and since mass and gravity are constant so it leaves us with just E=hi/hf 
which 1.5/2.0 x100= 75% so we see that we lost 25% of the energy or 0.25 in fraction 

b) Here use the equation vf^2=vi^2+2gd 

<span>where g is gravity, vf is the final velocity and vi is the initial velocity while d is the distance travelled

so in here we are looking for the vi so let us isolate that variable 
we know that at maximum height or peak, the velocity is 0 so vf is 0 

therefore,</span></span>

vi =sqrt(-2gd) <span>
vi =sqrt(-2x-9.81x1.5) </span>
<span>vi =5.4 m/s

<span>c) The energy was converted to heat due to friction with the air and the ground.</span></span>

6 0
3 years ago
A proton and an alpha particle are momentarily at rest at adistance r from each other. They then begin to move apart.Find the sp
Arte-miy333 [17]

Answer:

The unknown quantities are:

E and F

The final velocity of the proton is:

√(8/3) k e^2/(m*r)

Explanation:

Hello!

We can solve this problem using conservation of energy and momentum.

Since both particles are at rest at the beginning, the initial energy and momentum are:

Ei = k (q1q2)/r

pi = 0

where k is the coulomb constant (= 8.987×10⁹ N·m²/C²)

and q1 = e and q2 = 2e

When the distance between the particles doubles, the energy and momentum are:

Ef = k (q1q2)/2r + (1/2)m1v1^2 + (1/2)m2v2^2

pf = m1v1 + m2v2

with m1 = m,   m2 = 4m,    v1=vf_p,    v2 = vf_alpha

The conservation momentum states that:

pi = pf      

Therefore:

m1v1 + m2v2 = 0

That is:

v2 = (1/4) v1

The conservation of energy states that:

Ei = Ef

Therefore:

k (q1q2)/r = k (q1q2)/2r + (1/2)m1v1^2 + (1/2)m2v2^2

Replacing

      m1 =  m, m2 = 4m, q1 = e, q2 = 2e

      and   v2 = (1/4)v1

We get:

(1/2)mv1^2 = k e^2/r + (1/2)4m(v1/4)^2 =  k e^2/r + (1/8)mv1^2

(3/8) mv1^2 = k e^2/r

v1^2 = (8/3) k e^2/(m*r)

3 0
3 years ago
Denise is conducting a physics experiment to measure the acceleration of a falling object when it slows down and comes to a stop
Butoxors [25]

Answer:

D!

Explanation:

Using the formula F = ma, you plug in 4.9 for F (force), and 0.5 for m (mass), then solve for a (acceleration).

4 0
3 years ago
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