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Aleks [24]
3 years ago
11

Una gota de lluvia (m = 3.54 x 10-5 Kg.) cae verticalmente a velocidad constante bajo la influencia de la gravedad y la resisten

cia del aire. Después de que la gota ha descendido 121 m, ¿cuál es el trabajo (mJ) realizado por la gravedad?
Physics
1 answer:
konstantin123 [22]3 years ago
4 0

Answer:

Hola no speak Espanol

Explanation:

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Suppose that the acceleration of a model rocket is proportional to the difference between 100 ft/sec and the rocket's velocity.
sp2606 [1]

Answer:

Explanation:

initial velocity, u = 0

final velocity, v = 80 ft/s

acceleration, a = 150 ft/s²

Let the time taken is t.

v = u + at

80 = 0 + 150 x t

t = 0.53 second

3 0
3 years ago
A child standing on a bus remains still when the bus is at rest. When the bus moves forward and then slows down, the child conti
pochemuha
B. Newton's First Law, I'm pretty sure. The first states that an object in motion stays in motion, and an object at rest stays at rest until an outside force is applied, and that seems pretty relevant.
4 0
3 years ago
Read 2 more answers
"A steel rotating-beam test specimen has an ultimate strength of 120 kpsi. Estimate the life of the specimen if it is tested at
MrRissso [65]

Answer:

life (N) of the specimen is 117000  cycles

Explanation:

given data

ultimate strength Su = 120 kpsi

stress amplitude σa = 70 kpsi

solution

we first calculate the endurance limit of specimen Se i.e

Se = 0.5× Su   .............1

Se = 0.5 × 120

Se = 60 kpsi

and we know strength of friction f  = 0.82

and we take endurance limit Se is = 60 kpsi

so here coefficient value (a) will be

a = \frac{(f\times Su)^2}{Se}     ......................1  

put here value and we get

a = \frac{(0.82\times 120)^2}{60}  

a = 161.4  kpsi

so coefficient value (b) will be

b = -\frac{1}{3}log\frac{(f\times Su)}{Se}  

b =  -\frac{1}{3}log\frac{(0.82\times 120)}{60}  

b = −0.0716

so here number of cycle N will be  

N =  (\frac{ \sigma a}{a})^{1/b}

put here value  and we get

N =  (\frac{ 70}{161.4})^{1/-0.0716}

N = 117000

so life (N) of the specimen is 117000  cycles

7 0
3 years ago
In July 2005, NASA's "Deep Impact" mission crashed a 372-kg probe directly onto the surface of the comet Tempel 1, hitting the s
eimsori [14]

Question:

(a) What change in the comet’s velocity did this collision produce? Would this change be noticeable? (b) Suppose this comet were to hit the earth and fuse with it. By how much would it change our planet’s velocity? Would this change be noticeable? (The mass of the earth is 5.97×1024kg.)

Answer:

The answers to the question are;

(a) The change in the comet’s velocity produced by the collision is 2.86 × 10⁻⁶km/h or 7.944 × 10⁻⁷ m/s

(b) It would change our planet’s velocity by  6.70× 10⁻⁸ km/h or 1.86× 10⁻⁸ m/s

Change is too small to be noticeable

Explanation:

We not that the question is about conservation of liner momentum

Therefore we have, by listing out the known parameters

m₁ = Mass of "Deep Impact" = 372 kg

m₂ = Mass of Tempel 1 comet  = (0.1 to 2.5) × 10¹⁴ kg,

v₁ = Vaelocity of "Deep Impact" = 37000 km/h

v₂ = Velocity of Tempel 1 comet = 40000 km/h

From the principle of linear momentum, we have, for both bodies moving in opposite direction;

m₁×v₁ + m₂×v₂ = m₁×v₃ + m₂×v₃ since it was a crash, it is assumed that they both have the same final velocity

This gives

372 kg ×37000 km/h  - 0.1 × 10¹⁴ kg × 40000 km/h = (372 kg + 0.1 × 10¹⁴ kg )×v₃

13764000 kg·km/h - 4.0 × 10¹⁷  kg·km/h = 10000000000372×v₃

v₃ = (-399999999986236000 kg·km/h)/10000000000372 = -39999.999997 km/h ≈ - 40000  km/h in the direction of Deep Impact

Change in comet velocity

= 40000  km/h - 39999.999997 km/h

= 2.86 × 10⁻⁶km/h = 7.944 × 10⁻⁷ m/s

(b) If the colission is with earth, we have

m₃ = Mass of earth

From the principle of conservation of linear momentum, we have

m₂v₂+m₃ v₃ = (m₂ + m₃) v₄

v₃ = Initial velocity of Earth = 0 km/h

m₃ = Mass of Earth = 5.97 × 10²⁴ kg

Therefore, pluggin in the vaalues gives

0.1 × 10¹⁴ kg × 40000 km/h + 5.97 × 10²⁴ kg × 0 km/h = (0.1 × 10¹⁴ kg + 5.97 × 10²⁴ kg) × v₄

Therefore,

v₄ = (4.0 × 10¹⁷  kg·km/h + 0 kg·km/h)/ (5970000000010000000000000 kg)

= 6.70× 10⁻⁸ km/h = 1.86× 10⁻⁸ m/s

Change is too small

5 0
3 years ago
The apparent displacement of an object as it is viewed from two different positions is known as
Zigmanuir [339]
That's called 'parallax'.

To see it happen, go outside, hold your arm straight out in front
of your face and stick your thumb up.  Close one eye, and notice
where your thumb appears against the houses across the street. 
Now, don't move your arm, but just close the first eye and open
the other one.  You'll see your thumb jump in front of a different
house.  The angle that your thumb appeared to move is its parallax.
6 0
3 years ago
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