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erik [133]
3 years ago
6

Why do raindrops fall with a constant speed during the later stages of their descents?

Physics
1 answer:
BartSMP [9]3 years ago
6 0
Firstly they have a acceleration downwards due the force downwards due they gravitational field acting on it's mass.
as it falls it gains speed, and as it gains speed the air Resistance which is a upward force actin on the drop increases, eventually the rain drop's upward and downward forces are balanced and hence there is no RESULTANT force therefore no acceleration, so the drops falls in constant speed (terminal verlocity is a better term)

Are you wondering that why is the raindrop still moving given that the forces are balanced? If so according to Newton's 1st law an object will keep moving or Remain at rest until a RESULTANT force acts on it.
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3. Imagine a 10kg block moving with a speed of 20m/s<br> calculate the kinetic energy of this block
MatroZZZ [7]
The formula of the kinetic energy is:
E_{k}  =  \frac{m v^{2} }{2}
where m is a mass of the object, v is speed of the object at the moment of time. So we have:
E_{k}  =  \frac{10* 20^{2} }{2}  = 2000J
The answer is 2000 Joules.
6 0
3 years ago
Explain how a voltage is produced in the secondary coil of the transformer.
Tema [17]

Answer:

the primary coil current produces a magnetic field, which changes as the current changes. the iron core increases the strength of the magnetic field. the changing magnetic field induces a changing potential difference (voltage) in the secondary coil.

4 0
2 years ago
Suppose a star has the same apparent brightness as Alpha Centauri A ( 2.7×10−8watt/m2 ) but is located at a distance of 300 ligh
iragen [17]

For  a star that has the same apparent brightness as Alpha Centauri A ( 2.7×10−8watt/m2 is mathematically given as

L=2.7*10^30w

<h3> What is its luminosity?</h3>

Generally, the equation for the luminosity  is mathematically given as

L=4*\pi^2*b

Therefore

L=4*\pi^2*b

L=4* \pi *(2.83*10^{18})*2.7*10^{-8}

L=2.7*10^30

In conclusion, the luminosity

L=2.7*10^30w

Read more about Light

brainly.com/question/25770676

6 0
2 years ago
A mass m attached to a horizontal massless spring with spring constant k, is set into simple harmonic motion. its maximum displa
Lesechka [4]
At the point of maximum displacement (a), the elastic potential energy of the spring is maximum:
U_i= \frac{1}{2} ka^2
while the kinetic energy is zero, because at the maximum displacement the mass is stationary, so its velocity is zero:
K_i =0
And the total energy of the system is
E_i = U_i+K= \frac{1}{2}ka^2

Viceversa, when the mass reaches the equilibrium position, the elastic potential energy is zero because the displacement x is zero:
U_f = 0
while the mass is moving at speed v, and therefore the kinetic energy is
K_f =  \frac{1}{2} mv^2
And the total energy is
E_f = U_f + K_f =  \frac{1}{2} mv^2

For the law of conservation of energy, the total energy must be conserved, therefore E_i = E_f. So we  can write
\frac{1}{2} ka^2 =  \frac{1}{2}mv^2
that we can solve to find an expression for v:
v= \sqrt{ \frac{ka^2}{m} }
6 0
3 years ago
PLEASE ANSWER ASAP!!!!!!!
Viefleur [7K]

Answer:

B. Mechanical energy= 50J+30J=80J

4 0
3 years ago
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