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3 years ago

What is the final velocity?

2 answers:

6 0

Initial velocity describes how fast an object travels when gravity first applies force on the object. On the other hand, the final velocity is a vector quantity that measures the speed and direction of a moving body after it has reached its maximum acceleration.

Reil [10]3 years ago

5 0

The final velocity is a vector quantity that measures the speed and direction of a moving body after it has reached its maximum acceleration

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HELP PLEASE 15 POINTS !!!

Nookie1986 [14]

B(I)84.0*10^5J
B(ll)1400W

3 0

2 years ago

A raindrop of mass 0.5 * 10^-4 kg is falling verctically under the influence of gravity. The air drag on the raindrop is fdrag =

Elina [12.6K]

Answer:

The displacement of the air drop after 3 second is 18.27 m.

Explanation:

Mass of the rain drop = m = $0.5\times 10^{-4} kg$

Weight of the rain drop = W

Duration of time = t = 3 seconds

$W=m\times g$

Drag force on rain drop = $D=0.2\times 10^{-5} v^2$

$W=0.5\times 10^{-4} kg\time 10 m/s^2=0.5\times 10^{-3} N$

Motion of the rain drop:

$F=m\times a$

Net force on the rain drop , F= W - D

$W-D=m\times a$

$0.5\times 10^{-3} N-0.2\times 10^{-5} v^2=0.5\times 10^{-4} kg\times a$

$0.5\times 10^{-3} kg m/s^2-0.2\times 10^{-5} v^2=0.5\times 10^{-4} kg\times \frac{v}{t}$

$0.006v^2+0.05v-1.5=0$

v = 12.18 m/s

Initial velocity of the rain drop = u = 0 (since, it is starting from rest)

v=u+at (First equation of motion)

$12.18 m/s=0m/s+a\times 3 s$

$a=4.06 m/s^2$

$s=ut+\frac{1}{2}at^2$ (second equation of motion)

$s=0\times 3s+\frac{1}{2}\times 4.06m/s^2\times (3 s)^2$

s = 18.27 m

The displacement of the air drop after 3 second is 18.27 m.

6 0

2 years ago

21. (a) Aircraft sometimes acquire small static charges. Suppose a supersonic jet has a 0.500-μC charge and flies due west at a

beks73 [17]

Answer:

Explanation:

The picture attached shows the full explanation and i hope it helps. Thank you

6 0

3 years ago

The distance between Pluto and the Sun is 39.1 times more than the distance between the Sun and Earth. Calculate the time taken

german

Answer:

Explanation:

Given

Distance between Pluto and sun is 39.1 times more than the distance between earth and sun

According to Kepler's Law

$T^2=kR^3$

where k=constant

T=time period

R=Radius of orbit

Suppose $R_1$ is the radius of orbit of earth and sun

so Distance between Pluto and sun is $R_2=39.1\cdot R_1$

$T_1$ and $T_2$ is the time period corresponding to $R_1$ and R_2[/tex]

$(T_1)^2=k(R_1)^3---1$

$(T_2)^2=k(R_2)^3---2$

divide 1 and 2

$(\frac{365}{T_2})^2=(\frac{R_1}{39.1})^3$

$T_2^2=365^2\times 39.1^3$

$T_2=89239.67\ Earth\ days$

7 0

2 years ago

What is the effective dose to a worker who receives uniform, whole-body dose of 8.4 mGy from gamma-rays and 1.2 mGy from 80-Kev

Vera_Pavlovna [14]

Given :

Whole-body dose of 8.4 mGy from gamma-rays and 1.2 mGy from 80-Kev neutrons.

To Find :

The effective dose to a worker.

Solution :

By the given information effective dose to a worker is given by :

E.D = ( 8.4 × 1.2 × 0.12 ) + ( 1.2 × 1 × 1 )

E.D = 1.2096 + 1.2

E.D = 2.4096

Therefore, the effective dose to a worker is 2.4096 .

3 0

2 years ago