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3 years ago

Scientists observe an approaching asteroid that is on a collision course with

Physics

1 answer:

nasty-shy [4]3 years ago

3 0

Answer:

The approximate velocity the rocket must have to stop the asteroid completely after the collision is;

C. -324 m/s

Explanation:

The parameters of the asteroid and the rocket are;

The mass of the asteroid, m₁ = 11,000 kg

The initial velocity with which the asteroid is approaching Earth, v₁ = 50 m/s

The mass of the rocket, m₂ = 1700 kg

The initial velocity of the rocket = v₂

The final velocity of the combined asteroid and rocket after the collision, v₃ = 0 m/s

By the law of conservation of linear momentum, we have;

The total initial momentum = The total final momentum

m₁·v₁ + m₂·v₂ = (m₁ + m₂)·v₃

Substituting the known values, we get;

11,000 kg × 50 m/s + 1,700 kg × v₂ = (11,000 kg + 1,700 kg) × 0 m/s

11,000 kg × 50 m/s + 1,700 kg × v₂ = 0

∴ 1,700 kg × v₂ = -11,000 kg × 50 m/s

v₂ = (-11,000 kg × 50 m/s)/(1,700 kg) = -323.529412 m/s ≈ -324 m/s

The approximate initial velocity the jet must have to completely stop the asteroid after the collision is -324 m/s.

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SHOW WORK

Helga [31]

Answer:

Follows are the solution to the given question:

Explanation:

For point a:

$T= 2\pi \sqrt{\frac{m}{k}}\\\\k = \frac{4 \pi^2 m}{T^2}\\\\= \frac{4 \times (3.14)^2 \times 3}{2^2}\\\\=29.578 \ \frac{N}{m}\\\\$

For Point b:

$E=\frac{1}{2} m a^2 w^2\\\\$

$=\frac{1}{2} \times m \times a^2 \times \frac{4\pi^2}{T^2}\\\\=\frac{1}{2} \times 3 \times (0.15)^2 \times \frac{4\times 3.14^2}{2^2}\\\\=0.332 \ J$

For Point C:

$V_{max}= a w$

$= (0.15) \times \frac{2\pi}{T}\\\\= (0.15) \times \frac{2\times 3.14}{2}\\\\=0.471 \frac{m}{s}$

For point D:

$X= a \sin (wt+ \phi)\\\\0.91=0.15 \sin(\frac{2\pi}{T} \times t+\phi)\\\\0.91=0.15 \sin(\frac{2\times 3.14}{2} \times 0.5+\phi)\\\\0.60 = \sin(3.14 \times 0.5+\phi)\\\\0.60 = \sin(1.57+\phi)\\\\1.57 +\phi =\sin^{-1} 60^{\circ}\\\\1.57 +\phi = 36.86^{\circ}\\\\=35.29^{\circ}\\\\So, X=15 \sin(3.14t+35.29^{\circ}) \ cm$

5 0

3 years ago

A car was moving at 14 m/s After 30 s, its speed increased to 20 m/s. What was the acceleration during this time ( need help fas

Arada [10]

Answer:let initial velocity u=14m/s

Final velocity v=20m/s

Time taken t=30

Acceleration =a

V=u +at

a= (20-14)/30

a=0.2m/s^2

Explanation:

Acceleration is the change in velocity with respect to time.

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3 years ago

What did solomon asch discover in his famous experiment on judging the lengths of lines?

BaLLatris [955]

Answer:

Asch (1956) found that group size influenced whether subjects conformed. The bigger the majority group (no of confederates), the more people conformed, but only up to a certain point.

Explanation:

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3 years ago

PLEASE HELPPPPP

Ganezh [65]

The only graph that accurately depict the given motion is graph D.

The given parameters;

initial position of the man = 0

direction of the man's first displacement = backward
time of first motion, t₁ = 6 seconds
velocity of this first displacement = v₁
time without any motion (zero movement) = 6 seconds
direction of the second displacement = forward
velocity of second displacement = 2v₁

Let the acceleration of the first displacement = a

Acceleration of the second displacement = 2a

From the given graphs we can eliminate every graph without initial decrease or motion towards the negative direction.

The only options with initial motion towards the negative direction are;

graph B, and

graph D.

The difference between graph B and D;

in graph B there is a uniform motion for 6 seconds

in graph D there is no motion for 6 seconds (this is obvious as the line fall directly on top of the horizontal axis maintaining a value of zero for 6 seconds).

Thus, the only graph that accurately depict the given motion is graph D.

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2 years ago

In a tape recorder, the tape is pulled past the read-and-write heads at a constant speed by the drive mechanism. Consider the re

Irina-Kira [14]

Answer:

Torque decreases .

Explanation:

The tape is pulled at constant speed , speed v is constant , so there is

v = ω r where ω is angular speed and r is radius , As radius decreases , angular speed ω increases , So there is angular acceleration .

Let it be α . Let I be moment of inertia of reel .

Reel is in the form of disc

I = 1/2 m r²

I x α = torque

1/2 m r² x α = torque

As the reel is untapped , its mass decreases , r also decreases , so torques also decreases .

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