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3 years ago

15

A space vehicle approaches a space station in orbit. The intent of the engineers is to have the vehicle slowly approach, reducin

g velocity, until a docking maneuver is completed and the vehicle is attached to the station. How does the total momentum of the docked vehicle and station compare to the momentum of each object before the docking maneuver? greater momentum same momentum less momentum

1 answer:

NeX [460]3 years ago

3 0

Answer:

c. same momentum

Explanation:

got it correct on edge2020

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a car accelerates from 4 meters/second to 16 meter/second in 4 seconds. The cars acceleration is how many meter/seconds.

Allushta [10]

Answer:

=3 metre per second ^2

Explanation:

Formula for acceleration is

V-U÷T

In the given information

V=16

U=4

T=4

Acceleration =16-4/4

=3 metre per second ^2

4 0

2 years ago

For a damped simple harmonic oscillator, the block has a mass of 1.2 kg and the spring constant is 9.8 N/m. The damping force is

ArbitrLikvidat [17]

Answer:

a) t=24s

b) number of oscillations= 11

Explanation:

In case of a damped simple harmonic oscillator the equation of motion is

m(d²x/dt²)+b(dx/dt)+kx=0

Therefore on solving the above differential equation we get,

x(t)=A₀ $e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)$

where A(t)=A₀ $e^{\frac{-bt}{2m}}$

A₀ is the amplitude at t=0 and

$w'$ is the angular frequency of damped SHM, which is given by,

$w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }$

Now coming to the problem,

Given: m=1.2 kg

k=9.8 N/m

b=210 g/s= 0.21 kg/s

A₀=13 cm

a) A(t)=A₀/8

⇒A₀ $e^{\frac{-bt}{2m}}$ =A₀/8

⇒ $e^{\frac{bt}{2m}}=8$

applying logarithm on both sides

⇒ $\frac{bt}{2m}=ln(8)$

⇒ $t=\frac{2m*ln(8)}{b}$

substituting the values

$t=\frac{2*1.2*ln(8)}{0.21}=24s(approx)$

b) $w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }$

$w'=\sqrt{\frac{9.8}{1.2}-\frac{0.21^{2}}{4*1.2^{2}}}=2.86s^{-1}$

$T'=\frac{2\pi}{w'}$ , where $T'$ is time period of damped SHM

⇒ $T'=\frac{2\pi}{2.86}=2.2s$

let $n$ be number of oscillations made

then, $nT'=t$

⇒ $n=\frac{24}{2.2}=11(approx)$

8 0

3 years ago

State the law of conservation of energy

storchak [24]

Answer:

In physics and chemistry, the law of conservation of energy states that the total energy of an isolated system remains constant; it is said to be conserved over time. ... For instance, chemical energy is converted to kinetic energy when a stick of dynamite explodes.

7 0

2 years ago

Breaks in the Earths crust called faults from where plates meet true or false

KonstantinChe [14]

False theyre called plate boundries

5 0

3 years ago

Read 2 more answers

You see a lightning bolt flash in the distant sky. You hear the corresponding thunder 6.8 seconds later. If the temperature is 2

Anna35 [415]

Answer:

$Distance = 2.4\ km$

Explanation:

Given

$Time (t) = 6.8s$

$Temperature = 28C$

Required

Determine the distance at which the lighting struck

First, we need to determine the speed at which the lighting struck because the peed of sound varies with temperature.

At about 28C, the speed of sound is 346m/s

So, we have the following:

$Speed = 346m/s$

$Time = 6.8s$

Distance is calculated as thus:

$Distance = Speed * Time$

$Distance = 346m/s * 6.8s$

$Distance = 2352.8$

Divide by 1000 to get distance equivalent in kilometers

$Distance = \frac{2352.8km}{1000}$

$Distance = 2.3528km$

$Distance = 2.4\ km$ ---- Approximated

7 0

3 years ago