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Art [367]
3 years ago
8

An astronaut in a spacecraft looks out her window and observes a comet travel in the opposite direction at a relative speed of 2

37 m/s. The velocity of the spacecraft is 114 m/s directly away from the sun. What is the comet's velocity relative to the sun? Assume that motion away from the sun is in the positive direction.
Physics
2 answers:
Sergeu [11.5K]3 years ago
7 0
The velocity of the spacecraft with respect to the Sun is
v_s = +114 m/s
the velocity of the comet in the reference system of the spacecraft is
v_c' = 237 m/s

So, the velocity of the comet in the reference system of the Sun will be
v_c = v_c'-v_s = -237 m/s - 114 m/s=-351 m/s
And the negative sign tells us that the comet is directed toward the Sun.
Umnica [9.8K]3 years ago
4 0

Answer: The relative velocity of the comet with respect to sun is '-351 m/s'.

Explanation:

Let the magnitude of the velocity of the spaceship moving away from the sun be |v_s|=114 m/s

Let the magnitude of the velocity of the comet moving towards the sun be |v_c| m/s

Assuming the direction away from the sun to be positive then :

The velocity of the space ship :v_s=+114 m/s

The  velocity of the comet :-v_c

v_r = relative velocity of spaceship with respect to comet

v_r=-237 m/s=v_s+(-v_c)=114 m/s-v_c

v_c=351 m/s

Velocity of the comet.v_c=351 m/s

v_R = Relative velocity of comet with respect to sun:

v_R=v_s+v_c=0 m/s+(-351 m/s)=-351 m/s

v_s= 0 m/s(Sun is fixed at one point)

The negative sign means that comet is moving towards the sun.The relative velocity of the comet with respect to sun is '-351 m/s'.

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Consider a river flowing toward a lake at an average velocity of 3 m/s at a rate of 500m3/sat a location 90 m above the lake sur
olga_2 [115]

Answer:

mechanical energy per unit mass is 887.4 J/kg

power generated is 443.7 MW

Explanation:

given data

average velocity = 3 m/s

rate = 500 m³/s

height h = 90 m

to find out

total mechanical energy and power generation potential

solution

we know that mechanical energy is sum of potential energy and kinetic energy

so

E = \frac{1}{2}×m×v² + m×g×h    .............1

and energy per mass unit is

E/m =  \frac{1}{2}×v² + g×h

put here value

E/m =  \frac{1}{2}×3² + 9.81×90

E/m = 887.4 J/kg

so mechanical energy per unit mass is 887.4 J/kg

and

power generated is express as

power generated = energy per unit mass ×rate×density

power generated = 887.4× 500× 1000

power generated = 443700000

so power generated is 443.7 MW

3 0
3 years ago
How can accuracy be limited?
lozanna [386]

Answer:

accuracy refers to the deviation of a measurement from a standard or true value of the quantity being measured

5 0
3 years ago
A 100-kg spacecraft is in a circular orbit about Earth at a height h = 2RE .
maria [59]

To solve this problem it is necessary to apply the concepts related to the conservation of the Gravitational Force and the centripetal force by equilibrium,

F_g = F_c

\frac{GmM}{r^2} = \frac{mv^2}{r}

Where,

m = Mass of spacecraft

M = Mass of Earth

r = Radius (Orbit)

G = Gravitational Universal Music

v = Velocity

Re-arrange to find the velocity

\frac{GM}{r^2} = \frac{v^2}{r}

\frac{GM}{r} = v^2

v = \sqrt{\frac{GM}{r}}

PART A ) The radius of the spacecraft's orbit is 2 times the radius of the earth, that is, considering the center of the earth, the spacecraft is 3 times at that distance. Replacing then,

v = \sqrt{\frac{(6.67*10^{-11})(5.97*10^{24})}{3*(6.371*10^6)}}

v = 4564.42m/s

From the speed it is possible to use find the formula, so

T = \frac{2\pi r}{v}

T = \frac{2\pi (6.371*10^6)}{4564.42}

T = 8770.05s\approx 146min\approx 2.4hour

Therefore the orbital period of the spacecraft is 2 hours and 24 minutes.

PART B) To find the kinetic energy we simply apply the definition of kinetic energy on the ship, which is

KE = \frac{1}{2} mv^2

KE = \frac{1}{2} (100)(4564.42)^2

KE = 1.0416*10^9J

Therefore the kinetic energy of the Spacecraft is 1.04 Gigajules.

8 0
3 years ago
PLEASE ANSWER, I NEED HELP
Scorpion4ik [409]

1) The gravitational force between Ellen and the moon is 1.56\cdot 10^{-3} N

2) The two forces are equal, while the acceleration of the bus is smaller than the acceleration of the bicycle.

Explanation:

1)

The magnitude of the gravitational force between two objects is given by

F=G\frac{m_1 m_2}{r^2}

where

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2} is the gravitational constant

m_1, m_2 are the masses of the two objects

r is the separation between them

In this problem, we have:

m_1 = 47 kg is the mass of Ellen

m_2 = 7.35\cdot 10^{22} kg is the mass of the moon

r=3.84\cdot 10^8 m is the distance between Ellen and the moon

Substituting, we find the gravitational force between Ellen and the moon:

F=(6.67\cdot 10^{-11})\frac{(47)(7.35\cdot 10^{22})}{(3.84\cdot 10^8)^2}=1.56\cdot 10^{-3} N

2)

We can analyze the forces acting in the collision between the bus and the bicycle by using Newton's third law of motion, which states that:

"When an object A exerts a force (called action) on an object B, then object B exerts an equal and opposite force (called reaction) on object A"

Applied to our problem, this means that the force exerted by the bus on the bicycle during the collision (action force) is equal (and opposite) to the force exerted by the bicycle on the bus (reaction force).

Now let's analyze the accelerations of the two vehicles. We can find the acceleration of each vehicle by using Newton's second law:

a=\frac{F}{m}

where

a is the acceleration

F is the force exerted on the vehicle

m is the mass of the vehicle

As we said previously, the force F exerted on each of the two vehicles: so, the acceleration only depends on the mass. In particular, the acceleration is inversely proportional to the mass: therefore, the larger the mass of the vehicle, the smaller the acceleration. This means that the acceleration of the bus is smaller than the acceleration of the bicycle.

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

And about Newton's third law:

brainly.com/question/11411375

#LearnwithBrainly

6 0
3 years ago
At a time t = 2.80 s , a point on the rim of a wheel with a radius of 0.230 m has a tangential speed of 51.0 m/s as the wheel sl
bonufazy [111]

Answer:

Part A) the angular acceleration is α= 44.347 rad/s²

Part B) the angular velocity is 195.13 rad/s

Part C)  the angular velocity is 345.913 rad/s

Part D ) the time is t= 7.652 s

Explanation:

Part A) since angular acceleration is related with angular acceleration through:

α = a/R = 10.2 m/s² / 0.23 m =   44.347 rad/s²

Part B) since angular acceleration is related

since

v = v0 + a*(t-t0) =  51.0 m/s + (-10.2 m/s²)*(3.4 s - 2.8 s) = 44.88 m/s

since

ω = v/R = 44.88 m/s/ 0.230 m = 195.13 rad/s

Part C) at t=0

v = v0 + a*(t-t0) =  51.0 m/s + (-10.2 m/s²)*(0 s - 2.8 s) = 79.56 m/s

ω = v/R = 79.56 m/s/ 0.230 m = 345.913 rad/s

Part D ) since the radial acceleration is related with the velocity through

ar = v² / R → v= √(R * ar) = √(0.23 m  * 9.81 m/s²)= 1.5 m/s

therefore

v = v0 + a*(t-t0) → t =(v -  v0) /a + t0 = ( 1.5 m/s - 51.0 m/s) / (-10.2 m/s²) + 2.8 s = 7.652 s

t= 7.652 s

4 0
3 years ago
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