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zmey [24]
3 years ago
8

Two astronauts of identical mass are connected by a taut cable of negligible mass, as shown in the figure above, and are initial

ly at rest with respect to a nearby space station. Astronaut Y pulls on the cable toward herself with considerable force. Which of the following describes the direction of the velocity of the center of mass of the two astronauts after Astronaut Y pulls on the cable?
Physics
1 answer:
PolarNik [594]3 years ago
5 0

Answer:

The right answer is "The center of mass doesn't move".

Explanation:

  • It generates a voltage throughout the cable while the astronaut falls on either the wire. At other ends of the spectrum or cable, the tension will be similar. As such, with both astronauts, there would be the same energy, although throughout the opposite way.
  • Thus, the net force seems to be essentially negative on the machine. And therefore the mass center stays stationary.
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8. What is the momentum of a 120-kg professional fullback running<br> across the line at 11.2 m/s?
Alchen [17]

Answer:

134r kgm^-1 or 1344 kg /m

Explanation:

Momentum is is given by:

p=mv

Where:

p is momentum, m is mass in kg and v is velocity in ms−1

p=120kg×11.2 m/ s= 1344 kgms=1344kgm^−1

5 0
3 years ago
Astronomers have observed a small, massive object at the center of our Milky Way Galaxy. A ring of material orbits this massive
Burka [1]

Answer:

1.91773\times 10^{37}\ kg

Explanation:

v = Orbital speed = 130 km/s

d = Diameter = 16 ly

r = Radius = \dfrac{d}{2}=\dfrac{16}{2}=8\ ly

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

1\ ly=9.461\times 10^{15}\ m

As the centripetal force balances the gravitational energy we have the following relation

\dfrac{GMm}{r^2}=\dfrac{mv^2}{r}\\\Rightarrow M=\dfrac{v^2r}{G}\\\Rightarrow M=\dfrac{130000^2\times 8\times 9.461\times 10^{15}}{6.67\times 10^{-11}}\\\Rightarrow M=1.91773\times 10^{37}\ kg

Mass of the the massive object at the center of the Milky Way galaxy is 1.91773\times 10^{37}\ kg

4 0
3 years ago
Gas a bG1 5.22 0.0289G2 1.05 0.0388G3 2.31 0.0467G4 4.05 0.0310Based on the given van der Waals constants for four hypothetical
inysia [295]

Answer:

Gas 2, Gas 3, Gas 4, Gas 5 is the order of decreasing strength of inter-molecular forces.

Explanation:

The strength increases as there is a decrease in the vanderwaals constant and vice versa.

3 0
3 years ago
Assuming a Chevy Cruze may be operated indefinitely at 30 m/s with no stops for provisions,
kramer

To solve this problem we will apply the linear motion kinematic equations. In which the speed is defined as the distance traveled in a certain period of time. In turn we must emphasize that 1 light year is equivalent to 9.461*10^{15}m. Mathematically the speed is described as,

v = \frac{x}{t}  \rightarrow t = \frac{x}{v}

Replacing,

t = \frac{9.461*10^{15}}{30}

t = 3.15366*10^{14}s (\frac{1 year}{3.154*10^7})

t \approx 10,000,000 years

Therefore the correct answer is D.

4 0
3 years ago
A projectile is fired at time t= 0.0 s from point 0 at the edge of a cliff, with initial velocity components of Vox = 30 m/s and
BARSIC [14]

Answer:

At t = 15.0 s the magnitude of the velocity is 58.31 m/s

Explanation:

The given parameters are;

V₀ₓ = 30 m/s

V_{0y} = 100 m/s

The time of flight of the projectile = 25 s

For projectile motion;

Vₓ = V₀ × cos(θ₀)

The magnitude of the velocity V = √(V₀ₓ² + V_{0y}²)

We have the magnitude of the initial velocity = √(30² + 100²) = 10·√109 m/s

cos(θ₀) = V₀ₓ/V₀ = 30/(10·√109) = 3/√109

θ₀ = cos⁻¹(3/√109) = 73.3°

The components of the velocity after time t is given by the relations;

Vₓ = V₀ × cos(θ₀) = 30 m/s

V_y =  V₀ × sin(θ₀) - g×t

When V_y = 0, we have;

0 =  V₀ × sin(θ₀) - g×t

g×t  =  V₀ × sin(θ₀)  = 10·√109×0.958 = 100 m/s

t = 100/g = 100/10 = 10 s

The time to reach maximum height = 10 s

At 15.0 seconds, we have;

V_y =  V₀ × sin(θ₀) - g×t = 10·√109×0.958  - 10×15 = -50 m/s

Therefore, the projectile is returning at 50 m/s

The magnitude of the velocity =√(30² + 50²) = 10·√34 m/s = 58.31 m/s.

5 0
3 years ago
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