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Rainbow [258]
3 years ago
9

which has a greater speed, a heron that travels 600 m in 60 seconds or a duck that travels 60 m in 5 seconds

Physics
2 answers:
nasty-shy [4]3 years ago
8 0
A duck has a greater speed
Gnom [1K]3 years ago
4 0
The duck will have the greater speed because 600 divided by 60 is 10 m in a second and 60 divided by 5 will be 12 m in a second also known as 12 mps
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An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km fr
mestny [16]

An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km from its center. The black hole is 5.00 times the mass of the sun and has a Schwarzschild radius of 15.0 km. The astronaut is positioned inside the spaceship such that one of her 0.030 kg ears is 6.0 cm farther from the black hole than the center of mass of the spacecraft and the other ear is 6.0 cm closer.

What is the tension between her ears?

Would the astronaut find it difficult to keep from being torn apart by the gravitational forces?

Answer:

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

Explanation:

Given that:

Orbital radius of the spacecraft (R) = 120 Km = 120 × 10³ m

Mass of the black hole (m) = 5 \ * (M \ _{sun})

where : M_{sun} = 1.99*10^{33} \ kg

Then; we have:

 m = 5*(1.99*10^{30} \ kg ) \\ = 9.95*10^{30} kg

Schwarzchild radius of the black hole

r - 15.0 km

Mass of each ear m_{ear} = 0.030 \ kg

Farther distance between one ear and the black hole (d) = 6.0 cm

= 0.06 m

Closer distance between the other ear and the black home is (d) 6.0 cm

= 0.6 cm

NOW, If we assume that the tension force should be T; then definitely the two ears will posses the same angular velocity .

The net force on the ear closer to the black hole will be:

\frac{GMm_{ear} }{(R-d)}- T = m_{ear} (R -  d) \omega^2

\frac{GMm_{ear} }{(R-d)^2}- \frac{T}{(R-d)} = m_{ear} \omega^2 \ ----> \ (1)

The net force on the ear farther to the black hole is :

\frac{GMm_{ear} }{(R+d)}- T = m_{ear} (R +  d) \omega^2

\frac{GMm_{ear} }{(R+d)^2}- \frac{T}{(R+d)} = m_{ear} \omega^2 \ ----> \ (2)

Equating equation (1) and (2) & therefore making (T) the subject of the formula; we have:

T = \frac{3GMm_{ear}d}{R^3}

T = \frac{3(6.67*10^{-11}N.m^2/kg^2)(1.95*10^{30}kg)(0.03kg)(0.06m)}{(120*10^3m)^3}

T = 2073.9 N\\T = 2.07 KN

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

3 0
3 years ago
The two-second rule applies to any speed __________ A. in all weather conditions. B. in ideal weather and road conditions. C. in
kumpel [21]

Answer:

The answer is A

Explanation:

A large risk of tailgating is the collision avoidance time being much less than the driver reaction time. Driving instructors advocate that drivers always use the "two-second rule" regardless of speed or the type of road. During adverse weather, downhill slopes, or hazardous conditions such as black ice, it is important to maintain an even greater distance.

5 0
3 years ago
Read 2 more answers
Are all magnetic fields created with electricity
Natasha_Volkova [10]
No not all, because not all electrical field attract and repel
3 0
3 years ago
How strong is naruto i need a in depth answer going off his AP and DC i prefer powerscalers since they do this but anyone can an
marin [14]

Answer:

Naruto is kinda strong

Explanation:

If were just talking about just him I think that he is strong for letting go of his past and moving along with his life. Bu t he also isn't strong if you think about it if Naruto didn't have the nine tails he wouldn't be special so he also is not strong there for his only power really comes from the nine tails.

6 0
3 years ago
A 2000 kg truck traveling north at 34 km/h turns east and accelerates to 58 km/h. (a) What is the change in the truck's kinetic
barxatty [35]

Explanation:

It is given that,

Mass of the truck, m = 2000 kg

Initial velocity of the truck, u = 34 km/h = 9.44 m/s

Final velocity of the truck, v = 58 km/h = 16.11 m/s

(a) Change in truck's kinetic energy, \Delta E=\dfrac{1}{2}m(v^2-u^2)

\Delta E=\dfrac{1}{2}\times 2000\ kg\times (16.11^2-9.44^2)

\Delta E=170418.5\ J

\Delta E=1.7\times 10^5\ J

(b) Change in momentum of the truck, \Delta p=m(v-u)

\Delta p=2000\ kg\times (16.11-9.44)

\Delta p=13340\ kg-m/s

Hence, this is the required solution.

6 0
3 years ago
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