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4 years ago

What happens if u slow water molecules down

1 answer:

7 0

The only way to slow them down is to COOL the water. When you do that, you take kinetic energy away from the molecules. When you take enough kinetic energy away from water molecules, the water freezes.

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Suppose a star the size of our Sun, but of mass 9.0 times as great, were rotating at a speed of 1.0 revolution every 17 days. If

Tanya [424]

Use the conservation of angular momentum; angular momentum at the beginning = angular momentum at the end
Conservation of angular momentum:
I1 w1 = I2 w2
Where I is the moment of inertia. For a sphere, I=2/5 m R^2. Substituting into the equation above we get
w2 = I1 w1 / I2 = w1 m1 R1^2 / (m2 R2^2)
w2 = w1 4 * (R1/R2)^2
= 4*(1)*(7E5/7.5)^2
= 3.48E10 revs/(17days)
= 2.04705882 x 10^9 revs/sec

4 0

3 years ago

Tidal Forces near a Black Hole. An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black

arlik [135]

Answer:

$833.4801043*10^6N$ on ear that is closer to Black hole.

$13.83803929*10^6N$ On ear that is farther from Black hole.

Explanation:

This problem can be solved as two masses that are at two different location from a bigger mass whose gravity affects both.

tension is an equal and opposite force that is exerted in response to applied force.

so on ear that is closer to black hole would have tension that is equal in magnitude and opposite in direction to gravitational force that ear experience due to the black hole at that location.

this true for ear that is further away from black hole as well.

(1) Force on ear that is closer to black hole.

$F =\frac{m_{1}*m_{2} }{r^2} G$

$m_{1}= 5*1.989*10^30kg$ is mass of Black hole.

$m_{2}= 0.030kg$ is mass of ear that is close to black hole.

$G = 6.679*10^-11m^3*kg^-1*s^-2$

r = $120km-\frac{6}{100000} km=119.99994km=119999.94m$

Note, we have subtracted because ear is closer to black hole.

plugging all this in formula gives.

$F = 833.4801043*10^6N$

That is tension of ear.

(2) Force on ear that is further from black hole.

$F =\frac{m_{1}*m_{2} }{r^2} G$

$m_{1}= 5*1.989*10^30kg$ is mass of Black hole.

$m_{2}= 0.030kg$ is mass of ear that is close to black hole.

$G = 6.679*10^-11m^3*kg^-1*s^-2$

this time r is further away from black hole so it would be.

$r = 120km+\frac{6}{100000} km = 120000.06m$

Plugging this all in we get

$F = 13.83803929N$

and that is tension on ear that is further from black hole.

Notice the tension difference, and order of magnitude of tension,it is enormous .

this astronaut is lethally close to black hole.

5 0

3 years ago

You drive 4 km at 30 km/h and then another 4 km at 50 km/h. What is your average speed for the whole 8-km trip?

Setler79 [48]

Answer:

option A

Explanation:

given,

drive 4 km at 30 km/h and

another 4 km at 50 km/h

average speed for the whole 8 Km trip = ?

time for the trip of first 4 km

distance = speed x time

$t_1 = \dfrac{d}{s}$

$t_1 = \dfrac{4}{30}$

t₁= 0.1333 hr

time from another 4 km trip

$t_2 = \dfrac{d}{s}$

$t_2= \dfrac{4}{50}$

t₂ = 0.08 hr

average speed

$s= \dfrac{d}{t}$

$s= \dfrac{8}{0.1333 + 0.08}$

$s= \dfrac{8}{0.2133}$

s = 37.50 Km/hr

Less than 40 Km/h

the correct answer is option A

6 0

3 years ago

A car speeds up from 13 m/s to 23 m/s in 30 seconds. What is the acceleration of the car?

ra1l [238]

Well, v = u + a×t is the equation.

v: final velocity, which is 23 m/s in this equation.
u: initialo velocity = 13 m/s 
a: acceleration = ? 
t: time = 30s

Your equation would be...

23 = 13 + a×30 
a = (23 - 13) / 30 
a = 1 / 3 
a = 0.333 m/s

7 0

3 years ago

A hawk flies in a horizontal arc of radius 12.0 m at constant speed 4.00 m/s. (a) Find its centripetal acceleration. (b) It cont

olya-2409 [2.1K]

Answer:

a) $a_c= 1.33 m/s^2$