Ask question

Ask question

All categories

2 years ago

12

Suppose a tidal basin is 6 m above the ocean at low tide and that the area of the basin is 5×107 m2. Estimate the gravitational

potential energy change if the water is released from the tidal basin to the low-tide ocean level. The density of water is 1000 kg/m3. (Hint: The level does not change by 6 m for all of the water.)

1 answer:

Neko [114]2 years ago

3 0

Answer:

Potential energy will be $176.58\times 10^{11}j$

Explanation:

We have given the height of the basin is h = 6 m

Area of the basin $A=5\times 10^7m^2$

Volume $V=area \times height=5\times 10^7\times 6=30\times 10^7m^3$

Density $\rho =1000kg/m^3$

We know that mass is given by $m=\rho V=1000\times 30\times 10^7=3\times 10^{11}kg$

We know that potential energy is given by $E=mgh=3\times 10^{11}\times 9.81\times 6=176.58\times 10^{11}j$

You might be interested in

i will give brainliest Which of the following structures are present in plant cells but absent in animal cells? Choose 1 answer:

Sonja [21]

Answer:

A. cell walls

Explanation:

Plants have cell walls but animals dont.

3 0

3 years ago

If a vector that is 3cm long represents 30 km/h, what velocity does a 5 cm long vector which is drawn using the same scale repre

FrozenT [24]

Answer:

False knowww yess ofcure

8 0

3 years ago

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

a 5.0 charge is placed at the 0 cm mark of a meterstick and a -4.0 charge is placed at the 50 cm mark. what is the electric fiel

Lesechka [4]

Answer:

-1748*10^N/C

Explanation:

See attached file

8 0

3 years ago

A 4.0 Ω resistor has a current of 3.0 A in it for 5.0 min. How many electrons pass 3. through the resistor during this time inte

musickatia [10]

Answer:

Number of electrons, $n=5.62\times 10^{21}$

Explanation:

It is given that,

Resistance, R = 4 ohms

Current, I = 3 A

Time, t = 5 min = 300 s

We need to find the number of electrons pass through the resistor during this time interval. Let the number of electron is n.

i.e. q = n e ...............(1)

And current, $I=\dfrac{q}{t}$

$I\times t=n\times e$

$n=\dfrac{It}{e}$

e is the charge of an electron

$n=\dfrac{3\ A\times 300\ s}{1.6\times 10^{-19}}$

$n=5.62\times 10^{21}$

So, the number of electrons pass through the resistor is $5.62\times 10^{21}$ . Hence, this is the required solution.

6 0

3 years ago