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4 years ago

Which gas giant has a rotation axis so tilted that the planet rotates like a bowling ball as it orbits the sun?

Physics

1 answer:

stellarik [79]4 years ago

6 0

C || Uranus!
I believe it's axial tilt approximately 98°, 4 times greater than earths!

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(100 POINTS!!!!!) What type of solar radiation does not reach the surface of the Earth?

Kobotan [32]

Gamma radiations do not reach the surface of the Earth

Explanation:

Sun radiates all of these radiations but one of them does not reach the surface of the Earth. Most of the radiation switch reaching the Earth's surface are visible and infrared radiations.

8 0

3 years ago

12456787980-1234567890

Levart [38]

Answer:11222220090

Explanation:

4 0

3 years ago

Read 2 more answers

There is a bell at the top of a tower that is 45m high. The bell weighs 190N. The bell has ___________________ energy. Calculate

Arte-miy333 [17]

Answer:

250mn

Explanation:

5 0

4 years ago

6. What is the mass of a boy who is standing on top of a 1.5 meter high

puteri [66]

Answer:

m = 50 [kg]

Explanation:

The potential energy can be calculated by means of the following equation.

$E_{p}=m*g*h$

where:

Ep = potential energy = 735 [J]

m = mass [kg]

g = gravity acceleration = 9.81 [m/s²]

h = elevation = 1.5 [m]

Now replacing:

$735=m*9.81*1.5\\m = 50 [kg]$

8 0

3 years ago

cylindrical water tank is 20.0 m tall, open to the atmosphere at the top, and is filled to the top. It is noticed that a small h

Naya [18.7K]

Answer:

The velocity is $v =17.98 \ m/s$

The diameter is $d = 0.00184m$

Explanation:

The diagram of the set up is shown on the first uploaded image

From the question we are told that

The height of the water tank is $h = 20.0 \ m$

The position of the hole $p_h = 16.5m$ below water level

The rate of water flow $\r V = 2.90 *10^{-3} m^3 /min = \frac{2.90 *10^{-3}}{60} = 0.048*10^{-3} m^3/s$

According to Bernoulli's theorem position of the hole

$\frac{P_o + h \rho g}{\rho} + \frac{1}{2} u^2 = \frac{P_o}{\rho } + \frac{1}{2} v^2$

Where u is the initial speed the water through the hole = 0 m/s

$P_o$ is the atmospheric pressure

$\frac{P_o }{\rho} + \frac{ h \rho g}{\rho} + 0 = \frac{P_o}{\rho } + \frac{1}{2} v^2$

$v = \sqrt{2gh}$

Substituting value

$v = \sqrt{2 * 9.8 * 16.5 }$

$v =17.98 \ m/s$

The Volumetric flow rate is mathematically represented as

$\r V = A * v$

Making A the subject

$A = \frac{\r V}{v}$

substituting value

$A = \frac{0.048 *10^{-3}}{17.98}$

$= 2.66*10^{-6}m^2$

Area is mathematically represented as

$A = \frac{\pi d^2}{4}$

making d the subject

$d = \sqrt{\frac{4*A}{\pi} }$

Substituting values

$d = \sqrt{\frac{4 * 2.67 *10^{-6}}{3.142} }$

$d = 0.00184m$

6 0

3 years ago