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3 years ago

Which major planet has the largest . . .

Physics

1 answer:

svlad2 [7]3 years ago

7 0

Answer:

Answered

Explanation:

The largest Semi major axis- Neptune( 30.0611 AU).

Largest avg. Orbital speed around the Sun- 47.9 km/s of mercury.

The Largest orbital period around the Sun- 164.79 years for Neptune by 1 year on Earth.

The largest Eccentricity of orbit- is Of Mercury (0.206)

Note: All data has been taken from the internet, hope it helps

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A rain gutter is to be constructed from a metal sheet of width 30 cm by bending up one-third of the sheet on each side through a

GarryVolchara [31]

Answer:

Explanation:

The volume of the water carried will be proportional to the cross-sectional area. The cross-section is a trapezoid with height 10sint, where t represents theta. The top of the trapezoid is 10+(2)(10cost), i.e. 10 + 20cost. The base of the trapezoid is 10.

Area of trapezoid = (average of bases) times (height)

= ([10 + (10 +20cost)] / 2 ) 10 sint

= (10 + 10cost)(10sint)

= 100sint + 100sintcost, call this A(t). You need to maximize. So differentiate and set equal to zero.

dA/dt = -100sin(t)^2+100cos(t)+100cos(t)^2 = 0, divide by 100:

-sin(t)^2 + cos(t) + cos(t)^2 = 0, replace sin^2 by 1-cos^2

2cos(t)^2 + cos(t) - 1 = 0, factor

(1+cos(t))(2cos(t)-1)=0, so

cos(t)=1, t=0 that give a min (zero area) not a max, or

cos(t) = 1/2, so t=60 degrees. This gives the max.

7 0

4 years ago

30 points answer ASAP please

djverab [1.8K]

Explanation:

Momentum of the system is a constant when there aren't external forces..

So,When first Balls velocity is went down to zero... its momentum is also zero and beacause momentum should be a constant...now second ball must have a momentum of 20 kg * m/s

4 0

3 years ago

Un hombre parado en el techo de un edificio, tira una bola verticalmente hacia arriba con una velocidad de 12m/s, la bola llega

zmey [24]

Respuesta:

24m

Explicación:

Según la ecuación de movimiento

v = u + en

Dado

Velocidad final v = 12 m / s

velocidad inicial u = 0 m / s

tiempo t = 4s

Sustituir

12 = 0 + 4a

a = 12/4

a = 3 m / s²

Lo siguiente es obtener la distancia;

S = ut + 1 / 2at²

S = 0 (4) + 1/2 (3) (4) ²

S = 3 * 16/2

S = 48/2

S = 24 m

Por lo tanto, la distancia requerida es de 24 m.

4 0

3 years ago

A Huge water tank is 2m above the ground if the water level on it is 4.9m high and a small opening is there at the bottom then t

lapo4ka [179]

Answer:

The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.

Explanation:

Let assume the existence of a line of current between the water tank and the ground and, hence, the absence of heat and work interactions throughout the system. If water is approximately at rest at water tank and at atmospheric pressure ( $P_{atm}$ ), then speed of efflux of the non-viscous water is modelled after the Bernoulli's Principle:

$P_{1} + \rho\cdot \frac{v_{1}^{2}}{2} + \rho\cdot g \cdot z_{1} = P_{2} + \rho\cdot \frac{v_{2}^{2}}{2} + \rho\cdot g \cdot z_{2}$

Where:

$P_{1}$ , $P_{2}$ - Water total pressures inside the tank and at ground level, measured in pascals.

$\rho$ - Water density, measured in kilograms per cubic meter.

$g$ - Gravitational acceleration, measured in meters per square second.

$v_{1}$ , $v_{2}$ - Water speeds inside the tank and at the ground level, measured in meters per second.

$z_{1}$ , $z_{2}$ - Heights of the tank and ground level, measured in meters.

Given that $P_{1} = P_{2} = P_{atm}$ , $\rho = 1000\,\frac{kg}{m^{3}}$ , $g = 9.807\,\frac{m}{s^{2}}$ , $v_{1} = 0\,\frac{m}{s}$ , $z_{1} = 6.9\,m$ and $z_{2} = 4.9\,m$ , the expression is reduced to this:

$\left(9.807\,\frac{m}{s^{2}} \right)\cdot (6.9\,m) = \frac{v_{2}^{2}}{2} + \left(9.807\,\frac{m}{s^{2}} \right)\cdot (4.9\,m)$

And final speed is now calculated after clearing it:

$v_{2} = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (6.9\,m-4.9\,m)}$

$v_{2} \approx 6.263\,\frac{m}{s}$

The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.

5 0

4 years ago

A person on a road trip drives a car at different constant speeds over several legs of the trip. She drives for 10.0 min at 50.0

irakobra [83]

<h2>The average speed for the entire trip is 47.5 m/s .</h2>

It is given that for different time span car have different speed and also the person spend $40\ min=\dfrac{40}{60}\ hrs =0.67\ hrs\ .$ in eating lunch and buying gas.

We know , average speed is total distance covered by total time taken .

Therefore , average speed , $v=\dfrac{total\ distance }{total\ times}$

$v=\dfrac{\dfrac{10}{60}\times 50+\dfrac{19}{60}\times 100+\dfrac{60}{60}\times 55}{\dfrac{10}{60}+\dfrac{10}{60}+\dfrac{60}{60}+ \dfrac{40}{60}}\\\\\\v=47.5\ m/s$

Hence, this is the required solution.

Learn More :

Average speed

https://brainly.in/question/12701198

7 0

3 years ago