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Temka [501]
4 years ago
15

The most massive planet in the solar system is

Physics
1 answer:
Natasha_Volkova [10]4 years ago
5 0

Answer:

Jupiter

Largest Planet: Jupiter. The largest planet in our solar system by far is Jupiter, which beats out all the other planets in both mass and volume. Jupiter's mass is more than 300 times that of Earth, and its diameter, at 140,000 km, is about 11 times Earth's diameter.

Explanation:

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You wish to lift a 12,000 lb stone by a vertical distance of 15 ft. Unfortunately, you can only generate a maximum pushing force
DochEvi [55]

Answer:

5,760,000 lb-ft²/s²

Explanation:

Work done, W = mgy where m = mass of stone = 12,000 lb, g = 32 ft/s² and y = vertical distance = 15 ft.

So, W = mgy = 12,000 lb × 32 ft/s² × 15 ft. = 5,760,000 lb-ft²/s²

6 0
3 years ago
A4.0 kg object is moving with speed 2.0 m/s. A 1.0 kg object is moving with speed 4.0 m/s. Both objects encounter the same const
Evgesh-ka [11]

Answer:

Both objects travel the same distance.

(c) is correct option

Explanation:

Given that,

Mass of first object = 4.0 kg

Speed of first object = 2.0 m/s

Mass of second object = 1.0 kg

Speed of second object = 4.0 m/s

We need to calculate the stopping distance

For first particle

Using equation of motion

v^2=u^2+2as

Where, v = final velocity

u = initial velocity

s = distance

Put the value in the equation

0= u^2-2as_{1}

s_{1}=\dfrac{u^2}{2a}....(I)

Using newton law

a=\dfrac{F}{m}

Now, put the value of a in equation (I)

s_{1}=\dfrac{8}{F}

Now, For second object

Using equation of motion

v^2=u^2+2as

Put the value in the equation

0= u^2-2as_{2}

s_{2}=\dfrac{u^2}{2a}....(I)

Using newton law

F = ma

a=\dfrac{F}{m}

Now, put the value of a in equation (I)

s_{2}=\dfrac{8}{F}

Hence, Both objects travel the same distance.

6 0
3 years ago
Work done by a force forum X to X
Ksivusya [100]

Answer:

W = 68 J

Explanation:

On a force vs displacement chart, work is the area under the curve.

The area under the curve can be divided into a rectangle and a triangle

W = Fd = (2 N)(12 - 0 m) + ½(13 - 2 N)(12 - 4 m) = 68 N•m = J

5 0
3 years ago
A ball falls off the desk and hits the floor. Which
Yuliya22 [10]

Answer:

The potential energy is transformed into kinetic energy

Explanation:

This particular case is defined as the principle of energy conservation since energy is not created or destroyed only transforms. When you have potential energy it can be transformed into kinetic energy or vice versa. In this problem, we have the case of a ball that sits on a desk and then falls to the ground. In this way the ground will be taken as a reference point, this is a point at which the potential energy will be equal to zero in such a way that when the ball is on the desktop that is above the reference line its potential energy will be maximum. As the ball drops its potential energy decreases, as the height relative to the ground (reference point) decreases. In contrast its kinetic energy increases and increases as it approaches the ground. So when it hits the ground it will have maximum kinetic energy and will be equal to the potential energy for when the ball was on the desk.

Therefore:

E_{p} = potential energy [J] = E_{k} = kinetic energy [J]where:\\E_{p} =m*g*h\\m =mass [kg]\\g=gravity[m/s^2]\\h=elevation[m]\\E_{k} = \frac{1}{2} *m*v^{2} \\where:\\v=velocity [m/s]\\\frac{1}{2}  *m*v^{2} = m*g*h

8 0
3 years ago
A tank, shaped like a cone has height 12 meter and base radius 1 meter. It is placed so that the circular part is upward. It is
Temka [501]

Answer:

376966.991 Joules

Explanation:

Given that :

the height = 12  m

Let assume the tank have a thickness  = dh

The radius of the tank by using the concept of similar triangle is :

\dfrac{1}{r} = \dfrac{12}{h}

r = \dfrac{h}{12}

The area of the tank = \mathbf{\pi r^2}

The area of the tank = \mathbf{\pi( \dfrac{h}{12})^2}

The area of the tank = \mathbf{ \dfrac{\pi}{144}h^2}

The volume of the tank is  = area × thickness

= \mathbf{ \dfrac{\pi}{144}h^2 \  dh}

Weight of the element = \rho_ g * volume

where;

\rho_g = density of water ; which is given as 10000 N/m³

So;

Weight of the element = \mathbf{ 10000 *\dfrac{\pi}{144}h^2 \  dh}

Weight of the element = \mathbf{69.44 \ \pi  \ h^2 \  dh}

However; the work required to pump this water = weight × height  rise

where the height rise = 12 - h

the work required to pump this water  = \mathbf{69.44 \ \pi  \ h^2 \  dh}(12 - h)

the work required to pump this water  = \mathbf{69.44 \pi (12h^2-h^3)dh}

We can determine the total workdone by integrating the work required to pump this water

SO;

Workdone = \mathbf{\int\limits^{12}_0 {69.44 \pi(12h^2-h^3)} dh}

= \mathbf{ 69.44 \pi \int\limits^{12}_0 {(12h^2-h^3)} dh}

=  \mathbf{ 69.44 \pi[ \frac{12h^3}{3}-  \frac{h^4}{4}]^{12}}_0} }

= \mathbf{69.44 \pi [ \frac{12^4}{3}-\frac{12^4}{4}]}

= \mathbf{69.44 \pi*12^4 [ \frac{4-3}{12}]}

= \mathbf{69.44 \pi*12^4 *\frac{1}{12}}

= 376966.991 Joules

6 0
3 years ago
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