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3 years ago

Mercury is 0.39 AU from the sun what is its distance from the sun in kilometers

Physics

2 answers:

marissa [1.9K]3 years ago

5 0

Answer:The distance of Mercury from the Sun in kilometer is $5.8344\times 10^7 km$ .

Explanation;

Distance between the Mercury and Sun = 0.39 AU

1 AU = $1.496\times 10^8 km$

So, 0.39 AU = $0.39\times 1.496\times 10^8 km=5.8344\times 10^7 km$

The distance of Mercury from the Sun in kilometer is $5.8344\times 10^7 km$ .

lutik1710 [3]3 years ago

3 0

58,000 --------------------------------------------------------------

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Convection currents produce the heat in the Earth’s interior.

DaniilM [7]

Answer:

False

Explanation:

Convection currents do not produce heat. In fact, convection current are a method of transfer of heat, not of production. Convection occurs when there is a fluid which is heated from bottom, from an external source of heat (such as a pot of boiling water over a flame): the bottom part of the fluid becomes warmer, and so less dense than the colder part, therefore it starts moving up, and it is replaced by the colder parts of the fluid, which go down. Later, these colder parts become warmer, so they start going up, being replaced by new colder parts, etc... in a cycle. This is known as convection current, but it requires an external source of heat, it does not produce heat by itself.

8 0

3 years ago

Read 2 more answers

The phenomenon of vehicle "tripping" is investigated here. The sport-utility vehicle is sliding sideways with speed v1 and no an

Mrrafil [7]

Answer:

$v_1 = 3.5 \ m/s$

Explanation:

Given that :

mass of the SUV is = 2140 kg

moment of inertia about G , i.e $I_G$ = 875 kg.m²

We know from the conservation of angular momentum that:

$H_1= H_2$

$mv_1 *0.765 = [I+m(0.765^2+0.895^2)] \omega_2$

$2140v_1*0.765 = [875+2140(0.765^2+0.895^2)] \omega_2$

$1637.1 v_1$ $= 3841.575 \omega_2$

$\omega_2 = \frac{1637.1 v_1}{3841.575}$

$\omega _2 = 0.4626 \ v_1$

From the conservation of energy as well;we have :

$T_2 +V_{2 \to 3} = T_3 \\ \\ \\ \frac{1}{2} I_A \omega_2^2 - mgh =0$

$[\frac{1}{2} [875+2140(0.765^2+0.895^2)](0.4262 \ v_1)^2 -2140(9.81)[\sqrt{0.76^2+0.895^2} -0.765]] =0$

$706.93 \ v_1^2 - 8657.49 =0$

$706.93 \ v_1^2 = 8657.49$

$v_1^2 = \frac{8657.49}{706.93 }$

$v_1 ^2 = 12.25$

$v_1 = \sqrt{ 12.25$

$v_1 = 3.5 \ m/s$

6 0

3 years ago

2. Three super strong teenagers pull a heavy crate across the floor. Dion pulls with a force of 18.5 N towards 0°. Shirley pulls

kogti [31]

Answer:

A) The resultant force is 43.4 [N]

B) The movement of the heavy crate is going to the right and in the negative direction on the y-axis

Explanation:

We need to make a sketch of the different forces acting on the heavy crate.

In the attached image we can see the forces and the sum of the vector with their respective angles.

Forces in the X-axis

$Fdionx=18.5N\\\\Fshix=16.5*cos(30)=14.29N\\Fjoanx=19.5*cos(60)=9.75N\\\\Forcex= 18.5 + 14.29 + 9.75 = 42.54 N$

Forces in the y-axis

$FDiony=0[N]\\Fshirley= 16.5*sin(30)=8.25[N]\\Fjoany=19.5*sin(60)=16.88 [N]\\\\Forcesy=0+8.25-16.88= -8.63[N]$

Using the Pythagorean theorem

$Tforce=\sqrt{(42.54)^{2} +(8.63)^{2} } \\\\Tforce= 43.4N$

The movement of the heavy crate is going to the right and in the negative direction on the y-axis, this can be easily seen in the graphical sum of vectors.