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fomenos
3 years ago
7

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     --------------------------------------------------------------                  
You might be interested in
A fixed mass of gas has a volume of 25 cm³. The pressure of the gas is 100 kPa.
tekilochka [14]

Answer:

67 kPa

Explanation:

Given that,

Initial volume, V₁ = 25 cm³

Initial pressure, P₁ = 100 kPa

Final volume, V₂ = 15 cm³

We need to find the change in pressure of the gas. The relation between the volume and pressure of a gas is given by :

P\propto \dfrac{1}{V}\\\\\dfrac{P_1}{P_2}=\dfrac{V_2}{V_1}\\\\P_2=\dfrac{P_1V_1}{V_2}\\\\P_2=\dfrac{100\times 25}{15}\\\\=166.66\ kPa

or

= 167 kPa

The change in pressure,

= P₂ - P₁

= 167 kPa - 100 kPa

= 67 kPa

Hence, the correct option is (a).

7 0
2 years ago
Which element accumulates in the environment due to the use of
marissa [1.9K]

To Find :

Which element accumulates in the environment due to the use of  fertilizers.

Solution :

Most of the fertilizers used in daily practice are nitrogen fertilizers .

When soil became waterlogged, soil organisms take the oxygen they need from nitrates, leaving the nitrogen in a gaseous form which escapes into the air.

Also, nitrogen is very stable and did not participate in most of the reactions.

So, nitrogen is accumulates in the environment .

Therefore, option C. is correct.

6 0
3 years ago
One of two 25-year-old identical twins begins a trip on a spaceship traveling at 0.8 c while her twin remains on Earth. The twin
borishaifa [10]

Answer:

This equation is based on twin paradox - a phenomena where one of the twin travels to space at a speed close to speed of light and the other remains on earth. the twin from the space on return discovers that the one on earth age faster.

Solution:

t_{o} = 10 years

v = 0.8c

c = speed of light in vacuum

The problem can be solved by time dilation equation:

t = \frac{t_{o}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}            (1)

where,

t = time observed from a different inertial frame

Now, using eqn (1), we get:

t = \frac{10}{\sqrt{1 - \frac{(0.8c)^{2}}{c^{2}}}}

t = 16.67 years

The age of the twin on spaceship according to the one on earth = 25+16.67 =41.66 years

8 0
3 years ago
A 0.25-kg ball attached to a string is rotating in a horizontal circle of radius 0.5 m twice every second, what is the tension i
Svetradugi [14.3K]

Answer:

T = 19.75 N

Explanation:

given,

mass of ball = 0.25 Kg

radius = 0.5 m

frequency = 2 s⁻¹

tension in the string = ?

angular velocity

ω = 2 π f

ω = 2 π x 2

ω = 12.57 rad/s

tension on the string is equal to the centripetal force

T = m ω² r

T = 0.25 x 12.57² x 0.5

T = 19.75 N

Tension in the string is equal to T = 19.75 N

3 0
3 years ago
Two massless bags contain identical bricks, each brick having a mass M. Initially, each bag contains four bricks, and the bags m
stepladder [879]

Answer: F_{2}=\frac{3}{4}F_{1}

Explanation:

According to Newton's law of universal gravitation:

F=G\frac{m_{1}m_{2}}{r^2}

Where:

F is the module of the force exerted between both bodies

G is the universal gravitation constant.

m_{1} and m_{2} are the masses of both bodies.

r is the distance between both bodies

In this case we have two situations:

1) Two bags with masses 4M and 4M mutually exerting a gravitational attraction F_{1} on each other:

F_{1}=G\frac{(4M)(4M)}{r^2}   (1)

F_{1}=G\frac{16M^2}{r^2}   (2)

F_{1}=16\frac{GM^2}{r^2}   (3)

2) Two bags with masses 2M and 6M mutually exerting a gravitational attraction F_{2} on each other (assuming the distance between both bags is the same as situation 1):

F_{2}=G\frac{(2M)(6M)}{r^2}   (4)

F_{2}=G\frac{12M^2}{r^2}   (5)

F_{2}=12\frac{GM^2}{r^2}   (6)

Now, if we isolate \frac{GM^2}{r^2} from (3):

\frac{F_{1}}{16}=\frac{GM^2}{r^2}   (7)

Substituting \frac{GM^2}{r^2}  found in (7) in (6):

F_{2}=12(\frac{F_{1}}{16})   (8)

F_{2}=\frac{12}{16}F_{1}   (9)

Simplifying, we finally get the expression for F_{2}  in terms of F_{1} :

F_{2}=\frac{3}{4}F_{1}  

5 0
3 years ago
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