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2 years ago

At its closest point, Mercury is approximately 46 million kilometers from the sun. What is this distance in AU?

2 answers:

5 0

1 astronomical unit 1 AU = 1.4960 * 10^11 meters
it is the average distance between earth and sun
mercury to sun distance is = 46,000,000 * 1000 meters
= 4.6 * 10^9 meters = 4.6 * 10^9 / 1.4960 * 10^11 AU
= 3.0.74 / 100 = 0.0374 AU

Radda [10]2 years ago

5 0

Explanation :

Astronomical unit is the unit to find the average distance between the earth and the sun.

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

So, $1\ Km=\dfrac{1}{1.496\times 10^8}\ AU$

46 million kilometers means, $46\times 10^6\ Km$

$46\times 10^6\ Km=\dfrac{46\times 10^6}{1.496\times 10^8}$

$46\ million\ Km=0.3074\ AU$

Hence, this is the required solution.

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2 years ago

Consider a particle launched at a horizontal velocity v0 from a height h above the ground. 1. Derive an expression for the time

Rina8888 [55]

Answer: $t=\sqrt{\frac{2h}{g} }$

Explanation: The time taken by the object to reach the ground depends up on the vertical velocity of the object.

The object horizontal velocity component only moves the object in forward direction and does not have any effect vertically

as the object is launched horizontally its initial vertical velocity is zero

$u_y=0$

the object experiences downward fall due to gravitational acceleration only since there is no air resistance

therefore a=g

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now the kinematic equation could be used

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2 years ago

Calculate the tension (in N) in a vertical strand of spiderweb if a spider of mass 5.00 ✕ 10-5 kg hangs motionless on it.

Mandarinka [93]

Answer:

4.9 x 10⁻⁴N

Explanation:

Given parameters:

Mass of the vertical strand of spiderweb = 5 x 10⁻⁵kg

Unknown:

Tension in the vertical strand = ?

Solution:

The tension is the vertical strand will be the weight of the strand.

Weight of strand = mg

m is the mass

g is the acceleration due to gravity = 9.8m/s²

Tension in the vertical strand = 5 x 10⁻⁵kg x 9.8m/s²

Tension in the vertical strand = 4.9 x 10⁻⁴N

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2 years ago

A 15 g bullet traveling horizontally at 865 m/s passes through a tank containing 13.5 kg of water and emerges with a speed of 53

blagie [28]

Answer:

0.0613°C

Explanation:

the given parameters are m=15gm=15×10⁻³ V₁=865m/s V₂=534m/s

the bullet moves with different kinetic energies before and after the penetration, therefore

Kinetic energy before - kinetic energy after = 1/2 × m × ( V₁² - V₂²)

= $\frac{1}{2}$ × 15×10⁻³ × (865² - 534²)

= 3.47 × 10⁻³J

this loss in energy is transferred to the water, therefore

change in temperature = $\frac{Q}{m C}$

where c = heat capacity of water = 4.19 x 10^3

m = mass of water = 13.5 kg

= {3.47 × 10⁻³} / {13.5 x 4.19 x 10^3 }

=0.0613°C

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2 years ago