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boyakko [2]
3 years ago
12

An astronomer observes an asteroid in the solar system. He notes that the asteroid is three times farther from the Sun than Eart

h is. How far away from the Sun is the asteroid in astronomical units?
Physics
2 answers:
Dvinal [7]3 years ago
5 0

Answer: is  3 AU

Explanation:

wlad13 [49]3 years ago
3 0
450 kilometers ! or 933 million miles

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Calculate the gravitational potential energy of a body of mass 40 kg at a vertical height of 10 m. ( g = 9.8 m/s2)
olganol [36]
Ep= mgh
Ep = 40 x 9.8 x 10
Ep = 3920J
Ep = 3900J (2sf)
8 0
3 years ago
6. If a drag racer wins the final round of herrace by going an average speed of 320 m/sin 4.5 seconds, what distance did he cove
Ivan

We want to calculate the distance covered by the drag racer. Recall, the formula for calculating distance is expressed as

Distance = speed x time

From the information given,

speed = 320 m/s

time = 4.5 s

By substituting these values into the formula, we have

Distance = 320 m/s x 4.5s

s cancels out. We are left with m. Thus,

Distance = 1440m

4 0
1 year ago
Why should the substage condenser not be included in computing the magnification?
Leni [432]
The reason as to why the substage condenser does not need to be included in computing the magnification and the only component needed is the ocular lens and the objective lenses is because the condenser is only responsible for gathering light and it does not contribute with the magnification of the object under the microscope.
5 0
3 years ago
Read 2 more answers
An aluminum bar 600mm long, with diameter 40mm long has a hole drilled in the center of the bar.The hole is 30mm in diameter and
Svetradugi [14.3K]

Answer:

Total contraction on the Bar  = 1.22786 mm

Explanation:

Given that:

Total Length for aluminum bar = 600 mm  

Diameter for aluminum bar  = 40 mm

Hole diameter  = 30 mm

Hole length = 100 mm

elasticity for the aluminum is 85GN/m² = 85 × 10³ N/mm²

compressive load P = 180 KN = 180  × 10³ N

Calculate the total contraction on the bar = ???

The relation used in  calculating the contraction on the bar is:

\delta L = \dfrac{P *L }{A*E}

The relation used in  calculating the total contraction on the bar can be expressed as :

Total contraction in the Bar = (contraction in part of bar without hole + contraction in part of bar with hole)

i.e

Total contraction on the Bar = \dfrac{P *L_1 }{A_1*E} +  \dfrac{P *L_2 }{A_2 *E}

Let's find the area of cross section without the hole and with the hole

Area of cross section without the hole is :

Using A = πd²/4

A = π (40)²/4

A = 1256.64 mm²

Area of cross section with the hole is :

A = π (40²-30²)/4

A = 549.78 mm²

Total contraction on the Bar = \dfrac{P *L_1 }{A_1*E} +  \dfrac{P *L_2 }{A_2 *E}

Total contraction on the Bar  = \dfrac{180 *10^3 \N  }{85*10^3 \ N/mm^2} [\dfrac{500}{1256.64}+ \dfrac{100}{549.78}]

Total contraction on the Bar  = 2.117( 0.398 + 0.182)

Total contraction on the Bar  = 2.117*(0.58)

Total contraction on the Bar  = 1.22786 mm

5 0
3 years ago
How are the concepts of impulse and momentum related?
butalik [34]

Answer:

The impulse experienced by the object equals the change in momentum of the object. In equation form, F • t = m • Δ v. In a collision, objects experience an impulse; the impulse causes and is equal to the change in momentum. ... The collision would change the halfback's speed and thus his momentum.

Explanation:

4 0
3 years ago
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