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Butoxors [25]
3 years ago
6

This distance from the Earth to the Sun given above is a standard for measuring other distances in the solar system and is calle

d an Astronomical Unit or AU. 1AU=_________miles. Write this as a ratio:_________
Physics
1 answer:
sineoko [7]3 years ago
5 0

Answer:

The Astronomical Unit (AU) is approximately 92.96\times 10^{6}\,mi, the ratio of Astronomical Units per mile is r = 1\,AU\,:\,92.96\times 10^{6}\,mi.

Explanation:

The distance from the Earth to the Sun is approximately 92.96\times 10^{6}\,mi, then the Astronomical Unit (AU) is approximately 92.96\times 10^{6}\,mi. The ratio can be described by the quantity of miles per Astronomical Unit, then we define:

r = 1\,AU\,:\,92.96\times 10^{6}\,mi.

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a lamp or bulb is marked 12 volt 240 Watts how many joules does it consume in one hour and what is the current that passes throu
valkas [14]

Answer:

1 watt is 1 Joule per second. 240 watt would then be 240 joules per second. How many seconds are there in an hour? Current formula is P= IV , P is 240(not the power in an hour because the current is calculated in seconds), and V is 12V. Then you will get I, the current.

5 0
3 years ago
A 4kg book sits on a table and your pet hamster puts his front paws on the book and pushes down with a force of 3N. What is the
Natali [406]
There are three forces acting on the book. 
1. Force due to gravity
2. Force exerted downward by the hamster
3. Normal Force in reaction to the downward forces

Since the book is not moving, the net force is zero. The summation of all forces must be zero. Then we could find the normal force which is unknown (denoted as x).

∑F = -(4 kg)(9.81 m/s2) - 3 N + x =0
∑F = -39.24 N - 3N + x =0
x = 42 N

Therefore, the normal force is 42 N.
4 0
3 years ago
I don’t understand I need help asap
hammer [34]
For the first question, you got them right, for the two you left blank, initial(beginning) velocity: 2 m/s the final velocity is: 12 m/s
3 0
3 years ago
) A steel guitar string with a diameter of 1.00 mm is stretched between supports 80.0 cm apart. The temperature is 0.0°C. (a) Fi
ladessa [460]

Answer: a. Mass per unit length =0.0245kg/m

b. Tension =2.45x10^-8N

C. Tension = 2.45 x10^-8N

Fundamental frequency =200Hz

Explanation:

7 0
3 years ago
Which applied force will allow a 7.65 kg block of ice to begin sliding on a sheet of ice? The block of ice has a kinetic coeffic
Anni [7]

Answer:

force for start moving is 7.49 N

force for moving constant velocity 2.25 N

Explanation:

given data

mass = 7.65 kg

kinetic coefficient of friction = 0.030

static coefficient of friction = 0.10

solution

we get here first weight of block of ice that is

weight of block of ice = mass  ×  g

weight of block of ice = 7.65 × 9.8 = 74.97 N

so here Ff = Fa

so for force for start moving is

Fa = weight × static coefficient of friction  

Fa = 74.97 × 0.10

Fa =  7.49 N

and

force for moving constant velocity is

Fa =  weight × kinetic coefficient of friction

Fa = 74.97 × 0.030

Fa = 2.25 N

7 0
3 years ago
Read 2 more answers
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