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Luden [163]
3 years ago
10

One light-hour is the distance that light travels in an hour. How far is this, in kilometers? (Recall that the speed of light is

300,000 km/s.) _______A) 18 million kmB) 1.08 billion kmC) 300,000 kmD) 9.46 trillion kmE) 100 million km
Physics
1 answer:
ycow [4]3 years ago
6 0

Answer:

B 1.08 BILLION

Explanation:

SEE ATTACHMENT

Download pdf
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Rocket-powered sleds are been used to test the responses of humans to acceleration. Starting from rest, one sled can reach a spe
Greeley [361]

Answer:cho  v₀ =0s  

α=Δv/Δt

Explanation:

\frac{0-495}{2,16-1,78}

=-1302,631579

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3 0
2 years ago
Which circuit would have the most electrical power?
aliya0001 [1]

the answer is c because it has the most volts

a

6 0
2 years ago
1) Freddy drives 4 miles east to his friend's house. He then travels 9 more miles east to the supermarket. Finally on his way ba
timurjin [86]

Answer:

<h3>B. 19miles</h3>

Explanation:

If Freddy drives 4 miles east to his friend's house. He then travels 9 more miles east to the supermarket. Finally on his way back home he out  of gas 6 miles after leaving the supermarket, the distance travel by fred will be the sup of all the distances he covered throughout the journey.

Distance covered by fred = 4miles + 9miles + 6miles

Distance covered by fred = 13miles + 6miles

Distance covered by fred  = 19miles

8 0
3 years ago
How much heat is required to convert 5.53 g of ice at -12.0 ∘C to water at 24.0 ∘C? (The heat capacity of ice is 2.09 J/(g⋅∘C),
fredd [130]

Answer:

2.55 × 10³ J =2.55 kJ

Explanation:

Specific heat capacity of ice =  37.8 J / mol °C

Specific heat capacity of water = 76.0 J/ mol °C

Ice at -12 °C is converted to ice at 0 °C by absorbing heat Q₁

Ice at 0°C melts to water at 0 °C. Let Heat  absorbed during this phase change be Q₂ .

Let heat  absorbed to raise the temperature of water from 0 C to 24°C be Q₃ .

Total heat = Q = Q₁ + Q₂ + Q₃

Q₁ = (37.8 j/mol C )(5.53 g /18.01532 g/ mol )( 0-(-12)) = 139.23749 j

Q₂ =(5.53 g/18.01532 g H₂O / mol ) (6.02 x10³ j) = 1847.905 j

Q₃ = (76 j/mol C) ( (5.53 g/18.01532 g H₂O / mol )(24-0) = 559.8968 j

Total Heat required = Q = 139.23749 j + 1847.905 j + 559.8968 j

= 2547.039 j = 2.55 × 10³ J =2.55 kJ

5 0
3 years ago
A skier starts from rest at the top of a hill that is inclined 10.5° with respect to the horizontal. The hillside is 200 m long,
Svetlanka [38]

Answer:

d) 289.31 m

Explanation:

Energy provided by potential energy = mgh = m x 9.8x 200 sin10.5 = 357.18m

Energy used by friction = μmgcos 10.5 x 200 = .075 x m x 9.8 x cos 10.5 x200 = 144.54 m .

Energy used by friction on plain surface = μmg x d.( dis distance covered on plain ) =.075x m x 9.8 xd = .735 m d

To equate

357.18 m -144.54 m = .735 m d

d = 289.31 m .

4 0
3 years ago
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