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11Alexandr11 [23.1K]
4 years ago
6

Energy that Is transferred from a warmer object to a cooler object is called what?

Physics
1 answer:
stealth61 [152]4 years ago
3 0

Answer:

The movement of heat from a warmer object to a cooler one is called heat transfer.

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PLEASE HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! IM DESPERATE!!!!!!!!!!!!!! EVEN IF YOU DONT KNOW , GIVE A GUESS. What would
dezoksy [38]
Well, A = T or U C = G G = C T or U = A So it would be like this; DNA Sequence: GCTAATTGCATCCGA The Complementary Sequence: CGATTAACGTAGGCT Hope this helped :)
3 0
3 years ago
The existence of the dwarf planet Pluto was proposed based on irregularities in Neptune's orbit. Pluto was subsequently discover
givi [52]

Answer:

Acceleration due to gravity, a=4.61\times 10^{-14}\ m/s^2

Explanation:

It is given that,

Mass of Pluto, m=1.4\times 10^{22}\ kg

Distance between Neptune and Pluto, r=4.5\times 10^{12}\ m

The force of gravity is balanced by the gravitational force between Neptune and Pluto. It is given by :

a=\dfrac{Gm}{r^2}

a=\dfrac{6.67\times 10^{-11}\times 1.4\times 10^{22}}{(4.5\times 10^{12})^2}

a=4.61\times 10^{-14}\ m/s^2

So, the acceleration due to gravity at Neptune due to Pluto is 4.61\times 10^{-14}\ m/s^2. Hence, this is the required solution.  

4 0
4 years ago
What is the time feather to fall to the surface of the moon ?
denpristay [2]
Mbccbnb jdkdmmdodbs ndkdkmdnd ndkdkfnndf njdkdmdnxjsbd. Djjdjdndnfk ndkdkfkd
8 0
4 years ago
A plane is flying east at 135 m/s. The wind accelerates it at 2.18 m/s^2 directly northeast. After 18 s, what is the magnitude o
Korolek [52]

When we say directly northeast that is equivalent to 45˚ north of east.

First let us determine the north and east components of the acceleration using cos and sin functions:<span>

North = 2.18 * sin 45 
East = 2.18 * cos 45 

<span>Then we set to determine the east component of the plane’s displacement by calculating using the formula:

d = vi * t + ½ * a * t^2 
d = 135 * 18 + ½ * 2.18 * cos 45 * 18^2 
<span>d = 2430 + 353.16 * cos 45 = 2679.72 m</span>

Calculating for the north component:
North = ½ * 2.18 * sin 45 * 18^2 </span></span>

North = 249.72 m 

 

Hence magnitude is:

Magnitude = sqrt (2679.72^2 + 249.72^2)

Magnitude = 2,691. 33 m<span>

</span>

Calculating for angle:

Tan θ = North ÷ East <span>
<span>Tan θ = 249.72 m  ÷ 2679.72 m</span></span>

θ = 5.32°<span>


</span>

So the plane was flying at 2,691. 33 m at 5.32<span>°</span>

4 0
3 years ago
The Hubble Space Telescope (HST) orbits 569 km above Earth’s surface. Given that Earth’s mass is 5.97 × 1024 kg and its radius i
Kipish [7]
V (<span>HST’s tangential speed</span>) =  7570 m/s

4 0
4 years ago
Read 2 more answers
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