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scoundrel [369]
3 years ago
15

What is the answer to 17 ‐ 12× ‐ 18 ‐ 5×

Physics
1 answer:
Nastasia [14]3 years ago
7 0
The answer is -17x-1. You have to combine like terms.
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A skateboarder drops in off the top of one side of the half pipe shown below. She does not push off and starts from rest. She st
solong [7]

Answer:

v

Explanation:

4 0
2 years ago
How is thermal energy transferred during convection?
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Conduction and <span>convection it involves particles.</span>
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3 years ago
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four children pull on the same stuffed toy at the same time , yet there is no net force on the toy.how is this possible?
Kay [80]
There was no net force on the stuffed toy, because the kids might have the same strength,  The same force is on both sides of it.  T<span>hey cancel each other out. They exert a force on the stuffed toy equal in strength but opposite in direction. The forces are balanced and the stuffed toy does not move.  </span>Its like a game of tug-o-war, but you and I have the same strength. the rope would be still and not moving. 
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3 years ago
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A 40.0 kg wagon is towed up a hill inclined at 18.5 degrees with respect to the horizontal. The tow rope is parallel to the incl
sp2606 [1]

Answer:

7.9m/s

Explanation:

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Mass of wagon=40 kg

\theta=18.5^{\circ}

Tension=140 N

Initial velocity of wagon=u=0

Displacement=s=80 m

Net force applied  on wagon=F=T-mgsin\theta=140-40(9.8)sin18.5=15.7 N

By using g=9.8m/s^2

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We know that

v^2-u^2=2as

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v^2=2\times 0.39\times 80

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5 0
3 years ago
If an atomic nucleus were the size of a dime how far away might one of its electrons be?
mihalych1998 [28]
The radius of a nucleus of hydrogen is approximately r_{n1}=1\cdot 10^{-15}m, while we can use the Borh radius as the distance of an electron from the nucleus in a hydrogen atom: r_{e1}=5.3 \cdot 10^{-11}m

The radius of a dime is approximately r_{n2} = 9\cdot 10^{-3}m: if we assume that the radius of the nucleus is exactly this value, then we  can find how far is the electron by using the proportion
r_{n1}:r_{e1}=r_{n2}:r_{e2}
from which we find
r_{e2}= \frac{r_{e1} r_{n2}}{r_{n1}}= \frac{(5.3 \cdot 10^{-11}m)(9\cdot 10^{-3}m)}{1 \cdot 10^{-15}m}=477 m

So, if the nucleus had the size of a dime, we would find the electron approximately 500 meters away.
6 0
3 years ago
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