The Kelvin scale has no negatives on it.
Zero Kelvin is 'Absolute Zero', and nothing can get colder than that.
Answer:
No one is right
Explanation:
John Case:
The function
is defined between -1 and 1, So it is not possible obtain a value
greater.
In addition, if you move the function cosine a T Value, and T is the Period, the function take the same value due to the cosine is a periodic function.
Larry case:
Is you have
, the domain of this is [0,2].
it is equivalent to adding 1 to the domain of the
, and its mean that the function
, in general, is not greater than
.
<span>Data:
mass =
110-g bullet
d = 0.636 m
Force =
13500 + 11000x - 25750x^2, newtons.
a) Work, W
W = ∫( F* )(dx) =∫[13500+ 11000x - 25750x^2] (dx) =
W = 13500x + 5500x^2 - 8583.33 x^3 ] from 0 to 0.636 =
W = 8602.6 joule
b) x= 1.02 m
</span><span><span>W = 13500x + 5500x^2 - 8583.33 x^3 ] from</span> 0 to 1.02
W = 10383.5
c) %
[W in b / W in a] = 10383.5 / 8602.6 = 1.21 => W in b is 21% more than work in a.
</span>
That depends on what "objects" refer to.
If "objects" refer to the ones on Earth, then it is TRUE. These objects are in motion relative to the sun same as with Earth itself.
<span>If "objects" refer to the ones outside earth, then it may be a TRUE or FALSE depending on how far or near they are from the sun.</span>
Answer:
Force exerted by car on truck will be 6000 N in opposite direction
Explanation:
It is given that mass of the car is one quarter of the mass of the truck
Force exerted by the truck on the car is 6000 N
We have to find the force exerted by car on the truck
According to newtons third law for any action there is equal and opposite reaction
So force exerted by car on the truck will be equal to 6000 N in opposite direction