<span>Answer:
Venus is nearer to the Sun, and gets relatively more power from it. Earth is 93 million miles from the Sun, by and large, while Venus is just 67.25 million (by and large) from the Sun. Since the force of the Sun's radiation diminishes with remove from it as 1 over r-squared, Venus gets (93/67.25) squared, or 1.91 times the power for each unit range that Earth gets, all things considered.
Since the emanating temperature of a confined body in space fluctuates as the fourth-foundation of the power occurrence upon it, by the Stefan-Boltzmann law, the transmitting temperature of Venus ought to be the fourth-base of 1.91 = 1.18 times that of the Earth. Moreover, since the environmental weight differs as the temperature, the temperature at any given weight level in the Venusian air ought to be 1.18 times the temperature at that same weight level in the Earth climate, INDEPENDENT OF ANY INFRARED ABSORPTION in the air. Specifically, the found the middle value of temperature at 1000 millibars on Earth is around 15ÂşC = 288K, so the comparing temperature on Venus, WITHOUT ANY GREENHOUSE EFFECT, ought to be 1.18 times that, or 339K. Yet, this is only 66ÂşC, the temperature we really find there from the temperature and weight profiles for Venus.
So there is no nursery impact. You have recently demonstrated that atmosphere science is totally wrong to suspect something. This is the embarrassment that such huge numbers of "specialists" in atmosphere science, and all the logical experts, won't confront. Tune in to the physicists that let you know there is no nursery impact; they know without going to the Venus information - and I am one of them. The proceeding with ineptitude on this fundamental point among such a significant number of researchers, for over a century, is stunning, and deplorable..</span>
Answer:
The extension of the second wire is 
Explanation:
From the question we are told that
The length of the wire is 
The elongation of the wire is 
The tension is 
The length of the second wire is 
Generally the Young's modulus(Y) of this material is
Where 
Where A is the area which is evaluated as
and 
So
Since the wire are of the same material Young's modulus(Y) is constant
So we have
Now the ration between the first and the second wire is
Since tension , radius are constant
We have
substituting values
Answer:
The distance s of how far the ball will go at the highest setting = 2.25m
Explanation:
Let consider x to be the representative of the compression and the distance to be s
Recall that:
By cross multiplying
Thus, for the first setting
x = 1 , s = 0.25
for the second setting
x = 2, s = 1
1 = 0.25A + B --- (1)
4 = A + B ----- (2)
From (1); let B = 1 - 0.25A and substitute it into (2)
4 = A + 1 - 0.25 A
4 - 1 = A - 0.25 A
3 = 0.75 A
A = 3/0.75
A = 4
From (2)
4 = A + B
4 = 4 + B
B = 4 - 4
B = 0
Therefore, for the highest setting, where x = 3
Then :
will be:
3² = 4s + 0
9 = 4s
s = 9/4
s = 2.25 m
∴
The distance s of how far the ball will go at the highest setting = 2.25m
Well from a non professional standpoint, an athlete-or any other individual, can stretch to relax his/her muscles. Another easy method of muscle relaxation, is icing. Icing your muscles can help release tension in your muscle ligaments.
Answer:
a.
b.1058 N
Explanation:
We are given that
Mass of each dog,M=18.5 kg
Mass of sled with rider,m=250 kg
a.Average force,F=185 N
By Newton's second law
b.By Newton's second law
Substitute the values
Hence, the force in the coupling between the dogs and the sled=1058 N
