Answer:
19.2*10^6 s
Explanation:
The equation for time dilation is:

Then, if it is observed to have a life of 6*10^6 s, and it travels at 0.95 c:

It has a lifetime of 19.2*10^6 s when observed from a frame of reference in which the particle is at rest.
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4. For this problem, we have to write and solve a proportion. We would set this proportion up as 12/15 = 8/x. This is because we're looking for the length of the shadow and we know the height of the items, so we line them up horizontally and x goes with 8, because we're looking for the shadow length. Let's cross multiply the values. 15 * 8 = 120. 12 * x = 12. You get 120 = 12x. Now, we must divide each side by 12 to isolate the "x". 120/12 is 10. x = 10. There. The cardboard box casts a shadow that is 10 ft long.
5. For this question, you do the same thing. This time, you're finding the height of the tower, so you would do 1.2/0.6 = x/7. Cross multiply the values in order to get 8.4 = 0.6x. Now, divide each side by 0.6x to isolate the "x". 8.4/0.6 is 14. x = 14. There. The tower is 14 m tall.
If you need more help on proportions and using proportions in real life situations, feel free to search on the internet to find more information about how you solve them.
Answer:
225 N
Explanation:
"Below the horizontal" means he's pushing down at an angle.
Draw a free body diagram of the box. There are three forces: normal force N pushing up, weight force mg pulling down, and the applied force F at an angle θ.
Sum of forces in the y direction:
∑F = ma
N − mg − F sin θ = 0
N = F sin θ + mg
Plug in values:
N = (50 N) (sin 30°) + (20.0 kg) (10 m/s²)
N = 225 N
Answer:
a
The speed of wave is 
b
The new speed of the two waves is 
Explanation:
From the question we are told that
The mass of the string is 
The length is 
The tension is 
Now the velocity of the first wave is mathematically represented as

Where
is the linear density which is mathematically represented as

substituting values


So


Now given that the Tension, mass and length are constant the velocity of the second wave will same as that of first wave (reference PHYS 1100 )