Answer:
Your friend is 2.143 blocks from the restaurant.
You are 2.857 blocks from the restaurant.
Explanation:
Let t be the time both you and your friend take to walk to the restaurant.
The distance (m) from your building to the restaurant is your walking time t times your speed v1
![s_1 = v_1t = 1.6t](https://tex.z-dn.net/?f=s_1%20%3D%20v_1t%20%3D%201.6t)
Similarly the distance (m) from your friend building to the restaurant:
![s_2 = v_2t = 1.2t](https://tex.z-dn.net/?f=s_2%20%3D%20v_2t%20%3D%201.2t)
Let b be the length (in m) of a block, the total distance of 5 blocks is 5b
![s_1 + s_2 = 5b](https://tex.z-dn.net/?f=s_1%20%2B%20s_2%20%3D%205b)
![1.6t + 1.2t = 5b](https://tex.z-dn.net/?f=1.6t%20%2B%201.2t%20%3D%205b)
![2.8t = 5b](https://tex.z-dn.net/?f=2.8t%20%3D%205b)
![t = 5b/2.8 = 25b/14](https://tex.z-dn.net/?f=t%20%3D%205b%2F2.8%20%3D%2025b%2F14)
![s_2 = 1.2t = 1.2(25b/14) = 2.143b](https://tex.z-dn.net/?f=s_2%20%3D%201.2t%20%3D%201.2%2825b%2F14%29%20%3D%202.143b)
So your friend are 2.143b meters from the restaurant, since each block is b meters long, 2.143b meters would equals to 2.143b/b = 2.143 blocks. And you are 5 - 2.143 = 2.857 blocks from the restaurant.
An example of a quantitative observation is measuring the surface of an oil painting and finding its dimensions to be 12 inches by 12 inches. A quantitative observation occurs when a researcher takes a measurement that is recorded in an objective number of units.
Explanation:
T1 = 201k
T2 = 537k
C1 = 1220 m / s
C1/C2 =√(T1/T2)
C2 = C1×√(T2/T2)
C2 = 1220×√(537/201)
C2 = 1220 × √(179/67)
C2 = (1220√179)/(√69)
C2 = (1220√11993)/67 ≈ 1994.10924
I hope it helps you
feel free to share if you have any doubts on my answer :3
Answer:
![a=190\ m/s^2](https://tex.z-dn.net/?f=a%3D190%5C%20m%2Fs%5E2)
Explanation:
Mass of a hockey puck, m = 0.17 kg
Force exerted by the hockey puck, F' = 35 N
The force of friction, f = 2.7 N
We need to find the acceleration of the hockey puck.
Net force, F=F'-f
F=35-2.7
F=32.3 N
Now, using second law of motion,
F = ma
a is the acceleration of the hockey puck
![a=\dfrac{F}{m}\\\\a=\dfrac{32.3}{0.17}\\\\a=190\ m/s^2](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7BF%7D%7Bm%7D%5C%5C%5C%5Ca%3D%5Cdfrac%7B32.3%7D%7B0.17%7D%5C%5C%5C%5Ca%3D190%5C%20m%2Fs%5E2)
So, the acceleration of the hockey puck is
.