Answer:
car B will be 30 Km ahead of car A.
Explanation:
We'll begin by calculating the distance travelled by each car. This is illustrated below:
For car A:
Speed = 40 km/h
Time = 3 hours
Distance =?
Speed = distance / time
40 = distance / 3
Cross multiply
Distance = 40 × 3
Distance = 120 Km
For car B:
Speed = 50 km/h
Time = 3 hours
Distance =?
Speed = distance / time
50 = distance / 3
Cross multiply
Distance = 50 × 3
Distance = 150 Km
Finally, we shall determine the distance between car B an car A. This can be obtained as follow:
Distance travelled by car B (D₆) = 150 Km
Distance travelled by car A (Dₐ) = 120 Km
Distance apart =?
Distance apart = D₆ – Dₐ
Distance apart = 150 – 120
Distance apart = 30 Km
Therefore, car B will be 30 Km ahead of car A.
Answer:
0,00123 = 1,2*10^{-3}
Explanation:
To write down correctly the number 0,00123 in scientific notation, you take into account that the scientific notation demands that there in only one number after the comma ( , ). Furthermore, it is necessary that you move the comma to the right of the first number different of zero, in this case the number 1. To do this you move the comma three positions.
Then, you have to multiply the expresion 1.23 by 10 with an exponential -3 (because of the movement of the comma in three positions). That is:
0,00123 = 1,23*10^{-3}
But it is mandatory that nly one number can stay after the comma, so, you approximate the number three. In this case, the number is lower than 5, hence, you approximate 3 to 0.
Finally, you have:
0,00123 = 1,2*10^{-3}
Work done is 0.442J
<u>Explanation:</u>
Given:
Spring constant, k = 33 N/m
Distance, x₁ = 0.15m
Additional distance, x₂ = 0.072 m
Total distance = 0.15 + 0.072 m
= 0.222 m
Work done, W = ?
We can calculate work done by the formula
On substituting the value we get:
![W = \frac{1}{2}k [(x_2)^2 - (x_1)^2]\\ \\W = \frac{1}{2} X 33[(0.222)^2 - (0.15)^2]\\ \\W = \frac{1}{2}X 33 [ 0.0493 - 0.0225]\\ \\W = 0.442 J](https://tex.z-dn.net/?f=W%20%3D%20%5Cfrac%7B1%7D%7B2%7Dk%20%5B%28x_2%29%5E2%20%20-%20%28x_1%29%5E2%5D%5C%5C%20%5C%5CW%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20X%2033%5B%280.222%29%5E2%20-%20%280.15%29%5E2%5D%5C%5C%20%5C%5CW%20%3D%20%5Cfrac%7B1%7D%7B2%7DX%2033%20%5B%200.0493%20-%200.0225%5D%5C%5C%20%5C%5CW%20%3D%200.442%20J)
Therefore, work done is 0.442J
I think that would be the moon
"too small to clear objects that are in its orbital path" , which means that it probably not a planet.
hope this helps
