Answer:
x = 5[km]
Explanation:
We must convert the time from minutes to hours.
![t=30[min]*\frac{1h}{60min}= 0.5[h]\\](https://tex.z-dn.net/?f=t%3D30%5Bmin%5D%2A%5Cfrac%7B1h%7D%7B60min%7D%3D%200.5%5Bh%5D%5C%5C)
We know that speed is defined as the relationship between space and time.

where:
x = space [m]
t = time = 0.5 [h]
v = velocity [m/s]
Now replacing:
![x = 10[\frac{km}{h} ]*0.5[h]\\x=5[km]](https://tex.z-dn.net/?f=x%20%3D%2010%5B%5Cfrac%7Bkm%7D%7Bh%7D%20%5D%2A0.5%5Bh%5D%5C%5Cx%3D5%5Bkm%5D)
Answer- There are two reasons that we know quotations have been used first is the use of of name of the person who quoted it and secondly the quotation is written inside the quotation marks.
Explanation- Quotation is nothing but using a line that has been already quoted by someone somewhere. Such sentences are normally written inside quotation marks. In the above given paragraph the name of the person who quotes the sentence is also given hence we know that our quotation has been used.
Answer:
3.28 m
3.28 s
Explanation:
We can adopt a system of reference with an axis along the incline, the origin being at the position of the girl and the positive X axis going up slope.
Then we know that the ball is subject to a constant acceleration of 0.25*g (2.45 m/s^2) pointing down slope. Since the acceleration is constant we can use the equation for constant acceleration:
X(t) = X0 + V0 * t + 1/2 * a * t^2
X0 = 0
V0 = 4 m/s
a = -2.45 m/s^2 (because the acceleration is down slope)
Then:
X(t) = 4*t - 1.22*t^2
And the equation for speed is:
V(t) = V0 + a * t
V(t) = 4 - 2.45 * t
If we equate this to zero we can find the moment where it stops and begins rolling down, that will be the highest point:
0 = 4 - 2.45 * t
4 = 2.45 * t
t = 1.63 s
Replacing that time on the position equation:
X(1.63) = 4 * 1.63 - 1.22 * 1.63^2 = 3.28 m
To find the time it will take to return we equate the position equation to zero:
0 = 4 * t - 1.22 * t^2
Since this is a quadratic equation it will have to answers, one will be the moment the ball was released (t = 0), the other will eb the moment when it returns:
0 = t * (4 - 1.22*t)
t1 = 0
0 = 4 - 1.22*t2
1.22 * t2 = 4
t2 = 3.28 s
Answer: <em>Powdered sugar</em>
Powdered sugar dissolves faster compare to the sugar cube. Because sugar cube has less surface area (the granules are tightly packed) compared to powdered sugar