Answer:
The frictional force needed to overcome the cart is 4.83N
Explanation:
The frictional force can be obtained using the following formula:
where 
is the coefficient of friction = 0.02
R = Normal reaction of the load = 
= 
= 
Now that we have the necessary parameters that we can place into the equation, we can now go ahead and make our substitutions, to get the value of F.
F = 4.83 N
Hence, the frictional force needed to overcome the cart is 4.83N
Answer:
Isabella will not be able to spray Ferdinand.
Explanation:
We'll begin by calculating the time taken for the water to get to the ground from the hose held at 1 m above the ground. This can be obtained as follow:
Height (h) = 1 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =.?
h = ½gt²
1 = ½ × 9.8 × t²
1 = 4.9 × t²
Divide both side by 4.9
t² = 1/4.9
Take the square root of both side
t = √(1/4.9)
t = 0.45 s
Next, we shall determine the horizontal distance travelled by the water. This can be obtained as follow:
Horizontal velocity (u) = 3.5 m/s
Time (t) = 0.45 s
Horizontal distance (s) =?
s = ut
s = 3.5 × 0.45
s = 1.58 m
Finally, we shall compare the distance travelled by the water and the position to which Ferdinand is located to see if they are the same or not. This is illustrated below:
Ferdinand's position = 10 m
Distance travelled by the water = 1.58 m
From the above, we can see that the position of the water (i.e 1.58 m) and that of Ferdinand (i.e 10 m) are not the same. Thus, Isabella will not be able to spray Ferdinand.
<span>The sparse area surrounding a spiral galaxy is called a
Halo
</span>
Compounds. two or more compounds.
Answer:
1.549 m
Explanation:
Given:
The radius of the circular board, r = 2 m
The probability of hitting the red is given as 0.6
Now, this probability of hitting the red can be conclude as
0.6 = (Area of red)/ (Total area of the board)
Total area of the board = πr² = π × 2²
let the radius of the red area be R
thus, area of red circle, = πR²
on substituting the value of the area, we have
0.6 = (πR²)/ (π × 2²)
or
R² = 2.4
or
R = 1.549 m
Thus, the radius of the red circle is 1.549 m
