The decimal point is placed after two digits starting from the end. For each decimal place, we can write the number divided by 100.
21.12 can be written as 
.
Divide the numerator and denominator by 2:
The numerator and denominator can be divided by 2 again:
There is no other common factor between numerator and denominator other than 1. Hence, it is the reduced form.
Answer:
5.02 m
Explanation:
Applying the formula of maximum height of a projectile,
H = U²sin²Ф/2g...................... Equation 1
Where H = maximum height, U = initial velocity, Ф = angle, g = acceleration due to gravity.
Given: U = 46 ft/sec = 14.021 m/s, Ф = 45°
Constant: g = 9.8 m/s²
Substitute these values into equation 1
H = (14.021)²sin²45/(2×9.8)
H = 196.5884×0.5/19.6
H = 5.02 m.
Hence the ball goes 5.02 m high
Answer:
m = 2.01[kg]
Explanation:
This problem can be solved using Newton's second law which tells us that the force applied on a body is equal to the product of mass by acceleration.
where:
F = force = 12.5 [N]
m = mass [kg]
a = acceleration = 6.2 [m/s²]
![12.5=m*6.2\\m = 2.01[kg]](https://tex.z-dn.net/?f=12.5%3Dm%2A6.2%5C%5Cm%20%3D%202.01%5Bkg%5D)
Answer:
<em>The net force acting on the object is 0 N</em>
Explanation:
<u>Newton's Second Law of Forces</u>
The net force acting on a body is proportional to the mass of the object and its acceleration.
The net force can be calculated as the sum of all the force vectors in each rectangular coordinate separately.
The image shows a free body diagram where four forces are acting: two in the vertical direction and two in the horizontal direction.
Note the forces in the vertical direction have the same magnitude and opposite directions, thus the net force is zero in that direction.
Since we are given the acceleration a =0, the net force is also 0, thus the horizontal forces should be in equilibrium.
The applied force of Fapp=10 N is compensated by the friction force whose value is, necessarily Fr=10 N in the opposite direction.
The net force acting on the object is 0 N
Answer:
It requires more tension to pull up the track
Explanation:
Net force must be zero to maintain constant velocity.
Weight force will always be pointed down the slope. Call it W
Friction force (Call it Ff) will be down slope when movement is up slope.
Friction force will be up slope when movement is down slope.
W and Ff are always positive numbers
Call the pulling force T
If Up slope is considered the positive direction
Moving up slope
Tu - Ff - W = 0
Tu = Ff + W
Moving down slope
Td + W - Ff = 0
Td = Ff - W
Ff + W > Ff - W therefore Tu > Td
