Answer: 11%
Explanation:
Given that
X = ab^2/C. Calculate percentage error in X, when percentage error in a,b,c are 4,2 and 3 respectively.
Percentage error of b = 2%
Percentage error of b^2 = 2 × 2 = 4
When you are calculating for percentage error that involves multiplication and division, you will always add up the percentage error values.
Percentage error of X will be;
Percentage error of a + percentage error of b^2 + percentage error of c
Substitute for all these values
4 + 4 + 3 = 11%
Therefore, percentage error of X is 11%
Answer:
D. Solution
Explanation:
Sugar dissolved in water is an example of solution.
A solution is a homogenous mixture of solutes and solvents.
In a solution the solute particles ae distributed uniformly in the solvents. The solute is the substance and it is the sugar here that is dissolved to make a solution.
The solvent is the water in this instance that helps to dissolve the solute.
The change in speed of this object is 3m/s
According to Newton's second law;
F = ma
F = mv/t
Given the following parameters
Force F = 8.0N
mass m = 16kg
time t = 4.0s
Required
speed v
Substitute the given parameters into the formula
v = Ft/m
v = 8 * 6/16
v = 48/16
v = 3m/s
Hence the change in speed of this object is 3m/s
Learn more here: brainly.com/question/19072061
The period of the pendulum is directly proportional to the square root of the length of the pendulum
Explanation:
The period of a simple pendulum is given by the equation
where
T is the period
L is the length of the pendulum
g is the acceleration of gravity
From the equation, we see that when the length of the pendulum increases, the period of the pendulum increases as the square root of L, 
. This means that
The period of the pendulum is directly proportional to the square root of the length of the pendulum
From the equation, we also notice that the period of a pendulum does not depend on its mass.
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Answer:
The frictional force 
6.446 N
The acceleration of the block a = 6.04 
Explanation:
Mass of the block = 3.9 kg
°
= 0.22
(a). The frictional force is given by
3.9 × 9.81 × 
29.3 N
Therefore the frictional force
0.22 × 29.3
6.446 N
(b). Block acceleration is given by
F = 30 N
= 6.446 N
= 30 - 6.446
= 23.554 N
The net force acting on the block is given by
23.554 = 3.9 × a
a = 6.04
This is the acceleration of the block.
