Explanation:
First, we need to determine the distance traveled by the car in the first 30 minutes, 
.
Notice that the unit measurement for speed, in this case, is km/hr. Thus, a unit conversion of from minutes into hours is required before proceeding with the calculation, as shown below
Now, it is known that the car traveled 40 km for the first 30 minutes. Hence, the remaining distance, 
, in which the driver reduces the speed to 40km/hr is
.
Subsequently, we would also like to know the time taken for the car to reach its destination, denoted by 
.
.
Finally, with all the required values at hand, the average speed of the car for the entire trip is calculated as the ratio of the change in distance over the change in time.
Therefore, the average speed of the car is 50 km/hr.
Have everything in control and in order and discuss about different issues.
Answer:
the moment of inertia of the merry go round is 38.04 kg.m²
Explanation:
We are given;
Initial angular velocity; ω_1 = 37 rpm
Final angular velocity; ω_2 = 19 rpm
mass of child; m = 15.5 kg
distance from the centre; r = 1.55 m
Now, let the moment of inertia of the merry go round be I.
Using the principle of conservation of angular momentum, we have;
I_1 = I_2
Thus,
Iω_1 = I'ω_2
where I' is the moment of inertia of the merry go round and child which is given as I' = mr²
Thus,
I x 37 = ( I + mr²)19
37I = ( I + (15.5 x 1.55²))19
37I = 19I + 684.7125
37I - 19 I = 684.7125
18I = 684.7125
I = 684.7125/18
I = 38.04 kg.m²
Thus, the moment of inertia of the merry go round is 38.04 kg.m²
Variables:
Source charge, Q = 3 micro C = 3 * 10^ - 6 C
E = electric field = 2.86 * 10 ^5 N/C
K = 8.99 * 10^9 N * m^2 / C
d = distance = ?
Formula:
E = K * Q / (d^2) => d^2 = K * Q / E
=> d^2 = 8.99 * 10^9 N * m^2 / C * 3 * 10^ -6 C / (2.86 * 10^ 5 N/C)
d^2 = 9.43 * 10 ^ -2 m^2
=> d = 3.07 * 10^-1 m
Answer: 0.307 m
Note: it is a long distance due to the Electric field is very low
Answer;
-Flammability
Explanation;
-Chemical properties are characteristics of a material that become evident when the material undergoes a chemical reaction or chemical change. It is a substances ability to change into a whole new substance.
-Examples of Chemical Property include flammability, ability to rot, reactivity, ability to tarnish, ability to rust. Iron, for example, combines with oxygen in the presence of water to form rust.
