A bird flies at a speed of 15 m/s for 10 seconds, 1,200 m/min for 0.17 minute, and 2,500 cm/s for 5,000 milliseconds. What is the bird’s average speed?

Explanation:

average speed=distance travelled/time taken

distance=speed*time

15*10=150 m

1200 m/min=20 m/s

0.17 minute=10.2 seconds

20*10.2=204 m

2500 cm/s=25 m/s

5000 milliseconds=5 seconds

25*5=125 m

125+150+204=479 m

10.2+5+10=25.2

479/25.2=19.0

15 m/s for 10 sec; distance covered = (15 x 10) = 150 m
Second segment:
20 m/s for 20 sec; distance covered = (20 x 20) = 400 m
Third segment:
30 m/s for 15 sec; distrance covered = (30 x 15) = 450 m
Total time = (10s + 20s + 15s) = 45 sec
Total distance = (150m + 400m + 450m) = 1,000 m (1 km)
Average speed = (1,000 m) / (45 sec) = 22.22 m/s ...

Explanation:

Balance Chemical Equation:

A balanced chemical equation is that in which the the number of the reactant atoms equal to the number of the product atom.

For example if the oxygen at the reactant side is 4, the number of oxygen should be 4 on the product side.

The complete answer is in attachment

Equation 1:

2 is the correct number

In the equation look for the number of Carbon on product side

C = 4 so it should be 4 on reactant side too

So,

by putting 2 in Coefficient location the number of C become equal on reactant as well as product side.

____________________

Equation 2:

12 on reactant side and 6 on product side is the correct number

In the equation look for the number of Hydrogen (H) on product side

it is H = 12 so it should be 12 on reactant side too

Then look for the number of Chlorine (Cl) on reactant side

C = 12 so it should be 12 on Product side too

So,

by putting 12 in Coefficient location of the reactant side and 6 on product side the number of H and Cl become equal on reactant as well as product side.

______________________

Equation 3:

5 is the correct number

In the equation look for the number of Oxygen on product side

O = 12 so it should be 12 on reactant side too

So,

by putting 5 in Coefficient location the number of O become equal on reactant as well as product side.

______________________

Equation 4:

3 on reactant side and 1 on product side is the correct number

In the equation look for the number of Chlorine (Cl) on product side

it is Cl = 6 so it should be 6 on reactant side too

Then look for the number of Calcium (Ca) on reactant side

Ca = 1 so it should be 1 on reactant side too

So,

by putting 3 in Coefficient location of the reactant side and 1 on product side the number of Cl and Ca become equal on reactant as well as product side.

____________________

Equation 5:

4 and 7 on reactant side is the correct number

In the equation look for the number of Sulfur (S) on product side

it is S = 4 so it should be 4 on reactant side too

Then look for the number of Oxygen (O) on product side

O = 14 so it should be 14 on reactant side too

So,

by putting 4 and 7 in Coefficient location of the reactant side so the number of S and O become equal on reactant as well as product side.