The bulk of earths atmospheric oxygen comes from phytoplankton, small flower like things in the ocean.
Answer:
0.71 kg
Explanation:
L = length of the steel wire = 3.0 m
d = diameter of steel wire = 0.32 mm = 0.32 x 10⁻³ m
Area of cross-section of the steel wire is given as
A = (0.25) πd²
A = (0.25) (3.14) (0.32 x 10⁻³)²
A = 8.04 x 10⁻⁸ m²
ΔL = change in length of the wire = 1.3 mm = 1.3 x 10⁻³ m
Y = Young's modulus of steel = 20 x 10¹⁰ Nm⁻²
m = mass hanging
F = weight of the mass hanging
Young's modulus of steel is given as


F = 6.968 N
Weight of the hanging mass is given as
F = mg
6.968 = m (9.8)
m = 0.71 kg
B force i belive becuase when is done it force
This is false. Current is the speed of the charge, 1 amp of current is 1 coulomb per second. So you can imagine the current of a circuit as the current of a river. In a parallel circuit, the river breaks into two separate streams. Some of the water goes down one river, some goes down the other. However, the total amount of water/coulombs never changes. This means that some of the total current will go down one river, and one the other. However, with less coulombs now the current will decrease.
Long story short, since there are two paths, the charge will split and depending on the resistance of each parallel stream a different amount of charge will go down each branch.
Answer:
9.81 N
Explanation:
Given that a baseball weighs 1 kg. The pitcher throws the ball at a velocity of 10 m/s straight up. Neglecting air resistance. The maximum height reached will be calculated by using the formula
V^2 = U^2 - 2gH
U = 10 m/s
g = 9.8m/s^2
At maximum height, V = 0
Substitute u and g into the formula
0 = 10^2 - 2 × 9.8 × H
19.6H = 100
H = 100/19.6
H = 5.1 m
The Kinetic energy on the ball will be
K.E = 1/2mv^2
K.E = 1/2 × 1 × 10^2
K.E = 1/2 × 100
K.E = 50 J
But energy = work done
WD = Force × distance (height)
The force that acts on the baseball when it is HALF WAY to the top of the path will be
F × 5.1/2 = 50
F = 100/5.1
F = 19.61 N
The weight acting downward will be
W = mg
W = 1 × 9.8
W = 9.8 N
The net force acting on the ball will be
Net force = F - W
Net force = 19.61 - 9.8
Net force = 9.81 N
Therefore, the net force that acts on the baseball when it is HALF WAY to the top of the path is 9.81 N