Recessive genotype I hope this helps
Correct matching:
1 acceleration -->
rate of change in velocity, which is the change in velocity divided by the change in time
2. speed --> the rate at which an object changes position when traveling in a certain direction
4. gravity --> force of attraction between all masses in the universe
5. Inertia --> an object´s resistance to a change in motion
3. friction --> force of resistance acting between objects in contact and tending to dampen their motion
6. velocity --> the rate at which an object changes position
Answer:
a)Distance traveled during the first second = 4.905 m.
b)Final velocity at which the object hits the ground = 38.36 m/s
c)Distance traveled during the last second of motion before hitting the ground = 33.45 m
Explanation:
a) We have equation of motion
S = ut + 0.5at²
Here u = 0, and a = g
S = 0.5gt²
Distance traveled during the first second ( t =1 )
S = 0.5 x 9.81 x 1² = 4.905 m
Distance traveled during the first second = 4.905 m.
b) We have equation of motion
v² = u² + 2as
Here u = 0, s= 75 m and a = g
v² = 0² + 2 x g x 75 = 150 x 9.81
v = 38.36 m/s
Final velocity at which the object hits the ground = 38.36 m/s
c) We have S = 0.5gt²
75 = 0.5 x 9.81 x t²
t = 3.91 s
We need to find distance traveled last second
That is
S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m
Distance traveled during the last second of motion before hitting the ground = 33.45 m
Answer:
0.035 N
Explanation:
Parameters given:
Charge q1 = -3.31x10^(-7) C
Charge q2 = -5.7x10^(-7) C.
Distance between them, R = 22 cm = 0.22 m
Electrostatic force between to particles is given as:
F = (k* q1 * q2) / R²
F = (9 * 10^9 * -3.31 * 10^(-7) * -5.7 * 10^(-7)) / 0.22²
F = 0.035 N
