Density = mass/volume
Mass = 35g
Volume = 5cm3
35/5 = 7g/cm3
So the answer is A to 1 decimal place
T = 4.25 ms = 4 x 10⁻³ s, the time for rebound
v₁ = 25.5 m/s, the impacting velocty
v₂ = -19.5 m/s, the rebounding velocity (n the opposite directon)
The change in velocity is
v₂ - v₁ = - (25.5+19.5) = -45 m/s
The acceleration is
a = (-45 m/s)/(4 s) = -11.25 m/s²
The negative sign indicates that the final velocity is opposiye to the impact velocty.
Answer: The magnitude of the acceleration is 11.25 m/s²
That would be the atom I believe.
Answer:
distance between school and home is 21 miles
Explanation:
given data
in rush hour speed s1 = 28 mph
less traffic speed s2 = 42 mph
time t = 1 hr 15 min = 1.25 hr
to find out
distance d
solution
we consider here distance home to school is d and t1 time to reach at school
we get here distance equation when we go home to school that is
distance = 28 × t1 .......................1
and when we go school to home distance will be
distance = 42 × ( t - t1 )
distance = 42 × ( 1.25 - t1 ) ...................2
so from equation 1 and 2
28 × t1 = 42 × ( 1.25 - t1 )
t1 = 0.75
so
from equation 1
distance = 28 × t1
distance = 28 × 0.75
distance = 21 miles
Answer:
Explanation:
Total distance covered = 25 + 15 = 40 m
Total time = 15 + 8 = 23 s
Average speed = total distance covered / total time
= 40 / 23
= 1.74 m / s
Total displacement = 25 - 15 = 10 m
Total time = 15 + 8 = 23 s
Average velocity = total displacement / total time
= 10 / 23
= .434 m / s to the right .
