Answer:
j
Explanation:
x = 4 t^2 - 2 t - 4.5
Position at t = 3 s
x = 4 (3)^2 - 2 (3) - 4.5 = 25.5 m
Velocity at t = 3 s
v = dx / dt = 8 t - 2
v ( t = 3 s) = 8 x 3 - 2 = 22 m/s
Acceleration at t = 3 s
a = dv / dt = 8
a ( t = 3 s ) = 8 m/s^2
When is the velocity = 0
v = 0
8 t - 2 = 0
t = 0.25 second
When is the position = 0
x = 0
4 t^2 - 2 t - 4.5 = 0
t = 1.4 second
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
<span># of protons + # of neutrons = atomic mass</span>
Answer:
B.will increase the maximum static friction between the box and the floor
Explanation:
Because static friction is the force that keeps an object at rest
