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:
0.8 x 10^-9 kg
Explanation:
Given,
Distance ( R ) = 10 m
Force ( F ) = 3.2 x 10^-9 N
Mass ( m1 ) = 40 kg
To find : Mass ( m2 ) = ?
Formula : -
F = m1.m2 / R^2
m2 = FR^2 / m1
= 3.2 x 10^-9 x 10 / 40
= 3.2 x 10^-9 / 4
= ( 3.2 / 4 ) x 10^-9
m2 = 0.8 x 10^-9 kg
Answer:
22m/s
Explanation:
To find the velocity we employ the equation of free fall: v²=u²+2gh
where u is initial velocity, g is acceleration due to gravity h is the height, v is the velocity the moment it hits the ground, taking the direction towards gravity as positive.
Substituting for the values in the question we get:
v²=2×9.8m/s²×25m
v²=490m²/s²
v=22.14m/s which can be approximated to 22m/s
Answer:
The maximum current, in amperes, that a conductor can carry continuously under the conditions of abuse without exceeding its temperature rating.
