Answer:
Explanation:
Given
Distance to grandmother's house=100 mi
it is given that during return trip Julie spend equal time driving with speed 30 mph and 70 mph
Let Julie travel x mi with 30 mph and 100-x with 70 mph

x=30 mi
Therefore
Julie's Average speed on the way to Grandmother's house
=42 mph
On return trip

Answer:
a = -4/5 m/s^2
Explanation:
Acceleration = change in velocity / time
change in velocity = final velocity - initial velocity
a = (20 m/s - 60 m/s) / 50 s
a = -40 m/s / 50 s
a = -4/5 m/s^2
hope this helps! <3
Answer:
Opposite to the direction that you are pulling
Explanation:
Static friction acts in the opposite direction to the acceleration.
Kinetic friction acts in the opposite direction to the velocity.
Answer:
a) x = 0.200 m
b)E = 3.84*10^{-4} N/C
Explanation:


DISTANCE BETWEEN BOTH POINT CHARGE = 0.5 m
by relation for electric field we have following relation

according to question E = 0
FROM FIGURE
x is the distance from left point charge where electric field is zero

solving for x we get

x = 0.200 m
b)electric field at half way mean x =0.25

E = 3.84*10^{-4} N/C
Answer:
C) 1.0 m
Explanation:
The component of the velocity parallel to the sidewalk is:
vₓ = v cos θ
vₓ = 0.1 m/s cos 45°
vₓ = 0.0707 m/s
The distance traveled after 14 seconds is:
d = vₓ t
d = (0.0707 m/s) (14 s)
d = 0.99 m
Closest answer is C) 1.0 m.