Answer:
D. 3 hours or more
Explanation:
The average 8- to 18-year-old spends at least D. 3 hours every day in front of a screen, performing little to no physical activity. This is because, instead of exercising and socializing with their peers, children and teenagers frequently talk, watch a lot of movies/shows, or play video games on their computers. Unfortunately, this is typically considerably more than three hours every day. Although some children still prefer physical activities over this, the bulk of the population does not.
Answer:four times
Explanation:
Given
mass of both cars A and B are same suppose m
but velocity of car B is same as of car A
Suppose velocity of car A is u
Velocity of car B is 2 u
A constant force is applied on both the cars such that they come to rest by travelling certain distance
using to find the distance traveled
where, v=final velocity
u=initial velocity
a=acceleration(offered by force)
s=displacement
final velocity is zero
For car A


For car B


divide 1 and 2 we get

thus 
distance traveled by car B is four time of car A
Fnet = (mass) (acceleration)
= 11 kg x 3.7m/s^2
= 41 N
Answer:
The helicopter was 1103.63 meters high when the package was dropped.
Explanation:
We consider positive speed as a downward movement
y: height (m)
t: time (s)
v₀: initial speed (m/s)
Δy = v₀t +
gt²
Δy= 15
×15 s +
×9.81
×(15 s)²
Δy= 1103.63 m
Answer:
a) During the reaction time, the car travels 21 m
b) After applying the brake, the car travels 48 m before coming to stop
Explanation:
The equation for the position of a straight movement with variable speed is as follows:
x = x0 + v0 t + 1/2 a t²
where
x: position at time t
v0: initial speed
a: acceleration
t: time
When the speed is constant (as before applying the brake), the equation would be:
x = x0 + v t
a)Before applying the brake, the car travels at constant speed. In 0.80 s the car will travel:
x = 0m + 26 m/s * 0.80 s = <u>21 m </u>
b) After applying the brake, the car has an acceleration of -7.0 m/s². Using the equation for velocity, we can calculate how much time it takes the car to stop (v = 0):
v = v0 + a* t
0 = 26 m/s + (-7.0 m/s²) * t
-26 m/s / - 7.0 m/s² = t
t = 3.7 s
With this time, we can calculate how far the car traveled during the deacceleration.
x = x0 +v0 t + 1/2 a t²
x = 0m + 26 m/s * 3.7 s - 1/2 * 7.0m/s² * (3.7 s)² = <u>48 m</u>