It holds A. Less water than when it is warmer.
Answer:
Her speed is 9.8 meter per second
Explanation:
Newton's second law states that acceleration (a) is related with force (F) by:
(1)
Here the only force acting on the firefighter is the weight F=mg so (1) is:
Solving for a:

Now with the acceleration we can use the Galileo's kinematic equation:
(2)
With Vf the final velocity, Vo the initial velocity and Δx the displacement, because the firefighter stars from rest Vo=0 so (2) is:

Solving for Vf



Hi pupil here's your answer ::
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Action and Reaction do not act on the same body !! If they acted on the same body, the resultant force will be zero and their could be never accelerated motion.
If both the forces acted on the same body, then if they are equal to opposite direction the object will remain stationary. If on of the forces is greater than other the object will move in the direction of greater force.
If both acted in the same direction there would be an accelrated motion.
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Hope this helps . . . . .
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>