Answer:
s^ -1 ( or 1/sec)
Explanation:
Velocity is given in units of displacement / sec
like feet /sec or m/sec
so b would have units of s^-1
(or perhaps a more general term would be time^-1)
1- You should always have a question for your experiment.
2- You need to conduct background research. It helps to write down your sources so you can cite your references.
3- Propose a hypothesis (educated guess on what you believe the outcome of the experiment will be)
4- Design and perform an experiment to test your hypothesis (include independent and dependent variable)
5- Record observations and analyze what the data means.
6- Conclude whether you need to accept or reject your hypothesis, which accepting means your hypothesis was right and rejected is if it was wrong.
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>
Answer:
The radiation pressure of the light is 3.33 x 10⁻⁶ Pa.
Explanation:
Given;
intensity of light, I = 1 kW/m²
The radiation pressure of light is given as;
I kW = 1000 J/s
The energy flux density = 1000 J/m².s
The speed of light = 3 x 10⁸ m/s
Thus, the radiation pressure of the light is calculated as;
Therefore, the radiation pressure of the light is 3.33 x 10⁻⁶ Pa.
We have: K.E. = 1/2 mv²
Here: m = 50 g = 0.05 Kg
v = 4 m/s
Substitute their values,
K.E. = 1/2 * 0.05 * 4²
K.E. = 1/2 * 0.05 * 16
K.E. = 0.4 J
In short, Your Answer would be 0.4 Joules
Hope this helps!
