For this use the formula:
d = Vo * t - (at^2) / 2
Clearing t:
t = d/(v + 0.5*a)
Replacing:
t = 5 m / (7.2 m/s + 0.5 * (-1.1 m/s²)
Resolving:
t = 5 m / (7.2 m/s + (-0.55 m/s²)
t = 5 m / 6.65 m/s
t = 0.75 s
Result:
The time will be <u>0.75 seconds.</u>
Answer:
a. 18.13m/s
b. 0.84m
c. 2.4m
Explanation:
a. to find the speed at which the ball was lunched, we use the horizontal component.Since the point distance from the base of the ball is 24m and it takes 2.20 secs to reach the wall,we can say that
t=distance /speed

Hence the speed at which ball was lunched is 18.13m/s
b. from the equation

the vertical distance at which the ball clears the wall is
y=8.14-7.3=0.84m
c. the time it takes the ball to reach the 6.2m vertically

the horizontal distance covered at this speed is

Answer:
orbital speed of the electrons in their orbit will increase
Explanation:
As we know that centripetal force for electrons will be due to electrostatic attraction force of electron.
So it is given as

so we have

now on the left side if the force of attraction will increase and hence there must be the change in that part of equation
So here at the same position the speed of the electron
So we can say that correct answer will be
orbital speed of the electrons in their orbit will increase
1) D
2) D.) Greater than 
Explanation:
1)
The phenomenon of total internal reflection occurs when a ray of light hitting the interface between two mediums is totally reflected back into the original medium, therefore no refraction into the second medium occurs.
This phenomenon occurs only if two conditions are satisfied:
- The index of refraction of the first medium is larger than the index of refraction of the 2nd medium
- The angle of incidence is greater than a certain angle called critical angle
In picture 1, we have 4 different diagrams. In the diagrams:
- The red arrow represents the incident ray
- The green arrow represents the refracted ray
- The blue arrow represents the reflected ray
Total internal reflection occurs when there is no refraction, therefore when there is no green arrow: this occurs only in figure D, so this is the correct option. (in figure C, there is a refracted ray but it is parallel to the interface: this condition occurs when the angle of incidence is exactly equal to the critical angle, however in this problem, the angle of incidence is greater than the critical angle, so the correct option is D)
2)
As we stated in problem 1), total internal reflection occurs when the angle of incidence is equal or greater than the critical angle. Therefore in this case, the angle of incidence must be
D.) Greater than 