Answer:
R = 0.0503 m
Explanation:
This is a projectile launching exercise, to find the range we can use the equation
R = v₀² sin 2θ / g
How we know the maximum height
² =
² - 2 g y
= 0
= √ 2 g y
= √ 2 9.8 / 15
= 1.14 m / s
Let's use trigonometry to find the speed
sin θ = / vo
vo = / sin θ
vo = 1.14 / sin 60
vo = 1.32 m / s
We calculate the range with the first equation
R = 1.32² sin(2 60) / 30
R = 0.0503 m
I believe your answer is: Switching Costs & Customer Loyalty
since airplane is thrown towards west with speed 6 m/s
while air is blowing with speed 8 m/s towards north
so here the net speed of air plane will be the resultant of airplane speed and wind speed always
SO here we can say it would be a combination of vector along west with must be of length 6 m/s and other vector is towards north with is of length 8 m/s
so correct answer must be 1st option
The three longest wavelengths for the standing waves on a 264-cm long string that is fixed at both ends are:
- 5.2 meters.
- 2.6 meters.
- 1.7meters.
Given data:
Length of the fixed string = 264cms = 2.64 meters
The wavelength for standing waves is given by:
λ = 2L/n
where,
- λ is the wavelength
- L is the length of the string
For n = 1,
= 5.2 meters
For n = 2,
= 2.6 meters
For n = 3,
= 1.7 meters
To learn more about standing waves: brainly.com/question/14151246
#SPJ4
Answer:
statement - 'The work done by friction is equal to the sum of the work done by the gravity and the initial push' is correct.
Explanation:
The statement ''The work done by friction is equal to the sum of the work done by the gravity and the initial push" is correct.
The above statement is correct because, the initial push will tend to slide down the block thus the work done by the initial push will be in the downward direction. Also, the gravity always acts in the downward direction. thus, the work done done by the gravity will also be in the downward direction
here, the downward direction signifies the downward motion parallel to the inclined plane.
Now we know that the work done by the friction is against the direction of motion. Thus, the friction force will tend to move the block up parallel to the inclined plane.
Hence, for the block to stop sliding the the above statement should be true.
