Answer:
![v_o = 40.56 m/s](https://tex.z-dn.net/?f=v_o%20%3D%2040.56%20m%2Fs)
Explanation:
Let the initial speed of the ball is given as
![v_i = v_o](https://tex.z-dn.net/?f=v_i%20%3D%20v_o)
then angle of inclination is given as 33 degree
now its two components of velocity is given as
![v_x = v_o cos33](https://tex.z-dn.net/?f=v_x%20%3D%20v_o%20cos33)
![v_y = v_o sin33](https://tex.z-dn.net/?f=v_y%20%3D%20v_o%20sin33)
now the two positions with time given as
![y = y_o + v_y t + \frac{1}{2}at^2](https://tex.z-dn.net/?f=y%20%3D%20y_o%20%2B%20v_y%20t%20%2B%20%5Cfrac%7B1%7D%7B2%7Dat%5E2)
![y = 1 + v_osin33 t - 4.9 t^2](https://tex.z-dn.net/?f=y%20%3D%201%20%2B%20v_osin33%20t%20-%204.9%20t%5E2)
![x = v_o cos33 t](https://tex.z-dn.net/?f=x%20%3D%20v_o%20cos33%20t)
now when it hit the wall so we will have
![x = 114 = v_o cos33 t](https://tex.z-dn.net/?f=x%20%3D%20114%20%3D%20v_o%20cos33%20t)
![y = 20 = 1 + v_o sin33 t - 4.9 t^2](https://tex.z-dn.net/?f=y%20%3D%2020%20%3D%201%20%2B%20v_o%20sin33%20t%20-%204.9%20t%5E2)
now from above two equations we will have
![19 = \frac{114}{v_o cos33}(v_o sin33) - 4.9 t^2](https://tex.z-dn.net/?f=19%20%3D%20%5Cfrac%7B114%7D%7Bv_o%20cos33%7D%28v_o%20sin33%29%20-%204.9%20t%5E2)
![19 = 74 - 4.9 t^2](https://tex.z-dn.net/?f=19%20%3D%2074%20-%204.9%20t%5E2)
![t = 3.35 s](https://tex.z-dn.net/?f=t%20%3D%203.35%20s)
now from above equations again
![114 = v_o cos33 (3.35)](https://tex.z-dn.net/?f=114%20%3D%20v_o%20cos33%20%283.35%29)
![v_o = 40.56 m/s](https://tex.z-dn.net/?f=v_o%20%3D%2040.56%20m%2Fs)
Answer:
wildlife reserves are not available for camping...but you can camp at a skatepark
Answer:
1.31×10⁻⁴ C or 131 μC
Explanation:
From the question,
Applying Coulomb's Law,
F = kqq'/r²..................... Equation 1
Where F = Force between the charges, q = first charge, q' = second charge, r = distance between the charges, k = coulomb's constant.
make q' the subject of the equation
q' = F×r²/(kq)
q' = Fr²/kq................ Equation 2
Given: F = 1240 N, r = 16.4 cm = 0.164 m, q = 28.4 μC = 2.84×10⁻⁵ C,
Constant: k = 8.98×10⁹ Nm/C²
Substitute these values into equation 2
q' = (1240×0.164²)/[(2.84×10⁻⁵)×(8.98×10⁹)
q' = (33.35104)/(25.5032×10⁴)
q' = 1.31×10⁻⁴ C
q' = 131 μC
Answer:
The velocity of a trip from City A to City B using the car is 50 miles per hour.
Explanation:
Given
200 mile trip by car takes 4 hours.
Distance between cities = 120 miles
Required:
What is the velocity of a trip from City A to City B?
To get the velocity of a trip from city A to B, we consider the parameter given for the car that travel 200 miles distance in 4 hours
Velocity = displacement ÷ time
Where displacement 200 miles
Time = 4 hours.
By substituton,
So, velocity= 200 miles ÷ 4 hours
Velocity = 50 miles ÷ hour
Velocity = 50 miles per hour
The velocity of a trip from City A to City B using the car is 50 miles per hour.