Average speed = total distance travelled ÷ total time taken
AS = (75km + 68km) ÷ (1hr + 2hr)
As = 143km ÷ 3hr
AS = 47.66667 km/hr
AS = 47.7 km/hr (3sf)
Answer:
option E
Explanation:
given,
diameter = 4 mm
shutter speed = 1/1000 s
diameter of aperture = ?
shutter speed = 1/250 s
exposure time to the shutter time
![E V = log_2(\dfrac{N^2}{t})](https://tex.z-dn.net/?f=E%20V%20%3D%20log_2%28%5Cdfrac%7BN%5E2%7D%7Bt%7D%29)
N is the diameter of the aperture and t is the time of exposure
now,
![log_2(\dfrac{N^2}{t_1})=log_2(\dfrac{N^2}{t_2})](https://tex.z-dn.net/?f=log_2%28%5Cdfrac%7BN%5E2%7D%7Bt_1%7D%29%3Dlog_2%28%5Cdfrac%7BN%5E2%7D%7Bt_2%7D%29)
![\dfrac{N_1^2}{t_1}=\dfrac{N_2^2}{t_2}](https://tex.z-dn.net/?f=%5Cdfrac%7BN_1%5E2%7D%7Bt_1%7D%3D%5Cdfrac%7BN_2%5E2%7D%7Bt_2%7D)
inserting all the values
![\dfrac{4^2}{1000}=\dfrac{N_2^2}{250}](https://tex.z-dn.net/?f=%5Cdfrac%7B4%5E2%7D%7B1000%7D%3D%5Cdfrac%7BN_2%5E2%7D%7B250%7D)
N₂² = 4
N₂ = 2 mm
hence , the correct answer is option E
The tension in the string corresponds to the gravitational attraction between the Sun and any planet.
Actually what the problem meant about the westward
component of the ball’s displacement is the horizontal component of the
displacement. To help us better understand the problem, I attached a figure of
the situation.
We can see from the figure that to solve for the value of
the horizontal component, we have to make use of the sin function. That is:
sin θ = side opposite to the angle / hypotenuse of the
triangle
sin 42 = x / 40 m
x = (40 m) sin 42
x = 26.77 m
Therefore the ball has a westward
displacement of about 26.77 m