This question is not complete.
The complete question is as follows:
One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a space station that spins about its center at a constant rate. This creates “artificial gravity” at the outside rim of the station. (a) If the diameter of the space station is 800 m, how many revolutions per minute are needed for the “artificial gravity” acceleration to be 9.80m/s2?
Explanation:
a. Using the expression;
T = 2π√R/g
where R = radius of the space = diameter/2
R = 800/2 = 400m
g= acceleration due to gravity = 9.8m/s^2
1/T = number of revolutions per second
T = 2π√R/g
T = 2 x 3.14 x √400/9.8
T = 6.28 x 6.39 = 40.13
1/T = 1/40.13 = 0.025 x 60 = 1.5 revolution/minute
Answer: a
Explanation: The color of a star is linked to its surface temperature. The hotter the star, the shorter the wavelength of light it will emit. The hottest ones are blue or blue-white, which are shorter wavelengths of light. Cooler ones are red or red-brown, which are longer wavelengths.
Answer:
The torque about his shoulder is 34.3Nm.
The solution approach assumes that the weight of the boy's arm acts at the center of the boy's arm length 35cm from the shoulder.
Explanation:
The solution to the problem can be found in the attachment below.
Answer: Hello mate!
lets define the north as the y-axis and east as the x-axis.
Using the notation (x,y) we can define the initial position of the car as (0,0)
then the car travells 13 mi east, so now the position is (13,0)
then the car travels Y miles to the north, so the position now is (13, Y)
and we know that the final position is 25° degrees north of east of the initial position. This angle says that the distance traveled to the north is less than 13 mi because this angle is closer to the x-axis (or east in this case).
This angle is measured from east to north, then the adjacent cathetus is on the x-axis, in this case, 13mi
And we want to find the distance Y, so we can use the tangent:
Tan(25°) = Y/13
tan(25°)*13 mi = Y = 6.06 mi.
