The weight of an object is the product of its mass and the acceleration of gravity.
If g[e] is the acceleration of gravity on earth, and g[M] the same for Mars and g[m] the same for the moon,
then m[M]=m[e]g[M]/g[e] and m[m]=m[e]g[m]/g[e] where m[ ] denotes mass. Note that weight=mg (measured in newtons) while mass is in kilograms.
If g[M]=g[e]/3 and g[m]=g[e]/6 approximately. Then the weight of an object on Mars will be about a third of what it is on earth, while on the moon it would be about a sixth of what it is on earth.
Answer:
Step-by-step explanation:
To write this equation with h isolated, we cancel everything else on the right hand side of the equals sign. This means we cancel π and r².
Since both of these are attached to h by multiplication, we divide by both to cancel. We can divide by both at the same time:
Answer:
5 as 5/6 of 30= 25 and 25+5 =30
Step-by-step explanation:
Is RS perpendicular to DF? Select Yes or No for each statement. R (6, −2), S (−1, 8), D (−1, 11), and F (11 ,4) R (1, 3), S (4,7
guajiro [1.7K]
I'll do the first one to get you started.
Find the slope of the line between R (6,-2) and S (-1,8) to get
m = (y2-y1)/(x2-x1)
m = (8-(-2))/(-1-6)
m = (8+2)/(-1-6)
m = 10/(-7)
m = -10/7
The slope of line RS is -10/7
Next, we find the slope of line DF
m = (y2 - y1)/(x2 - x1)
m = (4-11)/(11-(-1))
m = (4-11)/(11+1)
m = -7/12
From here, we multiply the two slope values
(slope of RS)*(slope of DF) = (-10/7)*(-7/12)
(slope of RS)*(slope of DF) = (-10*(-7))/(7*12)
(slope of RS)*(slope of DF) = 10/12
(slope of RS)*(slope of DF) = 5/6
Because the result is not -1, this means we do not have perpendicular lines here. Any pair of perpendicular lines always has their slopes multiply to -1. This is assuming neither line is vertical.
I'll let you do the two other ones. Let me know what you get so I can check your work.
Answer:
A. 9.8 ft
Step-by-step explanation:
first convert the degrees to radians:
80° x (pi/180) ≈ 1.396
arc length = r x θ (always in radians)
= 7 x 1.396
= 9.772 ≈ 9.8 ft
