Answer:
303.29N and 1.44m/s^2
Explanation:
Make sure to label each vector with none, mg, fk, a, FN or T
Given
Mass m = 68.0 kg
Angle θ = 15.0°
g = 9.8m/s^2
Coefficient of static friction μs = 0.50
Coefficient of kinetic friction μk =0.35
Solution
Vertically
N = mg - Fsinθ
Horizontally
Fs = F cos θ
μsN = Fcos θ
μs( mg- Fsinθ) = Fcos θ
μsmg - μsFsinθ = Fcos θ
μsmg = Fcos θ + μsFsinθ
F = μsmg/ cos θ + μs sinθ
F = 0.5×68×9.8/cos 15×0.5×sin15
F = 332.2/0.9659+0.5×0.2588
F =332.2/1.0953
F = 303.29N
Fnet = F - Fk
ma = F - μkN
a = F - μk( mg - Fsinθ)
a = 303.29 - 0.35(68.0 * 9.8- 303.29*sin15)/68.0
303.29-0.35( 666.4 - 303.29*0.2588)/68.0
303.29-0.35(666.4-78.491)/68.0
303.29-0.35(587.90)/68.0
(303.29-205.45)/68.0
97.83/68.0
a = 1.438m/s^2
a = 1.44m/s^2
Answer:
a = 12.8 m/s^2
Explanation:
To find the centripetal acceleration you use the following formula:
(1)
v: tangential speed of the rock = 17.90mph
r: radius of the orbit = 1000cm/2 = 500cm = 0.5m
You first change the units of the tangential velocity in order to replace the values of v and r in the equation (1):

Then, you use the equation (1):

hence, the centripetal acceleration is 12.8 m/s^2
I'd go for D here. It also fits in with the idea of thermal expansion - as something is heated up, molecules vibrate and maybe collide. they vibrate with bigger amplitudes, so taking up more space, so expanding. maybe
Ok so the first thing you need to do is im not gonna help you
Answer:
C.
Explanation:
If there are the same number of protons (+) as electrons (-) they will cancel out and make the atom neutral.