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
The answer is B, luminous. Because of that fact that absorbing is taking light and Illuminated means having light because of a different light source. Luminous means bright and having lots of light. Hope this helps :)
Answer:
increases by a factor of 
Explanation:
First we need to find the initial velocity for it to stop at the distance 2d using the following equation of motion:
where v = 0 m/s is the final velocity of the package when it stops, 
is the initial velocity of the package when it, a is the deceleration, and 
is the distance traveled.
So the equation above can be simplified and plug in Δs = d, 
for the 1st case
(1)
For the 2nd scenario where the ramp is changed and distance becomes 2d, 
(2)
let equation (2) divided by (1) we have:
So the initial speed increases by . If the deceleration a stays the same and time is the ratio of speed over acceleration a
The time would increase by a factor of
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Answer:
84 protons and 128 neutron
