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
frequency=4Hz
wavelength=5m
amplitude=1/2×2=1m
period=1/frequency
1/4=0.25seconds.
velocity=wavelength×frequency
=5×4
=20m/s
Explanation:
Resting potential is the static membrane potential of the dormant cells. It is opposite of the action potential or the graded membrane potential. Na+/K+ ATPase pumps helps offset the ion diffusion through leak channels on neurons and helps return the membrane potential to its resting voltage.
Answer:
26.56°
Explanation:
given,
horizontal distance travel by the ant = 2 m
vertical distance travel by the ant = 1 m
angle of the ant above the initial point = ?
using trigonometric function
θ is the angle of the ant form initial position
h is the height on the wall
x is the horizontal distance covered by the ant
hence, the angle made by the ant from the initial position is equal to 26.56°
Inertia is the tendency of an object to resist a change in its motion.
