Answer:
64.5 cm
Explanation:
30 * 80 + x * 110 = 50* 190 => x = 64.5
KE = 2000 J
Explanation:
KE = (1/2)mv^2
= (1/2)(0.100 kg)(200 m/s)^2
= 2000 J
Given from the problem :mass m = 413 kg;coefficient of friction u = 0.0163;acceleration due to gravity g = 9.8 m/s2;inclined angle @1 = 14.3;inclined angle @2 = 4.69;distance travelled d = 175 m;applied fore F = 410 N; the component of the force from the donkey in the direction of motion isF2 = F1
[email protected]= 397.2964498768165 N
Fy = N - mg
[email protected] = 0N = mg
[email protected] = 4037.964151113007 NFx = F2 - mg
[email protected] - f = mahere f = u N=65.8188156631420141
F2 - mg
[email protected] - f = maa = F2 - mg
[email protected] - f/ m=0.31923412183075155 m/s^2
work done by donkeyW = F2 d=69526.8787284428875 J
Answer:
103.5 meters
Explanation:
Given that a stunt person has to jump from a bridge and land on a boat in the water 22.5 m below. The boat is cruising at a constant velocity of 48.3 m/s towards the bridge. The stunt person will jump up at 6.45 m/s as they leave the bridge.
The time the person will jump to a certain spot under the bridge can be calculated by using the formula below:
h = Ut + 1/2gt^2
since the person will fall under gravity, g = 9.8 m/s^2
Also, let assume that the person jump from rest, then, U = 0
Substitute h, U and g into the formula above
22.5 = 1/2 * 9.8 * t^2
22.5 = 4.9t^2
22.5 = 4.9t^2
t^2 = 22.5/4.9
t^2 = 4.59
t = 
t = 2.143 seconds
From definition of speed,
speed = distance /time
Given that the boat is cruising at a constant velocity of 48.3 m/s towards the bridge, substitute the speed and the time to get the distance.
48.3 = distance / 2.143
distance = 48.3 * 2.143
distance = 103.5 m
Therefore, the boat should be 103.5m away from the bridge at the moment the stunt person jumps?
Answer:
1.8x10⁻³m
Explanation:
From the question above, the following information was used to solve the problem.
wavelength λ = 4.5x10⁻⁷m
Length L = 2.0 meters
distance d = 5 x 10₋⁴m
ΔY = λL/d
= 4.5x10⁻⁷m (2) / 5 x 10₋⁴m
= 0.00000045 / 0.0005
= 0.0000009/0.0005
= 0.0018
= 1.8x10⁻³m
from the solution above The separation between two adjacent bright fringes is most nearly 1.8x10⁻³m
thank you!