100cos30° + 80cos120° + 40cos233° + Dx = 0
<span>Dx = -22.53 N </span>
<span>y components: </span>
<span>100sin30° + 80sin120° + 40sin233° + Dy = 0 </span>
<span>Dy = -87.34 N </span>
<span>magnitude of D: </span>
<span>sqrt[(-22.53)² + (-87.34)²] </span>
<span>90.2 N </span>
<span>direction of D: </span>
<span>arctan[(-87.34)/(-22.53)] </span>
<span>75.5° ref, but since Dx and Dy are both negative, we know this vector is in QIV: </span>
<span>360 - 75.5° = 284.5°</span>
Answer:
Refractive index of unknown liquid = 1.56
Explanation:
Using Snell's law as:
Where,
is the angle of incidence ( 65.0° )
is the angle of refraction ( 53.0° )
is the refractive index of the refraction medium (unknown liquid, n=?)
is the refractive index of the incidence medium (oil, n=1.38)
Hence,
Solving for ,
Refractive index of unknown liquid = 1.56
His average speed is 45 miles an hour
Answer:
7.15 m/s
Explanation:
We use a frame of reference in which the origin is at the point where the trucck passed the car and that moment is t=0. The X axis of the frame of reference is in the direction the vehicles move.
The truck moves at constant speed, we can use the equation for position under constant speed:
Xt = X0 + v*t
The car is accelerating with constant acceleration, we can use this equation
Xc = X0 + V0*t + 1/2*a*t^2
We know that both vehicles will meet again at x = 578
Replacing this in the equation of the truck:
578 = 24 * t
We get the time when the car passes the truck
t = 578 / 24 = 24.08 s
Before replacing the values on the car equation, we rearrange it:
Xc = X0 + V0*t + 1/2*a*t^2
V0*t = Xc - 1/2*a*t^2
V0 = (Xc - 1/2*a*t^2)/t
Now we replace
V0 = (578 - 1/2*1.4*24.08^2) / 24.08 = 7.15 m/s
Answer:
The effect of gravity on object makes objects stable and not floating in the air . It also makes any object thrown up to return back
