Answer:
<h3>50.23m</h3>
Explanation:
The distance talks about how far an object has travelled.
Given
9 = -9.8m/s²
t = 3.2s
to get the distance Δx, we will use the formula;
S = ut+1/2gt²
S = 0(3.2)+1/2(-9.8)(3.2)²
S = 0-4.905(10.24)
S = -50.23m
Hence the can of tuna drop by 50.23m
Answer:
0.15A
Explanation:
The parameters given are;
R=20.0 Ω
C= 2.50 μF
V= 3.00 V
f= 2.48×10^-3 Hz
Xc= 1/2πFc
Xc= 1/2×3.142 × 2.48×10^-3 × 2.5 ×10^-6
Xc= 25666824.1
Z= 1/√(1/R)^2 +(1/Xc)^2
Z= 1/√[(1/20)^2 +(1/25666824.1)^2]
Z= 1/√(2.5×10^-3) + (1.5×10^-15)
Z= 20 Ω
But
V=IZ
Where;
V= voltage
I= current
Z= impedance
I= V/Z
I= 3.00/20
I= 0.15A
Answer:
The relative size of an object serves as an important monocular cue for depth perception. It works like this: If two objects are roughly the same size, the object that looks the largest will be judged as being the closest to the observer. This applies to three-dimensional scenes as well as two-dimensional images.
Explanation:
Answer: v= 160ft/s
a=32ft/s^2 constant
Explanation:
s(t)=400-16t^2 derivative of position is velocity v(t) and derivative of velocity is acceleration a(t) so let s(t)=0 to find the time of flight to reach the ground and take the two derivatives and use the time found and solve. Also acceleration is a constant as it’s gravity.
0=400-16t^2
400=16t^2
25=t^2
t=5s
ds/dt=v(t)=0-32t
dv/dt=a(t)=-32 constant(gravity)
v(t)=-32(5s)= -160ft/s negative sign is only showing direction
