Answer:
professional communication style is expected to be formal while the casual is informal.
Explanation:
A 59 kg sprinter, starting from rest, runs 47 m in 7.0 s at constant acceleration.?
What is the sprinter's power output at 2.0 s, 4.0 s, and 6.0 s?
Instantaneous Power is the force times velocity
P = Fv
Because the acceleration is constant, the force will be constant as well
F = ma
P = mav
for constant acceleration, the velocity at each time is found using
v = at
P = ma(at) = ma²t
find the acceleration using kinematic equation
s = ½at²
a = 2s/t²
a = 2(47) / 7.0²
a = 1.918 m/s²
P(2.0) = 59(1.918²)2.0 = 434.25 W = 0.43 kW
P(4.0) = 59(1.918²)4.0 = 868.51 W = 0.87 kW
P(6.0) = 59(1.918²)6.0 = 1302.76 W = 1.3 kW
I hope this helped.
Answer:
0.5639m
Explanation:
For a young double slit experiment the expression below gives the angular separation for m dark fringe having slit width d and wavelength λ
=sin⁻¹(mλ/d)
mλ /d =y/L
for the first order,
y= mλL/d
For ratio separation y₀/yD=1 and d= 1
y₀/yD= [mλ ₀L₀/d]/[mλD.LD./d]
1=λ ₀L₀/λD.LD.
λD.LD= λ ₀L₀
L₀= λD.LD/ λ ₀..............(1)
Then substitute the given values into (1) we have
L₀=471 *0.497/611
= 0.3831m
Distance by which the screen has to be moved towards the slit is
LD- Lo
0.947-0.3831= 0.5639m
Answer:
(1) 2.25m/s^2
(2) 45.6m
Explanation:
(1) A car accelerates uniformly from 12m/s to 39m/s in 12 seconds
Therefore the average acceleration can be calculated as follows
a = 39-12/12
a = 27/12
a= 2.25m/s^2
(2) A butterfly is flying at 4m/s , it accelerates uniformly at 1.2 m/s for 6 seconds
u= 4
a= 1.2
t= 6
Therefore the distance can be calculated as follows
S= ut + 1/2at^2
= 4×6 + 1/2 × 1.2 × 6^2
= 24 + 1/2 × 1.2 × 36
= 24 + 1/2 × 43.2
= 24 + 21.6
S = 45.6m
Hence the butterfly travels at 45.6m
