(a) The work done by the person is 18.83465 kJ.
(b) The average power performed by the person during the walk is 51.4 W.
(c) The amount of food calories burnt is 4.5 Cal.
<h3>Work done by the person</h3>
The work done by the person is calculated as follows;
W = Fd
W = mgh
W = (89.2 x 9.8) x (0.162 x 133)
W = 18,834.65 J
W = 18.83465 kJ
<h3>Average power of the person</h3>
P = Fv
where;
v = (d)/t
v = (133 x 0.162)/(6 x 60 + 6)
v = (133 x 0.162)/(366)
v = 0.0588 m/s
P = (89.2 x 9.8) x 0.0588
P = 51.4 W
<h3>Amount of food calories burnt</h3>
4.1868 kJ = 1 Cal
18.83465 kJ = ?
= 4.5 Cal
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Answer:
v = 1.69 m/s
Explanation:
Given that,
Displacement of the student is 304 m due North and it takes 180 s.
We need to find the student's average velocity. Using formula of velocity.
Velocity = displacement/time
Hence, the student's average velocity is 1.69 m/s.
Answer:
Explanation:
a ) x ( t ) = t³ / 3 - t² + t
v = dx / dt = 3 t² / 3 - 2 t + 1 = t² - 2 t + 1
b ) lizard is at rest , v( t ) = 0
t² - 2 t + 1 = 0
( t - 1 )² = 0
t = 1
c )
velocity is positive when
t² - 2 t + 1 > 0
( t - 1 ) ² > 0
Here we see that LHS is a square so it is always positive whatever be the value of t
So velocity is always positive or lizard is always moving in positive x direction .
d ) It never moves in negative x direction .
e )
a ( t ) = dv / dt = 2t - 2
t = 1
so it has zero acceleration at t = 0 .
Answer:
Explanation:
Given that
Total race distance is 400m
Her initial velocity was 0m/s²
At the 100m mark, after she has travelled 100m, her final velocity was v=12m/s²
Using equation of motion
Let determine her constant acceleration
v²=u²+2as
12²=0²+2×a×100
144=0+200a
144=200a
a=144/200
a=0.72m/s²
Then we want to know her position after another 10second
So total time is 10+12=22seconds
Then, using equation of motion
Let determine his postion
S=ut+½at²
S=0•t+½×0.72×22²
S=0+174.24
S=174.24 m
Her position will be 174.24m
Answer:
3.62 m and - 1.4 m
Explanation:
Consider a location towards the positive side of x-axis beyond the location of charge Q₂
x = distance of the location from charge Q₂
d = distance between the two charges = 2 m
For the electric field to be zero at the location
E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location
x = 1.62 m
So location is 2 + 1.62 = 3.62 m
Consider a location towards the negative side of x-axis beyond the location of charge Q₁
x = distance of the location from charge Q₁
d = distance between the two charges = 2 m
For the electric field to be zero at the location
E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location
x = - 1.4 m
