Answer:
Explanation:
How long does the football need to rise?
4.70/3 = 2.35 s
What height will the football reach?
h = ½(9.81)2.35² = 27.1 m
With what speed does the punter need to kick the football?
vy = g•t = 9.81(2.35) = 23.1 m/s
vx = d/t = 56.0/4.70 = 11.9 m/s
v = √(vx²+vy²) = 26.0 m/s
At what angle (θ), with the horizontal, does the punter need to kick the football?
θ = arctan(vy/vx) = 62.7°
Answer:
La Educación Ambiental es un proceso educativo continuo que persigue hacer sensible, formar y modificar actitudes de forma objetiva sobre la realidad global del medio, tanto natural como social, con este trabajo consideramos que la asignatura de Educación Física en sí,
Explanation:
The kayaker has velocity vector
<em>v</em> = (2.50 m/s) (cos(45º) <em>i</em> + sin(45º) <em>j</em> )
<em>v</em> ≈ (1.77 m/s) (<em>i</em> + <em>j</em> )
and the current has velocity vector
<em>w</em> = (1.25 m/s) (cos(315º) <em>i</em> + sin(315º) <em>j</em> )
<em>w</em> ≈ (0.884 m/s) (<em>i</em> - <em>j</em> )
The kayaker's total velocity is the sum of these:
<em>v</em> + <em>w</em> ≈ (2.65 m/s) <em>i</em> + (0.884 m/s) <em>j</em>
That is, the kayaker has a velocity of about ||<em>v</em> + <em>w</em>|| ≈ 2.80 m/s in a direction <em>θ</em> such that
tan(<em>θ</em>) = (0.884 m/s) / (2.65 m/s) → <em>θ</em> ≈ 18.4º
or about 18.4º north of east.
Answer:
12552 J or 3000 calories
Explanation:
Q = m × c × ∆T
Where;
Q = amount of heat energy (J)
m = mass of water (g)
c = specific heat capacity (4.184 J/g°C)
∆T = change in temperature
For 50mL of water, there are 50g, hence, m = 50g, c = 4.184 J/g°C, initial temperature = 0°C, final temperature = 60°C.
Q = m × c × ∆T
Q = 50 × 4.184 × (60 - 0)
Q = 209.2 × 60
Q = 12552 J
Hence, the amount of heat energy used to heat the water is 12552 J or 3000 calories
Answer:
Explanation:
Remark
The only thing that might trip you up is what to do with the angle. The vertical component of the 15 degrees does no work against anything. So the 15 degrees limits the horizontal force.
The formula is
Work = F * d * cos(15)
The givens are
F = 2000 N
d = 30 m
Cos(15) = 0.9659
Solution
Work = 2000 * 30 * cos(15)
Work = 57,955
Rounded to two places would be 5.8 * 10^4
C
