Poder = (resistencia) x (corrente)²
Poder = (10 ohms) x (5 A)²
<em>Poder = 250 watts </em>(250 Joule por segundo)
2 horas = 7,200 segundos
Energia = (250 joule/seg) x (7,200 seg)
<em>Energia = 1,800,000 Joules</em>
Answer
Assuming
east is the positive x direction
north is the positive y direction
initial velocity , u = 19 j m/s
a)
acceleration , a = 1.6 j m/s^2
Using first equation of motion
v = u + a × t
v = 19 + 5.6× 1.6
v = 28 j m/s
the velocity of the car after 5.6 s is 28 m/s north
b)
acceleration , a = -1.5 j m/s^2
Using first equation of motion
v = u + a × t
v = 19 - 5.6 ×1.5
v = 10.6 j m/s
the velocity of the car after 5.6 s is 10.6 m/s north
Explanation:
The buoyant force must be greater to float, otherwise it would sink, its like a barrel in water, the more water weight in it the more it sinks, the more air weight the more it rises.
Hope this helps a little
initial distance up = 2
initial velocity component up = 9 sin 60 = 7.79
v = 9 sin 60 - 9.8 t
when v = 0, we are there
9.8 t = 7.79
t = .795 seconds to top
h = 2 + 7.79(.795) - 4.9(.795^2)
Answer:
Explanation:
To find out the angular velocity of merry-go-round after person jumps on it , we shall apply law of conservation of ANGULAR momentum
I₁ ω₁ + I₂ ω₂ = ( I₁ + I₂ ) ω
I₁ is moment of inertia of disk , I₂ moment of inertia of running person , I is the moment of inertia of disk -man system , ω₁ and ω₂ are angular velocity of disc and man .
I₁ = 1/2 mr²
= .5 x 175 x 2.13²
= 396.97 kgm²
I₂ = m r²
= 55.4 x 2.13²
= 251.34 mgm²
ω₁ = .651 rev /s
= .651 x 2π rad /s
ω₂ = tangential velocity of man / radius of disc
= 3.51 / 2.13
= 1.65 rad/s
I₁ ω₁ + I₂ ω₂ = ( I₁ + I₂ ) ω
396.97 x .651 x 2π + 251.34 x 1.65 = ( 396.97 + 251.34 ) ω
ω = 3.14 rad /s
kinetic energy = 1/2 I ω²
= 3196 J
