<span>The current is 6 miles per hour.
Let's create a few equations:
Traveling with the current:
(18 + c)*t = 16
Traveling against the current:
(18 - c)*t = 8
Let's multiply the 2nd equation by 2
(18 - c)*t*2 = 16
Now subtract the 1st equation from the equation we just doubled.
(18 - c)*t*2 = 16
(18 + c)*t = 16
(18 - c)*t*2 - (18 + c)*t = 0
Divide both sides by t
(18 - c)*2 - (18 + c) = 0
Now solve for c
(18 - c)*2 - (18 + c) = 0
36 - 2c - 18 - c = 0
36 - 2c - 18 - c = 0
18 - 3c = 0
18 = 3c
6 = c
So the current is 6 mph.
Let's verify that.
(18 + 6)*t = 16
24*t = 16
t = 16/24 = 2/3
(18 - 6)*t = 8
12*t = 8
t = 8/12 = 2/3
And it's verified.</span>
You get the net force acting on it ... the sum of the strengths and directions
of all the individual forces there may be.
He feels a 10 N to the left force moves. Yes ,he moves.
Answer:
t = 444.125 sec
Explanation:
Given data:
V = 24 volt
I = 0.1 ampere
mass of water mw = 51 gm
cr = 4.18 J/gm degree K^-1
mass of resistor = 8 gm
cr = 3.7 J/gm degree K^-1
we know that power is given as
Power P = VI
But P =E/t
so equating both side we have
solving for t
t = 444.125 sec
Answer:
K_a = 8,111 J
Explanation:
This is a collision exercise, let's define the system as formed by the two particles A and B, in this way the forces during the collision are internal and the moment is conserved
initial instant. Just before dropping the particles
p₀ = 0
final moment
p_f = m_a v_a + m_b v_b
p₀ = p_f
0 = m_a v_a + m_b v_b
tells us that
m_a = 8 m_b
0 = 8 m_b v_a + m_b v_b
v_b = - 8 v_a (1)
indicate that the transfer is complete, therefore the kinematic energy is conserved
starting point
Em₀ = K₀ = 73 J
final point. After separating the body
Em_f = K_f = ½ m_a v_a² + ½ m_b v_b²
K₀ = K_f
73 = ½ m_a (v_a² + v_b² / 8)
we substitute equation 1
73 = ½ m_a (v_a² + 8² v_a² / 8)
73 = ½ m_a (9 v_a²)
73/9 = ½ m_a (v_a²) = K_a
K_a = 8,111 J
