C.) Mitochondria is the answer...
The net force on the barge is 8000 N
Explanation:
In order to find the net force on the badge, we have to use the rules of vector addition, since force is a vector quantity.
In this problem, we have two forces:
- The force of tugboat A,
, acting in a certain direction - The force of tugboat B,
, also acting in the same direction
Since the two forces act in the same direction, this means that we can simply add their magnitudes to find the net combined force on the barge. Therefore, we get
and the direction is the same as the direction of the two forces.
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Answer:
=3.5 m/s
Explanation:
y = x tanθ - 1/2 g x² / (u²cos²θ )
y = 0.25 , x = 0.5, θ = 40°
.25 = .50 tan40 - .5 x 9.8x x²/ u²cos²40
.25 = .42 - 2.0875/u²
u = 3.5 m / s.
Answer:
32000 N
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 40 m/s
Distance (s) = 10 m
Final velocity (v) = 0 m/s
Mass (m) of car = 400 Kg
Force (F) =?
Next, we shall determine the acceleration of the the car. This can be obtained as follow:
Initial velocity (u) = 40 m/s
Distance (s) = 10 m
Final velocity (v) = 0 m/s
Acceleration (a) =?
v² = u² + 2as
0² = 40² + (2 × a × 10)
0 = 1600 + 20a
Collect like terms
0 – 1600 = 20a
–1600 = 20a
Divide both side by –1600
a = –1600 / 20
a = –80 m/s²
The negative sign indicate that the car is decelerating i.e coming to rest.
Finally, we shall determine the force needed to stop the car. This can be obtained as follow:
Mass (m) of car = 400 Kg
Acceleration (a) = –80 m/s²
Force (F) =?
F = ma
F = 400 × –80
F = – 32000 N
NOTE: The negative sign indicate that the force is in opposite direction to the motion of the car.
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
