Answer:
there yah go that's the answer
<span>The person is dragging
with a force of 58 lbs at an angle of 27 degrees relating to the ground. We
want to use cosine function to look for the horizontal force component. And
then we can compute for W = (Horizontal Force) x (Distance). We want the
horizontal force component since that is the component that is parallel to the
direction the cart is moving. </span><span>
(cos 27 degrees)(58 lbs) = 51.69 lbs (This is the horizontal
force component.)
W = (51.69 lbs) x (70 ft) = 3618.3 ft*lbs</span>
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
Answer:
The object will travel at the speed of 16 m/s.
Explanation:
Given
To determine
How fast is the object traveling?
<u>Important Tip:</u>
The product of the mass and velocity of an object — momentum.
Using the formula
where
Thus, in order to determine the speed of the object, all we need to do is to substitute p = 64 and m = 4 in the formula
switch the equation
divide both sides by 4
simplify
m/s
Therefore, the object will travel at the speed of 16 m/s.
If net external force acting on the system is zero, momentum is conserved. That means, initial and final momentum are same → total momentum of the system is zero.
