The cart travelled a distance of 14.4 m
Explanation:
The work done by a force when pushing an object is given by:
![W=Fd cos \theta](https://tex.z-dn.net/?f=W%3DFd%20cos%20%5Ctheta)
where:
F is the magnitude of the force
d is the displacement
is the angle between the direction of the force and the displacement
In this problem we have:
W = 157 J is the work done on the cart
F = 10.9 N is the magnitude of the force
, assuming the force is applied parallel to the motion of the cart
Therefore we can solve for d to find the distance travelled by the cart:
![d=\frac{W}{F cos \theta}=\frac{157}{(10.9)(cos 0)}=14.4 m](https://tex.z-dn.net/?f=d%3D%5Cfrac%7BW%7D%7BF%20cos%20%5Ctheta%7D%3D%5Cfrac%7B157%7D%7B%2810.9%29%28cos%200%29%7D%3D14.4%20m)
Learn more about work:
brainly.com/question/6763771
brainly.com/question/6443626
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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:
it have two answers a and c
Explanation:
please mark me as brainlyst
If it is stationary, its not moving. there is no movement