Answer:
KE = 1/2mv^2
KE = 1/2(24)(3^2)
KE = 12(9)
KE = 108 J
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Answer:
W = 112.58 N-unit
Explanation:
Given:
- Force F = 10 N
- Angle Q of force with x axis: 30 degrees
- distance to be moved d = 13 units along + x axis
Find:
Work Done by the force F:
Solution:
The work by force in positive x direction can only be done if the both the direction of distance traveled and direction of force are parallel vectors. Hence we compute the component of Force F in x direction F_x:
F_x = F*cos(Q)
F_x = 10*cos(30)
F_x = 8.66 N
Hence,
Work Done by force
W = F_x * d
W = 8.66 * 13
W = 112.58 N-unit
Answer: the mass of the second ball is 2.631 kg
Explanation:
Given that;
m1 = 0.877 kg
Initial velocity = V0
Initial momentum = m1 × V0
final velocity of m1 is u1, final velocity of m2 is u2 = v0/2
now final momentum = m1 × u1 + m2 × u2
using momentum conservation;
m1×V0 = m1×u1 + m2×v0/2
m1×(v0 - u1) = m2×V0/2 ----- let this be equation 1
Now, for elastic collision;
m1×v0²/2 = m1×u1²/2 + m2×(v0/2)²/2
m1×(v0² - u1²) = m2×(v0/2)² --------- let this be equation 2
now; equation 2 / equation 1
: V0 + u1 = v0/2
2V0 + 2u1 = V0
2u1 = V0 - 2V0
u1 = -V0/2
now we insert in equ 1
m1×3V0/2= m2×V0/2
m1 × 3 = m2
m2 = 0.877 × 3
m2 = 2.631 kg
Therefore, the mass of the second ball is 2.631 kg
Between g and h the object is not moving.