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The potential energy of the block is given by:
V = m*g*h
m mass
g = 9.81m/s²
h height
The potential energy of a spring is given by:
V = 0.5 * k * x²
k spring constant
x compression of the spring
If the block starts from rest it has potential energy, but no kinetic energy. As it slides down the incline potential energy is converted into kinetic energy. When the block hits the spring the kinetic energy is converted into spring's potential energy. If the spring is fully compressed and the block is at rest again, the block has transferred all its energy into the spring. No energy is lost. So we can write:
m * g * h = 0.5 * k * x²
m = 0.5 kg
g = 9.81 m/s²
h = 2.5m * sin 37° = 1,5 m
x = 0,6 m
Solve for k.
k = 2 * m * g * h / x² = 40.8 N/m
Speed = Distance ÷ Time
Speed = 116 ÷ 29
Speed = 4

Hi pupil Here's Your answer :::
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Student's justification is not correct. Two equal and opposite force cancel each other if the act on the same body. According to the third law of motion action and reaction forces are equal and opposite but they both act on different bodies. Hence, they cannot cancel each other.
When we push a message track, then the applied force on the truck is not sufficient to overcome the force of friction between the tyres of truck and ground, hence, truck does not move.
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Hope this helps .
Answer:
Answered
Explanation:
a) What is the work done on the oven by the force F?
W = F * x
W = 120 N * (14.0 cos(37))
<<<< (x component)
W = 1341.71
b) 

= 29.4 N


W_f= 328.72 J = 329 J
c) increase in the internal energy
U_2 = mgh
= 12*9.81*14sin(37)
= 991 J
d) the increase in oven's kinetic energy
U_1 + K_1 + W_other = U_2 + K_2
0 + 0 + (W_F - W_f ) = U_2 + K_2
1341.71 J - 329 J - 991 J = K_2
K_2 = 21.71 J
e) F - F_f = ma
(120N - 29.4N ) / 12.0kg = a
a = 7.55m/s^2
vf^2 = v0^2 + 2ax
vf^2 = 2(7.55m/s)(14.0m)
V_f = 14.5396m/s
K = 1/2(mv^2)
K = 1/2(12.0kg)(14.5396m/s)
K = 87.238J