Answer:
Explanation:
Newton's second law of motion is F = ma, or force is equal to mass times acceleration.
( can u give me brainlist plz)
The correct answer is <span>3)
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In fact, the total energy of the rock when it <span>leaves the thrower's hand is the sum of the gravitational potential energy U and of the initial kinetic energy K:
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<span>As the rock falls down, its height h from the ground decreases, eventually reaching zero just before hitting the ground. This means that U, the potential energy just before hitting the ground, is zero, and the total final energy is just kinetic energy:
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But for the law of conservation of energy, the total final energy must be equal to the tinitial energy, so E is always the same. Therefore, the final kinetic energy must be
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Refer to the diagram shown below.
The angle is in quadrant 4, therefore cos θ is positive.
The length of the hypotenuse is determined from the Pythagorean theorem as
c² = 3² + 4² = 25
c = 5
Calculate cosine of the reference angle.
cos θ = 3/5
Answer: 3/5
Answer:
0.56 m/s
Explanation:
The speed of the head at the end of the interval in each case is the area under the acceleration curve. Then the difference in speeds is the difference in areas.
We can find the area geometrically, using formulas for the area of a triangle and of a trapezoid.
A = 1/2bh . . . . area of a triangle
A = 1/2(b1 +b2)h . . . . area of a trapezoid
If h(t) is the acceleration at time t for a helmeted head, the area under that curve will be (in units of mm/s) ...
Vh = 1/2(h(3)·3) +1/2(h(3) +h(4))·1 +1/2(h(4) +h(6))·2 +1/2(h(6))·1
Vh = 1/2(4h(3) +3h(4) +3h(6)) = 1/2(4·40 +3·40 +3·80) = 260 . . . mm/s
If b(t) is the acceleration for a bare head, the area under that curve in the same units is ...
Vb = 1/2(b(2)·2 +1/2(b(2) +b(4))·2 +1/2(b(4) +b(6))·2 +1/2(b(6)·1)
Vb = 1/2(4b(2) +4b(4) +3b(6)) = 1/2(4·120 +4·140 +3·200) = 820 . . . mm/s
Then the difference in speed between the bare head and the helmeted head is ... (0.820 -0.260) m/s = 0.560 m/s