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Igoryamba
4 years ago
11

A 0.100-kilogram apple falls from a height of 1.50 meters to 1.00 meter. Ignoring frictional effects, the total mechanical energ

y of the apple at this height is _____. 0.00 J less than 1.47 J but more than 0.00 J exactly 1.47 J more than 1.47 J
Physics
2 answers:
d1i1m1o1n [39]4 years ago
5 0

Answer:

less than 1.47 J but more than 0.00 J

Explanation:

VashaNatasha [74]4 years ago
3 0
The total mechanical energy is the sum of the kinetic energy and the potential energy. It is easier to solve for the potential energy in this case because we are given with the position of the apple. 
                               PE = mgh
Substituting the mass, acceleration due to gravity and the height,
                              PE = (0.1 kg)(9.8 m/s²)(1 m) = 0.98 J
Then, we solve for the velocity at that point. The distance it had traveled is 0.5 meters.
                                      d = v²/g ;    0.5 = v²/9.8 m/s²
                                                       v = 2.213 m/s
Then, solving for KE
                                                KE = 0.5mv²
Substituting,
                                                KE = 0.5(0.1 kg)(2.213 m/s)² = 0.245 J
Adding the two mechanical forces will give us an answer of 1.225 J. Thus, the answer is the first choice. 
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fredd [130]

Answer:

F = 2349.6 N

Explanation:

We can solve this exercise using the relationship of momentum and momentum

         I = Δp

         I = F t

As the woman accelerates at a distance of 29.1 m to go from rest to 56.8 m / s, we can use the kinematics to find the acceleration

       v² = v₀² + 2 a x

       v₀ = 0

       a = v / 2x

       a = 56.8 2/2 29.1

       a = 55.43 m / s²

Let's look for the time you need to get this speed

      v = v₀ + a t

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      t = 56.8 / 55.43

      t = 1,025 s

Let's clear the average force momentum from the momentum

      F t = m v- m v₀

      F = mv / t

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3 0
3 years ago
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7 0
4 years ago
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Find the position vector of a particle that has the given acceleration and the specified initial velocity and position. a(t) = 1
kondor19780726 [428]

Answer:

Explanation:

Given

Acceleration a(t)=14t\hat{i]+\sin (t)\hat{j}+\cos (2t)\hat{k}[/tex]

and v(0)=\hat{i}

r(0)=\hat{j}

we know a=\frac{\mathrm{d} v}{\mathrm{d} t}

\int dv=\int adt

v(t)=\int (14t\hat{i}+\sin (t)\hat{j}+\cos (2t)\hat{k})dt

v(t)=7t^2\hat{i}-\cos t\hat{j}+\frac{\sin (2t)\hat{k}}{2}+c

at t=0

v(0)=0-1\cdot \hat{j}+0+c

c=\hat{i}+\hat{j}

v(t)=(7t^2+1)\hat{i}+(1-\cos t)\hat{j}+\frac{\sin (2t)\hat{k}}{2}

and \frac{\mathrm{d} r}{\mathrm{d} t}=v(t)

\int dr=\int vdt

r(t)=\int ((7t^2+1)\hat{i}+(1-\cos t)\hat{j}+\frac{\sin (2t)\hat{k}}{2})dt

r(t)=(\frac{7}{2}t^3+t)\hat{i}+(t-\sin (t))\hat{j}+\frac{1}{2}\times (-\frac{1}{2}\cos 2t)\hat{k}+c_2

at t=0

r(0)=\hat{j}

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4 0
3 years ago
While the negatively charged rod is near the disk without touching it, a hand briefly touches the end of the post. Then the nega
Paraphin [41]

Answer:

that initially the weather vane was at rest, by this load that remained on the pole it would begin to move.

Explanation:

Let us carefully analyze the situation, when the bar is facing the index post a load of equal magnitude, but opposite sign on its surface, these two charges are in balance; When the hand touches the pole, it creates a path to the ground where the charges that were induced on the pole can be balanced with the charge coming from the ground, leaving a zero charge on the pole.

 

   Now if the hand is removed, there can be no exchange of charges with the earth. When the bar is removed, the induced loads are redistributed in the post, but the excess loads that came from the earth that have the same value and are of a sign opposite to the induced ones remain, you want to sign that they are of the same sign as the charges of the bar.

   In summary, after the process, the post has a load of equal magnitude and sign (negative) that of the bar.

   If we assume that initially the weather vane was at rest, by this load that remained on the pole it would begin to move.

4 0
3 years ago
How much force is needed to accelerates at a rate of 12.5 m/s after a force of 24N is applied ?
Artist 52 [7]

Answer:

F = ma so 12.5 * 24 is answer

Explanation:

4 0
4 years ago
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