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Sergio039 [100]
2 years ago
10

If you used 1000 J of energy to throw a ball, would it travel faster if you threw the ball (ignoring air resistance)

Physics
1 answer:
wolverine [178]2 years ago
6 0

To solve this problem it is necessary to apply the kinematic equations of Energy for which the rotation of a circular body is described as

KE = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2

Where,

m = Mass of the Vall

v = Velocity

I = Moment of inertia abouts its centre of mass

\omega = Angular speed

Basically the two sums of energies is the consideration of translational and rotational kinetic energy.

a. so that it was also rotating?

The ball is rotating means that it has some angular speed:

KE = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2

1000J = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2

When there is a little angular energy (and not linear energy to travel faster), translational energy will be greater than the 1000J applied.

1000J > \frac{1}{2}mv^2

The ball will not go faster.

c. so that it wasn't rotating?

For the case where the angular velocity does not rotate it is zero therefore

KE = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2

1000J = \frac{1}{2}mv^2+\frac{1}{2}I(0)^2

1000J = \frac{1}{2}mv^2

All energy is transoformed into translational energy so it is possible to go faster. This option is CORRECT.

b. It makes no difference.

Although the order presented is different, I left this last option because as we can see with the previous two parts if there is an affectation regarding angular movement, therefore it is not correct.

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Answer:

Minimum capacitance = 200 μF

Explanation:

From image B attached, we can calculate the current flowing through the capacitors.

Thus;

Since V=IR; I = V/R = 5/500 = 0.01 A

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So minimum capacitance will be determined from;

i(t) = C(dv/dt)

So, C = i(t)(Δt/Δv) = 0.01[0.001/0.05]

C = 0.01 x 0.0002 = 200 x 10^(-6) F = 200 μF

8 0
2 years ago
The eiffel tower has a mass of 7.3 million kilograms and a height of 324 meters. its base is square with a side length of 125 me
uranmaximum [27]

Since the tower base is square with a side length of  125 m,

Therefore,

(125\ m)^2+ (125\ m)^2=31250 m^2

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The volume of cylinder,

=\pi r^2h=3.14(88.38834765 m)^2\times 324 m =7948168.803\ m^3

Thus, the mass of the air in the cylinder,

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Hence, the mass of the air in the cylinder is this more  than the mass of the tower.


4 0
2 years ago
A suspicious-looking man runs as fast as he can along a moving sidewalk from one end to the other, taking 2.20 s. Then security
Brilliant_brown [7]

Answer: The ratio is 1.54

Explanation:

Firs, we have to find the relative speed of the man moving forward and backward.

Forward:

vf = vs + vm

vm = the man's speed

vs = the sidewalk's speed

vf = relative velocity moving forward

Because we don't know how much the man moved,

vf = distance (meters) / time (seconds)

vf = x / 2.20s

Backward:

vb = -vm + vs

vb = relative velocity moving backward

vf = distance (meters) / time (seconds)

vf = -x / 10.30s

We now divide the relative speeds

vf / vb = (x / 2.20) / (-x / 10.30)

We cancel the x

vf / vb = -10.3s / 2.2s = -4.68

vf = -4.68 . vb

We now substitute this in the equation we used for the forward travel

-4.68vb = vm + vs

Subtracting this from the backward travel equation

vb - (-4.68vb) = -vm - vm + vs -vs

5.68vb = -2vm

vb = -2vm / 5.68

Now, adding to the backward travel equation

vb + (-4.68vb) = -vm + vm + vs + vs

-3.68vb = 2vs

Using the two resulting equations

-3.68 . (-2 / 5.68) vm = 2vs

7.36 / 5.68 vm = 2vs

vm / vs = 1.54

7 0
2 years ago
The first and second coils have the same length, and the third and fourth coils have the same length. They differ only in the cr
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Answer:

\frac{R_2}{R_1}=\frac{A_1}{A_2}\\\frac{R_4}{R_3}=\frac{A_3}{A_4}

Explanation:

The resistance of a conductor is directly proportional to its length and is inversely proportional to its cross-sectional area, this dependence is given by:

R=\frac{\rho L}{A}

\rho is the material's resistance, L is the legth and A is the cross-sectional area.

For the first and second coils, we have:

R_1=\frac{\rho L}{A_1}\\R_2=\frac{\rho L}{A_2}\\\rho L=R_1A_1\\\rho L=R_2A_2\\R_1A_1=R_2A_2\\\frac{R_2}{R_1}=\frac{A_1}{A_2}

For the third and fourth coils, we have:

R_3=\frac{\rho L'}{A_3}\\R_4=\frac{\rho L'}{A_4}\\\rho L'=R_3A_3\\\rho L'=R_4A_4\\R_3A_3=R_4A_4\\\frac{R_4}{R_3}=\frac{A_3}{A_4}

6 0
2 years ago
You are riding on a roller coaster that starts from rest at a height of 25.0 m and moves along a frictionless track. however, af
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I attached the missing picture.
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A: mgh_1+\frac{mv_1^2}{2}
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W_f=F_f\cdot L
So, we know that the car is approaching the point B with the following amount of energy:
mgh_1+\frac{mv_1^2}{2}- F_fL
The law of conservation of energy tells us that this energy must the same as the energy at point B. 
The energy at point B is the sum of car's kinetic and potential energy:
B: mgh_2+\frac{mv_2}{2}
As said before this energy must be the same as the energy of a car approaching the loop:
mgh_2+\frac{mv_2}{2}=mgh_1+\frac{mv_1^2}{2}- F_fL
Now we solve the equation for v_1:
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v_1^2=39.23\\
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4 0
2 years ago
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