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2 years ago

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-A coconut falls out of a tree 12.0 m above the ground and hits a bystander 3.00 m tall on the top of the head. It bounces back

up 1.50 m before falling to the ground. If the mass of the coconut is 2.00 kg, calculate the Eg of the coconut relative to the ground at each of the following sites: Assume no loss due to friction or air resistance.
a.Eg while it is still in the tree
b.Eg when it hits the bystander on the head
c.Eg when it bounces up to its maximum height
d. At what point will it have the most Ek?

1 answer:

Llana [10]2 years ago

3 0

Explanation:

Eg = mgh

a. Eg = (2.00 kg) (9.8 m/s²) (12.0 m)

Eg = 235 J

b. Eg = (2.00 kg) (9.8 m/s²) (3.00 m)

Eg = 58.8 J

c. Eg = (2.00 kg) (9.8 m/s²) (4.50 m)

Eg = 88.2 J

d. When the coconut hits the bystander:

Ek = 235 J − 58.8 J = 176 J

When the coconut hits the ground:

Ek = 88.2 J − 0 J = 88.2 J

Ek is the greatest when the coconut hits the bystander.

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2 years ago

The drawing shows a person (weight W = 588 N, L1 = 0.838 m, L2 = 0.398 m) doing push-ups. Find the normal force exerted by the f

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Complete Question

The complete question is shown on the first uploaded image

Answer:

Force on each hand is 196.22 N

Force on each foot is 95.8 N

Explanation:

In order to get a better understanding of this question let us explain some concepts

Normal Force:

We can define normal force Fn as that type of force which makes a 90 degree angle with the surface on which it is exerted.

Torque:

We can define torque as the moment of forces that tends to produce or cause rotation

From the question we are given that

Weight of body is (W) = 584 N

The normal force on both hands (Ha) = ?

The normal force on both legs (Lg) = ?

Looking at the diagram the person is at equilibrium so

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an also this mean that torques acting on the body is balanced

So, 0.410 Ha = 0.840 Lg

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Considering the first equation and replacing Lg with this recent equation we have

584 = Ha + 0.4881 Ha

Therefore Ha = 392.44 N

This value obtained is for both hands for each hand we divide by 2

Therefore we have for each hand $= 392.44/2$ =196.55 N

Since we have been able to get the force on both hands we can substitute it in to the equation where we made Lg the subject of formula and we have

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2 years ago

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Answer:

Explanation:

The mass flow rate of benzene from the leak in the pipeline containing benzene is:

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Here, $Q_m$ is the mass flow rate through the leak of the pipeline. $A$ is the area of the hole, $C_o$ is the discharge rate, $\rho$ is the fluid density, $g_c$ is the gravitational constant and $P_g$ is the constant gauge pressure within the process unit.

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The specific gravity of benzene is 0.8794. Specific gravity is the ratio of th density of a substance to the density of a reference substance.

Specific gravity of benzene = density of benzenee/denity of reference substance

Rewrite the expression in terms of density of benzene.

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Take the reference substance as water. Density of water is $62.4\frac{Ib_m}{ft^3}$ . Calculate density of benzene.

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Upstream x distance from upstream pressure end

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Calculate the distance from the downstream pressure end. The distance from upstream pressure end is 43 ft. Total of the pipe is 100 ft.

Distance from the downstream pressure end = Total length of the pipe - Distance from the upstream pressure end

The distance from upstream pressure end is 43 ft. Total length of the pipe is 100 ft. Substitute the values in the equation.

Distance from the downstream pressure end = Total length of the pipe - Distance from the upstream pressure end

= 100ft - 43ft = 57 ft

Substitute 50 psig for upstream, 43 ft fr distance from the upstream pressure end, 40 psig for downstream pressure, 57 ft for distance from the downstream pressure end, and 100 ft for the total length of the horizontal pipeline and calculate $P_g$ .

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2 years ago