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avanturin [10]
2 years ago
11

-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?
Physics
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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La altura de un tornillo de banco respecto a la superficie es de 80 cm expresar dicha medida en pies..
Andrej [43]

Answer:

this measurement if feet is: 2.624672 ft

Explanation:

Notice that 80 cm can be expressed as 0.8 meters, and In order to convert from meters to feet, one needs to multiply the meter measurement times 3.28084. Therefore:

0.80 m can be written in feet as: 0.80 * 3.28084 feet = 2.624672 feet

3 0
2 years ago
You are provided with three polarizers with filters making angles of (A) 90 ​∘ ​​ , (B) 180 ​∘ ​​ and (C) −45 ​∘ ​​ with respect
irinina [24]

Answer:

Order of maximum transmission of the polarizer is A, C and B.

Solution:

As per the question:

For the first polarizer, the angle is quite insignificant:

(A) 90^{\circ}:

The light intensity after passing through the first polarizer is I_{o} and this intensity does not depend on the angle of the polarizer.

Consider 90^{\circ} with the vertical, the intensity is given by:

I = I_{o}cos^{2}90^{\circ}

I = I_{o}cos(2(45^{\circ})) = I_{o}(\frac{1+cos90^{\circ})}{2} = \frac{I_{o}}{2}

(B) 180^{\circ}:

Suppose the second polarizer is  45^{\circ} with the vertical.

Now, intensity through the second polarizer:

I' = Icos^{2}(\theta_{2} - \theta_{1}) = \frac{I_{o}}{2}cos^{2}(- 45 - 90)

I' =  \frac{I_{o}}{2}cos^{2}135^{\circ} = \frac{I_{o}}{4}

Now, if we consider the second polarizer to be 180^{\circ},

I' = \frac{I_{o}}{2}cos^{2}180^{\circ} = \frac{I_{o}}{2}cos^{2}(180^{\circ} - 90^{\circ}) = 0

(C) - 45^{\circ}:

Now,

Intensity through the third polarizer, if it is 180^{\circ} with the vertical:

I' = Icos^{2}(\theta_{2} - \theta_{1}) = \frac{I_{o}}{2}cos^{2}(180 - (- 45))

I' = \frac{I_{o}}{8}

5 0
2 years ago
A man is dragging a trunk up the loading ramp of a mover’s truck. The ramp has a slope angle of 20.0°, and the man pulls upward
AleksandrR [38]

Answer:

(a)  104 N

(b) 52 N

Explanation:

Given Data

Angle of inclination of the ramp: 20°

F makes an angle of 30° with the ramp

The component of F parallel to the ramp is Fx = 90 N.  

The component of F perpendicular to the ramp is Fy.

(a)  

Let the +x-direction be up the incline and the +y-direction by the perpendicular to the surface of the incline.  

Resolve F into its x-component from Pythagorean theorem:  

Fx=Fcos30°

Solve for F:  

F= Fx/cos30°  

Substitute for Fx from given data:  

Fx=90 N/cos30°

   =104 N

(b) Resolve r into its y-component from Pythagorean theorem:

     Fy = Fsin 30°

   Substitute for F from part (a):

     Fy = (104 N) (sin 30°)  

          = 52 N  

5 0
2 years ago
The magnetic field at the earth's surface can vary in response to solar activity. During one intense solar storm, the vertical c
Flauer [41]

Answer:

EMF = 33880 Volts

Explanation:

As per Faraday's law of Electromagnetic induction we know that

Rate of change in magnetic flux will induce EMF in the closed conducting loop

so we have

\phi = B.A

now we have

A = (110 \times 10^3)(110 \times 10^3)

A = 1.21 \times 10^{10}

now we have

\phi = B(1.21 \times 10^{10})

now the induced EMF through this loop is given as

EMF = (\frac{dB}{dt})(1.21 \times 10^{10})

EMF = (2.8 \times 10^{-6})(1.21 \times 10^{10})

EMF = 33880 Volts

5 0
2 years ago
The 8 kg block is then released and accelerates to the right, toward the 2 kg block. The surface is rough and the coefficient of
natita [175]

Answer:

3.258 m/s

Explanation:

k = Spring constant = 263 N/m (Assumed, as it is not given)

x = Displacement of spring = 0.7 m (Assumed, as it is not given)

\mu = Coefficient of friction = 0.4

Energy stored in spring is given by

U=\dfrac{1}{2}kx^2\\\Rightarrow U=\dfrac{1}{2}\times 263\times 0.7^2\\\Rightarrow U=64.435\ J

As the energy in the system is conserved we have

\dfrac{1}{2}mv^2=U-\mu mgx\\\Rightarrow v=\sqrt{2\dfrac{U-\mu mgx}{m}}\\\Rightarrow v=\sqrt{2\dfrac{64.435-0.4\times 8\times 9.81\times 0.7}{8}}\\\Rightarrow v=3.258\ m/s

The speed of the 8 kg block just before collision is 3.258 m/s

7 0
2 years ago
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