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3 years ago

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A hiker walks 2.00 km north and then 3.00 km east, all in 2.50 hours. Calculate the magnitude and direction of the hiker’s (a) d

isplacement (in km) and (b) average velocity (in km/h) during those 2.50 hours. (c) What was her average

1 answer:

tino4ka555 [31]3 years ago

5 0

Answer:

Incomplete third question

I think it should be

C. What was her average speed

Explanation:

Check attachment for solution

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what is the potential energy of the ball when it gets to its maximum height just before falling back to the ground

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Answer:

I will say that the the potential energy will be at its maximum.

Explanation:

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2 years ago

A 645 g block is released from rest at height h0 above a vertical spring with spring constant k = 530 N/m and negligible mass. T

Nina [5.8K]

Answer:

a)5.88J

b)-5.88J

c)0.78m

d)0.24m

Explanation:

a) W by the block on spring is given by

W= $\frac{1}{2}$ kx² = $\frac{1}{2}$ (530)(0.149)² = 5.88 J

b) Workdone by the spring = - Workdone by the block = -5.88J

c) Taking x = 0 at the contact point we have U top = U bottom

So, mg $h_o$ = $\frac{1}{2}$ kx² - mgx

And, $h_o$ = ( $\frac{1}{2}$ kx² - mgx )/(mg) = $[\frac{1}{2} (530)(0.149^2)-(0.645)(9.8)(0.149)$ ]/(0.645x9.8)

$h_o$ = 0.78m

d) Now, if the initial initial height of block is 3 $h_o$

$h_o$ = 3 x 0.78 = 2.34m

then, $\frac{1}{2}$ kx² - mgx - mg $h_o$ =0

$\frac{1}{2}$ (530)x² - [(0.645)(9.8)x] - [(0.645)(9.8)(2.34) = 0

265x² - 6.321x - 14.8 = 0

a=265

b=-6.321

c=-14.8

By using quadratic eq. formula, we'll have the roots

x= 0.24 or x=-0.225

Considering only positive root:

x= 0.24m (maximum compression of the spring)

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2 years ago

The electromagnetic wave that delivers a cellular phone callto

spin [16.1K]

Answer:

$E=2.41\cdot 10^{-5} J$

Explanation:

The intensity of an electromagnetic wave can be expressed in terms of the magnetic field using the next relationship:

$I_{average}=\frac{cB_{0}^{2}}{2\mu_{0}}$ (1)

c is the speed of light (3*10⁸ m/s)
μ₀ is the permeability of free space (in vacuum ) (1.26*10⁻⁶ N/A²)
B₀ is the magnetic field

$I_{average}=\frac{3\cdot 10^{8}(1.5\cdot 10^{-10})^{2}}{2\cdot 1.26\cdot 10^{-6}}$

$I_{average}=2.68\cdot 10^{-6} W/m^{2}$

Now, let's define the relationship between power (P) and average intensity (I).

$I_{average}=\frac{P}{A}$

P is the power
A is the area crossed

So we can calculate the power.

$P=I_{average}\cdot A=2.68\cdot 10^{-6}\cdot 0.20=5.37\cdot 10^{-7} W$

Finally, energy is the product of P times time, so:

$E=P\cdot t=5.37\cdot 10^{-7} \cdot 45=2.41\cdot 10^{-5} J$

I hope it helps you!

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2 years ago