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2 years ago

Hii I need help ASAP with this physics question

1 answer:

4 0

Between the top of the first and the top of the second loop, the coaster has lost potential energy = mgh, where h = 22.2 - 15 = 7.2m

This energy would have converted to Kinetic. Write out an equation and the masses will cancel out. Does that hint help you to find the solution? If not, I will give you another hint.

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A TV satellite dish is designed to receive radio waves of wavelength

Verizon [17]

Answer:

4.7 GHz

Explanation:

Applying,

v = λf................. Equation 1

Where v = velocity of the radio wave, λ = wavelength, f = frequency

make f the subject of the equation

f = v/λ.............. Equation 2

Note: A radio wave is an electromagnetic wave, as such it moves with a velocity of 3.00 x 10⁸ m/s

From the question,

Given: λ = 0.0644 meters

Constant: v = <em>3.00 x 10⁸ m/s</em>

Substitute these values into equation 2

f = (3.00 x 10⁸)/0.0644

f = 4.66×10⁹ Hz

f = 4.7 GHz

3 0

2 years ago

You’ve made the finals of the science Olympics. As one of your tasks you’re given 1.0 g of copper and asked to make a cylindrica

Pani-rosa [81]

Answer:

Length = 2.92 m

Diameter = 0.11 mm

Explanation:

We have $m = dl D \ \ \& \ \ \ R = \frac{\rho l}{A}$ , where:

$l$ is the length

$m = 1.0 g = 1 \times 10^{-3} \ kg\\R = 1.3 \ \Omega\\\rho = 1.7 \times 10^{-8} \Omega m\\d = 8.96 \ g/cm^3 = 8960 kg/m^3$

We divide the first equation by the second equation to get:

$\frac{m}{R} = \frac{d A^2}{\rho}$

$A^2 = \frac{m \rho}{dR} \\\\A^2 = \frac { 1 \times 10^{-3} \times 1.7 \times 10^{-8}}{8960 \times 1.3}\\\\A^2 = 1.5 \times 10^{-15}\\\\ A= 3.8 \times 10^{-8} \ m^2$

Using this Area, we find the diameter of the wire:

$D = \sqrt{\frac{4A}{\pi}}$

$D = \sqrt{\frac{4 \times 3.8 \times 10^{-8} }{\pi}}$

$D = 0.00011 \ m = 1.1 \times 10^ {-4} = 0.11 \ mm$

To find the length, we multiply the two equations stated initially:

$mR = d\rho l^2\\\\l^2 = \frac{mR}{d\rho} \\\l^2 = \frac {1.0 \times 10^{-3} \times 1.3}{8960 \times 1.7\times 10^{-8}}$

$l^2 = 8.534\\l = 2.92 \ m$

8 0

2 years ago

Read 2 more answers

Assume that the position vector of A is r=i+j+k . Determine the moment about the origin O if the force F=(1)i+(0)j+(5)k . The mo

ddd [48]

Answer:

M₀ = 5i - 4j - k

Explanation:

Using the cross product method, the moment vector(M₀) of a force (F) is about a given point is equal to cross product of the vector A from the point (r) to anywhere on the line of action of the force itself. i.e

M₀ = r x F

From the question,

r = i + j + k

F = 1i + 0j + 5k

Therefore,

M₀ = (i + j + k) x (1i + 0j + 5k)

M₀ = $\left[\begin{array}{ccc}i&j&k\\1&1&1\\1&0&5\end{array}\right]$

M₀ = i(5 - 0) -j(5 - 1) + k(0 - 1)

M₀ = i(5) - j(4) + k(-1)

M₀ = 5i - 4j - k

Therefore, the moment about the origin O of the force F is

M₀ = 5i - 4j - k

3 0

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)

4 0

2 years ago

An electron is traveling with initial kinetic energy K in a uniform electric field. The electron comes to rest momentarily after

KatRina [158]

Answer:

$E = kQs/r.r$

Explanation:

Where E is the magnitude of electric field...

k is called Columb's Constant. It has a value of 8.99 x 109 N m2/C2.

Qs is the magnitude of the source charge...

and r is the magnitude of distance between source and target...

(When electron comes to rest Δt the magnitude of Electric field E become zero momentarily but later achieves the maximum value...)

5 0

2 years ago