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geniusboy [140]

3 years ago

5

A 26-foot ladder is placed against a wall. If the top of the ladder is sliding down the wall at -2 feet per second (note that th

e rate is negative because the height is decreasing). At what rate is the bottom of the ladder moving away from the wall when the bottom of the ladder is 10 feet away from the wall?

1 answer:

RSB [31]3 years ago

6 0

Answer:

Dx/dt = 4,8 f/s

Explanation:

The ladder placed against a wall, and the ground formed a right triangle

with x and h the legs and L the hypothenuse

Then

L² = x² + h² (1)

L = 26 f

Taking differentials on both sides of the equation we get

0 = 2x Dx/dt + 2h Dh/dt (1)

In this equation

x = 10 distance between the bottom of the ladder and the when we need to find, the rate of the ladder moving away from the wall

Dx/dt is the rate we are looking for

h = ? The height of the ladder when x = 10

As L² = x² + h²

h² = L² - x²

h² = (26)² - (10)²

h² = 676 - 100

h² = 576

h = 24 f

Then equation (1)

0 = 2x Dx/dt + 2h Dh/dt

2xDx/dt = - 2h Dh/dt

10 Dx/dt = - 24 ( -2 ) ( Note the movement of the ladder is downwards)

Dx/dt = 48/10

Dx/dt = 4,8 f/s

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tester [92]

Radius of nuclei is given by formula

$R = R_oA^{1/3}$

now we can say volume of the nuclei is given as

$V = \frac{4}{3}\pi R_o^3* A$

now the density is given as

density = mass / volume

mass of nuclei = mass of neutron + mass of protons

$m = z*m_p + (A- z)*m_n$

$m_p = m_n = 1.008u$

$m = A*1.008u$

Now density is given as

$\rho = \frac{A*1.008u}{\frac{4}{3}\pi R_0^3* A}$

here we know that

$R_0$ = 1.2 fm

$\rho = \frac{1.008u}{\frac{4}{3}\pi*(1.2*10^{-15})^3}$

$\rho = 2.31 * 10^{17} kg/m^3$

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5 0

3 years ago

A circular coil 14.0 cm in diameter and containing nine loops lies flat on the ground. The Earth's magnetic field at this locati

sp2606 [1]

Answer:

a) $T = 2.9*10^{-5} N-m$

b) north edge will rise up

Explanation:

torque on the coil is given as

$T = NIABsin\theta$

where N is number of loop = 9 loops

i is current = 7.80 A

-B -earth magnetic field = $5.00*10^{-5} T$

A- area of circular coil

$A = \frac{\pi d^{2}}{4}$

$A =\frac{\pi .14^{2}}{4}$

A =0.015 m2

PUTITNG ALL VALUE TO GET TORQUE

$T = 9*7.8*0.015*5*10^{-5} sin{90-56}$

$T = 2.9*10^{-5} N-m$

b) north edge will rise up

3 0

3 years ago

Read 2 more answers

a ball is thrown straight up into the air with a speed of 13 m/s. if the ball has a mass of 0.25 kg, how high does the ball go?

evablogger [386]

<h2>Hello!</h2>

The answer is: 8.62m

<h2>Why?</h2>

There are involved two types of mechanical energy: kinetic energy and potential energy, in two different moments.

<h2>First moment:</h2>

Before the ball is thrown, where the potential energy is 0.

<h2>Second moment: </h2>

After the ball is thrown, at its maximum height, the Kinetic Energy turns to 0 (since at maximum height,the speed is equal to 0) and the PE turns to its max value.

Therefore,

$E=PE+KE$

Where:

$PE=m.g.h$

$KE=\frac{1*m*v^{2}}{2}$

<em>E</em> is the total energy

<em>PE</em> is the potential energy

<em>KE</em> is the kinetic energy

<em>m</em> is the mass of the object

<em>g</em> is the gravitational acceleration

<em>h </em>is the reached height of the object

<em>v</em> is the velocity of the object

Since the total energy is always constant, according to the Law of Conservation of Energy, we can write the following equation:

$KE_{1}+PE_{1}=KE_{2}+PE_{2}$

Remember, at the first moment the PE is equal to 0 since there is not height, and at the second moment, the KE is equal to 0 since the velocity at maximum height is 0.

$\frac{1*m*v^{2}}{2}+m.g.(0)=\frac{1*m*0^{2}}{2}+m.g.h\\\frac{1*m*v_{1} ^{2}}{2}=m*g*h_{2}$

So,

$h_{2}=\frac{1*m*v_{1} ^{2}}{2*m*g}\\h_{2}=\frac{1*v_{1} ^{2}}{2g}=\frac{(\frac{13m}{s})^{2} }{2*\frac{9.8m}{s^{2}}}\\h_{2}=8.62m}$

Hence,

The height at the second moment (maximum height) is 8.62m

Have a nice day!

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3 years ago

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zlopas [31]

Answer:

Explanation:

we know that

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given data

v=3*10^8

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1 minute=60 seconds

so

10 minutes =10*60/1 =600 seconds

now putting the value of v and t we can find the value of s

s=vt

s=3*10^8*600

s=1.8*10^11m

i hope this will help you

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3 years ago

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katrin [286]

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There are several distinct sorts of fault lines, and they are called based on the fault's movement and the fault plane's own direction. These faults may be found all over the world, but the most active and earthquake-producing ones are in the Pacific Ocean's Ring of Fire region.

When you gaze along the length of a fault line, the left side will move toward you while the right side moves away from you. This is referred to as a left-lateral fault.

To know more about left lateral fault visit : brainly.com/question/14136640

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3 0

1 year ago