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Rina8888 [55]
3 years ago
9

A 16 foot ladder is leaning against a wall. If the top of the ladder slides down the wall at a rate of 3 feet per second, how fa

st is the bottom of the ladder moving along the ground when the bottom of the ladder is 4 feet from the wall? Leave your answer in exact form, simplified as much as possible. Do not include units.
Physics
1 answer:
JulsSmile [24]3 years ago
7 0

Answer:

11.625

Explanation:

L = length of the ladder = 16 ft

v_{y} = rate at which top of ladder slides down = - 3 ft/s

v_{x} = rate at which bottom of ladder slides

y = distance of the top of ladder from the ground

x = distance of bottom of ladder from wall = 4 ft

Using Pythagorean theorem

L² = x² + y²

16² = 4² + y²

y = 15.5 ft

Also using Pythagorean theorem

L² = x² + y²

Taking derivative both side relative to "t"

0 = 2x\frac{\mathrm{d} x}{\mathrm{d} t} + 2y\frac{\mathrm{d} y}{\mathrm{d} t}

0 = x v_{x} + y v_{y}

0 = 4 v_{x} + (15.5) (- 3)

v_{x} = 11.625 ft/s

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A slit has a width of W1 = 4.4 × 10-6 m. When light with a wavelength of λ1 = 487 nm passes through this slit, the width of the
Vitek1552 [10]

Answer:

The width of the central bright fringe on the screen is observed to be unchanged is 4.48*10^{-6}m

Explanation:

To solve the problem it is necessary to apply the concepts related to interference from two sources. Destructive interference produces the dark fringes.  Dark fringes in the diffraction pattern of a single slit are found at angles θ for which

w sin\theta = m\lambda

Where,

w = width

\lambda =wavelength

m is an integer, m = 1, 2, 3...

We here know that as sin\theta as w are constant, then

\frac{w_1}{\lambda_1} = \frac{w_2}{\lambda_2}

We need to find w_2, then

w_2 = \frac{w_1}{\lambda_1}\lambda_2

Replacing with our values:

w_2 = \frac{4.4*10^{-6}}{487}496

w_2 = 4.48*10^{-6}m

Therefore the width of the central bright fringe on the screen is observed to be unchanged is 4.48*10^{-6}m

3 0
3 years ago
The molecular weight of carbon dioxide (CO2) is 44.00 a.m.u., and the molecular weight of propane gas (C3H8) is 44.10 a.m.u. Thu
Nina [5.8K]

The molecular weight of carbon dioxide (CO2) is 44.00 a.m.u., and the molecular weight of propane gas (C3H8) is 44.10 a.m.u. Thus CO2 diffuses_______ C3H8.

Slower Than

Faster Than

The Same As

Slower than

6 0
3 years ago
Read 2 more answers
Un objeto se suelta desde determinada altura y emplea un tiempo t en caer al suelo. Si se cuadruplica la altura desde la cual se
blondinia [14]

When an object falls from a h height, you should work with the uniformly accelerated linear movement equations:

y=½*a*t²+Vo*t+yo

You should consider:

a=-g=-10m/s²

yo=h

If it’s a freefall, it means it starts from rest, which means it has no initial velocity:

Vo=0

Replacing that information in the equation:

y=½*(-10m/s²)*t²+0*t+h=-5m/s²*t²+0+h=-5m/s²*t²+h

So this is the

Besides, if you want to find out how long it takes for it to get to the floor, you should put the height of the floor as final height, which would be 0 (assuming the initial height has been measured from there):

y=0

0=-5m/s²*t²+h

5m/s²*t²=h

t²=h/(5m/s²)

t=√(h/(5m/s²))

t=√(hs²/(5m))

t=(√(h/(5m)))s

<span>If we <span>quadruple </span>h:</span>

t2=(√(h2/(5m)))s=(√(4*h1/(5m)))s=(√4)*(√h1/(5m)))s=2*(√h1/(5m)))s=2*t1

This 4 goes inside the square root, so then it converts to 2. So the new time is twice as much the previous time.

Concerning velocity, you have to use the other equation:

v=at+vo

As I said before, a is gravity and vo is zero.

v=-10m/s²*t+0=-10m/s²*t

Final velocity is directly related to time, so if time is doubled, so is velocity.

v2=-10m/s²*t2=-10m/s²*(2*t1)=2*(-10m/s²*t1)=2*v1

<span>So the correct answer is A, and the other ones are false.</span>

8 0
3 years ago
A 50.0 kg object is moving at 18.2 m/s when a 200 N force
gayaneshka [121]

Answer:

distance = 21.56 m

Explanation:

given data

mass = 50 kg

initial velocity  = 18.2 m/s

force = -200 N ( here force applied to opposite direction )

final velocity = 12.6 m/s

solution

we know here acceleration will be as

acceleration a  = force ÷ mass

a = \frac{-200}{50}   =  -4 m/s²

we get here now required time that is

required time = \frac{V_{(final)} - V_{(initial)}}{a}     ...............1

put here value

required time = \frac{12.6-18.2}{-4}  

so distance will be

distance = \frac{V_{(final)}^2 - V_{(initial)}^2}{2a}    ........2

distance = \frac{12.6}^2 -{18.2}^2}{2\times (-4)}  

distance = 21.56 m

7 0
3 years ago
In the diagram, R1 = 40.0 ,
Nostrana [21]

Answer:

51 Ω.

Explanation:

We'll begin by calculating the equivalent resistance of R₁ and R₃. This can be obtained as follow:

Resistor 1 (R₁) = 40 Ω

Resistor 3 (R₃) = 70.8 Ω

Equivalent Resistance of R₁ and R₃ (R₁ₙ₃) =?

Since the two resistors are in parallel connection, their equivalent can be obtained as follow:

R₁ₙ₃ = R₁ × R₃ / R₁ + R₃

R₁ₙ₃ = 40 × 70.8 / 40 + 70.8

R₁ₙ₃ = 2832 / 110.8

R₁ₙ₃ = 25.6 Ω

Finally, we shall determine the equivalent resistance of the group. This can be obtained as follow:

Equivalent Resistance of R₁ and R₃ (R₁ₙ₃) = 25.6 Ω

Resistor 2 (R₂) = 25.4 Ω

Equivalent Resistance (Rₑq) =?

Rₑq = R₁ₙ₃ + R₂ (series connection)

Rₑq = 25.6 + 25.4

Rₑq = 51 Ω

Therefore, the equivalent resistance of the group is 51 Ω.

4 0
3 years ago
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