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3 years ago

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A 6.0 kg mass is placed on a 20º incline which has a coefficient of friction of 0.15. What is the acceleration of the mass down

the incline? show all work please

1 answer:

Leona [35]3 years ago

6 0

Answer:

Explanation:

The form of Newton's 2nd Law that we use for this is:

F - f = ma where F is the Force pulling the mass down the ramp forward, f is the friction trying to keep it from moving forward, m is the mass and a is the acceleration (and our unknown).

We know mass and we can find f, but we don't have F. But we can solve for that by rewriting our main equation to reflect F:

$wsin\theta-\mu F_n=ma$ That's everything we need.

w is weight: 6.0(9.8). Filling in:

6.0(9.8)sin20 - .15(6.0)(9.8) = 6.0a and

2.0 × 10¹ - 8.8 = 6.0a and

11 = 6.0a so

a = 1.8 m/s/s

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The maximum wavelength that an electromagnetic wave can have and still eject electrons from a metal surface is 459 nm. What is t

spin [16.1K]

Answer:

2.7067 eV

Explanation:

h = Planck's constant = $6.626\times 10^{-34}\ m^2kg/s$

c = Speed of light = $3\times 10^8\ m/s$

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$W_0=\frac{hc}{\lambda_0}\\\Rightarrow W_0=\frac{6.626\times 10^{-34}\times 3\times 10^8}{459\times 10^{-9}}\\\Rightarrow W_0=4.33072\times 10^{-19}\ J$

Converting to eV

$1\ J=\frac{1}{1.6\times 10^{-19}}\ eV$

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The work function W0 of this metal is 2.7067 eV

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3 years ago

Check all choices below that are correct. Increasing the frequency increases the current. Changing the frequency does not affect

Annette [7]

Answer:

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Explanation:

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3 years ago

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A "moving sidewalk" in an airport terminal moves at 1.0 m/s and is 35.0 m long. If a woman steps on at one end and walks at 1.5

pishuonlain [190]

Answer:

a.14 s

b.70 s

Explanation:

a.Let the sidewalk moving in positive x- direction.

Speed of sidewalk relative to ground= $v_s=1m/s$

Speed of women relative to sidewalk=v=1.5m/s

The speed of women relative to the ground

$v_w=v_s+v=1+1.5=2.5m/s$

Distance=35 m

Time= $\frac{distance}{speed}$

Using the formula

Time taken by women to reach the opposite end if she walks in the same direction the sidewalk is moving= $\frac{35}{v_w}=\frac{35}{2.5}=14s$

b.If she gets on at the end opposite the end in part (a)

Then, we take displacement negative.

Speed of sidewalk relative to ground= $v_s=1m/s$

Speed of women relative to sidewalk=v=-1.5 m/s

The speed of women relative to the ground= $v_w=v_s+v=1-1.5=-0.5m/s$

Time= $\frac{-35}{-0.5}=70 s$

Hence, the women takes 70 s to reach the opposite end if she walks in the opposite direction the sidewalk is moving.

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Answer:

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2 years ago

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