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3 years ago

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A college student likes to take her books to class by placing them in a box and pulling the box around behind her. She pulls on

the box with a force of 90.0 N at an angle of 30.0⁰ relative to the horizontal. The box of books has a mass of 20.0 kg and the coefficient of kinetic friction between the bottom of the box and the sidewalk is 0.500. What is the net force acting horizontally and what is the acceleration of the box as she pulls it around?

1 answer:

just olya [345]3 years ago

8 0

Answer:

It would result an a negatice answer.

Explanation:

The accelarion should be pulled as a kite not a box :) columbus said that musical stuff no just no

Hope this helps!!

- Katty queen

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A uranium-238 atom can break up into a thorium-234 atom and a particle called an alpha particle, αα-4. The numbers indicate the

alexdok [17]

Answer: E = 5.80*10^-13 J

Explanation:

Given

We use the law of conservation of momentum to solve this

Momentum before breakup = momentum after breakup

0 = m1v1 + m2v2

0 = 238m * -2.2*10^5 + 4m * v2

0 = -523.6m m/s + 4m * v2

v2 * 4m = 523.6m m/s

v2 = 523.6 m m/s / 4m

v2 = 130.9*10^5 m/s

v2 = 1.31*10^7 m/s

Using this speed in the energy equation, we have

E = 1/2m1v1² + 1/2m2v2²

E = 1/2 * (238 * 1.66*10^-27) * -2.2*10^5² + 1/2 * (4 * 1.66*10^-27) * 1.31*10^7²

E = [1/2 * 3.95*10^-25 * 4.84*10^10] + [1/2 * 6.64*10^-27 * 1.716*10^14]

E = (1/2 * 1.911*10^-14) + (1/2 * 1.139*10^-12)

E = 9.56*10^-15 + 5.7*10^-13

E = 5.80*10^-13 J

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Calculate the pressure exerted by 11.1 moles of neon gas in a volume of 5.45 L at 25°C using (a) the ideal gas equation and (b)

Ilia_Sergeevich [38]

Answer:

49.82414 atm

50.74675 atm

Explanation:

P = Pressure

V = Volume = 5.45 L

R = Gas constant = 0.08205 L atm/mol K

T = Temperature = 25°C

a = 0.211 atm L²/mol²

b = 0.0171 atm L²/mol²

From ideal gas law we have

$PV=nRT\\\Rightarrow P=\dfrac{nRT}{V}\\\Rightarrow P=\dfrac{11.1\times 0.08205(273.15+25)}{5.45}\\\Rightarrow P=49.82414\ atm$

The pressure is 49.82414 atm

From Van der Waals equation we have

$\left(P+\frac{an^2}{V^2}\right)\left(v-nb\right)=nRT\\\Rightarrow P=\dfrac{nRT}{V-nb}-\dfrac{an^2}{V^2}\\\Rightarrow P=\dfrac{11.1\times 0.08205\times (273.15+25)}{5.45-(11.1\times 0.0171)}-\dfrac{0.211\times 11.1^2}{5.45^2}\\\Rightarrow P=50.74675\ atm$

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3 years ago

A river 500 ft wide flows with a speed of 8 ft/s with respect to the earth. A woman swims with a speed of 4 ft/s with respect to

White raven [17]

Answer:

1) $\Delta s=1000\ ft$

2) $\Delta s'=998.11\ ft.s^{-1}$

3) $t\approx125\ s$

$t'\approx463.733\ s$

Explanation:

Given:

width of river, $w=500\ ft$

speed of stream with respect to the ground, $v_s=8\ ft.s^{-1}$

speed of the swimmer with respect to water, $v=4\ ft.s^{-1}$

<u>Now the resultant of the two velocities perpendicular to each other:</u>

$v_r=\sqrt{v^2+v_s^2}$

$v_r=\sqrt{4^2+8^2}$

$v_r=8.9442\ ft.s^{-1}$

<u>Now the angle of the resultant velocity form the vertical:</u>

$\tan\beta=\frac{v_s}{v}$

$\tan\beta=\frac{8}{4}$

$\beta=63.43^{\circ}$

Now the distance swam by the swimmer in this direction be d.

so,

$d.\cos\beta=w$

$d\times \cos\ 63.43=500$

$d=1118.034\ ft$

Now the distance swept downward:

$\Delta s=\sqrt{d^2-w^2}$

$\Delta s=\sqrt{1118.034^2-500^2}$

$\Delta s=1000\ ft$

2)

On swimming 37° upstream:

<u>The velocity component of stream cancelled by the swimmer:</u>

$v'=v.\cos37$

$v'=4\times \cos37$

$v'=3.1945\ ft.s^{-1}$

<u>Now the net effective speed of stream sweeping the swimmer:</u>

$v_n=v_s-v'$

$v_n=8-3.1945$

$v_n=4.8055\ ft.s^{-1}$

<u>The component of swimmer's velocity heading directly towards the opposite bank:</u>

$v'_r=v.\sin37$

$v'_r=4\sin37$

$v'_r=2.4073\ ft.s^{-1}$

<u>Now the angle of the resultant velocity of the swimmer from the normal to the stream</u>:

$\tan\phi=\frac{v_n}{v'_r}$

$\tan\phi=\frac{4.8055}{2.4073}$

$\phi=63.39^{\circ}$

Now let the distance swam in this direction be d'.

$d'\times \cos\phi=w$

$d'=\frac{500}{\cos63.39}$

$d'=1116.344\ ft$

<u>Now the distance swept downstream:</u>

$\Delta s'=\sqrt{d'^2-w^2}$

$\Delta s'=\sqrt{1116.344^2-500^2}$

$\Delta s'=998.11\ ft.s^{-1}$

3)

Time taken in crossing the rive in case 1:

$t=\frac{d}{v_r}$

$t=\frac{1118.034}{8.9442}$

$t\approx125\ s$

Time taken in crossing the rive in case 2:

$t'=\frac{d'}{v'_r}$

$t'=\frac{1116.344}{2.4073}$

$t'\approx463.733\ s$

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c.

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andriy [413]

Answer:

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