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Murljashka [212]

3 years ago

10

A box full of books rests on a wooden floor. The normal force the floor exerts on the box is 234.0 N. (a) You push horizontally

on the box with a force of 123.0 N, but it refuses to budge. What can you say about the coefficient of static friction between the box and the floor?

1 answer:

katen-ka-za [31]3 years ago

3 0

Answer:

µ>=0.526.

Explanation:

The friction force is opposite to the horizontal force applied and is given by:

$F_k=\µ*Fn$

The force applied horizontally need to be bigger so the box can be moved, but in this case, the force friction is bigger so:

$F=\frac{F}{F_k}\\\\\µ=\frac{123.0}{234.0}\\\\\µ>=0.526$

So the coefficient of static friction is bigger or equal to 0.526.

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Determine the electric field (in N/C) required to give the maximum possible deviation angle. (Enter the magnitude.)

Zigmanuir [339]

Answer:

Maximum angle = 3.43⁰

Explanation:

Say that you are given the following information:

vertical distance between the charge plate = 0.03 m

length of the plate = 0.5 m

velocity of the electrons = 5 × 10⁶ ms⁻¹

the maximum angle is given by the formula:

$tan\theta _{max} = \frac{d}{l}$

where d = vertical distance between the charge plate

l = length of the plate

substituting the values l and d gives:

$tan\theta _{max} = \frac{0.03}{0.5}$

maximum angle, $\theta _{max} = \frac{0.03}{0.5}$

$\theta _{max} = tan^{-1} (\frac{0.03}{0.5} )$

= 3.43⁰

3 0

3 years ago

A horizontal circular platform (m = 119.1 kg, r = 3.23m) rotates about a frictionless vertical axle. A student (m = 54.3kg) walk

Murrr4er [49]

Answer:

$\omega_2=5.1rad/s$

Explanation:

Since there is no friction angular momentum is conserved. The formula for angular momentum thet will be useful in this case is $L=I\omega$ . If we call 1 the situation when the student is at the rim and 2 the situation when the student is at $r_2=1.39m$ from the center, then we have:

$L_1=L_2$

Or:

$I_1\omega_1=I_2\omega_2$

And we want to calculate:

$\omega_2=\frac{I_1\omega_1}{I_2}$

The total moment of inertia will be the sum of the moment of intertia of the disk of mass $m_D=119.1 kg$ and radius $r_D=3.23m$ , which is $I_D=\frac{m_Dr_D^2}{2}$ , and the moment of intertia of the student of mass $m_S=54.3kg$ at position $r$ (which will be $r_1=r=3.23m$ or $r_2=1.39m$ ) will be $I_{S}=m_Sr_S^2$ , so we will have:

$\omega_2=\frac{(I_D+I_{S1})\omega_1}{(I_D+I_{S2})}$

or:

$\omega_2=\frac{(\frac{m_Dr_D^2}{2}+m_Sr_{S1}^2)\omega_1}{(\frac{m_Dr_D^2}{2}+m_Sr_{S2}^2)}$

which for our values is:

$\omega_2=\frac{(\frac{(119.1kg)(3.23m)^2}{2}+(54.3kg)(3.23m)^2)(3.1rad/s)}{(\frac{(119.1kg)(3.23m)^2}{2}+(54.3kg)(1.39m)^2)}=5.1rad/s$

6 0

3 years ago

PLEASE HELPPP!!! I'LL GIVE BRAINLEST FOR CORRECT ANSWERS

babymother [125]

Answer:

Explanation:

Weight will be highest at the pole where there is no tangential velocity requiring centripetal acceleration. Also, due to the slight bulge of the earth at the equator, the distance from the surface to the center of mass of earth is slightly shorter there meaning gravity is slightly stronger. Fg = GmM/R²

Weight will be lowest at the equator where the object is moving about the earth axis at an angular velocity of 2π/(24(3600)) rad/s ( about 7.27e-5 rad/s)

This means that some of the (already weaker, see above) gravity there is required to supply the needed centripetal acceleration to keep the object on the ground.

4 0

3 years ago

NEED HELP!! ASAP JUST QUESTIONS 22, 23, 24, & 25

poizon [28]

22. reduction
25. Le Chatelier's principle

8 0

4 years ago

The inductor in the RLC tuning circuit of an AM radio has a value of 250 mHmH . You may want to review (Pages 857 - 860) . Part

ANEK [815]

Answer:

The value of variable capacitor is $1.89 \times 10^{-13}$ F

Explanation:

Given :

Inductance $L = 250 \times 10^{-3}$ H

Frequency $f = 731 \times 10^{3}$ Hz

According to the cutoff frequency,

$f = \frac{1}{2\pi \sqrt{LC} }$

Now we find the value of capacitance,

$C = \frac{1}{4\pi ^{2} f^{2} L }$

$C = \frac{1}{4\times 9.85 \times (731 \times 10^{3} )^{2} \times 250 \times 10^{-3} }$

$C = 1.89 \times 10^{-13}$ F

Therefore, the value of variable capacitor is $1.89 \times 10^{-13}$ F

6 0

3 years ago