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3 years ago

7

A large flat block of mass 7 m rests on a level, smooth tabletop (µk ≈ 0). On top of it is a much smaller block of mass m, with

a coefficient of static friction between the two blocks of µs. If the small block is pulled horizontally with constant force F, we want both blocks to slide together across the table without slipping on each other.
What is the largest acceleration that the system can be given in terms of µs and g?
1. F − µs g/ 8
2. F /8 m
3. µs g /7
4. µs g
5. 7 µs g /8
6. 8 µs g

1 answer:

mariarad [96]3 years ago

8 0

Answer:

3. µs g /7

Explanation:

The largest Force appear when the maximal friction Force is required.

Second Newton law for the small block:

$F_friction=u_s*N=u_s*(mg)$

$F-F_friction=ma$

$F-u_s*(mg)=ma$

Second Newton law for the Big Block:

$F_friction=7ma$

$u_s*(mg)=7ma$

$a=u_s*g/7$

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Answer:

The permittivity of rubber is $\epsilon = 8.703 *10^{-11}$

Explanation:

From the question we are told that

The magnitude of the point charge is $q_1 = 70 \ nC = 70 *10^{-9} \ C$

The diameter of the rubber shell is $d = 32 \ cm = 0.32 \ m$

The Electric field inside the rubber shell is $E = 2500 \ N/ C$

The radius of the rubber is mathematically evaluated as

$r = \frac{d}{2} = \frac{0.32}{2} = 0.16 \ m$

Generally the electric field for a point is in an insulator(rubber) is mathematically represented as

$E = \frac{Q}{ \epsilon } * \frac{1}{4 * \pi r^2}$

Where $\epsilon$ is the permittivity of rubber

=> $E * \epsilon * 4 * \pi * r^2 = Q$

=> $\epsilon = \frac{Q}{E * 4 * \pi * r^2}$

substituting values

$\epsilon = \frac{70 *10^{-9}}{2500 * 4 * 3.142 * (0.16)^2}$

$\epsilon = 8.703 *10^{-11}$

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3 years ago

if quasars often resemble little blue stars, what was it about them that so surprised astronomers when they were discovered?

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1 year ago

Similarity between mass and charge

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I think there is only one.
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3 years ago

A garden hose has a radius of 0.0120 m, and water initially comes out at a speed of 2.88m/s. Dasha puts her thumb over the end ,

creativ13 [48]

Answer:

v = 12.4 [m/s]

Explanation:

With the speed and Area information, we can determine the volumetric flow.

$V=v*A\\A=\pi *r^{2}$

where:

r = radius = 0.0120 [m]

v = 2.88 [m/s]

$A=\pi *(0.0120)^{2} \\A=4.523*10^{-4} [m]\\$

Therefore the flow is:

$V=2.88*4.523*10^{-4} \\V=1.302*10^{-3} [m^{3}/s ]$

Despite the fact that you cover the inlet with the finger, the volumetric flow rate is the same.

$v=V/A\\v=1.302*10^{-3} /1.05*10^{-4} \\v=12.4[m/s]$

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3 years ago

A photovoltaic panel of dimension 2 m × 4 m is installed on the roof of a home. The panel is irradiated with a solar flux of GS

Flura [38]

Answer:

(a) the electrical power generated for still summer day is 1013.032 W

(b)the electrical power generated for a breezy winter day is 1270.763 W

Explanation:

Given;

Area of panel = 2 m × 4 m, = 8m²

solar flux GS = 700 W/m²

absorptivity of the panel, αS = 0.83

efficiency of conversion, η = P/αSGSA = 0.553 − 0.001 K⁻¹ Tp

panel emissivity , ε = 0.90

Apply energy balance equation to determine he electrical power generated;

transferred energy + generated energy = 0

(radiation + convection) + generated energy = 0

$[\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - \eta \alpha_s G_s = 0$

$[\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - (0.553-0.001T_p)\alpha_s G_s$

(a) the electrical power generated for still summer day

$T_s = T_{\infty} = 35 ^oC = 308 \ k$

$[0.83*700-0.9*5.67*10^{-8}(T_p_1^4-308^4)]-10(T_p_1-308) - (0.553-0.001T_p_1)0.83*700 = 0\\\\3798.94-5.103*10^{-8}T_p_1^4 - 9.419T_p_1 = 0\\\\Apply \ \ iteration \ method \ to \ solve \ for \ T_p_1\\\\T_p_1 = 335.05 \ k$

$P = \eta \alpha_s G_s A = (0.553-0.001 T_p_1)\alpha_s G_s A \\\\P = (0.553-0.001 *335.05)0.83*700*8 \\\\P = 1013.032 \ W$

(b)the electrical power generated for a breezy winter day

$T_s = T_{\infty} = -15 ^oC = 258 \ k$

$[0.83*700-0.9*5.67*10^{-8}(T_p_2^4-258^4)]-10(T_p_2-258) - (0.553-0.001T_p_2)0.83*700 = 0\\\\8225.81-5.103*10^{-8}T_p_2^4 - 29.419T_p_2 = 0\\\\Apply \ \ iteration \ method \ to \ solve \ for \ T_p_2\\\\T_p_2 = 279.6 \ k$

$P = \eta \alpha_s G_s A = (0.553-0.001 T_p_2)\alpha_s G_s A \\\\P = (0.553-0.001 *279.6)0.83*700*8 \\\\P = 1270.763 \ W$

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2 years ago