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seropon [69]
2 years ago
5

A 3.0kg weight W is initially at rest on incline AB, which is raised 40° above the horizontal. The effective coefficient is

Physics
1 answer:
lakkis [162]2 years ago
3 0

(a) The acceleration of the system is determined as 1.58 m/s².

(b) The relative weight of P is pounds is determined as 0.14 lb.

<h3>Acceleration of the system</h3>

The acceleration of the system is calculated as follows;

W - T = m₂a --- (1)

T = m₁a ----(2)

μmgsinθ - m₁a = m₂a

(0.3 x 3 x 9.8 x sin40) - (0.4 + 0.2)a = 3a

5.67 - 0.6a = 3a

5.67 = 3.6a

a = 5.67/3.6

a = 1.58 m/s²

<h3> Relative Weight of P</h3>

W = ma

W = 0.4 x 1.58

W = 0.632 N = 0.14 lb

Learn more about weight here: brainly.com/question/2337612

#SPJ1

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Answer:

Scenario 1 is the correct answer.

Explanation:

The sound of the drumstick hitting the metal bar will get to me in a shorter amount of time in Scenario 1 . The sound wave will travel faster in the metal bar than through the air because the speed of sound waves in solid is faster than it is in gases.

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An extension cord is used with an electric weed trimmer that has a resistance of 17.9 Ω. The extension cord is made of copper (r
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Answer:

(a) R_{c}=0.87ohms

(b) V_{T}=114.44V

Explanation:

Part (a)

The total length of copper cord L=86.3 m

The cross sectional area A=1.71×10⁻⁶m²

The resistivity of copper p=1.72×10⁻⁸Ω

Thus the resistance of extension cord is

R_{c}=p\frac{L}{A}\\R_{c}=(1.72*10^{-8} )\frac{86.3}{1.71*10^{-6}}\\R_{c}=0.87ohms

Part (b)

The resistance of trimmer Rt=17.9 ohms

When voltage of 120V is applied then the current I is passing through series circuit is

I=\frac{120V}{R_{c} +R_{T} }\\I=\frac{120V}{0.87 +17.9 } \\I=6.4A

Thus the voltage across the trimmer is:

V_{T}=IR_{T}\\V_{T}=(6.4)*(17.9)\\V_{T}=114.44V

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Two isolated, concentric, conducting spherical shells have radii R1 = 0.500 m and R2 = 1.00 m, uniform charges q1=+2.00 µC and q
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Complete Question

The diagram for this question is shown on the first uploaded image  

Answer:

a E =1.685*10^3 N/C

b E =36.69*10^3 N/C

c E = 0 N/C

d V = 6.7*10^3 V

e   V = 26.79*10^3V

f   V = 34.67 *10^3 V

g   V= 44.95*10^3 V

h    V= 44.95*10^3 V

i    V= 44.95*10^3 V

Explanation:

From the question we are given that

       The first charge q_1 = 2.00 \mu C = 2.00*10^{-6} C

       The second charge q_2 =1.00 \muC = 1.00*10^{-6}

      The first radius R_1 = 0.500m

      The second radius R_2 = 1.00m

 Generally \ Electric \ field = \frac{1}{4\pi\epsilon_0}\frac{q_1+\ q_2}{r^2}

And Potential \ Difference = \frac{1}{4\pi \epsilon_0}   [\frac{q_1 }{r}+\frac{q_2}{R_2} ]

The objective is to obtain the the magnitude of electric for different cases

And the potential difference for other cases

Considering a

                      r  = 4.00 m

           E = \frac{((2+1)*10^{-6})*8.99*10^9}{16}

                = 1.685*10^3 N/C

Considering b

           r = 0.700 m \ , R_2 > r > R_1

This implies that the electric field would be

            E = \frac{1}{4\pi \epsilon_0}\frac{q_1}{r^2}

             This because it the electric filed of the charge which is below it in distance that it would feel

            E = 8*99*10^9  \frac{2*10^{-6}}{0.4900}

               = 36.69*10^3 N/C

   Considering c

                      r  = 0.200 m

=>   r

 The electric field = 0

     This is because the both charge are above it in terms of distance so it wont feel the effect of their electric field

       Considering d

                  r  = 4.00 m

=> r > R_1 >r>R_2

Now the potential difference is

                  V =\frac{1}{4\pi \epsilon_0} \frac{q_1 + \ q_2}{r} = 8.99*10^9 * \frac{3*10^{-6}}{4} = 6.7*10^3 V

This so because the distance between the charge we are considering is further than the two charges given  

          Considering e

                       r = 1.00 m R_2 = r > R_1

                V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2}  ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{1.00} \frac{1.00*10^{-6}}{1.00} ] = 26.79 *10^3 V

          Considering f

              r = 0.700 m \ , R_2 > r > R_1

                      V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2}  ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.700} \frac{1.0*10^{-6}}{1.00} ] = 34.67 *10^3 V

          Considering g

             r =0.500\m , R_1 >r =R_1

   V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2}  ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V

          Considering h

                r =0.200\m , R_1 >R_1>r

  V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2}  ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V

           Considering i    

   r =0\ m \ , R_1 >R_1>r

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