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Anna11 [10]
3 years ago
13

Two large books are stacked on top of each other on a table . The mass of each book is 6.5Kg Given that the coefficient of stati

c friction between bottom book and table is 0.15, what is the lowest the coefficient of static friction between the books can be in order to apply force and move both books?
Physics
1 answer:
Setler [38]3 years ago
3 0
<span>If you apply force (push) both books with the same energy at the same constant rate then the friction between them doesn't matter as both will move. If you push on the bottom book only, the friction between the books needs to be sufficient that the top book is carried along on the bottom book.</span>
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Tarzan swings back and forth on a
WARRIOR [948]

Answer:

Considering Tarzan as a point mass at point L and using the equation for a simple pendulum:

P = 2 pi (L / g)1/2

P^2 = 4 pi^2 L / g

L = g P^2 / (4 pi^2) = 9.8 m/s^2 * 7.27^2 s^2 / (4 pi^2) = 13.1 m

4 0
3 years ago
Two particles, each with charge 55.3 nC, are located on the y axis at y 24.9 cm and y -24.9cm (a) Find the vector electric field
ser-zykov [4K]

Answer:

Ex = kq 2x / ∛ (x² + y²)²  and  Ex = 2008 N / C

Explanation:

a)   The electric field is a vector quantity, so we must find the field for each particle and add them vectorially, as the whole process is on the X axis,

The equation for the electric field produced by a point charge is

         E = k q / r²

With r the distance between the point charge and the positive test charge

We look for each electric field

Particle 1.  Located at y = 24.9 m, let's use Pythagoras' theorem to find the distance

          r² = x² + y²

          E1 = k q / (x² + y²)

Particle 2.   located at x = -24.9 m

          r² = x² + y²

          E2 = k q / (x² + y²)

We can see that the two fields are equal since the particles have the same charge and coordinate it and that is squared.

In the attached one we can see that the Y components of the electric fields created by each particle are always the same and it is canceled, so we only have to add the X components of the electric fields. Let's use Pythagoras' theorem to find

Let's measure the angle from axis X

     cos θ = CA / H = x / (x2 + y2) ½

     E1x = E1 cos θ

      E2x = = E1 cos θ

The resulting field

      Ey = 0

      Ex = E1x + E2x 2 E1x

      Ex = 2 k q / (x² + y²) cos θ) = 2 k q / (x² + y²) x / √(x² + x²)

      Ex = kq 2x / ∛ (x² + y²)²

b) For this part we substitute the numerical values

      Ex = 8.99 10⁹ 55.3 10⁻⁹ x / (x² + 0.249 2) ³/₂

      Ex = 497.15   x / (x² + 0.062)  ³/₂  

Point where can the value of the electric field x = 38.1 cm = 0.381 m

       Ex = 497.15 0.381 / (0.381² + 0.062)  ³/₂  

       Ex = 497.15 0.381 / (0.1452 + 0.062) 3/2 = 189.41 / 0.2072 3/2

       Ex= 189.41 /0.0943

       Ex = 2008 N / C

c)  E = 1.00 kN / C = 1000 N / C

To solve this part we must find x in the equation

       Ex = 497.15 x / (x² + 0.062)  ³/₂  

Let's use some arithmetic

       Ex / 497.15 = x / (x² + 0.062)  ³/₂  

       [Ex / 497.15] ²/₃ = [x / (x² + 0.062) 3/2] ²/₃

       ∛[Ex / 497.15]² = (∛x²) / (x² + 0.062)                 (1)

The roots of this equation are the solution to the problem,

     

For Ex = 1.00 kN / C = 1000 N / C

 

      [Ex / 497.15] 2/3 = 1000 / 497.15) 2/3 = 1,312

       1.312 = (∛x² ) / (x² + 0.062)

       1.312 (x² + 0.062) = ∛x²

       1.312 X² - ∛x² + 1.312 0.062 = 0

       1.312 X² - ∛x² + 0.0813 = 0

We need used computer

4 0
3 years ago
Hellppppp&amp;/&amp;/&amp;/&amp;//&amp;/&amp;/&amp;/&amp;/&amp;&amp; will give brainiest
Komok [63]

Explanation:

i really didn't understand the question

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3 0
3 years ago
Given that y is the vertical displacement of a wave at anytime and A is the amplitude of the wave, w is the angular speed of the
denis-greek [22]

Explanation:

The general equation describing a wave is:

y(x,t) = A sin(kx - wt)

Let's say that for a particular wave on a string the equation is:

y(x,t) = (0.9 cm) sin[(1.2 m-1)x - (5.0 s-1)t]

(a) Determine the wave's amplitude, wavelength, and frequency.

(b) Determine the speed of the wave.

(c) If the string has a mass/unit length of m = 0.012 kg/m, determine the tension in the string.

(d) Determine the direction of propagation of the wave.

(e) Determine the maximum transverse speed of the string.

Solutions

Part (a): The wave's amplitude, wavelength, and frequency can be determined from the equation of the wave:

y(x,t) = (0.9 cm) sin[(1.2 m-1)x - (5.0 s-1)t]

The amplitude is whatever is multiplying the sine.

A = 0.9 cm

The wavenumber k is whatever is multiplying the x:

k = 1.2 m-1The wavelength isl=2pk= 5.2 m

The angular frequency w is whatever is multiplying the t.

w = 5.0 rad/sf=w2p= 0.80 Hz

Part (b): The wave speed can be found from the frequency and wavelength:

v = f l = 0.80 * 5.2 = 4.17 m/s

Part (c): With m = 0.012 kg/m and the wave speed given by:v=(Tm)½

This gives a tension of T = m v2 = 0.012 (4.17)2 = 0.21 N.

Part (d): To find the direction of propogation of the wave, just look at the sign between the x and t terms in the equation. In our case we have a minus sign:

y(x,t) = (0.9 cm) sin[(1.2 m-1)x - (5.0 s-1)t]

A negative sign means the wave is traveling in the +x direction.

A positive sign means the wave is traveling in the -x direction.

Part (e): To determine the maximum transverse speed of the string, remember that all parts of the string are experiencing simple harmonic motion. We showed that in SHM the maximum speed is:

vmax = Aw

In this case we have A = 0.9 cm and w = 5.0 rad/s, so:

vmax = 0.9 * 5.0 = 4.5 cm/s

maybe this should help

6 0
3 years ago
An electric heater has a Nichrome heating element with a resistance of 9 Ω at 20oC. When 112 V are applied, the electric current
maw [93]

Answer:

975.28 W.

Explanation:

Using,

R' = R(1+αΔt)....................... Equation 1

Where R' = Resistance at the final temperature, R = Resistance at the initial temperature, α = temperature coefficient of resistivity of Nichorome, Δt = Temperature rise.

Given: R = 9 Ω, α = 0.0004/°C, Δt = 1090-20 = 1070 °C

Substitute into equation 1

R' = 9(1+0.0004×1070)

R' = 9(1.428)

R' = 12.862  Ω.

Note: Operating wattage of the heater means the operating power of the heater

The power of the heater is given as,

P = V²/R'...................... Equation 2

Where P = Operating wattage of the heater, V = Voltage, R' = Operating resistance.

Given: V = 112 V, R' = 12.862 Ω

Substitute into equation 2

P = 112²/12.862

P = 975.28 W.

3 0
3 years ago
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