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Alenkinab [10]
4 years ago
5

Se dispara una bala con una velocidad de v = 300[m/s] contra un cascarón esférico de papel que gira con MCU respecto a un eje ve

rtical. Sabiendo que el radio del cascarón es de 10[cm]. Calcular la rapidez angular mínima que deberá girar el cascarón para que el proyectil haga únicamente un agujero. La dirección del movimiento de la bala pasa por el centro de la esfera.
Physics
1 answer:
Sladkaya [172]4 years ago
7 0

Answer:

 w = 4.712 10⁻³ rad / s

Explanation:

For this exercise, the time it takes for the bullet to travel the distance of 2R must be equal to the time that the hole must travel half a circle.

Let's start by calculating the time it takes for the bullet, which is going at constant speed.

         v = x / t

         t = x / v

         t = 2R / v

         t = 2 0.10 / 300

         t = 6.666 10⁻⁴ s

As they ask that a single hole is formed in this time, it must be rotated half a circle, that is, θ =π rad, for which we use the angular scientific relations, where the shell has constant angular velocity

          w = θ / t

          w = π / 6,666 10⁻⁴

          w = 4.712 10⁻³ rad / s

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Two 13 cm -long thin glass rods uniformly charged to +11nC are placed side by side, 4.0 cm apart. What are the electric field st
DedPeter [7]

Answer:

E1  = 10.15 * 10^4 N/C

E2 = 0

E3 = 10.15 *10^4 N/C

Explanation:

Given data:

Two 13 cm-long thin glass rods ( L ) = 0.13 m

charge (Q)  = +11nC

distance between thin glass rods   = 4 cm .

<u>Calculate the electric field strengths </u>

electric charge due to a single glass rod in the question ( E ) = \frac{Q}{2\pi e_{0}rL }

equation 1 can be used to determine E1, E2 and E3 because the points lie within the two rods hence the net electric field produced will be equal to the difference in electric fields produced

applying equation 1 to determine E1

E1 = \frac{Q}{2\pi e_{0}rL } ( \frac{1}{0.01} - \frac{1}{0.03} )    ( distance from 1 rod is 0.01 m and from the other rod is 0.03 )

   = \frac{11*10^{-9} }{2*3.14*8.85*10^{-12}*0.13 } ( 66.67 )

   = 10.15 * 10^4 N/C

applying equation 1 to determine E2

E2 = \frac{Q}{2\pi e_{0}rL }( \frac{1}{0.02} - \frac{1}{0.02} )

therefore E2 = 0

E1 = E3

hence E3 = 10.15*10^4 N/C

4 0
3 years ago
The end of a horizontal rope is attatched to a prong of an electricity driven tuning fork that vibrates at 100hz. The other end
Darina [25.2K]

here since string is attached with a mass of 2 kg

so here tension force in the rope is given as

T = mg

here we will have

T = 2(9.8) = 19.6 N

now we will have speed of wave given as

v = \sqrt{\frac{T}{\mu}}

here we will have

v = \sqrt{\frac{19.6}{0.75\times 10^{-2}}}

v = 16.33 m/s

now we know that frequency is given as

F = 100 Hz

now wavelength is given as

\lambda = \frac{v}{F}

\lambda = \frac{16.33}{100} = 0.16 m

so wavelength will be 0.16 m

5 0
3 years ago
Please help with this
sertanlavr [38]

Answer:I have to say 56

Explanation: because it is going up by 8

6 0
3 years ago
A projector is placed on the ground 22 ft. away from a projector screen. A 5.2 ft. tall person is walking toward the screen at a
Stella [2.4K]

Answer:

y = 67.6 feet,   y = 114.4/ (22 - 3t)

Explanation:

For this exercise let's use that light travels in a straight line and some trigonometric relationships, the symbols are in the attached diagram

Large triangle Projector up to the screen

         tan θ = y / L

For the small triangle. Projector up to the person

         tan θ = y₀ / (L-d)

The angle is the same, so we equate the two equations

         y₀ / (L -d) = y / L

         y = y₀  L / (L-d)

The distance from the screen (d), we look for it with kinematics

         v = d / t

        d = v t

we replace

         y = y₀ L / (L - v t)

         y = 5.2 22 / (22 - 3 t)

         y = 114.4 (22 - 3t)⁻¹

This is the equation of the shadow height change as a function of time

For the suggested distance the shadow has a height of

           y = 114.4 / (22-13)

           y = 67.6 feet

7 0
3 years ago
an object is placed 15.8 cm in front of a thin converging lens with an unknown focal length. if a real image forms behind the le
storchak [24]
<span>On what:

f (is the focal length of the lens) = ? 
p (is the distance from the object to the lens) =15.8 cm
p' (is the distance from the image to the spherical lens) = 4.2 cm

</span><span>Using the Gaussian equation, to know where the object is situated (distance from the point).
</span>
\frac{1}{f} = \frac{1}{p} + \frac{1}{p'}
\frac{1}{f} = \frac{1}{15.8} + \frac{1}{4.2}
\frac{1}{f} = \frac{2.1}{33.18} + \frac{7.9}{33.18}
\frac{1}{f} =  \frac{10}{33.18}
Product of extremes equals product of means:
10*f = 1*33.18
10f = 33.18
f =  \frac{33.18}{10}
\boxed{\boxed{f = 3.318\:cm}}\end{array}}\qquad\quad\checkmark

6 0
4 years ago
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