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QveST [7]
3 years ago
6

A 7 kg object attached to a horizontal string moves with constant speed in a circle on a frictionless horizontal surface. The ki

netic energy of the object is 36 J and the tension in the string is 326 N. Find the radius of the circle. Answer in units of m.
Physics
1 answer:
Marizza181 [45]3 years ago
8 0

Answer:

Explanation:

Given

mass of object m=7 kg

kinetic Energy k=36\ J

Tension in string T=326\ N

mass is moving in a horizontal circle so tension is providing the centripetal acceleration

therefore T=\frac{mv^2}{r}----1

where r=radius of circle

kinetic energy of particle k=\frac{1}{2}mv^2----2

divide 1 and 2 we get

\frac{T}{k}=\frac{\frac{mv^2}{r}}{\frac{1}{2}mv^2}

\frac{T}{k}=\frac{2}{r}

r=\frac{2k}{T}=\frac{2\times 36}{326}

r=0.2208\ m

   

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Δλ = 3*10⁻³ m.

Explanation:

  • At any wave, there exists a fixed relationship between the speed of  the wave, the wavelength, and the frequency, as follows:

       v = \lambda* f  (1)

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  • Rearranging terms, we can get λ from the other two parameters, as follows:

       \lambda = \frac{v}{f}  (2)

  • Since v is constant for sound at 343 m/s, we can find the different wavelengths at different frequencies, as follows:

        \lambda_{1} =\frac{v}{f_{1}} = \frac{343m/s}{440(1/s)} = 0.779 m  (3)

        \lambda_{2} =\frac{v}{f_{2}} = \frac{343m/s}{442(1/s)} = 0.776 m  (4)

  • The difference between both wavelengths, is just the difference between (3) and (4):

       \Delta \lambda = \lambda_{1} - \lambda_{2} = 0.779 m - 0.776m = 3e-3 m (5)

       ⇒ Δλ = 3*10⁻³ m.

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The national high magnetic field laboratory holds the world record for creating the strongest magnetic field. for brief periods
max2010maxim [7]

Newton’s 2nd law states that Force is equal to the product of mass (m) and acceleration (a):

F = m a                                  ---> 1

While in magnetic forces, force can also be expressed as:

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v = velocity = 45 cm / s = 0.45 m / s

B = the magnetic field = 85 T

First we solve for the total charge, q:

q = 3.8 × 10^-23 g (1 mol / 23 g) (6.022 × 10^23 electrons / mol) (1.602 × 10^-19 C / electron)

q = 1.594 × 10^-19 C

 

We equate equations 1 and 2 then solve for acceleration a:

m a = q v B

a = q v B / m

a = [1.594 × 10^-19 C * 0.45 m / s * 85 T] / 3.8 × 10-26 kg

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Therefore the maximum acceleration of Na ions is about 160 × 10^6 m/s^2.

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3 years ago
Read 2 more answers
What is the frequency of light with a wavelength of 7.9 x 10^-9 m? ( the speed of light is 3.00 x 10^8)
nlexa [21]
Using the formula v=f times lambada
then v=the speed of light.
and f=what’s we’re looking for
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3 0
3 years ago
A 0.10 g honeybee acquires a charge of +23 pC while flying.
kari74 [83]

Answer:

a) \frac{F}{w} =2.347\times 10^{-6}\ N

b) E=4.2609\times 10^7\ N.C^{-1} parallel to the earth surface.

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Explanation:

Given:

mass of the bee, m=10^{-4}\ kg

charge acquired by the bee, q_2=23\times 10^{-12}\ C

a.

Electrical field near the earth surface, E=100\ N.C^{-1}

Now the electric force on the bee:

we know:

F=\frac{1}{4\pi.\epsilon_0} \times \frac{q_1.q_2}{r^2}

F=E.q_2

F=100\times 23\times 10^{-12}

F=23\times 10^{-10}\ N

The weight of the bee:

w=m.g

w=10^{-4}\times 9.8

w=9.8\times10^{-4}\ N

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\frac{F}{w} =\frac{23\times 10^{-10}}{9.8\times10^{-4}}

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b.

The condition for the bee to hang is its weight must get balanced by the electric force acing equally in the opposite direction.

So,

F=9.8\times10^{-4}\ N

E.q_2=9.8\times10^{-4}\ N

E\times 23\times 10^{-12}=9.8\times10^{-4}\ N

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3 0
3 years ago
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