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taurus [48]
3 years ago
10

A 4.60 gg coin is placed 19.0 cmcm from the center of a turntable. The coin has static and kinetic coefficients of friction with

the turntable surface of μs = 0.820 and μk = 0.440. The turntable very slowly speeds up.1. What is the angular speed in rpm when the coin slides off?
Physics
1 answer:
Komok [63]3 years ago
5 0

Answer:

62.64 RPM.

Explanation:

Given that

m= 4.6 g

r= 19 cm

μs = 0.820

μk = 0.440.

The angular speed of the turntable = ω rad/s

Condition just before the slipping starts

The maximum value of the static friction force =Centripetal force

\mu_s\ m g=m\ \omega^2\ r\\ \omega^2=\dfrac{\mu_s\ m g}{m r}\\ \omega=\sqrt{\dfrac{\mu_s\  g}{ r}}\\ \omega=\sqrt{\dfrac{0.82\times 10}{ 0.19}}\ rad/s\\\omega= 6.56\ rad/s

\omega=\dfrac{2\pi N}{60}\\N=\dfrac{60\times \omega}{2\pi }\\N=\dfrac{60\times 6.56}{2\pi }\ RPM\\N=62.64\ RPM

Therefore the speed in RPM will be 62.64 RPM.

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An observer measures the length (L), width (w), and height (h) of a box while stationary relative to the box. The observer then
Elenna [48]

Answer:

b. less than w.

Explanation:

In this question, the application of length contraction is what helps us come to our conclusion. When an object moves very fast (relative to the observer), the length of the object seems to be smaller than it actually is (again, for the observer).

This is supported by the length contraction equation below:

L = L_0\sqrt{1-\frac{v^2}{c^2} }

Here, L is the observed length

L_0 is the original length of the object

v is the relative speed between the object and the observer

and c is the speed of light

Using this equation, we can see that as the speed between the object and the observer is increased to be close to that of light, the square root in the equation gives us values less than 1.0

This effectively decreases the length that is observed.

8 0
3 years ago
Urgent! please help!
kap26 [50]

_{94}^{239}Pu+ _{0}^{1}n--->_{40}^{100}Zr+_{x}^{y}Element+2_{0}^{1}n\\ \\ 94+0=40+x+0, x= 54,\\ \\ 239+1 = 100+y+2*1, y=138\\ \\_{54}^{138}Element= _{54}^{138}Xe

Answer is A.

8 0
3 years ago
The unit of electrical current is the ampere (A).
IrinaVladis [17]

Answer:

option B

Explanation:

Current unit is ampere (A)

Ampere will be equivalent to = ?

we know,

Current can be define as the charge per unit time

I = \dfrac{Q}{t}

unit of charge(Q) is coulomb which is equal to C.

unit of time(t) is equal to 's'.

now,

I = \dfrac{C}{s}

unit of I = C/s

ampere(A) is equivalent to C/s

The correct answer is option B

4 0
3 years ago
***PLEASE HELP WITH ANSWER AND EXPLANATION: Imagine the current in a current-carrying wire is flowing into the screen. What is t
aalyn [17]

Magnetic field direction is given by right hand thumb rule.

If we put our thumb in the direction of current then curl of fingers will show the magnetic field direction around the wire.

Now here since current is going into the screen so we will put our thumb into the screen and then the curl of fingers is clockwise around it.

The magnetic field is clockwise.

So this would be the direction of magnetic field

5 0
3 years ago
Read 2 more answers
A solid ball of radius rb has a uniform charge density rho.
Oksana_A [137]

Note: question B is incomplete.

Complete Question

A solid ball of radius rb has a uniform charge density ρ.

a.  What is the magnitude of the electric field E(r) at a distance r>rb from the center of the ball?  Express your answer in terms of ρ, rb, r, and ϵ0.

b.   What is the magnitude of the electric field E(r) at a distance r<rb from the center of the ball?  Express your answer in terms of ρ, r, rb, and ϵ0.

c.   Let E(r) represent the electric field due to the charged ball throughout all of space. Which of the following statements about the electric field are true?

1. E(0) = 0.

2. E(rb) = 0

3. lim E(r) = 0.

4. The maximum electric field occurs when r = 0.

5. The maximum electric field occurs when r = rb.

6. The maximum electric field occurs as r to infinity.

Answer:

a) the magnitude of E(r)= ρr³/3 ε₀r²

b) the magnitude at distance r from the centre E(r)= ρr/3 ε ₀ ( if r<rb)

c) statements 1(E(0) = 0), 3(E(0) = 0) and 5(The maximum electric field occurs when r = rb.) are true

Explanation:

given

charge density = ρ ,  ε

Volume of sphere , V = (⁴/₃)πr³

a) charge density = charge/volume

ρ = q ÷ V

make charge the subject of the formula

∴q = ρ × V=  ρ× (⁴/₃)πr³

where r³ = rb³(at distance rb³)

recall

E= q/4πε₀r²

E=  ρ × (⁴/₃)πrb³/4πε₀r²

∴E(r)= ρrb³/3 ε ₀r²

(b)  The Gaussian surface is inside the ball, therefore, surface only encloses a portion of ball's charge .

The net charge enclosed by the Gaussian surface is different to the of net charge enclosed in (a)

Recall

E= q/4πε₀r²

V= (⁴/₃)πr³

E=  ρ × (⁴/₃)πr³/4πε₀r²

∴E(r)= ρr/3 ε₀

(c)  E(0)= 0

limr-----∝

E(r)= 0

The maximum electric field occurs when r=rb.

4 0
3 years ago
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