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AlladinOne [14]
3 years ago
13

An athlete swings a ball, connected to the end of a chain, in a horizontal circle. The athlete is able to rotate the ball at the

rate of 8.13 rev/s when the length of the chain is 0.600 m. When he increases the length to 0.900 m, he is able to rotate the ball only 6.04 rev/s. (a) Which rate of rotation gives the greater speed for the ball? 6.04 rev/s 8.13 rev/s (b) What is the centripetal acceleration of the ball at 8.13 rev/s? m/s2 (c) What is the centripetal acceleration at 6.04 rev/s? m/s2
Physics
1 answer:
garik1379 [7]3 years ago
5 0

(a) 6.04 rev/s

The speed of the ball is given by:

v=\omega r

where

\omega is the angular speed

r is the distance of the ball from the centre of the circle

In situation 1), we have

\omega=8.13 rev/s \cdot 2\pi = 51.0 rad/s

r = 0.600 m

So the speed of the ball is

v=(51.0 rad/s)(0.600 m)=30.6 m/s

In situation 2), we have

\omega=6.04 rev/s \cdot 2\pi = 37.9 rad/s

r = 0.900 m

So the speed of the ball is

v=(37.9 rad/s)(0.900 m)=34.1 m/s

So, the ball has greater speed when rotating at 6.04 rev/s.

(b) 1561 m/s^2

The centripetal acceleration of the ball is given by

a=\frac{v^2}{r}

where

v is the speed

r is the distance of the ball from the centre of the trajectory

For situation 1),

v = 30.6 m/s

r = 0.600 m

So the centripetal acceleration is

a=\frac{(30.6 m/s)^2}{0.600 m}=1561 m/s^2

(c) 1292 m/s^2

For situation 2 we have

v = 34.1 m/s

r = 0.900 m

So the centripetal acceleration is

a=\frac{v^2}{r}=\frac{(34.1 m/s)^2}{0.900 m}=1292 m/s^2

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olchik [2.2K]

Answer:

The reactance of the capacitor

Explanation:

In an AC circuit containing different elements (capacitors, resistors and inductors), we cannot simply calculate the equivalent resistance of the circuit, so another quantity is used, which is called reactance.

For a capacitor, the reactance is given by:

X=\frac{1}{2 \pi f C}

where:

f is the frequency of the AC current in the circuit

C is the capacitance of the capacitor

The reactance has a similar meaning to that of the resistance for a DC current. In fact, we notice that:

- When f=0 (which means we are in regime of DC current, because the current never changes direction), the reactance is infinite. This is correct: in a DC circuit, the capacitor does not let current pass through it, so it like it has infinite resistance (=infinite reactance)

- When f tends to infinite, the reactance becomes zero: in such situation, the current in the circuit changes direction so quickly that the capacitor has no enough time to "block" the current in the circuit, so it like it has almost zero resistance (zero reactance).

4 0
3 years ago
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A very good insulator has an R-value of 29. The material's heat transfer coefficient is
Fittoniya [83]

Answer:

b. 0.034

Explanation:

The heat transfer coefficient of a material (U-value) is equal to the reciprocal of its R-value, therefore:

U = \frac{1}{R}

where

R is the R-value of the material

For the insulator in this problem,

R = 29

Substituting into the equation, we find the heat transfer coefficient:

U=\frac{1}{29}=0.034

4 0
3 years ago
A plastic rod that has been charged to − 15 nC touches a metal sphere. Afterward, the rod's charge is − 5.0 nC.
Natali5045456 [20]

Answer:

B) electrons transferred from sphere to rod.

(2) 1.248 x 10¹¹ electrons were transferred

Explanation:

Given;

initial charge on the plastic rod, q₁ = 15nC

final charge on the plastic rod, q₂ = - 5nC

let the charge acquired by the plastic rod = q

q + 15nC = -5nC

q = -5nC - 15nC

q = -20 nC

Thus, the plastic rod acquired excess negative charge from the metal sphere.

Hence, electrons transferred from sphere to rod

B) electrons transferred from sphere to rod.

2) How many charged particles were transferred?

1.602 x 10⁻¹⁹ C = 1 electron

20 x 10⁻⁹ C = ?

= 1.248 x 10¹¹ electrons

Thus,1.248 x 10¹¹ electrons were transferred

7 0
3 years ago
Red light of wavelength 630 nm passes through two slits and then onto a screen that is 1.4 m from the slits. The center of the 3
VARVARA [1.3K]

Answer:

Part a)

f = 4.76 \times 10^{14} Hz

Part b)

d = 3.48 \times 10^{-4} m

Part c)

\theta = 0.311 degree

Explanation:

Part a)

As we know that the speed of light is given as

c = 3 \times 10^8 m/s

\lambda = 630 nm

now the frequency of the light is given as

f = \frac{c}{\lambda}

so we have

f = \frac{3 \times 10^8}{630 \times 10^{-9}}

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Part b)

Position of Nth maximum intensity on the screen is given as

y_n = \frac{n\lambda L}{d}

so here we know for 3rd order maximum intensity

y_3 = 0.76 cm

n = 3

L = 1.4 m

0.76 \times 10^{-2} = \frac{3(630 \times 10^{-9})(1.4)}{d}

d = 3.48 \times 10^{-4} m

Part c)

angle of third order maximum is given as

d sin\theta = 3 \lambda

3.48 \times 10^{-4} sin\theta = 3(630 \times 10^{-9})

\theta = 0.311 degree

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bazaltina [42]

Answer:

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