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brilliants [131]

3 years ago

14

What is a centripetal acceleration of a point on a bicycle wheel of a radius of 0.70 m when a bike is moving 8.0 m/s

1 answer:

Furkat [3]3 years ago

3 0

Answer:

The acceleration of a point on the wheel is 11.43 m/s² acting radially inward.

Explanation:

The centripetal acceleration acts on a body when it is performing a circular motion.

Here, a point on the bicycle is performing circular motion as the rotation of the wheel produces a circular motion.

The centripetal acceleration of a point moving with a velocity $v$ and at a distance of $r$ from the axis of rotation is given as:

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

Here, $v=8\ m/s,r=0.70\ m$

∴ $a=\frac{8}{0.70}=11.43\ m/s^2$

Therefore, the acceleration of a point on the wheel is 11.43 m/s² acting radially inward.

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If 128 v is required to push 4.00 A of current though a resistor, what is the resistance

prisoha [69]

Answer:

32.0 Ω

Explanation:

Ohm's law:

V = IR

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R = 32.0 Ω

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2 years ago

(a) The Sun orbits the Milky Way galaxy once each 2.60 x 108y , with a roughly circular orbit averaging 3.00 x 104 light years i

PilotLPTM [1.2K]

Answer:

Part a)

$a_c = 1.67 \times 10^{-10} m/s^2$

Part b)

$v = 2.18 \times 10^5 m/s$

Explanation:

Time period of sun is given as

$T = 2.60 \times 10^8 years$

$T = 2.60 \times 10^8 (365 \times 24 \times 3600) s$

$T = 8.2 \times 10^{15} s$

Now the radius of the orbit of sun is given as

$R = 3.00 \times 10^4 Ly$

$R = 3.00 \times 10^4 (3\times 10^8)(365 \times 24 \times 3600)m$

$R = 2.84 \times 10^20 m$

Part a)

centripetal acceleration is given as

$a_c = \omega^2 R$

$a_c = \frac{4\pi^2}{T^2} R$

$a_c = \frac{4\pi^2}{(8.2\times 10^{15})^2}(2.84 \times 10^{20})$

$a_c = 1.67 \times 10^{-10} m/s^2$

Part b)

orbital speed is given as

$v = \frac{2\pi R}{T}$

$v = \frac{2\pi (2.84 \times 10^{20})}{8.2 \times 10^{15}}$

$v = 2.18 \times 10^5 m/s$

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2 years ago

Four point charges are individually brought from infinity and placed at the corners of a square. Each charge has the identical v

Brut [27]

To solve this problem we will apply the concept of voltage given by Coulomb's laws. From there we will define the charges and the distance, and we will obtain the total value of the potential difference in the system.

The length of diagonal is given as

$l = 2a$

The distance of the center of the square from each of the corners is

$r = \frac{2a}{2}= a$

The potential electric at the center due to each cornet charge is

$V_1 = \frac{kQ_1}{r_1}$

$V_2 = \frac{kQ_2}{r_2}$

$V_3 = \frac{kQ_3}{r_3}$

$V_4 = \frac{kQ_4}{r_4}$

The total electric potential at the center of the given square is

$V = V_1+V_2+V_3+V_4$

$V = \frac{kQ_1}{r_1}+ \frac{kQ_2}{r_2}+\frac{kQ_3}{r_3}+\frac{kQ_4}{r_4}$

Al the charges are equal, and the distance are equal to a, then

$V = \frac{kQ}{a}+ \frac{kQ}{a}+\frac{kQ}{a}+\frac{kQ}{a}$

$V = \frac{4kQ}{a}$

Therefore the correct option is E.

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3 years ago