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3 years ago

5

A wheel with a tire mounted on it rotates at the constant rate of 2.73 revolutions per second. Find the radial acceleration of a

tack stuck in the tire at a distance of 37.7 cm from the rotation axis.

1 answer:

Lostsunrise [7]3 years ago

4 0

Answer:

110.9 m/s²

Explanation:

Given:

Distance of the tack from the rotational axis (r) = 37.7 cm

Constant rate of rotation (N) = 2.73 revolutions per second

Now, we know that,

1 revolution = $2\pi$ radians

So, 2.73 revolutions = $2.73\times 2\pi=17.153\ radians$

Therefore, the angular velocity of the tack is, $\omega=17.153\ rad/s$

Now, radial acceleration of the tack is given as:

$a_r=\omega^2 r$

Plug in the given values and solve for $a_r$ . This gives,

$a_r=(17.153\ rad/s)^2\times 37.7\ cm\\a_r=294.225\times 37.7\ cm/s^2\\a_r=11092.28\ cm/s^2\\a_r=110.9\ m/s^2\ \ \ \ \ \ \ [1\ cm = 0.01\ m]$

Therefore, the radial acceleration of the tack is 110.9 m/s².

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A potential difference of 24 V is applied to a 150-ohm resistor. How much current flows through the resistor?

Lady_Fox [76]

Given :- A resistor of 150 ohm, hence Resistance (R) = 150 ohm

Potential Difference (v) = 24 V

Current (I) = ?

V = IR

24 = I × 150

I = 24/150

I = 0.16 ampere

hope it helps!

3 0

3 years ago

The driver of a car traveling at 22.8 m/s applies the brakes and undergoes a constant deceleration of 2.95 m/s 2 . How many revo

a_sh-v [17]

Answer:

70 revolutions

Explanation:

We can start by the time it takes for the driver to come from 22.8m/s to full rest:

$t = \Delta v/a = (22.8 - 0)/2.95 = 7.73 s$

The tire angular velocity before stopping is:

$\omega_0 = v/r = 22.8 / 0.2 = 114 rad/s$

Also its angular decceleration:

$\alpha = a / r = 2.95/0.2 = 14.75 rad/s^2$

Using the following equation motion we can findout the angle it makes during the deceleration:

$\omega^2 - \omega_0^2 = 2\alpha\Delta \theta$

where $\omega$ = 0 m/s is the final angular velocity of the car when it stops, $\omega_0$ = 114rad/s is the initial angular velocity of the car $\alpha$ = 14.75 rad/s2 is the deceleration of the can, and $\Delta \theta$ is the angular distance traveled, which we care looking for:

$-114^2 = 2*(-14.75)*\Delta \theta$

$\Delta \theta = 440rad$ or 440/2π = 70 revelutions

4 0

3 years ago

After your school's team wins the regional championship, students go to the dorm roof and start setting off fireworks rockets. T

oksian1 [2.3K]

Answer:

required distance is 233.35 m

Explanation:

Given the data in the question;

Sound intensity $I$ = 1.62 × 10⁻⁶ W/m²

distance r = 165 m

at what distance from the explosion is the sound intensity half this value?

we know that;

Sound intensity $I$ is proportional to 1/(distance)²

i.e

$I$ ∝ 1/r²

Now, let r² be the distance where sound intensity is half, i.e $I$ ₂ = $I$ ₁/2

Hence,

$I$ ₂/ $I$ ₁ = r₁²/r₂²

1/2 = (165)²/ r₂²

r₂² = 2 × (165)²

r₂² = 2 × 27225

r₂² = 54450

r₂ = √54450

r₂ = 233.35 m

Therefore, required distance is 233.35 m

6 0

3 years ago

This is 100 points. When i find out how i will put the first person to answer as brainiest.

grin007 [14]

Answer:

Thank you so much! Have a great day!

3 0

2 years ago

In an electric shaver, the blade moves back and forth over a distance of 2.0 mm in simple harmonic motion, with frequency 124 Hz

bagirrra123 [75]

Answer:

(a) A = 1 mm

(b) $V_{max}=0.77872 m/s$

(c) $a_{max}=606.4 m/s^{2}/tex]Explanation:Distance moved back and forth = 2 mm Frequency, f = 124 HzSo, amplitude is the half of the distance traveled back and forth. (a) So, amplitude, A = 1 mm(b) Angular frequency, ω = 2 π f = 2 x 3.14 x 124 = 778.72 rad/s The formula for the maximum speed is given by [tex]V_{max}=\omega \times A$

$V_{max}=778.72 \times 0.001$

$V_{max}=0.77872 m/s$

(c) The formula for the maximum acceleration is given by

$a_{max}=\omega ^{2}A$

$a_{max}=778.72 ^{2}\times 0.001$

[tex]a_{max}=606.4 m/s^{2}/tex]

8 0

3 years ago