Solve this question plz

A wall clock has a minute hand with a length of 0.55 m and an hour hand with a length of 0.26 m. Take the center of the clock as

Lemur [1.5K]

Answer:

The magnitude of the acceleration of the tip of the minute hand of the clock $1.675\times10^{-6}\ m/s^2$ .

Explanation:

Given that,

Length of minute hand = 0.55 m

Length of hour hand = 0.26 m

The time taken by the minute hand to complete one revelation is

$T= 3600\ sec$

We need to calculate the angular frequency

Using formula of angular frequency

$\omega=\dfrac{2\pi}{T}$

Put the value into the formula

$\omega=\dfrac{2\pi}{3600}$

$\omega=0.001745\ rad/s$

We need to calculate the magnitude of the acceleration of the tip of the minute hand of the clock

Using formula of acceleration

$a=r\omega^2$

Put the value into the formula

$a=0.55\times(0.001745)^2$

$a=1.675\times10^{-6}\ m/s^2$

Hence, The magnitude of the acceleration of the tip of the minute hand of the clock $1.675\times10^{-6}\ m/s^2$ .

3 0

3 years ago

A spring stretches by 0.0208 m when a 3.39-kg object is suspended from its end. How much mass should be attached to this spring

sashaice [31]

Answer:

COMPLETE QUESTION

A spring stretches by 0.018 m when a 2.8-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 3.0 Hz?

Explanation:

Given that,

Extension of spring

x = 0.0208m

Mass attached m = 3.39kg

Additional mass to have a frequency f

Let the additional mass be m

Using Hooke's law

F= kx

Where F = W = mg = 3.39 ×9.81

F = 33.26N

Then,

F = kx

k = F/x

k = 33.26/0.0208

k = 1598.84 N/m

The frequency is given as

f = ½π√k/m

Make m subject of formula

f² = ¼π² •(k/m

4π²f² = k/m

Then, m4π²f² = k

So, m = k/(4π²f²)

So, this is the general formula,

Then let use the frequency above

f = 3Hz

m = 1598.84/(4×π²×3²)

m = 4.5 kg

4 0

3 years ago

Suppose that a public address system emits sound uniformly in all directions and that there are no reflections . The intensity a

Marina86 [1]

Answer:

$I_{2}=2.39*10^{-5} W/m^{2}$

Explanation:

The definition of the intensity in terms of power is given by:

$I=\frac{P}{A}$

Where:

P is the power
A is the area

If the sound emits uniformly in all directions and that there are no reflections, we can assume the geometry of the wave sound is spherical.

Let's recall the area of a sphere is $A = 4\pi R^{2}$

To the first location we have:

$I_{1}=3*10^{-4} W/m^{2}=\frac{P}{4\pi 22^{2}}$

and to the second location we have:

$I_{2}=\frac{P}{4\pi 78^{2}}$

Now, we can divide each intensity to find the second intensity.

$\frac{I_{2}}{I_{1}}=\frac{\frac{P}{4\pi 78^{2}}}{\frac{P}{4\pi 22^{2}}}$

$I_{2}=I_{1}* \frac{22^{2}}{78^{2}}$

$I_{2}=3*10^{-4}\frac{22^{2}}{78^{2}}$

$I_{2}=2.39*10^{-5} W/m^{2}$

I hope it helps you!

5 0

3 years ago

Read 2 more answers

Which energy transformation converts energy from the sun’s core to light energy needed by plants?

Assoli18 [71]

Nuclear to electromagnetic

3 0

3 years ago

What is the stretch when you pull with a force of 25 N on a spring with a spring constant of 8 N/m? *

Pani-rosa [81]

Hooke's Law

$\tt F=k.\Delta x$

k = spring constant

x = stretch

F = force

Input the value

$\tt \Delta x=\dfrac{F}{k}=\dfrac{25}{8}=3.125\rightarrow 3.13\:m$

7 0

2 years ago