All categories

3 years ago

If you increase the length ? of a pendulum by a factor of 9, how will the period t increase?

Physics

2 answers:

NeX [460]3 years ago

4 0

T = 2π√(L/g)

T⇒ √L

If L goes up by a factor of 9, T goes up by a factor of √9.

garik1379 [7]3 years ago

3 0

T = 2 * pie √(L/g)

so, if length is increased by 9

then time period is increased by √9 = 3

hope it helped :)

You might be interested in

A flute player hears four beats per second when she compares her note to an 880 Hz tuning fork (note A). She can match the frequ

ludmilkaskok [199]

Answer:

884Hz

Explanation:

Beats is the absolute difference between two frequencies therefore

Beats = f1-f2

4=f1-880

F1=880+4

F1=884Hz

7 0

3 years ago

(20 points) You are at the center of a boat and have been rowing the boat for a long time. You weigh only 80 kg and your 120 kg

valina [46]

Answer:

Explanation:

From the given information:

Let the first weight be $m_ 1$ = 80 kg

The weight of the buddy be $m_2$ = 120 kg

The weight of Bubba be $m_3$ = 60 kg

Also, since you and Budda are a distance of 4m to each other, then the length to which both meet buddy will be:

$x_1 = x_3 = \dfrac{4}{2} \\ \\ = 2$

The length of the boat be $x_2$ = 4 m

∴

We can find the center of mass of the system by using the formula:

$X_{CM} = \dfrac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3} \\ \\ X_{CM} = \dfrac{(80 \times 2)+(120\times4)+(60\times2)}{80+120+60} \\ \\ X_{CM} = \dfrac{160+480+120}{260} \\ \\ \mathbf{X_{CM} = 2.923}$

4 0

3 years ago

Problems with solar energy include _____.

german

First choice: the inability of current technology to capture
large amounts of the Sun's energy

Well, it's true that large amounts of it get away ... our 'efficiency' at capturing it is still rather low. But the amount of free energy we're able to capture is still huge and significant, so this isn't really a major problem.

Second choice: the inability of current technology to store
captured solar energy

No. We're pretty good at building batteries to store small amounts, or raising water to store large amounts. Storage could be better and cheaper than it is, but we can store huge amounts of captured solar energy right now, so this isn't a major problem either.

Third choice: inconsistencies in the availability of the resource

I think this is it. If we come to depend on solar energy, then we're
expectedly out of luck at night, and we may unexpectedly be out
of luck during long periods of overcast skies.

Fourth choice: lack of demand for solar energy

If there is a lack of demand, it's purely a result of willful manipulation
of the market by those whose interests are hurt by solar energy.

4 0

3 years ago

Read 2 more answers

To develop muscle tone, a woman lifts a 2.50 kg weight held in her hand. She uses her biceps muscle to flex the lower arm throug

Romashka [77]

To solve this problem we will use the concepts related to Torque as a function of the Force in proportion to the radius to which it is applied. In turn, we will use the concepts of energy expressed as Work, and which is described as the Torque's rate of change in proportion to angular displacement:

$\tau = Fr$

Where,

F = Force

r = Radius

Replacing we have that,

$\tau = Fr$

$\tau = 21cm (\frac{1m}{100cm})* 550N$

$\tau = 11.55Nm$

The moment of inertia is given by 2.5kg of the weight in hand by the distance squared to the joint of the body of 24 cm, therefore

$I = 0.25Kg\cdot m^2 +(2.5kg)(0.24m)^2$

$I = 0.394kg\cdot m^2$

Finally, angular acceleration is a result of the expression of torque by inertia, therefore

$\tau = I\alpha \rightarrow \alpha = \frac{\tau}{I}$

$\alpha = \frac{11.55}{0.394}$

$\alpha = 29.3 rad/s^2$

PART B)

The work done is equivalent to the torque applied by the distance traveled by 60 °° in radians $(\pi / 3)$ , therefore

$W = \tau \theta$

$W = 11.5* \frac{\pi}{3}$

$W = 12.09J$

4 0

3 years ago

Problem 4: A uniform flat disk of radius R and mass 2M is pivoted at point P A point mass of 1/2 M is attached to the edge of th

brilliants [131]

From the case we know that:

The moment of inertia Icm of the uniform flat disk witout the point mass is Icm = MR².
The moment of inerta with respect to point P on the disk without the point mass is Ip = 3MR².
The total moment of inertia (of the disk with the point mass with respect to point P) is I total = 5MR².

Please refer to the image below.

We know from the case, that:

m = 2M

r = R

m2 = 1/2M

distance between the center of mass to point P = p = R

Distance of the point mass to point P = d = 2R

We know that the moment of inertia for an uniform flat disk is 1/2mr². Then the moment of inertia for the uniform flat disk is:

Icm = 1/2mr²

Icm = 1/2(2M)(R²)

Icm = MR² ... (i)

Next, we will find the moment of inertia of the disk with respect to point P. We know that point P is positioned at the arc of the disk. Hence:

Ip = Icm + mp²

Ip = MR² + (2M)R²

Ip = 3MR² ... (ii)

Then, the total moment of inertia of the disk with the point mass is:

I total = Ip + I mass

I total = 3MR² + (1/2M)(2R)²

I total = 3MR² + 2MR²

I total = 5MR² ... (iii)

Learn more about Uniform Flat Disk here: brainly.com/question/14595971

#SPJ4

8 0

1 year ago