8. You are heating a substance in a test tube. Always point the open end of the ne

A nylon guitar string is fixed between two lab posts 2.00 m apart. The string has a linear mass density of μ=7.20 g/m\mu=7.20~\t

vladimir2022 [97]

Answer:

4.6 m

Explanation:

First of all, we can find the frequency of the wave in the string with the formula:

$f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}$

where we have

L = 2.00 m is the length of the string

T = 160.00 N is the tension

$\mu =7.20 g/m = 0.0072 kg/m$ is the mass linear density

Solving the equation,

$f=\frac{1}{2(2.00 m)}\sqrt{\frac{160.00 N}{0.0072 kg/m}}=37.3 Hz$

The frequency of the wave in the string is transmitted into the tube, which oscillates resonating at same frequency.

The n=1 mode (fundamental frequency) of an open-open tube is given by

$f=\frac{v}{2L}$

where

v = 343 m/s is the speed of sound

Using f = 37.3 Hz and re-arranging the equation, we find L, the length of the tube:

$L=\frac{v}{2f}=\frac{343 m/s}{2(37.3 Hz)}=4.6 m$

4 0

3 years ago

Sort the forces as producing a torque of positive, negative, or zero magnitude about the rotational axis identified in part

Fantom [35]

a) Angular acceleration: $17.0 rad/s^2$

b) Weight: conterclockwise torque, reaction force: zero torque

Explanation:

a)

In this problem, you are holding the pencil at its end: this means that the pencil will rotate about this point.

The only force producing a torque on the pencil is the weight of the pencil, of magnitude

$W=mg$

where m is the mass of the pencil and g the acceleration of gravity.

However, when the pencil is rotating around its end, only the component of the weight tangential to its circular trajectory will cause an angular acceleration. This component of the weight is:

$W_p =mg sin \theta$

where $\theta$ is the angle of the rod with respect to the vertical.

The weight act at the center of mass of the pencil, which is located at the middle of the pencil. So the torque produced is

$\tau = W_p \frac{L}{2}=mg\frac{L}{2} cos \theta$

where L is the length of the pencil.

The relationship between torque and angular acceleration $\alpha$ is

$\tau = I \alpha$ (1)

where

$I=\frac{1}{3}mL^2$

is the moment of inertia of the pencil with respect to its end.

Substituting into (1) and solving for $\alpha$ , we find:

$\alpha = \frac{\tau}{I}=\frac{mg\frac{L}{2}sin \theta}{\frac{1}{3}mL^2}=\frac{3 g sin \theta}{2L}$

And assuming that the length of the pencil is L = 15 cm = 0.15 m, the angular acceleration when $\theta=10^{\circ}$ is

$\alpha = \frac{3(9.8)(sin 10^{\circ})}{2(0.15)}=17.0 rad/s^2$

b)

There are only two forces acting on the pencil here:

- The weight of the pencil, of magnitude $mg$

- The normal reaction of the hand on the pencil, R

The torque exerted by each force is given by

$\tau = Fd$

where F is the magnitude of the force and d the distance between the force and the pivot point.

For the weight, we saw in part a) that the torque is

$\tau =mg\frac{L}{2} cos \theta$

For the reaction force, the torque is zero: this is because the reaction force is applied exctly at the pivot point, so d = 0, and therefore the torque is zero.

Therefore:

- Weight: counterclockwise torque (I have assumed that the pencil is held at its right end)

- Reaction force: zero torque

8 0

3 years ago

Somebody please help I need exuse to not go to pe because I injured my self anybody know what to say

algol13

Answer:

Say you feel like you twisted ur ankle or can move your foot or something in that category

3 0

2 years ago

9As a student walks downhill at constant speed, his gravitational potential energy

Minchanka [31]

Answer:

jfsfjsjfsjfsjgsjfskra

Explanation:

sgnsfnNCzgnzncxmhshdgNdfjBFzktajfskgskgzngzngzgnzngzngzfNFzngzngznfzngzngzngzng

4)sfjskgdgxgkzhldlhdhkdkhdgkdkgzgksfnsksgkskfsufsursjfsjrsjfBFHDahdJFzbfzkgzndcxdnvxekvxlhecihckheclheclhexhiecohceoheclhecohecohecohceohceohecohceihcelhevlhecohcehlcegicgei.

4 0

3 years ago

An oscillating block - spring system has a mechanical energy of 1.10 j, and amplitude of 11.0 cm, and a maximum speed of 1.7 m/s

ioda

The total mechanical energy of the block-spring system is given by the sum of the potential energy and the kinetic energy of the block:
$E=U+K= \frac{1}{2}kx^2 + \frac{1}{2}mv^2$
where
k is the spring constant
x is the elongation/compression of the spring
m is the mass of the block
v is the speed of the block

At the point of maximum displacement of the spring, the velocity of the block is zero: v=0, so the kinetic energy is zero and the mechanical energy is just potential energy of the spring:
$E= \frac{1}{2}kA^2$ (1)
where we used x=A, the amplitude (which is the maximum displacement of the spring).
Since we know
A = 11.0 cm= 0.11 m
E = 1.10 J
We can re-arrange (1) to find the spring constant:
$k= \frac{2E}{A^2} = \frac{2 \cdot 1.10 J}{(0.11 m)^2}=181.8 N/m$

3 0

2 years ago