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1 year ago

Q|C A sinusoidal wave in a string is described by the wave function

Physics

1 answer:

creativ13 [48]1 year ago

4 0

The maximum transverse force on a 1.00cm segment of the string is 0.045N.

The maximum transverse force on the segment,

l=1.00cm=1.00cm×1m/100cm=0.01m is,

F=μla

=(12.0×10^−3kg/m)×(0.01m)×(375m/s^2)

=0.045N

Hence the maximum force is 0.045N

Learn more about transverse force here:

brainly.com/question/13618693

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A spring is 6.0cm long when it is not stretched, and 10cm long when a 7.0N force is applied. What force is needed to make it 20c

Artist 52 [7]

Answer:

Approximately $25\; {\rm N}$ (assuming that this spring is ideal.)

Explanation:

The displacement of a spring is the new length of the spring relative to the original length.

For example:

When the $6.0\; {\rm cm}$ -spring in this question is stretched to $10\; {\rm cm}$ , the displacement is $x = (10\; {\rm cm} - 6.0\; {\rm cm})$ .
Likewise, if this spring is stretched to $20\; {\rm cm}$ , the displacement would be $(20\; {\rm cm} - 6\; {\rm cm})$ .

If this spring is ideal, the force on the spring would be proportional to the displacement of the spring. In other words, if a force of $F_{\text{a}}$ displaces this spring by $x_{\text{a}}$ , while a force of $F_{\text{b}}$ displaces this spring by $x_{\text{b}}$ , then:

$\displaystyle \frac{F_{\text{a}}}{x_{\text{a}}} = \frac{F_{\text{b}}}{x_{\text{b}}}$ .

In this question, it is given that a force of $F_{\text{a}} = 7.0 \; {\rm N}$ would stretch this spring by $x_{\text{a}} = (10\; {\rm cm} - 6.0\; {\rm cm})$ . Thus, the force $F_{\text{b}}$ required to stretch this spring by $x_{\text{a}} = (20\; {\rm cm} - 6.0\; {\rm cm})$ would satisfy:

$\displaystyle \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}}= \frac{F_{\text{b}}}{20\; {\rm cm} - 6.0\; {\rm cm}}$ .

Rearrange and solve for $F_{\text{b}}$ :

$\begin{aligned} F_{\text{b}} &= \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}} \, (20\; {\rm cm} - 6.0\; {\rm cm}) \\ &\approx 25\; {\rm N}\end{aligned}$ .

7 0

1 year ago

An object is moving to the west at a constant speed. three forces are exerted on the object. one force is 10 n directed due nort

SVETLANKA909090 [29]

If we want the object to continue to move at constant speed, it means that the resultant of the forces acting on the object must be zero. So far, we have:
- force F1 with direction north, of 10 N
- force F2 with direction west, of 10 N
The third force must balance them, in order to have a net force of zero on the object.

The resultant of the two forces F1 and F2 is
$F_{12} = \sqrt{F_1^2+F_2^2}= \sqrt{(10 N)^2+(10 N)^2}= \sqrt{200}=14.1 N$
with direction at $45^{\circ}$ north-west. This means that F3 must be equal and opposite to this force: so, F3 must have magnitude 14.1 N and its direction should be $45^{\circ}$ south-east.

6 0

3 years ago

Using this formula a = F/m What acceleration results from exerting a 125N force on a 0.65kg

Softa [21]

Answer:

Acceleration = 192.3 m/s² (Approx.)

Explanation:

Given:

Force = 125 N

Mass of ball = 0.65 kg

Find:

Acceleration

Computation:

We know that;

Acceleration = Force / Mas

So,

Acceleration = 125 / 0.65

Acceleration = 192.3 m/s² (Approx.)

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3 years ago

2 ways in which the effects of friction can be minimised in the experiment

Vlad [161]

Answer:

Friction can be minimized by using lubricants like oil and grease and by using ball bearing between machine parts. A substance that is introduced between two surfaces in contact, to reduce friction, is called a lubricant. Fluid friction can be minimized by giving suitable shapes to the objects moving in the fluids.

Explanation:

hope it helps

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2 years ago

Use the motion map to answer the question, describe the position and velocity of the object based moo the motion map?

drek231 [11]

Do you have an image?

answer:

3 0

3 years ago