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Ilia_Sergeevich [38]

3 years ago

14

A dramatic demonstration, called "singing rods," involves a long, slender aluminum rod held in the hand near the rod's midpoint.

the rod is stroked with the other hand. with a little practice, the rod can be made to "sing," or emit a clear, loud, ringing sound.

1 answer:

Korvikt [17]3 years ago

8 0

More vibration = higher frequency

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(answer the question or get reported) Please solve it with the steps thanks :)

DaniilM [7]

Answer:

Fx = 35.36 N

Fy = 35.36 N

Explanation:

From the question,

The X component of the force is

Fx = Fcos∅.................. Equation 1

Where Fx = X component of the force, F = Force, ∅ = Angle to the horizontal.

Give: F = 50 N, ∅ = 45°

Substitute into equation 1

Fx = 50(cos45°)

Fx = 50(0.7071)

Fx = 35.36 N

Similarly,

For Y component

Fy = Fsin∅

Where F y = Y component

Fy = 50(sin45°)

Fy = 50(0.7071)

Fy = 35.36 N

4 0

2 years ago

A metal disk of radius 6.0 cm is mounted on a frictionless axle. Current can flow through the axle out along the disk, to a slid

Galina-37 [17]

Answer:

0.09 N

Explanation:

We are given that

Radius of disk,r=6 cm= $\frac{6}{100}=0.06 m$

1 m=100 cm

B=1 T

Current,I=3 A

We have to find the frictional force at the rim between the stationary electrical contact and the rotating rim.

$dF=IBdr$

$dF=IBdr$

$\tau=rdF=IBrdr$

$\tau=\int_{0}^{R}IBr dr$

$\tau=IB(\frac{R^2}{2}$

Torque due to friction

$\tau=R\times F$

Where friction force=F

$R\times F=\frac{IBR^2}{2}$

$F=\frac{IBR}{2}$

Substitute the values

$F=\frac{3\times 1\times 0.06}{2}$

$F=0.09 N$

7 0

3 years ago

A 1500-kg car locks its brakes and skids to a stop on a slippery horizontal road, leaving skid marks that are 15 m long. How muc

Harman [31]

Answer:

$E=88200\ J$

Explanation:

Given:

mass of car, $m=1500\ kg$

distance of skidding after the application of brakes, $d=15\ m$

coefficient of kinetic friction, $\mu_k=0.4$

<u>So, the energy dissipated during the skidding of car:</u>

<em>Frictional force:</em>

$f=\mu_k.N$

where N = normal reaction by ground on the car

$f=0.4\ties 1500\times 9.8$

$f=5880\ N$

<em>Now from the work-energy equivalence:</em>

$E=f.d$

$E=5880\times 15$

$E=88200\ J$ is the dissipated energy.

3 0

2 years ago

Can someone help me :)

zhenek [66]

Answer:

undergone a chemical change

Explanation:

6 0

2 years ago

Read 2 more answers

A uniform electric field of strength E points to the right. An electron is fired with a velocity v0 to the right and travels a d

zvonat [6]

Answer:

<u /> $D_l=d$ <u />

Explanation:

From the question we are told that:

The Electric field of strength direction =Right

The Velocity of The First Electron=V_0

The Velocity of The Second Electron=V_0

Therefore

$V_{e1}=V_{e2}$

Generally, the equation for the Horizontal Displacement of electron is mathematically given by

$D=\frac{at^2}{2}$

Where

Acceleration is given as

$a=\frac{V_o}{2d}$

And

Time

$T=\frac{d}{v_0}$

Therefore horizontal displacement towards the left is

$D_l=\frac{(\frac{V_o}{2d})(\frac{d}{v_0})^2}{2}$

<u /> $D_l=d$ <u />

5 0

2 years ago