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Ivan
3 years ago
7

Transverse waves travel with a speed of 20 m/s on a string under a tension 0f 6.00 N. What tension is required for a wave speed

of 30.0 m/s on the same string?
Physics
1 answer:
Aleksandr [31]3 years ago
5 0

Answer:

T_2=13.5\ N

Explanation:

Given that,

Speed of transverse wave, v₁ = 20 m/s

Tension in the string, T₁ = 6 N

Let T₂ is the tension required for a wave speed of 30 m/s on the same string. The speed of a transverse wave in a string is given by :

v=\sqrt{\dfrac{T}{\mu}}........(1)

T is the tension in the string

\mu is mass per unit length

It is clear from equation (1) that :

v\propto\sqrt{T}

\dfrac{v_1}{v_2}=\sqrt{\dfrac{T_1}{T_2}}

T_2=T_1\times (\dfrac{v_2}{v_1})^2

T_2=6\times (\dfrac{30}{20})^2

T_2=13.5\ N

So, the tension of 13.5 N is required for a wave speed of 30 m/s. Hence, this is the required solution.

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The volume of 2.0 kg of helium in a piston-cylinder device is initially 7 m3. Now the helium is compressed to 5 m3 while its pre
IrinaVladis [17]

Answer:

A) T1 = 269.63 K

T2 = 192.59 K

B) W = -320 KJ

Explanation:

We are given;

Initial volume: V1 = 7 m³

Final Volume; V2 = 5 m³

Constant Pressure; P = 160 KPa

Mass; m = 2 kg

To find the initial and final temperatures, we will use the ideal gas formula;

T = PV/mR

Where R is gas constant of helium = R = 2.0769 kPa.m/kg

Thus;

Initial temperature; T1 = (160 × 7)/(2 × 2.0769) = 269.63 K

Final temperature; T2 = (160 × 5)/(2 × 2.0769) = 192.59 K

B) world one is given by the formula;

W = P(V2 - V1)

W = 160(5 - 7)

W = -320 KJ

6 0
3 years ago
A net force of 0.7 N is applied on a body. What happens to the acceleration of the body in a second trial if half of the net for
Dmitriy789 [7]

Answer:

The answer  is The acceleration is double its original value.

Explanation:

<h2><u>It is because of the second trial of accelaration. Because of this, an object's acceleration doubles from its original value.</u></h2><h2><u></u></h2>

Hope this helps....

Have a nice day!!!!

6 0
3 years ago
Read 2 more answers
When you irradiate a metal with light of wavelength 433 nm in an investigation of the photoelectric effect, you discover that a
sergij07 [2.7K]

Answer:

The right solution is:

(a) 2.87 eV

(b) 1.4375 eV

Explanation:

Given:

Wavelength,

= 433 nm

Potential difference,

= 1.43 V

Now,

(a)

The energy of photon will be:

E = \frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}

  = 4.59\times 10^{-19} \ J

or,

  = \frac{4.59\times 10^{-19}}{1.6\times 10^{-19}}

  = 2.87 \ eV

(b)

As we know,

⇒ Vq=\frac{hc}{\lambda}-\Phi_0

By substituting the values, we get

⇒ 1.43\times 1.6\times 10^{19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}-\Phi_0

⇒                       \Phi_0=2.3\times 10^{-19} \ J

or,

⇒                            =\frac{2.3\times 10^{-19}}{1.6\times 10^{-19}}

⇒                            =1.4375 \ eV

5 0
3 years ago
A force of 15 newtons is used to push a box along the floor a distance of 3 meters. How much work was done?
jeka57 [31]

Answer:

<h3>The answer is 45 J</h3>

Explanation:

The work done by an object can be found by using the formula

<h3>workdone = force × distance</h3>

From the question

distance = 3 meters

force = 15 newtons

We have

workdone = 15 × 3

We have the final answer as

<h3>45 J</h3>

Hope this helps you

7 0
3 years ago
A 0.153 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.700 m/s. It has a head-on col
DedPeter [7]

Answer:

3.1216 m/s.

Explanation:

Given:

M1 = 0.153 kg

v1 = 0.7 m/s

M2 = 0.308 kg

v2 = -2.16 m/s

M1v1 + M2v2 = M1V1 + M2V2

0.153 × 0.7 + 0.308 × -2.16 = 0.153 × V1 + 0.308 × V2

= 0.1071 - 0.66528 = 0.153 × V1 + 0.308 × V2

0.153V1 + 0.308V2 = -0.55818. i

For the velocities,

v1 - v2 = -(V1 - V2)

0.7 - (-2.16) = -(V1 - V2)

-(V1 - V2) = 2.86

V2 - V1 = 2.86. ii

Solving equation i and ii simultaneously,

V1 = 3.1216 m/s

V2 = 0.2616 m/s

8 0
3 years ago
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