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sleet_krkn [62]
3 years ago
11

Calculate the frequency of the 3rd normal mode of a guitar string of length 40.0cm and mass 0.5g when stretched with a tension o

f 80N.
Physics
1 answer:
Alexandra [31]3 years ago
7 0

Answer:

Explanation:

For third normal mode of vibration

l = \frac{3\lambda}{2}  , λ is wavelength , l is length of string .

.4 = \frac{3\lambda}{2}

λ = .267 m

velocity = \sqrt{\frac{T}{m} }

T is tension and m is mass unit length

m = .5 x 10⁻³ / 40 x 10⁻²

= .00125 kg / m

Putting the values

velocity = \sqrt{\frac{80}{.00125} }

= 253 m /s

frequency

= velocity / λ

= 253 / .267

= 947.5 Hz .

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A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t
Yuliya22 [10]

Answer:

a) W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

b) W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

c) V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

d)  d_1 =0.183m or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

Part b

For this case first we can convert the spring constant to N/m like this:

2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}

And the work donde by the spring on this case is given by:

W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i

And if we solve for the initial velocity we got:

V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

-1/2mV^2_i = W_g + W_{spring}

And replacing we got:

-1/2mV^2_i =mg d_1 -1/2 k d^2_1

And we can put the terms like this:

\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0

If we multiply all the equation by 2 we got:

k d^2_1 -2 mg d_1 -m V^2_i =0

Now we can replace the values and we got:

200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0

200 d^2_1 -2.058 d_1 -6.3525=0

And solving the quadratic equation we got that the solution for d_1 =0.183m or 18.3 cm because the negative solution not make sense.

5 0
3 years ago
4.00 kg glass coffee cup is 30.0°C at room temperature. It is then plunged into hot dishwater at a temperature of 90.0°C, as sho
statuscvo [17]

Answer:

Q = 200800 Joules.

Explanation:

Given the following data;

Mass = 4kg

Initial temperature = 30.0°C

Final temperature = 90.0°C

Specific heat capacity of glass = 837 J/kg°C

To find the quantity of heat absorbed;

Heat capacity is given by the formula;

Q = mcdt

Where;

Q represents the heat capacity or quantity of heat.

m represents the mass of an object.

c represents the specific heat capacity of water.

dt represents the change in temperature.

dt = T2 - T1

dt = 90 - 30

dt = 60°C

Substituting the values into the equation, we have;

Q = 4*837*60

Q = 200800 Joules.

Therefore, the amount of heat absorbed is 200800 Joules.

8 0
3 years ago
What are particles of carbon called?
Ainat [17]
Co2 or greenhouse gas
8 0
3 years ago
Power selection feature for resistors to become water modules 10 liters of water at 25°C to đến
Mila [183]

Answer:

P = 2439.5 W = 2.439 KW

Explanation:

First, we will find the mass of the water:

Mass = (Density)(Volume)

Mass = m = (1 kg/L)(10 L)

m = 10 kg

Now, we will find the energy required to heat the water between given temperature limits:

E = mCΔT

where,

E = energy = ?

C = specific heat capacity of water = 4182 J/kg.°C

ΔT = change in temperature = 95°C - 25°C = 70°C

Therefore,

E = (10 kg)(4182 J/kg.°C)(70°C)

E = 2.927 x 10⁶ J

Now, the power required will be:

Power = P = \frac{E}{t}

where,

t = time = (20 min)(60 s/1 min) = 1200 s

Therefore,

P = \frac{2.927\ x\ 10^6\ J}{1200\ s}

<u>P = 2439.5 W = 2.439 KW</u>

7 0
3 years ago
What is the middle layer of earth called
irinina [24]
The middle or centre of the Earth is the core. However the middle of the layers from the surface to the centre of the Earth is known as mantle. 
6 0
3 years ago
Read 2 more answers
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