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Vesna [10]
3 years ago
15

If a metallic wire is stretched regularly so that its length increased to the double,

Physics
1 answer:
Travka [436]3 years ago
5 0

Answer:

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Explanation:

scjigbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbefvuabddddddddaaetaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaatttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttgbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbuyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy

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Find τf, the torque about point p due to the force applied by the achilles' tendon.
Luba_88 [7]
The formula for the torque is
<span>τf = p F
where
</span><span>τf is the torque
p is the distance where the force is applied by the tendon
F is force applied by the tendon

If there are given values, substitute in the equation and solve for the torque.</span>
3 0
4 years ago
How much heat is required to raise the temperature of 5kg of iron from 50°c to250°c​
Marianna [84]

Answer:

462000J

Explanation:

Quantity of heat= mass x specific heat capacity of iron x change in temp

specific heat capacity of iron is 462J/Kg/K

change in temp = 250-50= 200°C

200°C is equivalent to 200K since 1°C is 1K

Q= mct

= 5x462x200

= 462000J

4 0
3 years ago
A violinist tuning her violin plays her A-string while sounding a tuning fork at A 415 Hz, and hears 5 beats per second. When sh
frutty [35]

Answer:410 Hz

Explanation:

It is given that frequency of tuning fork is 415 Hz and hears a beat of 5 beat per second

so untuned frequency must be either 410 Hz or 420 Hz because beat frequency is the difference in frequency of two notes.  

When she tightens the string the beat frequency decrease

and velocity v=\nu \lambda

where \nu =frequency

\lambda =wavelength

v=velocity

v=\sqrt{\frac{T}{\mu }}

where T=tension

it T increase v also increases and from 1 st equation frequency is also increasing therefore untuned frequency must be 410 Hz because beat frequency is decreasing.                    

3 0
4 years ago
A very long uniform line of charge has charge per unit length 4.54 μC/m and lies along the x-axis. A second long uniform line of
kozerog [31]

Answer:

The magnitude of the net electric field is 6.57\times10^{5}\ N/C

Explanation:

Given that,

Charge density \lambda = 4.54\ \mu C/m

Charge density \lambda' = -2.58\ \mu C/m

Distance y_{1}= 0.384\ m

Distance y_{2}= 0.204\ m

We need to calculate the magnitude of the net electric field

Using formula of electric field

E=E_{1}+E_{2}

E=\dfrac{1}{2\pi\epsilon_{0}}(\dfrac{\lambda}{r}+\dfrac{\lambda'}{r'})

Put the value into the formula

E=\dfrac{1}{2\pi\times8.85\times10^{-12}}(\dfrac{4.54\times10^{-6}}{0.204}+\dfrac{2.58\times10^{-6}}{0.384-0.204})

E=6.57\times10^{5}\ N/C

Hence, The magnitude of the net electric field is 6.57\times10^{5}\ N/C

8 0
4 years ago
Which would have the highest frequency of vibration? (Prove mathematically.) Pendulum A with a 200 g mass on a 1.0 m string Pend
Lera25 [3.4K]

Answer:

Pendulum B

Explanation:

The time period of a pendulum is given by :

T=2\pi\sqrt{\dfrac{L}{g}}

Case 1.

Mass, m = 200 g = 0.2 kg

Length of string, l = 1 m

Time, T_1=2\pi\sqrt{\dfrac{1\ m}{9.8\ m/s^2}}

T₁ = 2.007 Seconds

Since, f=\dfrac{1}{T_1}

f_1=\dfrac{1}{2.007}

f₁ = 0.49 Hz

Case 2.

Mass, m = 400 g = 0.4 kg

Length of string, l = 0.5 m

Time, T_2=2\pi\sqrt{\dfrac{0.5\ m}{9.8\ m/s^2}}

T₂ = 1.41 seconds

f₂=\dfrac{1}{T_2}

f₂=\dfrac{1}{1.41}

f₂ = 0.709 seconds

Hence, pendulum B have highest frequency of vibration.

3 0
4 years ago
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