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3 years ago

C) An identical spring is pulled with a force or 75 N The elastie limit of the spring is 72N

Physics

1 answer:

tiny-mole [99]3 years ago

8 0

Answer:

Spring cannot return to its original, since a part of its deformation is plastic, not elastic.

Explanation:

Physically speaking, stress is equal to the axial force divided by effective transversal area of spring. In addition, springs have usually a linear relationship between stress and strain in elastic region, since they are made of ductile materials. Axial force is directly proportional to axial stress, which is also directly proportional to axial strain.

Then, if force is greater than force associated with elastic limit of the spring, then spring cannot return to its original, since a part of its deformation is plastic, not elastic.

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What would happen when two or more capacitors are connected in parallel across a potential difference?

Varvara68 [4.7K]

Answer:

The Current Iₜ = I₁ + I₂ + I₃

Charge Qₜ = Q₁ + Q₂ + Q₃

Potential difference Vₜ = V₁ = V₂ = V₃

The total capacitance Cₜ = C₁ + C₂ + C₃

Explanation:

According to the attached image;

For parallel arrangements of capacitors, the current flowing through each of the capacitors sums up to the total current flowing through the circuit;

Iₜ = I₁ + I₂ + I₃

Also the charge storage by each capacitor sums up to give the total charge stored;

Qₜ = Q₁ + Q₂ + Q₃

The potential difference across each of the capacitors are the same and equal to the total voltage across the circuit;

Vₜ = V₁ = V₂ = V₃

The total capacitance equals the sum of the capacitances of each of the capacitors;

Cₜ = C₁ + C₂ + C₃

6 0

3 years ago

A 0.25-m string, vibrating in its sixth harmonic, excites a 0.96-m pipe that is open at both ends into its second overtone reson

Andrews [41]

Answer:

option D

Explanation:

given,

length of the pipe, L = 0.96 m

Speed of sound,v = 345 m/s

Resonating frequency when both the end is open

$f = \dfrac{nv}{2L}$

n is the Harmonic number

2nd overtone = 3rd harmonic

so, here n = 3

now,

$f = \dfrac{3\times 345}{2\times 0.96}$

f = 540 Hz

The common resonant frequency of the string and the pipe is closest to 540 Hz.

the correct answer is option D

7 0

3 years ago

A 70.0 kg ice hockey goalie, originally at rest, has a 0.110 kg hockey puck slapped at him at a velocity of 31.5 m/s. Suppose th

NISA [10]

Answer

given,

mass of the goalie(m₁) = 70 kg

mass of the puck (m₂)= 0.11 kg

velocity of the puck = 31.5 m/s

elastic collision

$v_1=\dfrac{m_2-m_1}{m_1+m_2}v_1+\dfrac{2m_2}{m_1+m_2}v_2$

$v_{pf}=\dfrac{0.11-70}{0.11+70}31.5+\dfrac{2m_2}{m_1+m_2}\times (0)$

$v_{pf}=-31.4\ m/s$

$v'_2 = \dfrac{2m_1v_1}{m_1+m_2}-\dfrac{(m_2-m_1)v_2}{m_2+m_1}$

$v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}-\dfrac{(0.11-70)\times 0}{m_1+m_2}$

$v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}$

$v_{gf} = 0.0988\ m/s$

4 0

3 years ago

A spaceship is traveling toward Earth while giving off a constant radio signal with a wavelength of 1 meter (m). What will the s

Nadya [2.5K]

Answer:

Less than 1 m

Explanation:

When objects are getting closer to each other there is a slight change in the wavelength that is being transmitted by either objects. This is known as the blue shift of waves. Here, the wavelength reduces.

In the opposite case the when objects are getting farther from each other there is a slight change in the wavelength that is being transmitted by either objects. This is known as the red shift. Here, the wavelength increases.

In this case the spaceship is getting close to Earth hence the wavelength will be lower than 1 m.

4 0

3 years ago

A stationary police car emits a sound of frequency 1240 HzHz that bounces off of a car on the highway and returns with a frequen

Tju [1.3M]

Answer

given,

frequency from Police car= 1240 Hz

frequency of sound after return = 1275 Hz

Calculating the speed of the car = ?

Using Doppler's effect formula

Frequency received by the other car

$f_1 = \dfrac{f_0(u + v)}{u}$ ..........(1)

u is the speed of sound = 340 m/s

v is the speed of the car

Frequency of the police car received

$f_2= \dfrac{f_1(u)}{u-v}$

now, inserting the value of equation (1)

$f_2= f_0\dfrac{u+v}{u-v}$

$1275=1240\times \dfrac{340+v}{340-v}$

1.02822(340 - v) = 340 + v

2.02822 v = 340 x 0.028822

2.02822 v = 9.799

v = 4.83 m/s

hence, the speed of the car is equal to v = 4.83 m/s

5 0

3 years ago