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4 years ago

Find the first three harmonics of a string of linear mass density 2.00 g/m and length 0.600 m when the tension in it is 50.0 n.

Physics

1 answer:

blsea [12.9K]4 years ago

8 0

As we know that Nth harmonic of the string is given by

$f = \frac{N}{2L}\sqrt{\frac{T}{m/L}}$

now here we will have

$m/L = mass density = 2 g/m$

$m/L = 0.002 kg/m$

$Length = L = 0.600 m$

$Tension = T = 50.0 N$

now from above formula we have

$f = \frac{N}{2(0.600)}\sqrt{\frac{50.0}{0.002}}$

$f = 131.8N$

now for first harmonic N = 1

$f_1 = 131.8 Hz$

for second harmonic N = 2

$f_2 = 263.5 Hz$

for third harmonic N = 3

$f_3 = 395.3 Hz$

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In a lab, four balls have the same velocities but different masses.

olya-2409 [2.1K]

Answer:

New Momentum of Ball B $=13.2 \frac{\mathrm{kgm}}{\mathrm{s}}$

Explanation:

Given:

Mass of Ball A=1kg

Mass of Ball B= 2kg

Mass of Ball C=5kg

Mass of Ball D=7kg

Velocities of A=B=C=D=2.2 $\frac{m}{s}$

Momentum of Ball A=2.2 $\frac{k g m}{s}$

Momentum of Ball B=4.4 $\frac{k g m}{s}$

Momentum of Ball C=11 $\frac{k g m}{s}$

Momentum of Ball D=15 $\frac{k g m}{s}$

To Find:

Change in Momentum When of Ball B gets tripled

Solution:

Though all balls have same velocity, thus we get

Velocities of A=B=C=D=2.2 $\frac{m}{s}$

Initial Momentum of Ball B=4.4 $\frac{k g m}{s}$

If the Mass of Ball B gets tripled;

We get New Mass of Ball B=3×Actual Mass of the ball

=3×2=6kg

Thus we get Mass of Ball B=6kg

According to the formula,

Change in momentum of Ball B $\Delta p=m \times \Delta v$

Where $\Delta p$ =change in momentum

m=mass of the ball B

$\Delta v$ =change in velocity ball B

And $\Delta v=v,$ since all balls, have same velocity

Thus the above equation, changes to

$\Delta p=m \times v$

Substitute all the values in the above equation we get

$\Delta p=6 \times 2.2$

$=13.2 \frac{\mathrm{kgm}}{\mathrm{s}}$

Result:

Thus the New Momentum of ball B $=13.2 \frac{\mathrm{kgm}}{\mathrm{s}}$

3 0

3 years ago

Read 2 more answers

how much time would it take for an airplane to reach its destination if it traveled at an average speed of 790 km/hr for a dista

NeTakaya

I think your question is incomplete because the distance between destination and departure point isn't given in the question

8 0

3 years ago

A train is accelerating at a rate of 2 km/hr/s. If its initial velocity is 20 km/hr, what is its velocity after 30 seconds?

Mademuasel [1]

Here it is. *WARNING* VERY LONG ANSWER

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11) If Galileo had dropped a 5.0 kg cannon ball to the ground from a height of 12 m, the change in PE of the cannon ball would have been product of mass(m),acceleration(g)and height(h) 

The change in PE =mgh=5*9.8*12=588 J 
______________________________________... 
12.) The 2000 Belmont Stakes winner, Commendable, ran the horse race at an average speed = v = 15.98 m/s. 

Commendable and jockey Pat Day had a combined mass =M= 550.0 kg, 

Their KE as they crossed the line=(1/2)Mv^2 

Their KE as they crossed the line=0.5*550*(15.98)^2 

Their KE as they crossed the line is 70224.11 J 

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13)Brittany is changing the tire of her car on a steep hill of height =H= 20.0 m 

She trips and drops the spare tire of mass = m = 10.0 kg, 

The tire rolls down the hill with an intial speed = u = 2.00 m/s. 

The height of top of the next hill = h = 5.00 m 

Initial total mechanical energy =PE+KE=mgH+(1/2)mu^2 

Initial total mechanical energy =mgH+(1/2)mu^2 

Suppose the final speed at the top of second hill is v 

Final total mechanical energy =PE+KE=mgh+(1/2)mv^2 

As mechanical energy is conserved, 

Final total mechanical energy =Initial total mechanical energy 

mgh+(1/2)mv^2=mgH+(1/2)mu^2 

v = sq rt [u^2+2g(H-h)] 

v = sq rt [4+2*9.8(20-5)] 

v = sq rt 298 

v =17.2627 m/s 

The speed of the tire at the top of the next hill is 17.2627 m/s 
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14.) A Mexican jumping bean jumps with the aid of a small worm that lives inside the bean. 

a.)The mass of bean = m = 2.0 g 

Height up to which the been jumps = h = 1.0 cm from hand 

Potential energy gained in reaching its highest point= mgh=1.96*10^-4 J or 1960 erg 

b.) The speed as the bean lands back in the palm of your hand =v=sq rt2gh =sqrt 0.196 =0.4427 m/s or 44.27 cm/s 
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15.) A 500.-kg horse is standing at the top of a muddy hill on a rainy day. The hill is 100.0 m long with a vertical drop of 30.0 m. The pig slips and begins to slide down the hill. 

The pig's speed a the bottom of the hill = sq rt 2gh = sq rt 2*9.8*30 =sq rt 588 =24.249 m/s 
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16.) While on the moon, the Apollo astronauts Neil Armstrong jumped up with an intitial speed 'u'of 1.51 m/s to a height 'h' of 0.700 m, 

The gravitational acceleration he experienced = u^2/2h = 2.2801 /(2*0.7) = 1.629 m/s^2 
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EDIT 
1.) A train is accelerating at a rate = a = 2.0 km/hr/s. 

Acceleration 

Initial velocity = u = 20 km/hr, 

Velocity after 30 seconds = v = u + at 

Velocity after 30 seconds = v = 20 km/hr + 2 (km/hr/s)*30s = 

Velocity after 30 seconds = v = 20 km/hr + 60 km/hr = 80 km/ hr 

Velocity after 30 seconds = v = 80 km/hr=22.22 m/s 
_______________________________- 
2.) A runner achieves a velocity of 11.1 m/s 9 s after he begins. 

His acceleration = a =11.1/9=1.233 m/s^2 

Distance he covered = s = (1/2)at^2=49.95 m

7 0

3 years ago

Over a short interval the coordinate of a car in the meters isgiven by x(t) = 27t - 4.0 t3 where time t is in seconds,at the end

wariber [46]

Answer:

5. -24 m/s²

Explanation:

Acceleration: This can be defined as the rate of change of velocity.

The S.I unit of acceleration is m/s².

mathematically,

a = dv/dt ............................ Equation 1

Where a = acceleration, dv/dt = is the differentiation of velocity with respect to time.

But

v = dx(t)/dt

Where,

x(t) = 27t-4.0t³...................... Equation 2

Therefore, differentiating equation 2 with respect to time.

v = dx(t)/dt = 27-12t²............. Equation 3.

Also differentiating equation 3 with respect to time,

a = dv/dt = -24t

a = -24t .................... Equation 4

from the question,

At the end of 1.0 s,

a = -24(1)

a = -24 m/s².

Thus the acceleration = -24 m/s²

The right option is 5. -24 m/s²

4 0

3 years ago

The barricade at the end of a subway line has a large spring designed to compress 2.00 m when stopping a 1.10 ✕ 105 kg train mov

Mrac [35]

Answer:

(a) k = 1684.38 N/m = 1.684 KN/m

(b) Vi = 0.105 m/s

Explanation:

(a)

First, we find the deceleration of the car. For that purpose we use 3rd equation of motion:

2as = Vf² - Vi²

a = (Vf² - Vi²)/2s

where,

a = deceleration = ?

Vf = final velocity = 0 m/s (since, train finally stops)

Vi = Initial Velocity = 0.35 m/s

s = distance covered by train before stopping = 2 m

Therefore,

a = [(0 m/s)² - (0.35 m/s)²]/(2)(2 m)

a = 0.0306 m/s²

Now, we calculate the force applied on spring by train:

F = ma

F = (1.1 x 10⁵ kg)(0.0306 m/s²)

F = 3368.75 N

Now, for force constant, we use Hooke's Law:

F = kΔx

where,

k = Force Constant = ?

Δx = Compression = 2 m

Therefore.

3368.75 N = k(2 m)

k = (3368.75 N)/(2 m)

k = 1684.38 N/m = 1.684 KN/m

(c)

Applying Hooke's Law with:

Δx = 0.6 m

F = (1684.38 N/m)(0.6 m)

F = 1010.62 N = 1.01 KN

(b)

Now, the acceleration required for this force is:

F = ma

1010.62 N = (1.1 kg)a

a = 1010.62 N/1.1 x 10⁵ kg

a = 0.0092 m/s²

Now, we find initial velocity of train by using 3rd equation of motion:

2as = Vf² - Vi²

a = (Vf² - Vi²)/2s

where,

a = deceleration = -0.0092 m/s² (negative sign due to deceleration)

Vf = final velocity = 0 m/s (since, train finally stops)

Vi = Initial Velocity = ?

s = distance covered by train before stopping = 0.6 m

Therefore,

-0.0092 m/s² = [(0 m/s)² - Vi²]/(2)(0.6 m)

Vi = √(0.0092 m/s²)(1.2 m)

Vi = 0.105 m/s

4 0

4 years ago