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Tanya [424]
2 years ago
13

Consider a string of length 1.0 meter, fixed at both ends, with mass 100 grams and tension 100 newtons. part a give the number o

f nodes and antinodes set up on the string if it is driven at a frequency of 95 hertz. be sure to include the ends of the string in your count. give the number of nodes followed by the number of antinodes, separated by a comma.
Physics
1 answer:
Bond [772]2 years ago
4 0
To answer the problem we would be using this formula which isv = sqrt(T/(m/L)) 
v = sqrt(100 N / [(0.100 kg)/(1.0 m)]) 
v = 31.62 m/s 
v = fλ 
31.62 m/s = (95 Hz)(λ) 
λ = 0.333 m 
For every wavelength along a string there will be 2 antinodes. 
1.0 m / 0.333 m = 3 
3 * 2 = 6 antinodes 
6 + 1 = 7 nodes
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A 1.15-kg mass oscillates according to the equation x = 0.650 cos(8.40t) where x is in meters and t in seconds. Determine (a) th
zheka24 [161]

Answer:

(a) A = 0.650 m

(b) f = 1.3368 Hz

(c) E = 17.1416 J

(d)  K = 11.8835 J

     U = 5.2581 J

Explanation:

Given

m = 1.15 kg

x = 0.650 cos (8.40t)

(a) the amplitude,

A = 0.650 m

(b) the frequency,

if we know that

ω = 2πf = 8.40    ⇒   f = 8.40 / (2π)

⇒   f = 1.3368 Hz

(c) the total energy,

we use the formula

E = m*ω²*A² / 2

⇒  E = (1.15)(8.40)²(0.650)² / 2

⇒  E = 17.1416 J

(d) the kinetic energy and potential energy when x = 0.360 m.

We use the formulas

K = (1/2)*m*ω²*(A² - x²)       (the kinetic energy)

and

U = (1/2)*m*ω²*x²              (the potential energy)

then

K = (1/2)*(1.15)*(8.40)²*((0.650)² - (0.360)²)

⇒  K = 11.8835 J

U = (1/2)*(1.15)*(8.40)²*(0.360)²

⇒  U = 5.2581 J

4 0
2 years ago
A sheet of gold leaf has a thickness of 0.125 micrometer. A gold atom has a radius of 174pm. Approximately how many layers of at
Olegator [25]

Answer:

719

Explanation:

Conversion

1 picometer (pm) is equivalent to 1 × 10^{-12} meter

1 micrometer is equivalent to 1 × 10^{-6} meter

To find the number of layers, we divide the overal leaf thickness by the thickness of one atom hence dividing tex]0.125 × 10^{-6}[/tex] meter by 174 × 10^{-12} meter we get that the number of sheets will be as follows

\frac {0.125× 10^{-6}}{174\times 10^{-12}}=718.3908045\approx 719

Therefore, they are approximately 719 sheets

7 0
2 years ago
A heavy frog and a light frog jump straight up into the air. They push off in such away that they both have the same kinetic ene
Ilia_Sergeevich [38]

Answer:

The lighter frog goes higher than the heavier frog.

The lighter frog is moving faster than the heavier frog

Explanation:

If both frogs have the same kinetic energy when they leave the ground, the following equality applies:

K(light) = K(heavy) = \frac{1}{2} *ml*vol^{2} = \frac{1}{2}*mh*voh^{2}

Now, if the only force acting on the frogs is gravity, when they reach to the maximum height, we can apply the following kinematic equation:

vf^{2} -vo^{2} = 2*a*hmax = vf^{2} -vo^{2} = 2*(-g)*hmax

When h= hmax, the object comes momentarily to an stop, so vf =0

Solving for hmax:

hmax =\frac{vo^{2} }{2*g}

As the lighter frog, in order to have the same kinetic energy than the heavier one, has a greater initial velocity, it will go higher than the other.

As a consequence of both having the same kinetic energy, the lighter frog will be moving faster than the heavier frog.

5 0
2 years ago
A person who weighs 625 N is riding a 98-N mountain bike. Suppose that the entire weight of the rider and bike is supported equa
kozerog [31]

Answer:

A = 4.76 x 10⁻⁴ m²

Explanation:

given,                                  

weight of the person = 625 N

weight of the bike = 98 N        

Pressure on each Tyre = 7.60 x 10⁵ Pa

Area of contact on each Tyre = ?          

total weight of the system = 625 + 98

                                             = 723 N

Let F be the force on both the Tyre

F + F = W                                    

2 F  = 723                                    

F = 361.5 N                                

F = P A                                            

A = \dfrac{F}{P}                          

A = \dfrac{361.5}{7.60 \times 10^5}

A = 4.76 x 10⁻⁴ m²

7 0
2 years ago
In an experiment you are performing, your lab partner has measured the distance a cart has traveled: 28.4inch. You need the dist
alexgriva [62]

As per the question the distance travelled by  a car is 28.4 inch.

we are asked to determine the  conversion factor in centimeter  which when multiplied with 28.4 inch will give a unit.

we know that one inch =2.54 centimeter.

Hence 28.4 inch = 2.54  ×28.4 cm

                            =72.136 cm.

Now we have to determine the conversion factor .The multiplication factor is calculated as    P =\frac{28.4 inch*2.54cm}{1 inch}

                         p= 72.136 cm        [p is the multiplication factor.]

Hence the multiplication factor is 72.137 cm which will give unit conversion when multiplied with 28.4 inch.

                 

                             


4 0
2 years ago
Read 2 more answers
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