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Masja [62]
3 years ago
12

A 0.49-kg cord is stretched between two supports,7.3m apart. When one support is struck by a hammer, a transverse wave travels d

own the cord and reaches the other support in 0.83s .
What is the tension in the cord? (Express your answer to two significant figures and include the appropriate units.)
Physics
1 answer:
Wewaii [24]3 years ago
3 0

Answer:

T= 5.18N

Explanation:

u = mass of chord / length of chord

u = 0.49/ 7.3

u = 0.067 kg/m

Velocity of sound waves (v) =length of chord / time taken for wave to travel

v = 7.3 / 0.83 = 8.795m/s

Tension is calculated below using the formula

T = v² * u

T = (8.795)² x 0.067

T= 5.18N

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An ideal gas, initially at a pressure of 11.2 atm and a temperature of 299 K, is allowed to expand adiabatically until its volum
Tju [1.3M]

Answer:

The pressure is  P_2  =  4.25 \ a.t.m

Explanation:

From the question we are told that

   The initial pressure is P_1  =  11.2\ a.t.m

   The  temperature is  T_1 =   299 \ K

   

Let the first volume be  V_1 Then the final volume will be  2 V_1

 Generally for a diatomic  gas

           P_1 V_1 ^r  =  P_2  V_2  ^r

Here r is the radius of the molecules which is  mathematically represented as

    r =  \frac{C_p}{C_v}

Where C_p \  and\   C_v are the molar specific heat of a gas at constant pressure and  the molar specific heat of a gas at constant volume with values

     C_p=7 \  and\   C_v=5

=>   r =  \frac{7}{5}

=>  11.2*( V_1 ^{\frac{7}{5} } ) =  P_2  *  (2 V_1 ^{\frac{7}{5} } )

=>   P_2  =  [\frac{1}{2} ]^{\frac{7}{5} } * 11.2

=>  P_2  =  4.25 \ a.t.m

8 0
3 years ago
Dos cargas iguales de 10^-7C estan separadas por una distancia de 2m. calcule la fuerza con que se repelen
m_a_m_a [10]

Answer:

F=2.25\times 10^{-5}\ N

Explanation:

According to question,

Charge 1 and charge 2 are 10^{-7}\ C

The distance between charges is 2 m

We need to find the force with which two positive charges repel. It is called electrostatic force of repulsion. It can be given by :

F=\dfrac{kq^2}{r^2}\\\\F=\dfrac{9\times 10^9\times (10^{-7})^2}{2^2}\\\\F=2.25\times 10^{-5}\ N

So, the electric force of repulsion is 2.25\times 10^{-5}\ N.

8 0
2 years ago
Data that shows consistent, regular, repetitive form displays a pattern.<br><br> True<br><br> False
anyanavicka [17]
I think the answer is True
6 0
3 years ago
Read 2 more answers
A 18.0-kg rock is sliding on a rough, horizontal surface at 7.10 m/s and eventually stops due to friction. the coefficient of ki
Bond [772]
A = .3*g = 2.94 m/s² 

<span>t = v/a = 9/2.94 = 3.061 sec </span>

<span>W = E/t = ½mv²/t = ½*40*9²/3.061 = 529.2 watts</span>
4 0
2 years ago
A 5.0 kg object moving at 5.0 m/s. KE = mv2 times 1/2
steposvetlana [31]

Answer: KE = 62.5J

Explanation:

Given that

Mass of object = 5kg

kinetic energy KE = ?

velocity of object = 5m/s

Since kinetic energy is the energy possessed by a moving object, and it depends on the mass (m) of the object and the velocity (v) by which it moves. Therefore, the object has kinetic energy.

i.e K.E = 1/2mv^2

KE = 1/2 x 5kg x (5m/s)^2

KE = 0.5 x 5 x 25

KE = 62.5J

Thus, the object has 62.5 joules of kinetic energy.

5 0
3 years ago
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