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mafiozo [28]
3 years ago
14

Another example of square-root relationships is the relation between the speed of a wave along a string under tension and the te

nsion itself. Suppose you hold one end of a string and attach the other end to a wall. If you hold the string taut, and wiggle the free end up and down, a wave travels along the string. If the tension in the string is 9 N, the wave travels along the string at 6 m/s; if the tension in the string is 36 N, the wave speed along the string is 12 m/s. If the tension in the string is increased to 81 N , how fast will you expect a wave to travel along the string if you wiggle its free end
Physics
1 answer:
klio [65]3 years ago
7 0

Answer:

The right answer is "18 m/s".

Explanation:

The given values are:

Velocity,

V = 6 m/s

Tension,

T = 9 N

As we know,

⇒  V=(\frac{T}{u})^{\frac{1}{2}}

On substituting the given values, we get

⇒  6=(\frac{9}{u} )^{\frac{1}{2}}

⇒  36=\frac{9}{u}

⇒  u=\frac{9}{36}

⇒     =\frac{1}{4}

Now,

For 81 N the velocity will be:

⇒  v=(\frac{81}{(\frac{1}{4} )} )^{\frac{1}{2} }

⇒     =18 \ m/s

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sergiy2304 [10]

Answer:

The answer is 493.05kJ

Explanation:

given

mass m= 3kg

t1=  10°C

t2= 60°C

Specific heat capacity of ham =3287 cal/kgºC

Required,

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Step two:

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substitute

Q= 3*3287(60-10)

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3 0
3 years ago
A 1990 kg car moving at 20.0 m/s collides and locks together with a 1540 kg car at rest at a stop sign. Show that momentum is co
Ilya [14]

Answer:

The momentum before collision and momentum after collision is equal in a frame of reference moving at 10.0 m/s in the direction of the moving car.

Explanation:

m_1 = Mass of moving car = 1990 kg

u_1 = Velocity of moving car in stationary frame = 20 m/s

m_2 = Mass of stationary car = 1540 kg

u_2 = Velocity of stationary car in stationary frame = 0 m/s

v = Combined velocity in stationary frame

Momentum conservation for stationary frame

m_1u_1+m_2u_2=(m_1+m_2)v\\\Rightarrow v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}\\\Rightarrow v=\dfrac{1990\times 20+0}{1990+1540}\\\Rightarrow v=\dfrac{3980}{353}\ \text{m/s}

In frame moving at 10 m/s the velocities change in the following ways

u_1=20-10\\\Rightarrow u_1=10\ \text{m/s}

u_2=0-10\\\Rightarrow u_1=-10\ \text{m/s}

v=\dfrac{3980}{353}-10\\\Rightarrow v=\dfrac{450}{353}\ \text{m/s}

Momentum before collision

m_1u_1+m_2u_2=1990\times 10+1540\times -10\\\Rightarrow m_1u_1+m_2u_2=4500\ \text{kg m/s}

Momentum after collision

(m_1+m_2)v=(1990+1540)\times \dfrac{450}{353}\\\Rightarrow (m_1+m_2)v=4500\ \text{kg m/s}

The momentum before collision and momentum after collision is equal. So, the momentum is conserved in a reference frame moving at 10.0 m/s in the direction of the moving car.

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