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alex41 [277]
2 years ago
12

2001240Determine the specific kinetic energy of a mass whose velocity is 40 m/s, in kJ/kg.

Physics
1 answer:
guapka [62]2 years ago
6 0

Answer:

The specific kinetic energy of a mass is 0.8 kJ/kg

Explanation:

Given that,

Velocity = 40 m/s

Specific kinetic energy is the kinetic energy per unit mass.

We need to calculate the specific kinetic energy

Using formula of specific kinetic energy

K.E=\dfrac{\dfrac{1}{2}mv^2}{m}

K.E=\dfrac{v^2}{2}

Put the value into the formula

K.E=\dfrac{40^2}{2}

K.E=800\ J/kg

We know that,

1 kJ = 1000 J

or, 1J=0.001 KJ

The specific energy is

K.E=800\times0.001

K.E=0.8\ kJ/kg

Hence, The specific kinetic energy of a mass is 0.8 kJ/kg

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A short current element dl⃗ = (0.500 mm)j^ carries a current of 5.40 A in the same direction as dl⃗ . Point P is located at r⃗ =
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Answer:

The magnetic field along x axis is

B_{x}=1.670\times10^{-10}\ T

The magnetic field along y axis is zero.

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Explanation:

Given that,

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B=\dfrac{\mu_{0}}{4\pi}\timesI\times\dfrac{\vec{dl}\times\vec{r}}{|\vec{r}|^3}

Put the value into the formula

B=10^{-7}\times5.40\times\dfrac{(0.5\times10^{-3})\times(-0.730)i+(0.390)k}{(0.827)^3}

We need to calculate the value of \vec{dl}\times\vec{r}

\vec{dl}\times\vec{r}=(0.5\times10^{-3})\times(-0.730)i+(0.390)k

\vec{dl}\times\vec{r}=i(0.350\times0.5\times10^{-3}-0)+k(0+0.730\times0.5\times10^{-3})

\vec{dl}\times\vec{r}=0.000175i+0.000365k

Put the value into the formula of magnetic field

B=10^{-7}\times5.40\times\dfrac{(0.000175i+0.000365k)}{(0.827)^3}

B=1.670\times10^{-10}i+3.484\times10^{-10}k

Hence, The magnetic field along x axis is

B_{x}=1.670\times10^{-10}\ T

The magnetic field along y axis is zero.

The magnetic field along z axis is

B_{z}=3.484\times10^{-10}\ T

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