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2 years ago

How many joules are in 3.5kJ?

Physics

1 answer:

jek_recluse [69]2 years ago

4 0

Answer:

3500 joules

Explanation:

Multiply the value by 1000.

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The leaning tower of Pisa is about 56 meters tall. A ball released from the top takes 3.4 seconds to reach the ground. The final

geniusboy [140]

S=56, u=0, v=33, a=?, t=3.4

v=u+at
33=3.4 a
a = 9.7m/s^2

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2 years ago

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Friction occurs when the ____ and ____ of two surfaces grind against each other.

monitta

Friction occurs when the surfaces and heat of two surfaces grind against each other.

I think these are the answers

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2 years ago

A beam of light, with a wavelength of 4170 Å, is shined on sodium, which has a work function (binding energy) of 4.41 × 10 –19 J

Lyrx [107]

Explanation:

Given that,

Wavelength of the light, $\lambda=4170\ A=4170\times 10^{-10}\ m$

Work function of sodium, $W_o=4.41\times 10^{-19}\ J$

The kinetic energy of the ejected electron in terms of work function is given by :

$KE=h\dfrac{c}{\lambda}-W_o\\\\KE=6.63\times 10^{-34}\times \dfrac{3\times 10^8}{4170\times 10^{-10}}-4.41\times 10^{-19}\\\\KE=3.59\times 10^{-20}\ J$

The formula of kinetic energy is given by :

$KE=\dfrac{1}{2}mv^2\\\\v=\sqrt{\dfrac{2KE}{m}} \\\\v=\sqrt{\dfrac{2\times 3.59\times 10^{-20}}{9.1\times 10^{-31}}} \\\\v=2.8\times 10^5\ m/s$

Hence, this is the required solution.

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2 years ago

In which situation is a person doing work on an object?

Marianna [84]

Answer : When a person applies force and covers distance then the person is doing work on an object.

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2 years ago

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A circular bird feeder 19.0 cm in radius has rotational inertia 0.130 kg·m2. It's suspended by a thin wire and is spinning slowl

soldi70 [24.7K]

Answer:

N₂=20.05 rpm

Explanation:

Given that

R= 19 cm

I=0.13 kg.m²

N₁ = 24.2 rpm

$\omega_1=\dfrac{2\pi \times 24.2}{60}\ rda/s$

ω₁= 2.5 rad/s

m= 173 g = 0.173 kg

v=1.2 m

Initial angular momentum L₁

L₁ = Iω₁ - m v r ( negative sign because bird coming opposite to motion of the wire motion)

Final linear momentum L₂

L₂= I₂ ω₂

I₂ = I + m r²

The is no any external torque that is why angular momentum will be conserve

L₁ = L₂

Iω₁ - m v r = I₂ ω₂

Iω₁ - m v r = ( I + m r²) ω₂

Now by putting the all values

Iω₁ - m v r = ( I + m r²) ω₂

0.13 x 2.5 - 0.173 x 1.2 x 0.19 = ( 0.13 + 0.173 x 0.19²) ω₂

0.325 - 0.0394 = 0.136 ω₂

ω₂ = 2.1 rad/s

$\omega_2=\dfrac{2\pi \times N_2}{60}$

N₂=20.05 rpm

3 0

3 years ago