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2 years ago

Can you answer one of the questions

Physics

1 answer:

bagirrra123 [75]2 years ago

4 0

Answer:

2.e

3.b

4.c conduction needs friction

5.a

1, d

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Which method is used to deliver heat in most central heating systems?

inessss [21]

The right answer is B. hope this helps you :)

4 0

2 years ago

Read 2 more answers

Phÿśïčš<br><br>write the relation between celsius and fahrenheit scale?

Aleksandr-060686 [28]

Answer:

C/100 = (F-32) / 180

or, C/5 = (F-32)/9

Explanation:

relation between any two scales is given by:

(X- lower fixed point ) / (upper fixed point -lower fixed point)

where X is any temperature

5 0

2 years ago

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

A slender rod is 80.0 cm long and has mass 0.390 kg . A small 0.0200-kg sphere is welded to one end of the rod, and a small 0.05

Keith_Richards [23]

Answer

given,

length of slender rod =80 cm = 0.8 m

mass of rod = 0.39 Kg

mass of small sphere = 0.0200 kg

mass of another sphere weld = 0.0500 Kg

calculating the moment of inertia of the system

$I = \dfrac{ML^2}{12}+\dfrac{m_1L^2}{4}+\dfrac{mL^2}{4}$

$I = \dfrac{0.39\times 0.8^2}{12}+\dfrac{0.02\times 0.8^2}{4}+\dfrac{0.05\times 0.8^2}{4}$

$I =0.032\ kg.m^2$

using conservation of energy

$\dfrac{1}{2}I\omega^2 = (m_1-m_2)g\dfrac{L}{2}$

$\omega=\sqrt{\dfrac{(m_1-m_2)gL}{I}}$

$\omega=\sqrt{\dfrac{(0.05-0.02)\times 9.8 \times 0.8}{0.032}}$

$\omega=2.71 \rad/s$

we know,

v = r ω

$v = \dfrac{L}{2} \times 2.71$

$v = \dfrac{0.8}{2} \times 2.71$

v = 1.084 m/s

3 0

3 years ago

Whar are the types of energy associated with the microwave

Naily [24]

Type motion examples and subtypes
electromagnetic radiation disturbance propagating through electric and magnetic fields (classical physics) or the motion of photons (modern physics) radio waves, microwaves, infrared, light, ultraviolet, x-rays, gamma rays

Hope this helps and please mark me as brainlest and like:)

8 0

2 years ago