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3 years ago

How is temperature and viscosity related?

Physics

1 answer:

Vlad1618 [11]3 years ago

6 0

Answer:

With an increase in temperature, there is typically an increase in the molecular interchange as molecules move faster in higher temperatures. The gas viscosity will increase with temperature. ... With high temperatures, viscosity increases in gases and decreases in liquids, the drag force will do the same.

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An ambulance with a siren emitting a whine at 1790 Hz overtakes and passes a cyclist pedaling a bike at 2.36 m/s. After being pa

Deffense [45]

Answer:

The speed of the ambulance is 4.30 m/s

Explanation:

Given:

Frequency of the ambulance, f = 1790 Hz

Frequency at the cyclist, f' = 1780 Hz

Speed of the cyclist, v₀ = 2.36 m/s

let the velocity of the ambulance be 'vₓ'

Now,

the Doppler effect is given as:

$f'=f\frac{v\pm v_o}{v\pm v_x}$

where, v is the speed of sound

since the ambulance is moving towards the cyclist. thus, the sign will be positive

thus,

$v_x=\frac{f}{f'}(v+v_o)-v$

on substituting the values, we get

$v_x=\frac{1790}{1780}(343+2.36)-343$

vₓ = 4.30 m/s

Hence, <u>the speed of the ambulance is 4.30 m/s</u>

6 0

3 years ago

Three resistors of resistances, R1=10Ω, R2=5Ω, R3=20Ω are connected in a circuit in such way that same amount of current flows t

Nezavi [6.7K]

Answer

Explanation:

As the three resistors are connected in series, the expression to be used for the

calculation of RT equivalent resistance

is:

RT = R1 + R2 + R3

We replace the data of the statement in the previous expression and it remains:

5 10 15 RT + R1 + R2 + R3 + +

We perform the mathematical operations that lead us to the result we are looking for:

RT - 30Ω

5 0

3 years ago

The earplug can reduce the sound level to about 18 decibels (dB). What percentage reduction is this intensity?

zlopas [31]

Answer:

1 x 10 -10 whisper at 1m distance.

Explanation:

Properly fitted ear plugs an reduce noise form 15-30db. Although they are better for low frequency

8 0

3 years ago

Read 2 more answers

A wheel initially spinning at wo = 50.0 rad/s comes to a halt in 20.0 seconds. Determine the constant angular acceleration and t

svetlana [45]

Answer:

(I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.

(II). The torque is 84.87 N-m.

Explanation:

Given that,

Initial spinning = 50.0 rad/s

Time = 20.0

Distance = 2.5 m

Mass of pole = 4 kg

Angle = 60°

We need to calculate the angular acceleration

Using formula of angular velocity

$\omega=-\alpha t$

$\alpha=-\dfrac{\omega}{t}$

$\alpha=-\dfrac{50.0}{20.0}$

$\alpha=-2.5\ rad/s^2$

The angular acceleration is -2.5 rad/s²

We need to calculate the number of revolution

Using angular equation of motion

$\theta=\omega_{0}t+\dfrac{1}{2}\alpha t$

Put the value into the formula

$\theta=50\times20-\dfrac{1}{2}\times2.5\times20^2$

$\theta=500\ rad$

The number of revolution is 500 rad.

(II). We need to calculate the torque

Using formula of torque

$\vec{\tau}=\vec{r}\times\vec{f}$

$\tau=r\times f\sin\theta$

Put the value into the formula

$\tau=2.5\times4\times 9.8\sin60$

$\tau=84.87\ N-m$

Hence, (I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.

(II). The torque is 84.87 N-m.

8 0

3 years ago

A 60 Watt light bulb runs for 120 seconds. How much energy does it use?

sattari [20]

ENERGY = POWER X TIME
=60 X 120=7200KWh

6 0

3 years ago