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3 years ago

Below is a physics question

Physics

1 answer:

Allisa [31]3 years ago

8 0

Explanation:

i1=1ampere

i1+i2=I

VA−VB=i1×3Ω=i2×6

⇒1×3=i2×6

2i=21ampere

Hence I=i1+i2=1+0.5=1.5ampere

Req=2+3+63×6=2+93×6=4Ω

equivalent circuit

Refer image .2

From KVL

1req=v

v=1.5×4

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A horizontal circular curve on a highway is designed for traffic moving at 60 km/h. If the radius of the curve is 150 m, and if

KengaRu [80]

Answer:

The value of the correct angle of banking for the road is $\theta = 67.76$ °

Explanation:

Given data

Velocity (v) = 60 $\frac{m}{s}$

Radius = 150 m

The velocity of the car in this case is given by

$v = \sqrt{r g \tan \theta}$

$v^{2} = r g \tan \theta$

$\tan \theta = \frac{v^{2} }{rg}$

Put all the values in above formula we get

$\tan \theta = \frac{60^{2} }{(150)(9.81)}$

$\tan \theta =$ 2.446

$\theta = 67.76$ °

Therefore the value of the correct angle of banking for the road is $\theta = 67.76$ °

4 0

3 years ago

The howler monkey is the loudest land animal and, under some circumstances, can be heard up to a distance of 8.9 km. Assume the

Airida [17]

Answer:

$\frac{I}{I_0}=113.68$

Explanation:

P = Acoustic power = 63 µW

r = Distance to the sound source = 210 m

Acoustic power

$P=IA\\\Rightarrow I=\frac{P}{A}\\\Rightarrow I=\frac{63\times 10^{-6}}{4\times \pi \times 210^2}$

Threshold intensity = $I_0=1\times 10^{-12}\ W/m^2$

Ratio

$\frac{I}{I_0}=\frac{\frac{63\times 10^{-6}}{4\times \pi \times 210^2}}{1\times 10^{-12}}\\\Rightarrow \frac{I}{I_0}=113.68$

Ratio of the acoustic intensity produced by the juvenile howler to the reference intensity is 113.68

6 0

3 years ago

A scooter is traveling at a constant speed v when it encounters a circular hill of radius r = 480 m. The driver and scooter toge

Grace [21]

Answer:

68.585m/sec , 779.1 N

Explanation:

To feel weightless, centripetal acceleration must equal g (9.8m/sec^2). The accelerations then cancel.

From centripetal motion.

F =( mv^2)/2

But since we are dealing with weightlessness

r = 480m

g = 9.8m/s^2

M also cancels, so forget M.

V^2 = Fr

V = √ Fr

V =√ (9.8 x 480) = 4704

= 68.585m/sec.

b) Centripetal acceleration = (v^2/2r) = (68.585^2/960) = 4704/960

= 4.9m/sec^2.

Weight (force) = (mass x acceleration) = 159kg x (g - 4.9)

159kg × ( 9.8-4.9)

159kg × 4.9

= 779.1N

6 0

3 years ago

Of gold, oxygen, iron, and sulfur, which is a renewable source?

Ulleksa [173]

Oxygen bc plants go through photosynthesis which keeps producing more oxygen

6 0

3 years ago

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On the planet Xenophous a 1.00 m long pendulum on a clock has a period of 1.32 s. What is the free fall acceleration on Xenophou

Vilka [71]

The period of the pendulum is given by the following equation

T = 2π * sqrt (L/g)

Where g is the gravity (free fall acceleration)

L is the longitude of the pendulum

T is the period.

We find g.............> (T /2π)^2 = L/g

g = L/(T /2π)^2...........> g = 22.657 m/s^2

3 0

4 years ago