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VashaNatasha [74]
3 years ago
12

. In the following three arrangements each resistor has the same resistance R. Rank the equivalent resistances of the arrangemen

ts in descending order (largest first) 15. The peak value of the ac current in a circuit is 6.0 A, and peak value of the ac voltage applied to the circuit is 4.1 V. What is the average power delivered to the circuit? 16. The temperature coefficient of resistivity or a material is 0.0035 (C^o)^-1. By how much AT does the temperature of a wire made from this material have to change, in order that the resistance of the wire doubles?
Physics
1 answer:
Natali5045456 [20]3 years ago
6 0

Answer:

Explanation:

15 ) Current_rms = Peak current / √2

= 6 / √2

= 4.24 A

Voltage_rms = 4.1 / √2

= 2.9 v

Average power = Voltage_rms x Current_rms

= 2.9 x 4.24

= 12.3 W

16 ) R₂ = 2R₁

Resistance of a material changes due to change in resistivity of material .

So resistivity ρ₂ will also be twice resistivity ρ₁

ρ₂  = 2ρ₁

2ρ₁ = ρ₁ ( 1 + .0035 t ) where t is rise in temperature.

1 = .0035 t

t = 1 / .0035

= 285.7 ⁰ C

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A shaving or makeup mirror is designed to magnify your face by a factor of 1.40 when your face is placed 20.0cm in front of it
Dennis_Churaev [7]

Answer:

(a) convex mirror

(b) virtual and magnified

(c) 23.3 cm

Explanation:

The having mirror is convex mirror.

distance of object, u = - 20 cm

magnification, m = 1.4

(a) As the image is magnified and virtual , so the mirror is convex in nature.

(b) The image is virtual and magnified.

(c) Let the distance of image is v.

Use the formula of magnification.

m =-\frac{v}{u}\\1.4=-\frac{v}{-20}\\v =28 cm

Use the mirror equation, let the focal length is f.

\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\\frac{1}{f}=\frac{1}{28}+\frac{1}{20}\\\frac{1}{f}=\frac{28+20}{560}\\f=11.67cm

Radius of curvature, R = 2 f = 2 x 11.67 = 23.3 cm

5 0
3 years ago
A speeder passes a parked police car at a constant speed of 23.3 m/s. At that instant, the police car starts from rest with a un
KonstantinChe [14]

Answer:

32s

Explanation:

We must establish that by the time the police car catches up to the speeder, both have travelled a certain distance during the same amount of time. However, the police car experiences accelerated motion whereas the speeder travels at a constant velocity. Therefore we will establish two formulas for distance starting with the speeder's distance:

x=vt=23.3\frac{m}{s}t

and the police car distance:

x=vt+\frac{at^{2}}{2}=0+\frac{2.75\frac{m}{s^{2}} t^{2}}{2}=0.73\frac{m}{s^{2}}

Since they both travel the same distance x, we can equal both formulas and solve for t:

0 = 0.73\frac{m}{s^{2}}t^{2}-23.3\frac{m}{s} t\\\\0=t(0.73\frac{m}{s^{2}}t-23.3\frac{m}{s} )\\\\

Two solutions exist to the equation; the first one being t=0

The second solution will be:

0.73\frac{m}{s^{2}}t=23.3\frac{m}{s}\\\\t=\frac{23.3\frac{m}{s}}{0.73\frac{m}{s^{2}}}=32s

This result allows us to confirm that the police car will take 32s to catch up to the speeder

7 0
3 years ago
Which of the following was NOT one of the top 10 reported Stressors for Teens?
olga55 [171]
I think it life changes
4 0
3 years ago
Read 2 more answers
A Person whose weight is 5.20 x 10^2 N is being pulledup
Gnoma [55]

Answer:

Explanation:

Given

Weight of Person W=5.20\times 10^{2} N

Cave is h=35.1 m deep

Breaking stress T=569 N

Net Force on Person

F_{net}=569-520=49 N

a_{net}=\frac{F_{net}}{\frac{W}{g}}

a_{net}=\frac{49}{\frac{520}{9.8}}

a_{net}=0.923 m/s^2

The shortest time such that the person can be taken out of cave

h=ut+\frac{1}{2}at^2

where

h=distance moved

t=time

a=acceleration

35.1=0+\frac{1}{2}(0.923)(t)^2

t^2=76.05

t=\sqrt{76.05}

t=8.72\ s                    

6 0
3 years ago
A basketball is held over head at a height of 2.4 m. The ball is lobbed to a teammate at 8 m/s at an angle of 40'. If the ball i
cupoosta [38]

Explanation:

since both the teammates are of the same height, their height won't matter. Because now the basketball won't cover any vertical distance.

We have to calculate its range the horizontal distance covered by it when tossed from one teammate to the other.

range can be calculated by the formula :-

\boxed{\mathfrak{range =  \frac{  u  {}^{2}   \sin 2\theta }{g} }}

u is the velocity during its take off and \theta is the angle at which its thrown

Given that

  • u = 8m/ s
  • \theta = 40°

calculating range using the above formula

= \frac{ {8}^{2} \sin2(40)  }{10}

=  \frac{64 \times  \sin(80) }{10}

value of sin 80 = 0. 985

=  \frac{64 \times 0.985}{10}

=  \frac{63.027}{10}

= 6.3027

Hence,

\mathfrak { \blue{the \: teammate \: is \:  \red{\underline{6.3027 \: meters} }\: away } }

7 0
3 years ago
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