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Elden [556K]
3 years ago
5

A complete series circuit consists of a 12.0 V battery, a 4.70 O resistor, and a switch. The internal resistance of the battery

is 0.30 O. The switch is open. What does an ideal voltmeter read when placed across the terminals of the switch?
12.0 V
9.40 V
2.40 V
zero
Physics
2 answers:
uysha [10]3 years ago
6 0

Answer:

12.0V is correct !!

Explanation:

RoseWind [281]3 years ago
4 0

With the switch open, there's no current in the circuit, and therefore
no voltage drop across any of the dissipative elements (the resistor
or the battery's internal impedance).  So the entire battery voltage
appears across the switch, and the voltmeter reads 12.0V .

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Jill does twice as much work as Jack does and in half the time. Jill's power output is Group of answer choices one-fourth as muc
Musya8 [376]

Answer:

Second Choice.

Explanation:

Jack's Power = W/t

Jill's Power = 2W/(0.5)*t

2/0.5 = 4

Jill's Power = 4*W/t

Jill's Power is 4 times greater than Jack's

Second Choice

3 0
3 years ago
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3 years ago
A piano wire with mass 2.60g and length 84.0 cm is stretched with a tension of 25.0 N. A wave with frequency 120.0 Hz and amplit
likoan [24]

Answer:

Power will be 0.2023 watt

And when amplitude is halved then power will be 0.0505 watt

Explanation:

We have given mass of the Piano wire m = 2.60 gram = 0.0026 kg

Length of wire l = 84 cm = 0.84 m

So mass density \mu =\frac{m}{l}=\frac{0.0026}{0.84}=0.0031kg/m

Tension in the wire T = 25 N

Frequency f = 120 Hz

So angular frequency \omega =2\pi f=2\times 3.14\times 120=753.6rad/sec

And amplitude A = 1.6 mm = 0.0016 m

We have to find the generated power

Power is given by P=\frac{1}{2}\sqrt{\mu T}\omega ^2A^2=\frac{1}{2}\times \sqrt{0.0031\times 25}\times 753.6^2\times 0.0016^2=0.2023watt

From the relation we can see that power P\ \propto\ A^2

So if amplitude is halved then power will be \frac{1}{4} times

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4 0
3 years ago
A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The
EleoNora [17]

Answer:

The tank is losing 4.976*10^{-4}  m^3/s

v_g = 19.81 \ m/s

Explanation:

According to the Bernoulli’s equation:

P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 +  \frac{1}{2}  \rho v_2^2 + \rho gh_2

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume  v_1 ≅ 0 ;

then v_2 can be determined as:\sqrt{[2g (h_1- h_2)]

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h₂ = 15 m

v_2 = \sqrt{[2*9.81*(20 - 15)]

v_2 = \sqrt{[2*9.81*(5)]

v_2= 9.9 \ m/s  as it leaves the hole at the base.

radius r = d/2  = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = A_1v_2

J = πr²v_2    

J =\pi *(2*10^{-3})^{2}*9.9

J =1.244*10^{-4}  m^3/s

b)

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : v_g = \sqrt{392.31}

v_g = 19.81 \ m/s

4 0
3 years ago
A trampoline spring has a force constant k = 800 N/m and is stretched exactly 17.5cm. What is the energy required to do this?
Artist 52 [7]

Answer:

the energy required for the extension is 12.25 J

Explanation:

Given;

force constant of trampoline spring, k = 800 N/m

extension of trampoline spring, x = 17.5 cm = 0.175 m

The energy required for the extension is calculated as;

E = ¹/₂kx²

E = 0.5 x 800 x 0.175²

E = 12.25 J

Therefore, the energy required for the extension is 12.25 J

6 0
3 years ago
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