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3 years ago

Could someone answer this?

Physics

1 answer:

Nadusha1986 [10]3 years ago

6 0

Answer:

Here the circuit in which a 4Ω resistor resistor is connected in series and two 8Ω resistor resistors are connected in parallel. Also, ammeter and voltmeter connected in series and parallel circuit respectively.

Now,

The maximum power of each resistance is 16 W

The 4Ω resistor is linked in series with the circuit.

so, P o w e r = I

two

R, here i is the current through the resistor resistor R

1 6 = I

two

∗ 4 Ω

i = 2A

Now 2A passes through parallel resistors of 8Ω resistance.

we know that, in parallel, the potential difference must be constant,

the current is divided into two parts, because the same resistance current in each resistance will be half. then the current through each resistor in parallel is

2 A

two

= 1 A

So finally the current through the 4Ω resistor = 2 A

current through each 8Ω resistor = 1 A

Explanation:

I hope this answer has helped you

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What is the force on an object that goes from 35 m/s to 85 m/s in 20 seconds and has a mass of 148 kg

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F=ma
F = 148×(85-35)÷20
F = 148×(50÷20)
F = 148×2.5
F = 370N

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3 years ago

What is the net torque on the square plate, with sides 0.2 m, from each of the three forces? F1=18 N, F2=26 N, and F3=14 N. Use

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Answer:

The net torque on the square plate is 2.72 N-m.

Explanation:

Given that,

Side = 0.2 m

Force $F_{1}=18\ N$

Force $F_{2}=26\ N$

Force $F_{3}=14\ N$

We need to calculate the torque due to force F₁

Using formula of torque

$\tau_{1}=-F_{1}d_{1}$

$\tau_{1}=-F_{1}\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{1}=-18\times\dfrac{0.2}{2}$

$\tau_{1}=-1.8\ N-m$

We need to calculate the torque due to force F₂

Using formula of torque

$\tau_{2}=F_{2}d_{2}$

$\tau_{2}=F_{2}\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{2}=26\times\dfrac{0.2}{2}$

$\tau_{2}=2.6\ N-m$

We need to calculate the torque due to force F₃

Using formula of torque

$\tau_{3}=F_{3}d_{3}$

$\tau_{3}=(F_{3}\sin45+F_{3}\cos45)\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{3}=0.1(14\sin45+14\cos45)$

$\tau_{3}=1.92\ N-m$

We need to calculate the net torque on the square plate

$\tau=\tau_{1}+\tau_{2}+\tau_{3}$

$\tau=-1.8+2.6+1.92$

$\tau=2.72\ N-m$

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3 years ago

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3 years ago

The amplitude of a lightly damped oscillator decreases by 3.42% during each cycle. What percentage of the mechanical energy of t

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Answer:

The percentage of the mechanical energy of the oscillator lost in each cycle is 6.72%

Explanation:

Mechanical energy (Potential energy, PE) of the oscillator is calculated as;

PE = ¹/₂KA²

During the first oscillation;

PE₁ = ¹/₂KA₁²

During the second oscillation;

A₂ = A₁ - 0.0342A₁ = 0.9658A₁

PE₂ = ¹/₂KA₂²

PE₂ = ¹/₂K (0.9658A₁)²

PE₂ = (0.9658²)¹/₂KA₁²

PE₂ = (0.9328)¹/₂KA₁²

PE₂ = 0.9328PE₁

Percentage of the mechanical energy of the oscillator lost in each cycle;

$Change \ in \ percent= \frac{PE_2 - PE_1}{PE_1} \\\\Change \ in \ percent= \frac{0.9328PE_1 -PE_1}{PE_1} *100\\\\Change \ in \ percent= \frac{-0.0672PE_1}{PE_1}*100 \\\\Change \ in \ percent= -0.0672*100\\\\Change \ in \ percent= -6.72 \ \% \\\\Loss \ in \ percent= 6.72 \ \%$

Therefore, the percentage of the mechanical energy of the oscillator lost in each cycle is 6.72%

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olga nikolaevna [1]

Lindsay has to fly this plane towards this direction [W 12.5° S] to get to Hamilton.

From this question, the plane is still up in the air.

We have wind blowing in [W 60° N ]

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$\frac{SinA}{a}=\frac{SinB}{b} =\frac{SinC}{c}$

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The next step is to cross multiply

50 x sin60° = 200Sinθ

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7 0

3 years ago

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