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goldfiish [28.3K]
3 years ago
12

Two appliances are connected in parallel to a 120-v battery and draw currents i1 = 3.0 a and i2 = 3.1

Physics
1 answer:
8090 [49]3 years ago
4 0
Initially they are connected in parallel, so they have the same voltage V=120 V at their ends. Therefore we can use Ohm's law to calculate the resistance of each appliance:
R_1 =  \frac{V}{I_1}= \frac{120 V}{3.0 A}=40 \Omega
R_2 =  \frac{V}{I_2}= \frac{120 V}{3.1 A}=38.7 \Omega

When they are connected in series, they are crossed by the same current I. The equivalent resistance of the circuit in this case is R_{eq}=R_1+R_2 = 78.7 \Omega, so we can use Ohm's law for the entire circuit to find the current in the circuit:
I= \frac{V}{R_{eq}}= \frac{120 V}{78.7 \Omega}=1.52 A
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Two forces, F⃗ 1F→1F_1_vec and F⃗ 2F→2F_2_vec, act at a point. F⃗ 1F→1F_1_vec has a magnitude of 9.20 NN and is directed at an a
marishachu [46]

Answer:

-9.46 N

Explanation:

In order to get the value of the x component of the resultant force, we need to get the value of the x component of each force.

This value will be the projection of the force vector, on the x-axis.

For F₁, as it is directed at an angle of 55.0º above the negative x axis, we can find F₁ₓ just applying the definition of cosine of an angle, as follows:

cos θ = \frac{x}{r}

In this case, x = F₁ₓ, and r = F₁

θ, measured from the positive x axis counterclockwise, is as follows:

θ= 180º-55º = 125º

⇒ F₁ₓ = F₁* cosθ = 9.2 N * cos 125º = -5.28 N

We can repeat the process for F₂, as follows:

For F₂, as it is directed at an angle of 53.3º below the negative x axis, we can find F₂ₓ just applying the definition of cosine of an angle, as follows:

cos θ = \frac{x}{r}

In this case, x = F₂ₓ, and r = F₂

θ, measured from the positive x axis counterclockwise, is as follows:

θ= 180º + 53.3º = 233.3º

⇒ F₂ₓ = F₂* cosθ = 7.00 N * cos 233.3º = -4.18 N

The total component of both forces along the x axis, can be found just adding both components, as follows:

Fₓ = F₁ₓ + F₂ₓ = -5.28 N + -4.18 N = -9.46 N

5 0
3 years ago
The force F⃗ pulling the string is constant; therefore the magnitude of the angular acceleration α of the wheel is constant for
abruzzese [7]

Answer:

The answer is "\boxed{\boxed{\omega = \sqrt{\frac{2fd}{kmr^2}}}}"

Explanation:

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Label these parts on the wave below: Amplitude, Wavelength, Crest, Trough, Rest Position
Leviafan [203]

Answer:

Wavelength is the distance between from one crest to another crest or from one trough to another trough. The amplitude is the distance from the midpoint to the crest or trough. Crest is the highest point of the or a wave. Tough is the lowest point of the or a wave. Rest position is the position where it lies on the midpoint line.

Explanation:

I need a diagram to label these parts.

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Andru [333]

There's a problem with the question as given. Even with a maximum projection angle of <em>θ</em> = 90°, the initial velocity is not large enough to get the ball up in the air 160 m. With angle 90°, the ball's height <em>y</em> at time <em>t</em> would be

<em>y</em> = (40 m/s) <em>t</em> - 1/2 <em>g t</em> ²

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From another perspective: recall that

<em>v </em>² - <em>v</em>₀² = 2<em>a </em>∆<em>y</em>

where

• <em>v</em>₀ = initial velocity

• <em>v</em> = final velocity

• <em>a</em> = acceleration

• ∆<em>y</em> = displacement

At its maximum height, the ball has zero vertical velocity, and ∆<em>y</em> = maximum height = 160 m. The ball is in free fall once it's launched, so <em>a</em> = -<em>g</em>.

So we have

0² - (40 m/s)² = -2<em>g </em>(160 m)

but this reduces to

(40 m/s)² = 2 (9.8 m/s²) (160 m)

1600 m²/s² ≠ 3136 m²/s²

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Explanation:

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