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2 years ago

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Given that R1 is 13 Ω, R2 is 10 Ω, and R3 is 4 Ω, what is the current coming out of the 9-V battery? What is the power dissipati

on of each resistor?

1 answer:

Kisachek [45]2 years ago

8 0

So sorry I need point good luck

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A 3500kg suv is driving north at 36m/s what the momentum

lianna [129]

Your answer is

<span>126000</span>

4 0

2 years ago

How much voltage (in terms of the power source voltage bV) will the capacitor have when it has started at zero volts potential d

Archy [21]

Answer:

The voltage is $V = 0.993V_b$

Explanation:

From the question we are told that

The time that has passed is $t = \frac{\tau}{2}$

Here $\tau$ is know as the time constant

The voltage of the power source is $V_b$

Generally the voltage equation for charging a capacitor is mathematically represented as

$V = V_b [1 - e^{- \frac{t}{\tau} }]$

=> $V = V_b [1 - e^{- \frac{\frac{\tau}{2}}{\tau} }]$

=> $V = V_b [1 - e^{- \frac{\tau}{2\tau} }]$

=> $V = V_b [1 - e^{- \frac{1}{2} }]$

=> $V = 0.993V_b$

5 0

2 years ago

Uncle Harry weighs 750 N. What is his mass in kg?

Elenna [48]

Answer:

W = MG

750 = M * 10

M = 750/10

M = 75 kg

8 0

2 years ago

Read 2 more answers

2. While standing near a bus stop, a student hears a distant horn beeping. The frequency emitted by the horn is 440 Hz. The bus

jarptica [38.1K]

Given Information:

Frequency of horn = f₀ = 440 Hz

Speed of sound = v = 330 m/s

Speed of bus = v₀ = 20 m/s

Answer:

Case 1. When the bus is crossing the student = 440 Hz

Case 2. When the bus is approaching the student = 414.9 Hz

Case 3. When the bus is moving away from the student = 468.4 Hz

Explanation:

There are 3 cases in this scenario:

Case 1. When the bus is crossing the student

Case 2. When the bus is approaching the student

Case 3. When the bus is moving away from the student

Let us explore each case:

Case 1. When the bus is crossing the student:

Student will hear the same frequency emitted by the horn that is 440 Hz.

f = 440 Hz

Case 2. When the bus is approaching the student

f = f₀ ( v / v+v₀ )

f = 440 ( 330/ 330+20 )

f = 440 ( 330/ 350 )

f = 440 ( 0.943 )

f = 414.9 Hz

Case 3. When the bus is moving away from the student

f = f₀ ( v / v+v₀ )

f = 440 ( 330/ 330-20 )

f = 440 ( 330/ 310 )

f = 440 ( 1.0645 )

f = 468.4 Hz

6 0

3 years ago

Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th

DiKsa [7]

Answer:

a. $\rm -1.49\ m/s^2.$

b. $\rm 50.49\ m.$

Explanation:

<u>Given:</u>

Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .

<h2>(a):</h2>

The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).$

At time t = 3 seconds,

$\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.$

<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>

<h2>(b):</h2>

The velocity of the particle at some is defined as the rate of change of the position of the particle.

$\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.$

For the time interval of 2 seconds,

$\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt$

The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,

$\Delta r=3\ \left (\dfrac{\sin(0.5\ t)}{0.05} \right )\limits^2_0\\=3\ \left (\dfrac{\sin(0.5\times 2)-sin(0.5\times 0)}{0.05} \right )\\=3\ \left (\dfrac{\sin(1.0)}{0.05} \right )\\=50.49\ m.$

It is the displacement of the particle in 2 seconds.

7 0

3 years ago