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2 years ago

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An 8.0-ohm resistor and a 6.0-ohm resistor are connected in series with a battery. The potential difference across the 6.0-ohm r

esistor is measured as 12-V. (a) Find the total resistance of the circuit. (b) Find the current in the 6.0-ohm resistor. (c) Find the potential difference across the battery.
Please show work! :)

1 answer:

ipn [44]2 years ago

6 0

Answer:

a) 14 Ω

b) 2.0 A

c) 28 V

Explanation:

a) The total resistance of resistors in series is the sum:

R = R₁ + R₂

R = 8.0 Ω + 6.0 Ω

R = 14 Ω

b) The current in the 6.0 Ω resistor can be found with Ohm's law:

V = IR

12 V = I (6.0 Ω)

I = 2.0 A

c) Since the resistors are in series, they have the same current. So the total voltage is:

V = IR

V = (2.0 A) (14 Ω)

V = 28 V

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Comparing helium atoms with nitrogen molecules at the same temperature, the helium atoms on average are moving _______ and have

mylen [45]

Answer:

Helium atoms compared to nitrogen atoms are moving faster and have a greater kinetic energy.

Explanation:

The molecular velocity of a gas at room temperature is inverse proportional to the square root of its molecular mass.

The greater the molecular mass of the gas the lesser the average speed of its molecules. Comparing the molecular masses of nitrogen and helium, helium is found to have a lower molecular mass and a corresponding greater velocity.

Hence helium moves faster than nitrogen and has a higher kinetic energy than nitrogen

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3 years ago

Black_prince [1.1K]

First off, you need to know the weight of the projectile, lift and drag coefficients something like a high Reynolds number is preferred, then use the gravitational constant of 9.8 meters per second squared those would be a good start to get closer to your goal

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3 years ago

A car horn emits a frequency of 400 Hz. A car traveling at 20.0 m/s sounds the horn as it approaches a stationary pedestrian. Wh

Temka [501]

Answer:

The observed frequency by the pedestrian is 424 Hz.

Explanation:

Given;

frequency of the source, Fs = 400 Hz

speed of the car as it approaches the stationary observer, Vs = 20 m/s

Based on Doppler effect, as the car the approaches the stationary observer, the observed frequency will be higher than the transmitted (source) frequency because of decrease in distance between the car and the observer.

The observed frequency is calculated as;

$F_s = F_o [\frac{v}{v_s + v} ] \\\\$

where;

F₀ is the observed frequency

v is the speed of sound in air = 340 m/s

$F_s = F_o [\frac{v}{v_s + v} ] \\\\400 = F_o [\frac{340}{20 + 340} ] \\\\400 = F_o (0.9444) \\\\F_o = \frac{400}{0.9444} \\\\F_o = 423.55 \ Hz \\$

F₀ ≅ 424 Hz.

Therefore, the observed frequency by the pedestrian is 424 Hz.

8 0

3 years ago

A running mountain lion can make a leap 10.0 m long, reaching a maximum height of 3.0 m.?a.What is the speed of the mountain lio

Arisa [49]

Answer:

What is the speed of the mountain lion as it leaves the ground?

9.98m/s

At what angle does it leave the ground?

50.16°

Explanation:

This is going to be long, so if you want to see how it was solved refer to the attached solution. If you want to know the step by step process, read on.

To solve this, you will need use two kinematic equations and SOHCAHTOA:

$d = v_it + \dfrac{1}{2}at^{2}\\\\vf = vi + at$

With these formulas, we can derive formulas for everything you need:

Things you need to remember:

A projectile at an angle has a x-component (horizontal movement) and y-component (vertical movement), which is the reason why it creates an angle.

Treat them separately.

At maximum height, the vertical final velocity is always 0 m/s going up. And initial vertical velocity is 0 m/s going down.

Horizontal movement is not influenced by gravity.
acceleration due to gravity (a) on Earth is constant at 9.8m/s

First we need to take your given:

10.0 m long (horizontal) and maximum height of 3.0m (vertical).

$d_x=10.0m\\d_y=3.0m$

What your problem is looking for is the initial velocity and the angle it left the ground.

Vi = ? Θ =?

Vi here is the diagonal movement and do solve this, we need both the horizontal velocity and the vertical velocity.

Let's deal with the vertical components first:

We can use the second kinematic equation given to solve for the vertical initial velocity but we are missing time. So we use the first kinematic equation to derive a formula for time.

$d_y=V_i_yt+\dfrac{1}{2}at^{2}$

Since it is at maximum height at this point, we can assume that the lion is already making its way down so the initial vertical velocity would be 0 m/s. So we can reduce the formula:

$d_y=0+\dfrac{1}{2}at^{2}$

$d_y=\dfrac{1}{2}at^{2}$

From here we can derive the formula of time:

$t=\sqrt{\dfrac{2d_y}{a}}$

Now we just plug in what we know:

$t=\sqrt{\dfrac{(2)(3.0m}{9.8m/s^2}}\\t=0.782s$

Now that we know the time it takes to get from the highest point to the ground. The time going up is equal to the time going down, so we can use this time to solve for the intial scenario of going up.

$vf_y=vi_y+at$

Remember that going up the vertical final velocity is 0m/s, and remember that gravity is always moving downwards so it is negative.

$0m/s=vi_y+-9.8m/s^{2}(0.782s)\\-vi_y=-9.8m/s^{2}(0.782s)\\-vi_y=-7.66m/s\\vi_y=7.66m/s$

So we have our first initial vertical velocity:

Viy = 7.66m/s

Next we solve for the horizontal velocity. We use the same kinematic formula but replace it with x components. Remember that gravity has no influence horizontally so a = 0:

$d_x=V_i_xt+\dfrac{1}{2}0m/s^{2}(t^{2})\\d_x=V_i_xt$

But horizontally, it considers the time of flight, from the time it was released and the time it hits the ground. Also, like mentioned earlier the time going up is the same as going down, so if we combine them the total time in flight will be twice the time.

T= 2t

T = 2 (0.782s)

<em>T = 1.564s</em>

<em>So we use this in our formula:</em>

<em> $d_x=V_i_xT\\\\10.0m=Vi_x(1.564s)\\\\\dfrac{10.0m}{1.564s}=V_i_x\\\\6.39m/s=V_i_x$ </em>

Vix=6.39m/s

Now we have the horizontal and the vertical component, we can solve for the diagonal initial velocity, or the velocity the mountain lion leapt and the angle, by creating a right triangles, using vectors (see attached)

To get the diagonal, you just use the Pythagorean theorem:

c²=a²+b²

Using it in the context of our problem:

$Vi^{2}=Viy^2+Vix^2\\Vi^2=(7.66m/s)^2+(6.39m/s)^2\\\sqrt{Vi}=\sqrt{(7.66m/s)^2+(6.39m/s)^2}\\\\Vi=9.98m/s$

The lion leapt at 9.98m/s

Using SOHCAHTOA, we know that we can TOA to solve for the angle, because we have the opposite and adjacent side:

$Tan\theta=\dfrac{O}{A}\\\\Tan\theta=\dfrac{V_i_y}{V_i_x}\\\\\theta=Tan^{-1}\dfrac{V_i_y}{V_i_x}\\\\\theta=Tan^{-1}\dfrac{7.66m/s}{6.39m/s}\\\\\theta=50.17$

The lion leapt at an angle of 50.16°.

6 0

2 years ago

Consider a variety of colors of visible light (say 400 nm to 700 nm) falling onto a pair of slits.

babymother [125]

Answer:

Explanation:

The relationship between angle and wavelength for maxima and minima in Young's double slit experiment is given by

For constructive interference

$d\sin \theta =m\lambda$

For Destructive interference

$d\sin \theta =(m+\frac{1}{2})\lambda$

where $\lambda =wavelength$

$d=slit\ width$

m=order of maxima and minima

for second order maxima i.e. $m=2$

For smallest separation taking $\lambda =400 nm, \theta =90^{\circ}$

$d\sin 90=2\times 400\times 10^{-9}$

$d=0.8\times 10^{-6}$

$d=0.8\mu m$

6 0

3 years ago