Physics Physics Physics

College students living off-campus frequently consume large amounts of ramen noodles and boxed macaroni and cheese. When they fi

tigry1 [53]

Answer:

Ramen Noodles and boxed Macaroni and Cheese are COMPLEMENTARY Goods

Explanation:

Complementary Goods are goods that are demanded jointly (together) to satisfy a particular want.

Ramen Noodles, Boxed Macaroni, Cheese are demanded together by off campus college students. So, they are complementary goods.

Also, their demand are directly related - demand of one good increases, demand of other good increases & demand of one good decreases , demand of other good decreases ; because they are demanded together.

After school end , consumption of all goods declining together - is also their characteristic pointing towards them being Complementary Goods .

4 0

3 years ago

A boat has a mass of 7660 kg. Its engines generate a drive force of 4080 N due west, while the wind exerts a force of 680 N due

makvit [3.9K]

Answer:

0.29 m/s due west.

Explanation:

According to newton's second law,

Net force acting on an object = mass×acceleration

From the question,

F+F₁+F₂ = ma................ Equation 1

Where F = The force generated from the engine, F₁ = Force exerted by the wind, F₂ = Force exerted due to the water, m = mass of the boat, a = acceleration of the boat.

Given: F = 4080 N , F₁ = -680 N(east), F₂ = -1160 N(east). m = 7660 kg

substitute into equation 1

4080-680-1160 = 7660(a)

2240 = 7660a

Therefore,

a = 2440/7660

a = 0.29 m/s due west.

8 0

2 years ago

A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t

givi [52]

To solve this problem we will apply the concepts related to the Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. Mathematically it can be described as,

$f = f_0 (\frac{v_0}{v_0-v})$

Here,

$f_0$ = Frequency of Source

$v_s$ = Speed of sound

f = Frequency heard before slowing down

f' = Frequency heard after slowing down

v = Speed of the train before slowing down

So if the speed of the train after slowing down will be v/2, we can do a system equation of 2x2 at the two moments, then,

The first equation is,

$f = f_0 (\frac{v_0}{v_0-v})$

$300 = f_0 (\frac{343}{343-v})$

$(300*343) - 300v = 343f_0$

Now the second expression will be,

$f' = f_0 (\frac{v_0}{v_0-v/2})$

$290 = (343)(\frac{v_0}{343-v/2})$

$290*343-145v = 343f_0$

Dividing the two expression we have,

$\frac{(300*343) - 300v}{290*343-145v} = 1$

Solving for v, we have,

$v = 22.12m/s$

Therefore the speed of the train before and after slowing down is 22.12m/s

6 0

3 years ago

What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500

padilas [110]

Answer:

I = 1.21x10^-5 A

Explanation:

You are missing the first part of the problem. This is an example, but it will give you the idea of how to solve yours with your data.

The first part is like this:

<em>A 4.0 cm diameter parallel plate capacitor has a 0.44 m m gap. What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000 V/s?</em>

Now with this, we can solve the problem.

In order to do this, we need to use the following expression:

q = CV (1)

Where:

C: Capacitance of a parellel capacitor (in Faraday)

q: charge of plate or capacitor (In coulombs)

V: voltage in Volts.

However, we need is the current, and we have data of potential difference, so, all we have to do is divide the expression between time so:

q/t = CV/t

And the current is q/t, thus:

I = C * V/t (2)

And finally, Capacitance C with two plates of area A separated by a distance d is:

C = Eo*A/d (3)

Where:

Eo = constant equals to 8.85x10^-12 F/m.

A = Area of the plate, in this case, πr²

d = gap of the capacitor.

Let's calculate first the Capacitance using equation (3):

C = 8.85x10^-12 * π * (0.04/2)² / 0.00046 = 2.42x10^-11 F

Now, it's time to use equation (2) and solve for I:

I = 2.42x10^-11 * 500,000

I = 1.21x10^-5 A

5 0

3 years ago

If she asks you to cut a strip of pie crust long around the outside of the pan, how long does it need to be? (Remember c/d =3.14

Nookie1986 [14]

I remember c/d. That's not a problem. But if you want 'c', you'll have to give me 'd'.

5 0

3 years ago

Physics Physics Physics - 10​

Physics Physics Physics - 10