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3 years ago

Can someone help me?

Physics

1 answer:

Leviafan [203]3 years ago

4 0

Car X traveled 3d distance in t time. Car Y traveled 2d distance in t time. Therefore, the speed of car X, is 3d/t, the speed of car Y, is 2d/t. Since speed is the distance taken in a given time.

In figure-2, they are at the same place, we are asked to find car Y's position when car X is at line-A. We can calculate the time car X needs to travel to there. Let's say that car X reaches line-A in t' time.

$V_x .t' = 3d\\ \frac{3d}{t} .t' = 3d\\ t'=t$

Okay, it takes t time for car X to reach line-A. Let's see how far does car Y goes.

$V_y.t = \frac{2d}{t} .t = 2d$

We found that car Y travels 2d distance. So, when car X reaches line-A, car Y is just a d distance behind car X.

You might be interested in

What are the characteristics of a magnetic force

fomenos

Answer:

Magnetic field lines never intersect each other.

They originates(from north pole) and terminates (to south pole) perpendicularly from/to the surface of magnet. Also magnetic field lines continues through the magnet.

Magnetic field is denser at the pole and becomes less dense as we move away from the poles.

Explanation:

8 0

3 years ago

A 0.18-kg turntable of radius 0.32 m spins about a vertical axis through its center. A constant rotational acceleration causes t

borishaifa [10]

Answer:

Angular acceleration will be $18.84rad/sec^2$

Explanation:

We have given that mass m = 0.18 kg

Radius r = 0.32 m

Initial angular velocity $\omega _i=0rev/sec$

And final angular velocity $\omega _f=24rev/sec$

Time is given as t = 8 sec

From equation of motion

We know that $\omega _f=\omega _i+\alpha t$

$24=0+\alpha \times 8$

$\alpha =3rev/sec^2=3\times 2\times \pi rad/sec^2=18.84rad/sec^2$

So angular acceleration will be $18.84rad/sec^2$

4 0

4 years ago

Fiber optics are an important part of our modern internet. In these fibers, two different glasses are used to confine the light

professor190 [17]

Complete Question

The complete question is shown on the first uploaded image

Answer:

$\theta_{max} =18.38^o$

New $n_{cladding} =1.491$

Explanation:

From the question we are told that

The refractive index of the core is $n_{core} = 1.497$

The refractive index of the cladding is $n_{cladding} = 1.421$

Generally according to Snell's law

$n_{core} * sin(90- \theta) = n_{cladding} * sin (90)$

Where $\theta_{max}$ is the largest angle a largest angle a ray will make with respect to the interface of the fiber and experience total internal reflection

$\theta_{max} = 90 - sin^{-1} [\frac{n_{cladding}}{n_{core}} ]$

$\theta_{max} = 90 - sin^{-1} [\frac{1.421}{1.497}} ]$

$\theta_{max} =18.38^o$

Given from the question the the largest angle is 5°

Generally the refraction index of the cladding is mathematically represented as

$n_{cladding} = n_{core} * sin (90 - 5)$

$n_{cladding} =1.491$

5 0

3 years ago

Capacitors C1 = 5.85 µF and C2 = 2.80 µF are charged as a parallel combination across a 250 V battery. The capacitors are discon

posledela

Answer:

Q1_new = 515.68 µC

Q2_new = 246.82 µC

Explanation:

Since the capacitors are charged in parallel and not in series, then both are at 250 V when they are disconnected from the battery.

Then it is only necessary to calculate the charge on each capacitor:

Q1 = 5.85 µF * 250 V = 1462.5 µC

Q2 = 2.8 µF * 250 V = 700 µC

Now, we will look at 1462.5 µC as excess negative charges on one plate, and 1462.5 µC as excess positive charges on the other plate. Now, we will use this same logic for the smaller capacitor.

When there is a connection of positive plate of C1 to the negative plate of C2, and also a connection of the negative plate of C1 to the positive plate of C2, some of these excess opposite charges will combine and cancel each other. The result is that of a net charge:

1462.5 µC - 700 µC = 762.5 µC

Thus,762.5 µC of net charge will remain in the 'new' positive and negative plates of the resulting capacitor system.

This 762.5 µC will be divided proportionately between the two capacitors.

Q1_new = 762.5 µC * (5.85/(5.85 + 2.8)) = 515.68 µC

Q2_new = 762.5 µC * (2.8/(5.85 + 2.8) = 246.82 µC

4 0

4 years ago

The burner on an electric stove has a power output of 2.0 kW. A 710 g stainless steel tea kettle is filled with 20∘C water and p

asambeis [7]

Answer:

The volume of water that was in the kettle is 1170 $cm^{3}$

Explanation:

Given:

Power, P = 2.0 kW = 2000 W, Mass of stainless steel, $m_{s}$ = 710 g = 0.71 kg at temperature of $20^{0} C$

Part A:

If it takes time, t = 3.5 minutes to reach boiling point of water $100^{0} C$ , then from conservation of energy,

Total energy supplied by the burner = Total heat gained by the water and the stainless steel to rise from $20^{0} C$ to $100^{0} C$

i.e. Pt = $m_{s}$ $c_{s}$ (100 - 20 ) + $m_{w}$ $c_{w}$ (100 - 20 )

$m_{w}$ = $\frac{pt - 80m_{s} C_{s} }{80c_{w} } = \frac{2000*3.5*60 - 80*0.71* 450}{80*4200}$

$m_{w}$ = 1.17 kg

where $c_{w}$ = 4200 J/Kgk (specific heat capacity of water), $c_{s}$ = 450 J/Kgk (specific heat capacity of steel)

But volume of water in the the kettle, v = $\frac{mass}{density} = \frac{1.17}{1000}= 1.17 *10^{-3}$ $m^{3}$

∴ v = 1170 $cm^{3}$

4 0

4 years ago

Can someone help me?​

Can someone help me?