Which of these substances has the greatest index of refraction?

Which circuit component is usually a device that uses electrical energy to to perform a certain function? A. wire B. power sourc

Ksenya-84 [330]

Answer:

A

Explanation:

3 0

3 years ago

Read 2 more answers

During which two time intervals does the particle undergo equal displacement?

san4es73 [151]

Answer:

BC and DE

Explanation:

In the given figure, the velocity time graph is shown. We know that the area under v-t curve gives the displacement of the particle.

Area under AB, $d_1=\dfrac{1}{2}\times 2\times 10=10\ m$

Area under BC, $d_2=\dfrac{1}{2}\times 2\times 4=4\ m$

Area under CD, $d_3=\dfrac{1}{2}\times 2\times 7=7\ m$

Area under DE, $d_4=\dfrac{1}{2}\times 2\times 4=4\ m$

Area under EF, $d_5=\dfrac{1}{2}\times 2\times 3=3\ m$

So, form above calculations it is clear that, during BC and DE undergo equal displacement. Hence, the correct option is (c) "BC and DE = 4 meters".

4 0

3 years ago

suppose a car manufacturer tested its cars for front end collsion by hauling them up on a crane and dropping them from a certain

IRINA_888 [86]

Initial height: 66.5 m

Explanation:

The problem can be solved by using the principle of conservation of energy.

If we neglect air resistance, the total mechanical energy of the car is conserved during the fall, therefore we can write:

$K_i + U_i = K_f + U_f$

where :

$K_i = 0$ is the kinetic energy of the car at the top (it starts from rest)

$U_i = mgh$ is the gravitational potential energy of the car at the top, with:

m = the mass of the car

g = the acceleration of gravity

h = the heigth of the car

$K_f = \frac{1}{2}mv^2$ is the kinetic energy of the car just before hitting the ground, with

v = 130 km/h final speed of the car

$U_f = 0$ is the gravitational potential energy of the car at the bottom

Re-arranging the equation, we find

$mgh=\frac{1}{2}mv^2$

and we have:

$g=9.8 m/s^2\\v = 130 km/h = 36.1 m/s$

Solving for h, we find the initial height of the car:

$h=\frac{v^2}{2g}=\frac{36.1^2}{2(9.8)}=66.5 m$

Learn more about kinetic energy and potential energy:

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5 0

3 years ago

A yellow train of mass 100 kg is moving at 8 m/s toward an orange train of mass 200 kg traveling in the opposite direction on th

Serhud [2]

The initial momentum of the yellow and the orange train is 1000kgm/s.

Momentum is the product of the mass and velocity of any object.

Momentum is denoted by P.

Momentum P = mv , where m = mass and v = velocity.

<h3>Given:</h3>

Mass of the orange train = 200kg

Velocity of the orange train = 1m/s

So, the momentum of the orange train will be,

∴ P = mv

P = 200 x 1

P = 200 kgm/s

∴ The initial momentum of the orange train is 200kgm/s.

Mass of the yellow train = 100kg

Velocity of the yellow train = 8m/s

So, the momentum of the yellow train will be,

∴ P = mv

P = 100 x 8

P = 800 kgm/s

∴ The initial momentum of the yellow train is 800kgm/s.

Therefore, the initial momentum of the yellow and the orange train is 1000kgm/s.

Learn more about momentum here:

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5 0

2 years ago

A low resistance light bulb and a high resistance light bulb are connected in parallel with each other. Which bulb is brighter i

sweet [91]

<h2>Answer:</h2>

The bulb with low resistance will be brighter.

<h2>Explanation:</h2>

The brightness of a bulb is a function of both the voltage across the bulb and current flowing through the bulb. The higher the voltage, the higher the current. Hence the brighter the bulb.

Now, according to the question, the bulbs (the high resistance bulb and the low resistance bulb) are connected in parallel with each other. This means that the same voltage passes across them.

Also, we know that according to Ohm's law, the voltage (V) and current (I) through a conductor are related by the following equation;

V = I x R -------------------(i)

Where;

R is the resistance of the conductor.

We can re-write equation (i) as follows;

I = V / R -----------------------(ii)

According to equation (ii), at fixed voltage (V), the current (I) will increase as the resistance (R) decreases.

Now, since the two bulbs have the same voltage, the bulb with the low resistance will allow a larger flow of current than the bulb with high resistance. Therefore, as said earlier that brightness is dependent on voltage and current, the bulb with the low resistance (and having larger current at some voltage) will be brighter than the bulb with the high resistance (having smaller current at same voltage).

6 0

3 years ago