A 750-kg automobile is moving at 26.2 m/s at a height of 5.00 m above the bottom of a hill when it runs out of gasoline. The car coasts down the hill and then continues coasting up the other side until it comes to rest. Ignoring frictional forces and air resistance, what is the value of h, the highest position the car reaches above the bottom of the hill?

Answers

Answer 1

Answer:

To solve this problem it is necessary to apply to the concepts related to energy conservation. For this purpose we will consider potential energy and kinetic energy as the energies linked to the body. The final kinetic energy is null since everything is converted into potential energy, therefore

Potential Energy can be defined as,

$PE = mgh$

Kinetic Energy can be defined as,

$K= (1)/(2) mv^2$

Now for Conservation of Energy,

$KE_i+PE_i = PE_f$

$(1)/(2)mv_i^2+mgh_1 = mgh_2$

$(1)/(2) (750kg) (26.2m/s)^2 + (750)(9.8)(5) = (750)(9.8)h_2$

$h_2 = 40.0224m$

Therefore the highets position the car reaches above the bottom of the hill is 40.02m

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Two capacitors give an equivalent capacitance of 9.42 pF when connected in parallel and an equivalent capacitance of 1.68 pF when connected in series. What is the capacitance of each capacitor?

Answers

Answer:

$C_1=7.23pF\ and\ C_2=2.19pF$

Explanation:

Let the two capacitance are $C_1\ and\ C_2$

It is given that when capacitors are connected in parallel their equvilaent capacitance is 9.42 pF

So $C_1+ C_2=9.2$ --------EQN 1

And when they are connected in series their equivalent capacitance is 1.68 pF

So $(1)/(C_1)+(1)/(C_2)=(1)/(1.68)$

$(C_1+C_2)/(C_1C_2)=(1)/(1.68)$

$C_1C_2=1.68* 9.42=15.8256pF$

$C_1-C_2=√((C_1+C_2)^2-4C_1C_2)=√(9.42^2-4* 15.8256)=5.0432pF$ -----EQN

On solving eqn 1 and eqn 2

$C_1=7.23pF\ and\ C_2=2.19pF$

The voltage difference between the AA and AAA batteries should be quite small. What then might be the difference between them?

Answers

Answer:

The major difference is the capacity of both batteries. The AA battery has a higher capacity (a higher current) than the AAA battery.

Explanation:

The AA batteries and the AAA batteries are very similar in their voltage; both of them have 1.5 V.

The difference between these two batteries is their size and also the current that they have. The AAA battery is smaller than the AA battery, which means that the amount of electrochemical material is lower, so the AA battery has a higher capacity (a higher current) than the AAA battery.Generally, AA battery has 2400 mAh capacity and AAA battery has a capacity of 1000mAh; this means that AA battery has almost three times the capacity of an AAA battery.

Furthermore, the size of the AA battery makes it more common than the AAA battery and therefore has higher commercial demand.

I hope it helps you!

Suppose there are 10,000 civilizations in the Milky Way Galaxy. If the civilizations were randomly distributed throughout the disk of the galaxy, about how far (on average) would it be to the nearest civilization?

Answers

Here is the full question

(Hint: Start by finding the area of the Milky Way's disk, assuming that it is circular and 100,000 light-years in diameter. Then find the average area per civilization, and use the distance across this area to estimate the distance between civilizations.)

Answer:

1000 light-years (ly)

Explanation:

If we go by the hint; The area of the disk can be expressed as:

$A = \pi ((D)/(2))^2$

where D = 100, 000 ly

Let's divide the Area by the number of civilization; if we do that ; we will be able to get 'n' disk that is randomly distributed; so ;

$d= (A)/(N) =(\pi ((D)/(2))^2 )/(10, 000)$

The distance between each disk is further calculated by finding the radius of the density which is shown as follows:

$d = \pi r^2 e$

$r^2_e= (d)/(\pi)$

$r_e = \sqrt{(d)/(\pi) }$

replacing d = $(\pi ((D)/(2))^2 )/(10, 000)$ in the equation above; we have:

$r_e = \sqrt{((\pi ((D)/(2))^2 )/(10, 000))/(\pi) }$

$r_e = \sqrt{(((D)/(2))^2 )/(10, 000)}$

$r_e = \sqrt{(((100,000)/(2))^2 )/(10, 000)}$

$r_e = 500 ly$

The distance (s) between each civilization = $2(r_e)$

= 2 (500 ly)

= 1000 light-years (ly)

Two narrow slits separated by 1.5 mm are illuminated by 514 nm light. Find the distance between adjacent bright fringes on a screen 5.0 m from the slits. Express your answer in meters using two significant figures.

Answers

The distance between the adjacent bright fringes is : 1.7 * 10⁻³ M

Given data :

separation between slits ( d ) = 1.5 x 10⁻³ m

wavelength of light ( λ ) = 514 * 10⁻⁹ m

Distance from narrow slit ( D ) = 5.0 m

Determine the distance between the adjacent bright fringes

we apply the formula below

w = D * λ / d ---- ( 1 )

where : w = distance between adjacent bright fringes

Back to equation ( 1 )

w = ( 5 * 514 * 10⁻⁹ ) / 1.5 x 10⁻³

= 1.7 * 10⁻³ M

Hence we can conclude that The distance between the adjacent bright fringes is : 1.7 * 10⁻³ M

Learn more about bright fringes calculations : brainly.com/question/4449144

Answer:

$1.7* 10^(-3)$ m

Explanation:

d = separation between the two narrow slits = 1.5 mm = 1.5 x 10⁻³ m

λ = wavelength of the light = 514 nm = 514 x 10⁻⁹ m

D = Distance of the screen from the narrow slits = 5.0 m

w = Distance between the adjacent bright fringes on the screen

Distance between the adjacent bright fringes on the screen is given as

$w = (D\lambda )/(d)$

$w = ((5.0)(514* 10^(-9)) )/(1.5* 10^(-3))$

$w = 1.7* 10^(-3)$ m

A truck traveling with an initial velocity of 44.1 m/s comes to a stop in 15.91 secs. What is theacceleration of the truck?

Answers

Answer:

a=-2.77 m/s^2

Explanation:

Assuming constant acceleration,

v=at + v_0

where v_0 is the initial velocity.

At rest, v=0, so

0=at+v_0

So solving the equation for a:

a=(-v_0)/t

Inserting the numbers yields

a=-2.77 m/s^2

Which is true about inelastic collisions: a. An inelastic collision does not obey conservation of energy. b. An inelastic collision conserves kinetic energy. c. Objects will stick together upon collision. d. Momentum is not conserved in inelastic collisions..

Answers

Answer:

Option c is correct

Explanation:

There are two types of collisions-elastic collision and inelastic collision.

In elastic collision, both kinetic energy and total momentum are conserved. On the other hand, in inelastic collision, total momentum is conserved but kinetic energy is not conserved. Thus, option b and d are incorrect.

Total energy is always conserved in both types. Thus, option a is incorrect.

In a perfectly inelastic collision, objects stick together. This happens because maximum kinetic energy is dissipated and used in bonding of the two objects. Thus, correct option is c.

Answer:

i believe its a?

Explanation:

In an inelastic collision, momentum is conserved

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