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3 years ago

I NEED HELP ASAP DUE TOMMORW

Physics

1 answer:

solniwko [45]3 years ago

8 0

The water was heavier since it was more concentrated

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A transformer consisting of two coils wrapped around an iron core is connected to a generator and a resistor (Resistor is connec

V125BC [204]

Answer:

The peak emf of the generator is 40.94 V.

Explanation:

Given that,

Number of turns in primary coil= 11

Number of turns in secondary coil= 18

Peak voltage = 67 V

We nee to calculate the peak emf

Using relation of number of turns and emf

$\dfrac{N_{1}}{N_{2}}=\dfrac{E_{1}}{E_{2}}$

$E_{1}=\dfrac{N_{1}}{N_{2}}\times E_{2}$

Where, N₁ = Number of turns in primary coil

N₂ = Number of turns in secondary coil

E₂ = emf across secondary coil

Put the value into the formula

$E_{1}=\dfrac{11}{18}\times67$

$E_{1}=40.94\ V$

Hence, The peak emf of the generator is 40.94 V.

4 0

2 years ago

A person hears a siren as a fire truck approaches and passes by. The frequency varies from 480Hz on approach to 400Hz going away

alekssr [168]

Answer:

31.2 m/s

Explanation:

$f_{app}$ = Frequency of approach = 480 Hz

$f_{aw}$ = Frequency of going away = 400 Hz

$V$ = Speed of sound in air = 343 m/s

$v$ = Speed of truck

Frequency of approach is given as

$f_{app} = \frac{Vf}{V - v}$ eq-1

Frequency of moving awayy is given as

$f_{aw} = \frac{Vf}{V + v}$ eq-2

Dividing eq-1 by eq-2

$\frac{f_{app}}{f_{aw}} = \frac{V + v}{V - v}$

$\frac{480}{400} = \frac{343 + v}{343 - v}$

$v$ = 31.2 m/s

7 0

3 years ago

What is the momentum of a 950 kg car moving at 10.0 m/s?

pentagon [3]

Answer:

<h3>The answer is 9500 kgm/s</h3>

Explanation:

The momentum of an object can be found by using the formula

<h3>momentum = mass × velocity</h3>

From the question

mass = 950 kg

velocity = 10.0 m/s

We have

momentum = 950 × 10

We have the final answer as

Hope this helps you

8 0

3 years ago

Read 2 more answers

Describe how one plays Dr.Dogeball

JulsSmile [24]

•To play Dr. Dodgeball you need to have 2 teams to verse each other.
•Next, select one person from each team to be the doctor (depending on the size of the teams you can have varying amounts of doctors)
•Continue to play dodgeball how you normally would
•When a player gets hit and is “out” they have to sit on the ground and wait for the doctor to “revive them” (this usually requires the doctor dragging,touching, or moving the player that is out to a “revival place” which is usually decided on by the advisor or person in charge.
•Finally, try to get all the doctors and players out from the other team. Get the doctors first, for they cannot revive themselves. Which means the other players are out after they get hit with a ball since the doctors are out. (Some games are played where if all doctors are out the game ends)
Hope this helped! Play on! And plz mark brainliest lol this was long to write :D

4 0

3 years ago

Compare the circular velocity of a particle orbiting in the Encke Division, whose distance from Saturn 133,370 km, to a particle

Ket [755]

Answer:

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

Explanation:

The circular velocity is define as:

$v = \frac{2 \pi r}{T}$

Where r is the radius of the trajectory and T is the orbital period

To determine the circular velocity of both particles it is necessary to know the orbital period of each one. That can be done by means of the Kepler’s third law:

$T^{2} = r^{3}$

Where T is orbital period and r is the radius of the trajectory.

Case for the particle in the Encke Division:

$T^{2} = r^{3}$

$T = \sqrt{(133370 Km)^{3}}$

$T = \sqrt{(2.372x10^{15} Km)}$

$T = 4.870x10^{7} Km$

It is necessary to pass from kilometers to astronomical unit (AU), where 1 AU is equivalent to 150.000.000 Km ( $1.50x10^{8} Km$ )

1 AU is defined as the distance between the earth and the sun.

$\frac{4.870x10^{7} Km}{1.50x10^{8}Km} . 1AU$

$T = 0.324 AU$

But 1 year is equivalent to 1 AU according with Kepler’s third law, since 1 year is the orbital period of the earth.

$T = \frac{0.324 AU}{1 AU} . 1 year$

$T = 0.324 year$

That can be expressed in units of days

$T = \frac{0.324 year}{1 year} . 365.25 days$

$T = 118.60 days$

<em>Circular velocity for the particle in the </em><em>Encke Division</em><em>:</em>

$v = \frac{2 \pi r}{T}$

$v = \frac{2 \pi (133370 Km)}{(118.60 days)}$

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

$118.60 days .\frac{86400 s}{1 day}$ ⇒ 10247040 s

$133370 Km .\frac{1000 m}{1 Km}$ ⇒ 133370000 m

$v = \frac{2 \pi (133370000 m)}{(10247040 s)}$

$v = 81.778 m/s$

Case for the particle in the D Ring:

For the case of the particle in the D Ring, the same approach used above can be followed

$T^{2} = r^{3}$

$T = \sqrt{(69000 Km)^{3}}$

$T = \sqrt{(3.285x10^{14} Km)}$

$T = 1.812x10^{7} Km$

$\frac{1.812x10^{7} Km}{1.50x10^{8}Km} . 1 AU$

$T = 0.120 AU$

$T = \frac{0.120 AU}{1 AU} . 1 year$

$T = 0.120 year$

$T = \frac{0.120 year}{1 year} . 365.25 days$

$T = 43.83 days$

<em>Circular velocity for the particle in </em><em>D Ring</em><em>:</em>

$v = \frac{2 \pi r}{T}$

$v = \frac{2 \pi (69000 Km)}{(43.83 days)}$

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

$43.83 days . \frac{86400 s}{1 day}$ ⇒ 3786912 s

$69000 Km . \frac{1000 m}{ 1 Km}$ ⇒ 69000000 m

$v = \frac{2 \pi (69000000 m)}{(3786912 s)}$

$v = 114.483 m/s$

$\frac{114.483 m/s}{81.778 m/s} = 1.399$

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

7 0

2 years ago