All categories

3 years ago

The engine in your car is sometimes called: A. A 2-stroke engine

Physics

1 answer:

Juli2301 [7.4K]3 years ago

7 0

Answer:A

Explanation:

Engines in car are 4 stroke engine . A 4-stroke engine is internal combustion engine which derives its power by four piston strokes . Internal combustion means combustion takes inside the engine i.e. is in cylinder.

There are process in 4 stroke engine

Intake: Intake of air
Compression:compression of intake air to a high pressure
Combustion:Fuel is injected and burned to get power
Exhaust:removal of exhaust gases after combustion

You might be interested in

An electron having 500ev energy enters at right angle to a uniform magnetic field of 10^-4 Tesla. If its specific charge is 1.75

gtnhenbr [62]

Answer:

The correct answer might be $r = 2.8^{27}$ meters.

Explanation:

<u>The Answer Given might not be correct, I just did what my brain said.</u>

As the angle is perpendicular so Θ=90.

Putting this in the equation to calculate the magnetic force as:

F = evBsinΘ

F= evBsin90 *sin90 = 1 so,

F= evB.

Now when the electron will start to move in a circle, The necessary force that makes the electron rotate in a circle is given by Centripetal force.

So,

Magneteic Force = Centripetal Force

evB = $\frac{mv^{2} }{r}$

r = $\frac{mv}{Be}$ ......(1)

Now the problem is, We don't know " v " so we need to calculate velocity first,

Calculation of Velocity:

In order to calculate the velocity of electron, We should know the potiential difference with which the electrons are accelerated which in our case is 500ev. If "V" is the potiential difference, the energy gained by electrons during accelreation will be Ve. This appear as kinectic enrgy of electrons as,

K.E = Ve

$\frac{1}{2}$ m $v^{2}$ = Ve

v = $\sqrt{\frac{2ve}{m} }$ ................(2)

Putting value of velocity in equation 2 from 1:

r = $\frac{mv}{Be}$ $\sqrt{\frac{2ve}{m} }$

$r = \sqrt{\frac{2mev}{Be}}$

$r = \sqrt{\frac{(2)(9.1^{-31})(500) (1.6^{-19} ) }{ (1.75^{11} ) (10^{-4} ) } }$

$r = 2.8^{27}$ meters.

8 0

3 years ago

A current of 0.8 A flows through the motor for 3 minutes. Calculate the total charge in this time.

mixas84 [53]

$▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪$

The given terms are :

Current = 0.8 Amperes

Time = 3 minutes = 3 × 60 = 180 seconds

we know that,

$current = \dfrac{charge}{time}$

now, let's solve for charge (q) :

$0.8 = \dfrac{q}{180}$

$q = 180 \times 0.8$

$q = 144 \:$

Total Charge (q) = 144 Coulombs

7 0

3 years ago

You are in charge of a cannon that exerts a force 11500 N on a cannon ball while the ball is in the barrel of the cannon. The le

IRISSAK [1]

Answer:

m = 7.48 kg

Explanation:

force (f) = 11,500 N

length of barrel (s) = 1.7 m

angle above the ground = 49.3 degrees

acceleration due to gravity (g) = 9.8 m/s^{2}

initial velocity (u) = 0 m/s

final velocity (v) = 72.3 m/s

mass (m) = ?

force = mass (m) x acceleration (a)

acceleration (a) = force / mass (m)

acceleration (a) = 11500 / m

applying the equation of motion $v^{2} = u^{2} + 2as$ , we can get the mass

$72.3^{2} = 0^{2} + (2 x \frac{11500}{m} x 1.7 )$

5227.3 = 0 + $\frac{39100}{m}$

m = $\frac{39100}{5227.3}$

m = 7.48 kg

5 0

3 years ago

________ is a measure of how many waves pass by in one second.

Sergio [31]

C. Frequency is a measure of how many waves pass by in one second.

4 0

3 years ago

Taking the density of air to be 1.29 kg/m3, what is the magnitude of the angular momentum (in kg · m2/s) of a cubic meter of air

Svet_ta [14]

The magnitude of the angular momentum of air will be 4.128 x 10^(-3) kg·m^2/s

The rotating equivalent of linear momentum in physics is called angular momentum. Because it is a conserved quantity—the total angular momentum of a closed system stays constant—it is significant in physics. Both the direction and the amplitude of angular momentum are preserved.

Given the density of air to be 1.29 kg/m3 and a wind speed of 73.0 mi/h

We have to find the magnitude of the angular momentum

Let,

ρ = Density of air = 1.29 kg/m^3

v = Speed of wind = 73.0 mi/h = 0.032 km/s

M = angular momentum of air

Let the volume of air be 1 m^3

Mass = Volume x ρ = 1 x 1.29 = 1.29 kg

Momentum = M = mass x velocity

Momentum = 1.29 x 0.0032

Momentum = 4.128 x 10^(-3) kg·m^2/s

Hence the magnitude of the angular momentum of air will be 4.128 x 10^(-3) kg·m^2/s

Learn more about angular momentum here:

brainly.com/question/7538238

#SPJ10

8 0

2 years ago