John and Suzie are trying to improve the speed of their race car by adjusting the angle of the rear spoiler. Which choice is the controlled variable(s)?the maximum speed and the angle of the rear spoiler

the type of tire and the type of fuel

only the maximum speed

only the angle of the rear spoiler

Answers

Answer 1

Answer:

Answer:

the type of tire and the type of fuel

Explanation:

Related Questions

Why are chemical processes unable to produce the same amount of energy flowing out of the sun as nuclear fusion?
In a cyclotron, the orbital radius of protons with energy 300 keV is 16.0 cm . You are redesigning the cyclotron to be used instead for alpha particles with energy 300 keV . An alpha particle has charge q=+2e and massm=6.64×10−27kg .If the magnetic field isnt changed, what will be the orbital radius of the alpha particles? Express your answer to three significant figures and include the appropriate units.
The predictions of Rutherford's scattering formula failed to correspond with experimental data when the energy of the incoming alpha particles exceeded 32MeV32MeV. This can be explained by the fact that the predictions of the formula apply when the only force involved is the electromagnetic force and will break down if the incoming particles make contact with the nucleus. Use the fact that Rutherford's prediction ceases to be valid for alpha particles with an energy greater than 32MeV32MeV to estimate the radius rrr of the gold nucleus.
The type of function that describes the amplitude of damped oscillatory motion is _______. The type of function that describes the amplitude of damped oscillatory motion is _______. quadratic sinusoidal inverse exponential linear
Answer plz answer plzzz I am a little confused with full time

A convex mirror with a focal length of 0.25 m forms a 0.080 m tall image of an automobile at a distance of 0.24 m behind the mirror. What is the magnification of the image? Where is the car located, and what is its height? Is the image real or virtual? Is the image upright or inverted? Draw a ray diagram to show where the image forms and how large it is with respect to the object

Answers

Answer:

The distance and height of the object is 6 m and 2 m.

The image is virtual and upright.

Explanation:

Given that,

Focal length = 0.25 m

Length of image = 0.080 m

Image distance = 0.24 m

We need to calculate the distance of the object

Using formula of lens

$(1)/(v)=(1)/(f)+(1)/(u)$

Put the value into the formula

$(1)/(0.24)=(1)/(0.25)+(1)/(u)$

$(1)/(u)=(1)/(0.24)-(1)/(0.25)$

$(1)/(u)=(1)/(6)$

$u=6\ m$

We need to calculate the magnification

Using formula of magnification

$m=-(v)/(u)$

Put the value into the formula

$m=-(0.24)/(-6)$

$m=0.04$

We need to calculate the height of the object

Using formula of magnification

$m=(h')/(h)$

$h=(0.080)/(0.04)$

$h=2\ m$

A convex mirror produce a virtual and upright image behind the mirror.

Hence, The distance and height of the object is 6 m and 2 m.

The image is virtual and upright.

Answer:

Distance of the object = 6 m

Height of the object = 2 m

Explanation:

Thinking process:

Given that,

Focal length = 0.25 m

Length of image = 0.080 m

Image distance = 0.24 m

We need to calculate the distance of the object

Therefore, using formula of lens:

$(1)/(u) = (1)/(f) + (1)/(u)$

$(1)/(u) = (1)/(6)$

solving, gives u = 6

The magnification is calculated as follows:

m = -0.24/-6

= 0.04

The height = 2 m

The diagram yields an image behind the mirror which is upright.

The intensity of a sound wave at a fixed distance from a speaker vibrating at 1.00 kHz is 0.750 W/m2. (a) Determine the intensity if the frequency is increased to 2.20 kHz while a constant displacement amplitude is maintained.(b) Calculate the intensity if the frequency is reduced to 0.250 kHz and the displacement amplitude is quadrupled.

Answers

Final answer:

The intensity of sounds is dependent on the square of the amplitude, not the frequency. Therefore, the intensity of sound remains the same when frequency is altered but the amplitude is constant. When the amplitude is quadrupled, the intensity of the sound becomes sixteen times greater.

Explanation:

In the field of physics, the intensity of a sound wave is defined as the power per unit area carried by the wave. This question involves calculating the change in sound wave intensity when the frequency and displacement amplitude of the source are altered.

(a) When the frequency is increased to 2.20 kHz while keeping the displacement amplitude constant, the intensity does not change, as the intensity in this case is not dependent on the frequency but on the square of the amplitude. Therefore, the intensity remains 0.750 W/m2.

(b) When the frequency is reduced to 0.250 kHz and the displacement amplitude is quadrupled, the intensity changes. Since the intensity of a sound wave is proportional to the square of the amplitude, by quadrupling the amplitude, the intensity will become 16 times greater (since 4 squared is 16). Hence, the new intensity will be 16 * 0.750 = 12 W/m2.

Learn more about Sound Wave Intensity here:

brainly.com/question/31851162

#SPJ12

The acceleration of a particle moving along a straight line is given by a = −kt2 m/s2 where k is a constant and time t is in seconds. The initial velocity of the particle at t = 0 is v0 = 12 m/s and the particle reverses it direction of motion at t = 6 s. Determine the constant k and the displacement of the particle over the same 6-second interval of motion. Ans: k = 1/6 m/s4, Δs = 54 m

Answers

Answer:

$X - Xo = 54m$

k = 1/18

Explanation:

Data:

a = -k $t^(2)$ $(m)/(s^(2) )$

to = 0s Vo = 12m/s

t = 6s the particle chage it's moviment, so v = 0 m/s

We know that acceleration is the derivative of velocity related to time:

$a = (dV)/(dT)$

rearranging...

$a*dT = dV$

Then, we must integrate both sides:

$\int\limits^f_i {dV} \, dV =-k \int\limits^f_i {t^(2) } \, dT$

$V - Vo = -k(t^(3) )/(3)$

V = 0 because the exercise says that the car change it's direction:

$0 - 12 = -k(6^(3) )/(3)$

k = 1/6

In order to find X - Xo we must integer v*dT = dX

$V - Vo = -k(t^(3) )/(3)$

so...

$(Vo -k(t^(3) )/(3))dT = dX$

$\int\limits^f_i {dX} \, dX = \int\limits^f_i {Vo -k(t^(3) )/(3) } \, dT$

integrating...

$X - Xo = Vot -k(t^(4) )/(12)$

$X - Xo = 12*6 -(1)/(6)* (6^(4) )/(12)$

X - Xo = 54m

URGENT!!!!!! Assume that a wire has 1.5 ohms of resistance. If the wire is connected to two batteries with a total voltage of 3.0 V, how much current will flow through the wire? 3.0 amps 2.3 amps 2.0 amps 1.0 amps

Answers

Answer:

2.0 amps

Explanation:

Current is the ratio of voltage to resistance:

I = V/R = (3.0)/(1.5) = 2.0

The current in the wire is 2.0 amps.

A 12.0 kg mass, fastened to the end of an aluminum wire with an unstretched length of 0.50 m, is whirled in a vertical circle with a constant angular speed of 120 rev/min. The cross-sectional area of thewire is 0.014 cm2. Calculate the elongation of the wire when the mass is a) at its lowest point of the path and b) at the highest point of its path

Answers

Answer:

a) the elongation of the wire when the mass is at its lowest point on the path = 0.5 cm

b) the elongation of the wire when the mass is at its highest point on the path = 0.42 cm

Explanation:

Given that;

the angular speed $\omega = 120 \ rev/min$

Then converting it to rad/s ; we have:

= $(120 \ rev/min )((2 \ \pi \ rad )/(1 \ rev) ) ((1 \ min )/(60 \ s) )$

= 12.57 rad/s

The cross-sectional area of the wire A = 0.014 cm²

A = (0.014 cm²) ( $(10^(-4) \ m^2)/(1 \ cm^3)$ )

A = $0.014*10^(-4) \ m^2$

mass (m) = 12.0 kg

R = 0.5 m

g = 9.8 m/s²

To calculate for the mass when its at the lowest point of the path; we use the Newton's second law of motion; which is expressed as:

$T - mg = ma_(rad)$

where;

$a_rad = ( radical \ acceleration ) = R \omega^2$

Now; we can rewrite our equation as;

$T -mg = m R \omega ^2$

$T = mR \omega^2 + mg$

$T = m( R \omega^2 + g)$

Replacing our given values ; we have:

$T = 12.0( 0.5(12.57)^2 + 9.8)$

$T = 12.0( 0.5(158.0049) + 9.8)$

$T = 12.0( 79.00245 + 9.8)$

$T = 12.0( 88.80245)$

T = 1065.6294 N

T ≅ 1066 N

Determining the elongation $\delta l$ in the wire by using the equation

Y = $(Tl)/(AY)$

Making $\delta l$ the subject of the formula; we have

$\delta l = (Tl)/(AY)$

where ;

l = length of the wire

T =Tension in the wire

A = cross - sectional area

Y = young's modulus

Then;

$\delta l = ((1066 N)(0.5m))/((0.014*10^(-4)m^2)(7.0*10^(10)Pa))$

$\delta l$ = $( 0.00544 m ) *((10 ^2 cm)/(1m) )$

$\delta l$ = 0.5 cm

Thus, the elongation of the wire when the mass is at its lowest point on the path = 0.5 cm

Using Newton's second law of motion also for the mass at its highest point of the path; we have:

$T +mg = m R \omega ^2$

$T = mR \omega^2- mg$

$T = m( R \omega^2 - g)$

Replacing our given values ; we have:

$T = 12.0( 0.5(12.57)^2 - 9.8)$

$T = 12.0( 0.5(158.0049)-9.8)$

$T = 12.0( 79.00245 - 9.8)$

$T = 12.0( 69.20245)$

T = 830.4294 N

T = 830 N

Determining the elongation $\delta l$ in the wire by using the equation

Y = $(Tl)/(AY)$

Making $\delta l$ the subject of the formula; we have

$\delta l = (Tl)/(AY)$

where ;

l = length of the wire

T =Tension in the wire

A = cross - sectional area

Y = young's modulus

Then;

$\delta l = ((830 N)(0.5m))/((0.014*10^(-4)m^2)(7.0*10^(10)Pa))$

$\delta l$ = $( 0.00424 m ) *((10 ^2 cm)/(1m) )$

$\delta l$ = 0.42 cm

Thus, the elongation of the wire when the mass is at its highest point on the path = 0.42 cm

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

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,

$PE = mgh$