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2 years ago

"when you double the vehicle's weight, you will __________ the vehicle's stopping distance. "

1 answer:

5 0

With same braking power you will be stopping faster on the original weight therefore the answer to fill the blank is increase. The stopping distance will increase as there'll be higher energy to dissipate than lighter cars applied with the braking force similar with that of the lighter car. Also the skid and drag will add to the distance as well as the inertia of the moving heavier vehicle would be greater as well.

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Explain whether or not there is any difference between a light ray emitted by a candle flame and one reflected off the cover of

nexus9112 [7]

Answer:

the difference between the two is that the candle forms an emission spectrum and the book an absorption spectrum.

the book it is observed in all directions so that its reflection has to be diffused

Explanation:

The ray of light emitted by a candle is the light generated by the temperature of the flame, which is made up of the emissions of a black body at this temperature plus the emissions of the chemical elements that make up the candle.

The Light reflected from the cover of a book is the same incident light spectrum minus the wavelengths that create transitions in the elements of the cover, these wavelengths will be seen as dark areas.

As a consequence of the above, the difference between the two is that the candle forms an emission spectrum and the book an absorption spectrum.

For the cover of the book form a specular reflection the incident rays are reflected in one direction and the rest would be dark, but in the book it is observed in all directions so that its reflection has to be diffused

6 0

2 years ago

(a) Aircraft sometimes acquire small static charges. Suppose a supersonic jet has a 0.500 - μC charge and flies due west at a sp

12345 [234]

(a) $2.64\cdot 10^{-8} N$ north

We can treat the aircraft as a single point charge moving in a magnetic field. In this case, the magnetic force exerted on the plane is

$F=qvB sin \theta$

where

$q=0.500 \mu C = 0.500\cdot 10^{-6} C$ is the charge on the plane

v = 660 m/s is the velocity

$B=8.00\cdot 10^{-5} T$ is the magnitude of the magnetic field

$\theta=90^{\circ}$ is the angle between the direction of motion of the jet and of the magnetic field

Substituting,

$F=(0.5\cdot 10^{-6})(660)(8.0\cdot 10^{-5})=2.64\cdot 10^{-8} N$

The direction can be found by using Fleming's left hand rule. We have:

- index finger: magnetic field direction (straight up)

- middle finger: velocity of the plane (due west)

- force: thumb --> north

(b) Not negligible

As we can see from part (a), the magnitude of the force is not really big, so the effects are negligible.

For instance, we can compare this force with the weight of a plane. If we take a Boeing 737, its mass is about 80,000 kg, so its weight is

$W=mg=(80000)(9.8)=784,000 N$

As we can see, this is several orders of magnitude bigger than the magnetic force calculated at point (a), so the effects of the magnetic force are negligible.

8 0

2 years ago

The difference between the two molar specific heats of a gas is 8000J/kgK. If the ratio of the two specific heats is 1.65, calcu

Serjik [45]

Answer:

sorry

Explanation:

pls search on google

4 0

2 years ago

The U.S. Department of Energy had plans for a 1500-kg automobile to be powered completely by the rotational kinetic energy of a

navik [9.2K]

Answer:

230

Explanation:

$\omega$ = Rotational speed = 3600 rad/s

I = Moment of inertia = 6 kgm²

m = Mass of flywheel = 1500 kg

v = Velocity = 15 m/s

The kinetic energy of flywheel is given by

$K=\dfrac{1}{2}I\omega^2\\\Rightarrow K=\dfrac{1}{2}6\times 3600^2\\\Rightarrow K=38880000\ J$

Energy used in one acceleration

$K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}1500\times 15^2\\\Rightarrow K=168750\ J$

Number of accelerations would be given by

$n=\dfrac{38880000}{168750}\\\Rightarrow n=230.4$

So the number of complete accelerations is 230

8 0

2 years ago

A passenger compartment of a rotating amusement park ride contains a bench on which a book of mass

Basile [38]

a) 120 s

b) v = 0.052R [m/s]

Explanation:

The period of a revolution in a simple harmonic motion is the time taken for the object in motion to complete one cycle (in this case, the time taken to complete one revolution).

The graph of the problem is missing, find it in attachment.

To find the period of revolution of the book, we have to find the time between two consecutive points of the graph that have exactly the same shape, which correspond to two points in which the book is located at the same position.

The first point we take is t = 0, when the position of the book is x = 0.

Then, the next point with same shape is at t = 120 s, where the book returns at x = 0 m.

Therefore, the period is

T = 120 s - 0 s = 120 s

The tangential speed of the book is given by the ratio between the distance covered during one revolution, which is the perimeter of the wheel, and the time taken, which is the period.

The perimeter of the wheel is:

$L=2\pi R$

where R is the radius of the wheel.

The period of revolution is:

$T=120 s$

Therefore, the tangential speed of the book is:

$v=\frac{L}{T}=\frac{2\pi R}{120}=0.052R$

8 0

2 years ago