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AleksandrR [38]
2 years ago
12

Two cars, with the same mass and traveling at the same speed, hit large trees head-on. One car has a rigid body that undergoes l

ittle or no deformation in the collision. The other has ``crumple zones'': portions of the body designed to crumple and deform in such a collision. How does this improve the chances that the driver of the second car will survive the event
Physics
1 answer:
nordsb [41]2 years ago
8 0

The crumple zones in the second car will improve the chance of survival of the driver because it will act as shock absorber, reducing the impact of the force on the driver.

<h3>Newton's third law of motion</h3>

According to Newton's third law of motion, action and reaction are equal and opposite.

The car with rigid body will exert maximum force to the driver while the car with crumple zone will exert lesser force to the driver since the crumple zone will act as shock absorber, reducing the impact of the force on the driver.

Thus, the crumple zones in the second car will improve the chance of survival of the driver because it will act as shock absorber, reducing the impact of the force on the driver.

Learn more about Newton's third law of motion here: brainly.com/question/25998091

#SPJ1

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Consider the hydrogen atom. How does the energy difference between adjacent orbit radii change as the principal quantum number i
Kisachek [45]

Answer:

the energy difference between adjacent levels decreases as the quantum number increases

Explanation:

The energy levels of the hydrogen atom are given by the following formula:

E=-E_0 \frac{1}{n^2}

where

E_0 = 13.6 eV is a constant

n is the level number

We can write therefore the energy difference between adjacent levels as

\Delta E=-13.6 eV (\frac{1}{n^2}-\frac{1}{(n+1)^2})

We see that this difference decreases as the level number (n) increases. For example, the difference between the levels n=1 and n=2 is

\Delta E=-13.6 eV(\frac{1}{1^2}-\frac{1}{2^2})=-13.6 eV(1-\frac{1}{4})=-13.6 eV(\frac{3}{4})=-10.2 eV

While the difference between the levels n=2 and n=3 is

\Delta E=-13.6 eV(\frac{1}{2^2}-\frac{1}{3^2})=-13.6 eV(\frac{1}{4}-\frac{1}{9})=-13.6 eV(\frac{5}{36})=-1.9 eV

And so on.

So, the energy difference between adjacent levels decreases as the quantum number increases.

5 0
3 years ago
A motorcycle has 100,000 J of kinetic energy and is traveling at 20 m/s. Find its mass.
tankabanditka [31]

Answer:

<h3>The answer is 500 kg</h3>

Explanation:

The mass of the object can be found by using the formula

m =  \frac{2KE}{ {v}^{2} }  \\

v is the velocity

KE is the kinetic energy

From the question we have

m =  \frac{2 \times 100000}{  {20}^{2}  }  =  \frac{200000 }{400} =   \frac{2000}{4}  \\

We have the final answer as

<h3>500 kg</h3>

Hope this helps you

7 0
3 years ago
A device that converts high voltages to lower voltages is a(n) ________.
AysviL [449]

Answer: A) Transformer

Explanation: A Transformer is an apparatus, device or a component in a system that is used to convert high voltage into low voltage. It is used to either increase or decrease the voltage of an alternating current.

A transformer uses the basic principle of electro magnetic induction, having two or more coils, the voltage is changed from one coil to another but with thesame frequency as alternating current energy passes through them.

7 0
3 years ago
A wave's frequency is 2Hz and its wavelength is 4 m. What is the wave's speed?
icang [17]

Wave speed = (wavelength) x (frequency)

                       =      (4 m)        x  (2 /sec)

                       =               8 m/sec

7 0
3 years ago
A long, thin solenoid has 450 turns per meter and a radius of 1.06 . The current in the solenoid is increasing at a uniform rate
kirill115 [55]

Answer:9.34 A/s

Explanation:

Given

radius of solenoid R=1.06 m

Emf induced E=8.50\times 10^{-6} V/m

no of turns per meter n=450

we know Induced EMF is given by

\int Edl=-\frac{\mathrm{d} \phi}{\mathrm{d} t}=-\frac{\mathrm{d} B}{\mathrm{d} t}A

Magnetic Field is given by

B=\mu _0ni

thus \frac{\mathrm{d} B}{\mathrm{d} t}=-\mu _0n\frac{\mathrm{d} i}{\mathrm{d} t}

Area of cross-section

A=\pi R^2 where

solving integration we get

E.\cdot 2\pi r=\mu _0n\frac{\mathrm{d} i}{\mathrm{d} t}\pi R^2

where r=distance from axis

R=radius of Solenoid

\frac{\mathrm{d} i}{\mathrm{d} t}=\frac{Er}{\mu _0nR^2}

\frac{\mathrm{d} i}{\mathrm{d} t}=\frac{8.50\times 10^{-6}\times 3.49\times 10^{-2}}{4\pi \times 10^{-7}\times 450\times 1.06^2}

\frac{\mathrm{d} i}{\mathrm{d} t}=9.34 A/s

4 0
3 years ago
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