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Debora [2.8K]
3 years ago
12

6. Your friend is going 30 mph, when she slams on the brakes. If you are not wearing a safety belt. Which way does your body mov

e? How fast will you be going when you slam into the windshield? Explain why you move this way in terms your inertia.
Physics
1 answer:
8090 [49]3 years ago
8 0

Answer:

We will go in the same direction where the car was was moving before applying brakes

Our speed will be same as that of speed of the car before applying brakes i.e. 30 mph

This is due to law of inertia as we are initially moving with speed 30 mph so the due to inertia of motion of our body after applying brakes we move in the same direction.

Explanation:

As we know by Newton's first law that every object moves in its state of motion or state of rest until and unless some external force will act upon it.

Here we know that car is initially moving with speed 30 mph so our body has state of motion so it will have tendency to move in same direction.

Here when brakes are applied in the car then also our body will move with same speed in same direction due to inertia of motion.

So our body will strike with windshield with the speed of 30 mph.

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un conductor de 0.358km de longitud tiene una seccion hexagonal de lado 1.25 milimetros y una resistencia de 25,75 ohmios, la re
dangina [55]

Answer:

Explanation:

he answer is 540 because you need to do the area of the wall times six. 15 times 6 times 6 equals 540

6 0
4 years ago
At what speed, as a fraction of c , is a particle's total energy twice its rest energy
WINSTONCH [101]
The rest energy of a particle is
E_0=m_0 c^2
where m_0 is the rest mass of the particle and c is the speed of light.

The total energy of a relativistic particle is
E=mc^2 =  \frac{m_0 c^2}{ \sqrt{1- \frac{v^2}{c^2} } }
where v is the speed of the particle.

We want the total energy of the particle to be twice its rest energy, so that
E=2E_0
which means:
\frac{m_0c^2}{ \sqrt{1- \frac{v^2}{c^2} } }=2m_0 c^2
\frac{1}{ \sqrt{1- \frac{v^2}{c^2} } }=2
From which we find the ratio between the speed of the particle v and the speed of light c:
\frac{v}{c}=  \sqrt{1- (\frac{1}{2})^2 }  =0.87
So, the particle should travel at 0.87c in order to have its total energy equal to twice its rest energy.
3 0
3 years ago
An earthquake produces longitudinal P waves that travel outward at 8000 m/s and transverse S waves that move at 4500 m/s. A seis
vivado [14]

Answer:

1234285.7 m or 1234.3 km

Explanation:

Let the distance be d, the time taken by P waves be t_P and the time taken by the S waves be t_S.

\text{Velocity}\dfrac{\text{Distance}}{\text{Time}}

\text{Time}\dfrac{\text{Distance}}{\text{Velocity}}

For the P waves,

t_P=\dfrac{d}{8000}

d=8000t_P

For the S waves,

t_S=\dfrac{d}{4500}

d=4500t_S

Equating the d,

8000t_P=4500t_S

Divide both sides of the equation by 500 to reduce the terms.

16t_P=9t_S

Since S waves arrive 2 minutes (= 120 seconds) after P waves,

t_S-t_P=120

t_S=120+t_P

Substitute this in the equation of the distance.

16t_P=9(t_P+120)

16t_P=9t_P+1080

7t_P=1080

t_P=\dfrac{1080}{7}

Substitute this in the equation for d involving t_P.

d=8000t_P

d=8000\times\dfrac{1080}{7}

d=1234285.7 \text{ m }= 1234.3 \text{ km}

4 0
3 years ago
Question:
iogann1982 [59]
The correct answer is:
<span>C: in the protons and neutrons of an atom

In fact, the nuclear energy refers to the binding energy of the nucleons (protons and neutrons) of an atom. The protons and the neutrons are held together by the strong nuclear interaction, one of the four fundamental forces of nature, and the energy associated to this interaction is called nuclear energy.

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7 0
3 years ago
An infinite conducting cylindrical shell of outer radius r1 = 0.10 m and inner radius r2 = 0.08 m initially carries a surface ch
irinina [24]

Answer:

a) \sigma_{\rm in} = -2.18~{\rm \mu C/m^2}

b) \sigma_{\rm out}= 1.12~{\rm \mu C/m^2}

c) E = \frac{\sigma(r_1 + r_2)}{\epsilon_0 r}

Explanation:

Before the wire is inserted, the total charge on the inner and outer surface of the cylindrical shell is as follows:

Q_{\rm in} = \sigma A_{\rm in} = \sigma(2\pi r_1 h) = (-0.35)(2\pi (0.08) h) = -0.175h~{\rm \mu C}

Q_{\rm out} = \sigma A_{\rm out} = \sigma(2\pi r_2 h) = (-0.35)(2\pi (0.1) h) = -0.22h~{\rm \mu C}

Here, 'h' denotes the length of the cylinder. The total charge of the cylindrical shell is -0.395h μC.

When the thin wire is inserted, the positive charge of the wire attracts the same amount of negative charge on the inner surface of the shell.

Q_{\rm wire} = \lambda h = 1.1h~{\rm \mu C}

a) The new charge on the inner shell is -1.1h μC. Therefore, the new surface charge density of the inner shell can be calculated as follows:

\sigma_2 = \frac{Q_{\rm in}}{2\pi r_1h} = \frac{-1.1h}{2\pi r_1 h} = \frac{-1.1}{2\pi(0.08)} = -2.18~{\rm \mu C/m^2}

b) The new charge on the outer shell is equal to the total charge minus the inner charge. Therefore, the new charge on the outer shell is +0.705 μC.

The new surface charge density can be calculated as follows:

\sigma_{\rm out}= \frac{Q_{\rm out}}{2\pi r_2h} = \frac{0.705h}{2\pi r_2 h} = \frac{0.705}{2\pi(0.1)} = 1.12~{\rm \mu C/m^2}

c) The electric field outside the cylinder can be found by Gauss' Law:

\int{\vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}

We will draw an imaginary cylindrical shell with radius r > r2. The integral in the left-hand side will be equal to the area of the imaginary surface multiplied by the E-field.

E(2\pi r h) = \frac{Q_{\rm enc}}{\epsilon_0}\\E2\pi rh = \frac{\sigma 2\pi (r_1 + r_2)h}{\epsilon_0}\\E = \frac{\sigma(r_1 + r_2)}{\epsilon_0 r}

4 0
3 years ago
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