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Lilit [14]
3 years ago
14

A person wearing a shoulder harness can survive a car crash if the acceleration is smaller than -300 m/s . assuming constant acc

eleration how far must the front end of the car collapse if 2 it crashes while going 101 km/hr?
Physics
1 answer:
mars1129 [50]3 years ago
3 0

To solve this problem, we use the equation:

<span>d = (v^2  - v0^2) / 2a</span>

 

where,

d = distance of collapse

v0 = initial velocity = 101 km / h = 28.06 m / s

v = final velocity = 0

a = acceleration = - 300 m / s^2

 

d = (-28.06 m / s)^2 / (2 * - 300 m / s^2)

<span>d = 1.31 m</span>

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Two particles have a mass of 8kg and 12kg, respectively. if they are 800mm apart, determine the force of gravity acting between
natima [27]
The magnitude of the force of gravity acting between the particles is:
F=G\frac{m_1.m_2}{d^2} 
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P=mg
Now let's plug in the numbers knowing that G=6.67\times10^{-11} , g=9.81, d=0.8 and m1 and m2 are already given in kilograms. We get then:
P_1=m_1.g=78.48N
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3 years ago
A proton moves along the x-axis in the laboratory with velocity uy = 0.6c. An observer moves with a velocity of v=0.8c along the
Dima020 [189]

Explanation:

Given that,

Velocity of the proton in lab frame u_{x} = 0.6c

Velocity of the observer v= 0.8c

We need to calculate the velocity of the proton with respect to the observer

Using formula of velocity

u'=\dfrac{v-u_{x}}{1-\dfrac{u_{x}v}{c^2}}

u'=\dfrac{0.8c-0.6c}{1-\dfrac{0.6c\times0.8c}{c^2}}

u'=c(\dfrac{0.8-0.6}{1-(0.8\times0.6)})

u'=0.385c

(a). We need to calculate the total energy of the proton in the lab frame

Using formula of kinetic energy

K.E=m_{0}c^2(\dfrac{1}{\sqrt{1-\dfrac{u_{x}^2}{c^2}}}-1)

Where, Proton mass energy = m₀c²

Put the value into the formula

K.E=938.28\times(\dfrac{1}{\sqrt{1-\dfrac{(0.6c)^2}{c^2}}}-1)

K.E=938.28\times(\dfrac{1}{\sqrt{1-(0.6)^2}}-1)

K.E=234.57\ MeV

(b). We need to calculate the kinetic energy of the proton in the observer

Using formula of kinetic energy

K.E=m_{0}c^2(\dfrac{1}{\sqrt{1-\dfrac{u'^2}{c^2}}}-1)

K.E=938.28\times(\dfrac{1}{\sqrt{1-\dfrac{(0.385)^2}{c^2}}}-1)

K.E=938.28\times(\dfrac{1}{\sqrt{1-(0.385)^2}}-1)

K.E=78.366\ MeV

(c). We need to calculate the momentum of the proton with respect to observer

Using formula of momentum

P_{obs}=\dfrac{m_{0}u'^2}{\sqrt{1-\dfrac{u'^2}{c^2}}}

We know that,

Proton mass energy = m₀c²

m_{0}=\dfrac{938.28}{c^2}

P_{obs}=\dfrac{\dfrac{938.28}{c^2}\times(0.385c)^2}{\sqrt{1-\dfrac{(0.385c)^2}{c^2}}}

P_{obs}=\dfrac{938.28\times(0.385)^2}{\sqrt{1-0.385^2}}

P_{obs}=150.69\ MeV/c

Hence, This is required solution.

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KiRa [710]

Answer:

0 - 6.9 --> Acidic

Explanation:

4 0
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