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Liono4ka [1.6K]
1 year ago
14

In a car crash, large accelerations of the head can lead to severe injuries or even death. A driver can probably survive an acce

leration of 50g that lasts for less than 30 ms, but in a crash with a 50g acceleration lasting longer than 30 ms, a driver is unlikely to survive. Imagine a collision in which a driver's head experienced a 50g acceleration. What is the shortest survivable distance over which the driver's head could have come to rest?
Physics
1 answer:
Mandarinka [93]1 year ago
8 0

Answer:

0.22 m

Explanation:

We are told that the driver can survive an acceleration of 50g only if the collision lasts no longer than 30 ms. So,

t = 30 ms = 0.030 s

The acceleration is

a=-50g = -50(9.8)=-490 m/s^2

where the negative sign is due to the fact that this is a deceleration, since the driver comes to a stop in the collision.

First of all, we can find what the initial velocity of the car should be in this conditions by using the equation:

v=u+at

And since the final velocity is zero, v=0, and solving for u,

u=-at=-490(0.030)=14.7 m/s

And now we can find the corresponding distance travelled using the equation:

d=ut+\frac{1}{2}at^2 = (14.7)(0.030)+\frac{1}{2}(-490)(0.030)^2=0.22 m

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When a resistor with resistance R is connected to a 1.50-V flashlight battery, the resistor consumes 0.0625 W of electrical powe
ch4aika [34]

Answer:

4.41 W

Explanation:

P = IV, V = IR

P = V² / R

Given that P = 0.0625 when V = 1.50:

0.0625 = (1.50)² / R

R = 36

So the resistor is 36Ω.

When the voltage is 12.6, the power consumption is:

P = (12.6)² / 36

P = 4.41

So the power consumption is 4.41 W.

5 0
2 years ago
Honeybees can see light in the ________ range of the electromagnetic spectrum.
konstantin123 [22]
Humans can see wavelengths in the visible part of the electromagnetic spectrum. That is the range of approximately 400 - 700 nm. Honeybees can see visible light and about 100 nm more in the ultraviolet part of the electromagnetic spectrum. That is approximately 300 - 700 nm. 
4 0
2 years ago
The Lyman series comprises a set of spectral lines. All of these lines involve a hydrogen atom whose electron undergoes a change
mihalych1998 [28]

Answer:

a) 1.2*10^-7 m

b) 1.0*10^-7 m

c) 9.7*10^-8 m

d) ultraviolet region

Explanation:

To find the different wavelengths you use the following formula:

\frac{1}{\lambda}=R_H(1-\frac{1}{n^2})

RH: Rydberg constant = 1.097 x 10^7 m^−1.

(a) n=2

\frac{1}{\lambda}=(1.097*10^7m^{-1})(1-\frac{1}{(2)^2})=8227500m^{-1}\\\\\lambda=1.2*10^{-7}m

(b)

\frac{1}{\lambda}=(1.097*10^7m^{-1})(1-\frac{1}{(3)^2})=9751111,1m^{-1}\\\\\lambda=1.0*10^{-7}m

(c)

\frac{1}{\lambda}=(1.097*10^7m^{-1})(1-\frac{1}{(4)^2})=10284375m^{-1}\\\\\lambda=9.7*10^{-8}m

(d) The three lines belong to the ultraviolet region.

8 0
2 years ago
The mass m1 enters from the left with velocity v0 and strikes a mass m2 > m1 which is initially at rest. The collision betwee
enot [183]

Answer:

1. False 2) greater than. 3) less than 4) less than

Explanation:

1)

  • As the collision is perfectly elastic, kinetic energy must be conserved.
  • The expression for the final velocity of the mass m₁, for a perfectly elastic collision, is as follows:

        v_{1f} = v_{10} *\frac{m_{1} -m_{2} }{m_{1} +m_{2}}

  • As it can be seen, as m₁ ≠ m₂, v₁f ≠ 0.

2)

  • As total momentum must be conserved, we can see that as m₂ > m₁, from the equation above the final momentum of m₁ has an opposite sign to the initial one, so the momentum of m₂ must be greater than the initial momentum of m₁, to keep both sides of the equation balanced.

3)    

  • The maximum energy stored in the in the spring is given by the following expression:

       U =\frac{1}{2} *k * A^{2}

  • where A = maximum compression of the spring.
  • This energy is always the sum of the elastic potential energy and the kinetic energy of the mass (in absence of friction).
  • When the spring is in a relaxed state, the speed of the mass is maximum, so, its kinetic energy is maximum too.
  • Just prior to compress the spring, this kinetic energy is the kinetic energy of m₂, immediately after the collision.
  • As total kinetic energy must be conserved, the following condition must be met:

       KE_{10} = KE_{1f}  + KE_{2f}

  • So, it is clear that KE₂f  < KE₁₀
  • Therefore, the maximum energy stored in the spring is less than the initial energy in m₁.

4)

  • As explained above, if total kinetic energy must be conserved:

        KE_{10} = KE_{1f}  + KE_{2f}

  • So as kinetic energy is always positive, KEf₂ < KE₁₀.
4 0
2 years ago
An astronaut on a small planet wishes to measure the local value of g by timing pulses traveling down a wire which has a large o
8_murik_8 [283]

Answer:

The gplanet is 0.193 m/s^2

Explanation:

The speed of the pulse is:

v=\frac{lengthofthewipe}{traveltime} =\frac{1.6}{0.0656} =15.24m/s

v=\sqrt{\frac{MgL}{m} } \\v^{2} =\frac{MgL}{m} \\g=\frac{mv^{2} }{ML}

where

m=mass of the wire=4 g= 4x10^-3 kg

M=mass of the object= 3 kg

Replacing values:

g=\frac{4x10^{-3}*15.24^{2}  }{3*1.6} =0.193 m/s^{2}

7 0
2 years ago
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