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alisha [4.7K]
3 years ago
15

You are driving at 25 m/s with your cruise control on when you see a fallen tree in the road. It takes you 0.30 s to put on the

brakes, your car decelerates at a rate of 5.0 m/s2, and the tree is 75 m ahead. Will you hit the tree? Choose the correct answer and reason below.
Physics
1 answer:
xxTIMURxx [149]3 years ago
3 0

Explanation:

Given data

velocity v= 25m/s

The time it takes to put on brake t= 0.3s

the distance covered when the  brake was put on is

v=s/t

s= v*t

s= 25*0.3s

s= 7.5m

hence the distance covered is 7.5m

Also the rate of decrease in aceleration is 5m/s^2

we can also calculate the distance covered at this rate

v^2=u^2+2as

25^2= 0+2*5*s

625=10s

divide both sides by 10

s=625/10

s= 62.5m

The total  distance covered between putting on the brakes and decelareation is  7.5+62.5= 70m

Given that the tree is 75m ahead, the car would not hit the tree

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Jobisdone [24]

Answer:

kinetic to thermal

Explanation:

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3 years ago
What do measurements of the number of neutrinos emitted by the Sun tell us about conditions deep in the solar interior?
OverLord2011 [107]

Answer:

The neutrinos are produced in the core of the sun by nuclear fusion and measuring their number helps us confirm that there are enough proton-proton chain reactions of each which produce a neutrino and going on in the Sun's core to explain the energy output of the Sun.

4 0
2 years ago
I need both parts please (a) Given a material with an attenuation coefficient (a) of 0.6/cm, what is the intensity of a beam (wi
Masteriza [31]

Answer:

<h3>a.</h3>
  • After it has traveled through 1 cm : I(1 \ cm) = 0.5488 I_0
  • After it has traveled through 2 cm : I(2 \ cm) = 0.3012 I_0
<h3>b.</h3>
  • After it has traveled through 1 cm : od( 1\ cm) =  0.2606
  • After it has traveled through 2 cm :  od( 2\ cm) =  0.5211

Explanation:

<h2>a.</h2>

For this problem, we can use the Beer-Lambert law. For constant attenuation coefficient \mu the formula is:

I(x) = I_0 e^{-\mu x}

where I is the intensity of the beam, I_0 is the incident intensity and x is the length of the material traveled.

For our problem, after travelling 1 cm:

I(1 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 1 cm}

I(1 \ cm) = I_0 e^{- 0.6}

I(1 \ cm) = I_0 e^{- 0.6}

I(1 \ cm) = 0.5488 \ I_0

After travelling 2 cm:

I(2 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 2 cm}

I(2 \ cm) = I_0 e^{- 1.2}

I(2 \ cm) = I_0 e^{- 1.2}

I(2 \ cm) = 0.3012 \ I_0

<h2>b</h2>

The optical density od is given by:

od(x) = - log_{10} ( \frac{I(x)}{I_0} ).

So, after travelling 1 cm:

od( 1\ cm) = - log_{10} ( \frac{0.5488 \ I_0}{I_0} )

od( 1\ cm) = - log_{10} ( 0.5488 )

od( 1\ cm) = - (  - 0.2606)

od( 1\ cm) =  0.2606

After travelling 2 cm:

od( 2\ cm) = - log_{10} ( \frac{0.3012 \ I_0}{I_0} )

od( 2\ cm) = - log_{10} ( 0.3012 )

od( 2\ cm) = - (  - 0.5211)

od( 2\ cm) =  0.5211

3 0
3 years ago
A 5.92 g object moving to the right at 17.1 cm/s makes an elastic head-on collision with an 11.84 g object that is initially at
Lelechka [254]

Answer:

v₁f = -5.7 cm/s

Explanation:

  • Assuming no external forces acting during the collision, total momentum must be conserved, as follows:

        m_{1} * v_{10} = m_{1} *v_{1f} + m_{2} * v_{2f} (1)

  • Rearranging terms, we have:

        m_{1} * (v_{10} - *v_{1f} ) = m_{2} * v_{2f} (2)

  • We also know that the collision is elastic, so total kinetic energy must be conserved , as follows:

        \Delta K = 0 \\ \\ \frac{1}{2} * m_{1} *v_{10} ^{2} = \frac{1}{2}* m_{1}  *v_{1f} ^{2}  + \frac{1}{2}* m_{2}  *v_{2f} ^{2} (3)

  • Rearranging , and simplifying common terms, we have:

        m_{1}* (v_{10} ^{2} -v_{1f} ^{2} ) = m_{2}  *v_{2f} ^{2} (4)

  • Replacing by the givens, doing some algebra and dividing (4) by (2), we find the following relationship:

        v_{10} + v_{1f} = v_{2f}

  • Replacing the expression above in (1), as m₂ = 2*m₁, we can find the value of v₁f, as follows:

       m_{1} * v_{10}  = m_{1} * v_{1f} +2*m_{1} * (v_{10} + v_{1f})\\ \\ -(m_{1} * v_{10}) = 3* m_{1} *v_{1f} \\ \\ v_{1f} = - \frac{v_{10} }{3}  = \frac{-17.1cm/s}{3} = -5.7 cm/s

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The roughness of the surface, the mass of the object, and the area of contact.
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