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Setler79 [48]
3 years ago
14

The young modulus of an elastic material is 2.5×10 raised to power 3 newton per meter square. What magnitude of load will cause

an extension of 200 cm in 20cm length of the material given that it has a diameter of 15m
the young modulus of an elastic material is  {2.5 \times 10}^{3}   nm ^{ - 2}  .  What magnitude of load will cause an extension of 200 cm in 20cm length of the material given that it has a diameter of 15m
​
Physics
1 answer:
katrin2010 [14]3 years ago
5 0

Answer:.

200

Explanation:

.

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A proton is moving at 425 m/s. (a) How much work must be done on it to stop it? (A proton has a mass of 1.67×10−27 kg.) (b) Assu
Tpy6a [65]

Answer:

a)1.51*10^-22joules b) 1.89*10^-7m

Explanation:

Work done to stop the proton = the kinetic energy of the proton = 1/2 mv^2 = 1/2* 1.67*10^-27* 425*425 = 1.51* 10 ^ -22 joules

b) net force acting to stop the proton = 8.01*10^-16

Work done needed to stop the proton = net force acting opposite the motion * distance

Distance covered = need work done/ net force

Distance = 1.51*10^-22/8.01*10^-16= 1.89*10^-7m

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Why are all galaxies moving away from us?
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A balloonist happens to drop his metal ballpoint pen from his balloon as he is taking notes on his flight. Because his pen has a
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So this is dealing with the conservation of energy. So you set kinetic energy equal to potential energy, so it looks like this:

1/2mv^2=mgh. The m's cancel out, so it is 1/2v^2=gh.

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h=0.5v^2/g. v=22m/s, g=9.81m/s^2, so h=(0.5)(22^2)/(9.81)=24.67m
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The Andromeda Galaxy (our nearest spiral neighbor) has spectral lines that show a blue shift. From this we may conclude that:
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The universe is the collection of galaxies, The Andromeda Galaxy has spectral lines with blue shift. The conclusion is that the Universe has stopped expanding.

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2 years ago
QUESTION 10
Elena L [17]

The maximum value of θ of such the ropes (with a maximum tension of 5,479 N) will be able to support the beam without snapping is:

\theta =37.01^{\circ}

We can apply the first Newton's law in x and y-direction.

If we do a free body diagram of the system we will have:

x-direction

All the forces acting in this direction are:

T_{1}sin(\theta)-T_{2}sin(\theta)=0    (1)

Where:

  • T(1) is the tension due to the rope 1
  • T(2) is the tension due to the rope 2

Here we just conclude that T(1) = T(2)

y-direction

The forces in this direction are:

T_{1}cos(\theta)+T_{2}cos(\theta)-W=0   (2)

Here W is the weight of the steel beam.

We equal it to zero because we need to find the maximum angle at which the ropes will be able to support the beam without snapping.

Knowing that T(1) = T(2) and W = mg, we have:

T_{1}cos(\theta)+T_{1}cos(\theta)-m_{steel}g=0

2T_{1}cos(\theta)-m_{steel}g=0

2T_{1}cos(\theta)=m_{steel}g

T(1) must be equal to 5479 N, so we have:

cos(\theta)=\frac{m_{steel}g}{2T_{1}}

cos(\theta)=\frac{892*9.81}{2*5479}

cos(\theta)=\frac{892*9.81}{2*5479}

cos(\theta)=0.80

Therefore, the maximum angle allowed is θ = 37.01°.

You can learn more about tension here:

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I hope it helps you!

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3 years ago
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