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Mumz [18]
3 years ago
11

3. A concrete highway is built of slabs 12 m long (20o C). How wide should the expansion cracks between the slabs be (at 20o C)

to prevent buckling if the range of temperature is -30o to +50o C? ( = 12 x 10-6 oC-1 for concrete)
Physics
1 answer:
lesantik [10]3 years ago
4 0

Answer:

\Delta x=2.304\times 10^{-2}\ m=2.304\ cm is the minimum gap between the slabs

Explanation:

Given:

  • length of the concrete highway, l=12\ m
  • coefficient of thermal expansion, \alpha=12\times 10^{-6}\ ^{\circ}C^{-1}
  • range of temperature variation, (-30^{\circ}C\ to\ 50^{\circ}C) \Rightarrow \Delta T=80^{\circ}C

<u>Now form the equation of thermal expansion:</u>

\Delta l=l\times \alpha\times \Delta T

\Delta l=12\times (12\times 10^{-6})\times 80

\Delta l=1.152\times 10^{-2}\ m

Since each slab of the highway expands by the above length so the minimum gap between the slabs to prevent buckling:

\Delta x=2\times \Delta l

\Delta x=2\times( 1.152\times 10^{-2})

\Delta x=2.304\times 10^{-2}\ m=2.304\ cm is the minimum gap between the slabs

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3 years ago
A bus travels north on some busy city streets for 2.5 km, and a trip
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Answer:

V = 4.63 m/s

V = 11.31 m/s

Explanation:

Given,

The distance traveled by the bus, towards north, d = 2.5 km

                                                                                     = 2500 m

The time taken by the trip is, t = 9 min

                                                  = 540 s

The velocity of the bus,

                                       V = d / t

                                           = 2500 / 540

                                          = 4.63 m/s

At another  point, the bus travels at a constant speed of v = 18 m/s

Therefore the velocity becomes

                                                V = (4.63 + 18)/2

                                                   = 11.31 m/s

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8 0
3 years ago
A ball is kicked at an angle of 35° with the ground.a) What should be the initial velocity of the ball so that it hits a target
stiks02 [169]

Answer:

a.18.5 m/s

b.1.98 s

Explanation:

We are given that

\theta=35^{\circ}

a.Let v_0 be the initial velocity of the ball.

Distance,x=30 m

Height,h=1.8 m

v_x=v_0cos\theta=v_0cos35

v_y=v_0sin\theta=v_0sin35

x=v_0cos\theta\times t=v_0cos35\times t

t=\frac{30}{v_0cos35}

h=v_yt-\frac{1}{2}gt^2

Substitute the values

1.8=v_0sin35\frac{30}{v_0cos35}-\frac{1}{2}(9.8)(\frac{30}{v_0cso35})^2

1.8=30tan35-\frac{6574.6}{v^2_0}

\frac{6574.6}{v^2_0}=21-1.8=19.2

v^2_0=\frac{6574.6}{19.2}

v_0=\sqrt{\frac{6574.6}{19.2}}=18.5 m/s

Initial velocity of the ball=18.5 m/s

b.Substitute the value then we get

t=\frac{30}{18.5cos35}

t=1.98 s

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Find the distance in nm between two slits that produces the first minimum for 410-nm violet light at an angle of 14.5°.
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Answer:

820 nm

Explanation:

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m=0

We know that for destructive interference

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Substitute the values

(0+\frac{1}{2})\times 410\times 10^{-9}=dsin 14.5

d=\frac{410\times 10^{-9}}{2\times sin 14.5}

d=820\times 10^{-9} m=820 nm

Hence, the distance between two slits that produces the first minimum=820 nm

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3 years ago
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