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3 years ago

On a very hot summer day, railroad tracks can buckle, causing train delays

2 answers:

5 0

The best reason for buckling of rail tracks on hot summer days is the thermal expansion or compressive forces acting on the rail tracks.

<u>Explanation:</u>

The rail tracks are composed of metals and they are rigidly connected with each other. If the materials used in construction of rail tracks are old, then there can be weakening of the track materials leading to decrease in their resistivity to thermal heat.

So this increase in temperature on hot days leads to compressive stress on the tracks which causes the tracks to expand and misaligned themselves.

Also the loosening of the ballast, ties and fasteners enhance the speed of buckling on hot summer days. Thus due to thermal expansion of the materials of the tracks, they buckle up as there is no space for desired expansion.

patriot [66]3 years ago

3 0

Answer:

The metal rails undergo thermal expansion APEX Verified

Explanation:

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You can use any coordinate system you like in order to solve a projectile motion problem. To demonstrate the truth of this state

posledela

Answer:

a) y₂ = 49.1 m , t = 1.02 s , b) y = 49.1 m , t= 1.02 s

Explanation:

a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero

$v_{y}$ ² = $v_{oy}$ ² - 2 g (y –yo)

The origin of the coordinate system is on the floor and the ball is thrown from a height

y-yo = $v_{oy}² /2 g
 y- 0 = 10.0²/2 9.8
 y - 0 = 5.10 m
 
The height from the ground is the height that rises from the reference system plus the depth of the ground from the reference system
 y₂ = 5.1 + 44
 y₂ = 49.1 m
Let's use the other equation to find the time
 [tex]v_{y}$ = $v_{oy}$ - g t

t = $v_{oy}$ / g

t = 10 / 9.8

t = 1.02 s

b) the maximum height

y- 44.0 = $v_{y}$ ² / 2 g

y - 44.0 = 5.1

y = 5.1 +44.0

y = 49.1 m

The time is the same because it does not depend on the initial height

t = 1.02 s

7 0

4 years ago

How does friction affect the distance of an object?

stiv31 [10]

It slows the object down so it cannot move well and evetually the object cannot be pushed and farther

4 0

3 years ago

Write a numerical expression for the emissive intensity (in W/m^2.sr) coming out of a tiny hole in an enclosure of surface tempe

stiks02 [169]

Answer:

6.0 × $10^{11}$ W/ $m^{2}$

Explanation:

From Wien's displacement formula;

Q = e A $T^{4}$

Where: Q is the quantity of heat transferred, e is the emissivity of the surface, A is the area, and T is the temperature.

The emissive intensity = $\frac{Q}{A}$ = e $T^{4}$

Given from the question that: e = 0.6 and T = 1000K, thus;

emissive intensity = 0.6 × $(1000)^{4}$

= 0.6 × 1.0 × $10^{12}$

= 6.0 × $10^{11}$ $\frac{W}{m^{2} }$

Therefore, the emissive intensity coming out of the surface is 6.0 × $10^{11}$ W/ $m^{2}$ .

3 0

3 years ago

A roller coaster is traveling at 13 m/s when it approaches a hill that is 400 m long. Heading down the hill, it accelerates at 4

SIZIF [17.4K]

Answer:

The value is $v = 47 \ m/s$

Explanation:

From the question we are told that

The initial speed of the roller coaster is $u = 13 \ m/s$

The length of the hill is $l = 400 \ m$

The acceleration of the roller coaster is $a=4.0 \ m/s^2$

Generally the acceleration is mathematically represented as

$a = \frac{ v - u}{ t_f - t_i }$

Here $t_i$ is the initial time which is equal to zero

$v_f$ is the final velocity which is mathematically represented as

$v_f = \frac{d}{ t_f}$

$a = \frac{ \frac{d}{d_f} - u }{ t_f - t_i}$

$4 = \frac{\frac{400}{ t_f} - 13}{t_f - 0}$

$4 = \frac{400 - 13t_f}{ t_f} * \frac{1}{t_f}$

$4t_f ^2 +13f + 400 =$

Solving this using quadratic formula we obtain

$t_f = 8.5 \ s$

$t_f = -11.8 \ s$

Generally time cannot be negative so

$t_f = 8.5 \ s$

Generally the final velocity is mathematically represented as

$v = \frac{400}{8.5}$

$v = 47 \ m/s$

5 0

3 years ago

You observe three carts moving to the left. Cart A moves to the left at nearly constant speed. Cart B moves to the left, gradual

Lady bird [3.3K]

Answer:cart B

Explanation:

For cart A speed is constant therefore there is no acceleration because acceleration is rate of change of velocity

thus there is no net force

For cart B there is change in velocity in the left direction , so there is net acceleration towards left

$Force=mass\times acceleration$

so there is net force in the left direction

For cart C there is decrease in velocity i.e. negative acceleration or deceleration . Therefore there is a net force towards right which opposes the motion

6 0

3 years ago