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rewona [7]
3 years ago
15

A steel railroad track has a length of 23 m when the temperature is 7◦C. What is the increase in the length of the rail on a hot

day when the temperature is 30 ◦C? The linear expansion coefficient of steel is 11 × 10−6 ( ◦C)−1 . Answer in units of m.
Physics
1 answer:
xxTIMURxx [149]3 years ago
7 0

Answer:

5.82 mm

Explanation:

length of the track, L = 23 m

Initial temperature, T = 7 °C

final temperature, T' = 30 °C

Coefficient of linear expansion, α = 11 x 10^-6 / °C

Let the increase in length of the track on the hot day is ΔL.

use the formula of the linear expansion

ΔL = L x α x ΔT

ΔL = L x α x (T' - T)

ΔL = 23 x 11 x 10^-6 x (30 - 7)

ΔL = 5.82 x 10^-3 m

ΔL = 5.82 mm

Thus, the increase in the length of the track is 5.82 mm.

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An electron is moving the east with a speed of 5.0 × 106 m/s. There is an electric field of
Tanzania [10]

The velocity of the electron after moving a distance of 1cm in the electric field is 5.95×10⁶m.

<h3>What is Electric field?</h3>

Electric field is the physical field that surrounds a charge.

<h3>How to find final velocity of the electron when it moves some distance in a certain electric field?</h3>
  • From Newton's second law, the acceleration the electron will be

a=F/m=qE/m

  • where q= charge of electron

E= electric field

m= mass of electron

=(−1.60×10^−19C)(3×10³N/C)/(9.11×10^-31kg)

=10¹⁵×0.526m/s²

  • The kinematics equation v²=v0²+2a(Δx)
  • where v=final velocity of the electron

v0=initial velocity of the electron =5×10⁶m/s

a=acceleration of the electron =10¹⁵×0.526m/s²

Δx=distance moved by the electron in east direction =1cm=10^-2m

  • Now v^2=(5×10⁶)²+2×10¹⁵×0.526×10^-2

=25×10¹²+10.52×10¹²

=35.52×10¹²

  • Now velocity of electron=5.95×10⁶m/s.

Thus , we can conclude that the velocity of the electron after moving a distance of 1cm in the electric field is 5.95×10⁶m.

Learn more about electric field here:

brainly.com/question/26199225

#SPJ1

5 0
2 years ago
7) List three (3) automobile safety features currently used to minimize the risk of injury to its passengers. Relate these
r-ruslan [8.4K]
Seatbelt- strapped in egg
air bag- cushion around the egg
brakes- parachute(bag that helps the egg go down slower)
3 0
3 years ago
A single mass (m1 = 3.5 kg) hangs from a spring in a motionless elevator. The spring constant is k = 278 N/m. 1)What is the dist
artcher [175]
<h2>Answer:</h2>

0.126m

<h2>Explanation:</h2>

According to Hooke's law, the force (F) acting on a spring to cause an extension or compression (e) is given by;

F = k x e            -------------------(i)

Where;

k = the spring's constant.

From the question, the force acting on the spring is the weight(W) of the mass. i.e

F = W               -----------------------(ii)

<em>But;</em>

W = m x g;

where;

m = mass of the object

g = acceleration due to gravity [usually taken as 10m/s²]

<em>From equation (ii), it implies that;</em>

F = W = m x g

<em>Now substitute F = m x g into equation(i) as follows;</em>

F = k x e

m x g = k x e      ------------------(iii)

<em>From the question;</em>

m = m1 = 3.5kg

k = 278N/m

<em>Substitute these values into equation (iii) as follows;</em>

3.5 x 10 = 278 x e

35 = 278e

<em>Now solve for e;</em>

e = 35/278

e = 0.126m

Therefore, the distance the spring is stretched from its unstretched length (which is the same as the extension of the spring) is 0.126m

3 0
3 years ago
You want to slide a 0.39 kg book across a table. If the coefficient of kinetic friction is .21, what force is required to move t
uranmaximum [27]
Find the force that would be required in the absence of friction first, then calculate the force of friction and add them together.  This is done because the friction force is going to have to be compensated for.  We will need that much more force than we otherwise would to achieve the desired acceleration:

F_{NoFric}=ma=0.39kg \times0.18 \frac{m}{s^2}  =0.0702N

The friction force will be given by the normal force times the coefficient of friction.  Here the normal force is just its weight, mg

F_{Fric}=0.39kg \times 9.8 \frac{m}{s^2} \times 0.21=0.803N

Now the total force required is:

0.0702N+0.803N=0.873N

5 0
3 years ago
A 1400kg car is moving at a speed of 25m/s. How much KE does the car have?
Nady [450]

Answer:

437500Joules

Explanation:

Kinetic energy=1/2mvsquare

1/2 x 1400 x 25 x25

kinetic energy= 437500Joules

6 0
4 years ago
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