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A piece of steel is 11.5cm long at 22C. It is heated to 1221C, close to its melting point. How long is it, in cm, at the high te

Nataly [62]

Answer:

The length at the final temperature is 11.7 cm.

Explanation:

We need to use the thermal expansion equation:

$\Delta L=\alpha L_{0}\Delta T$

Where:

L(0) is the initial length
ΔT is the differential temperature, final temperature minus initial temperature (T(f)-T(0))
ΔL is the final length minus the initial length (L(f)-L(0))
α is the coefficient of linear expantion of steel (12.5*10⁻⁶ 1/°C)

So, we have:

$L_{f}-L_{0}=\alpha L_{0}(T_{f}-T_{0})$

$L_{f}=L_{0}+\alpha L_{0}(T_{f}-T_{0})$

$L_{f}=0.115+(12.5*10^{-6})(0.115)(1221-22)$

$L_{f}=0.117\: m$

Therefore, the length at the final temperature is 11.7 cm.

I hope it helps you!

7 0

2 years ago

Which types of geometrical symmetry does a sphere have?.

Helen [10]

A sphere has reflection symmetry across any plane through its center.

8 0

2 years ago

A golf ball is dropped from rest from a height of 9.50m. It hits the pavement, then bounces back up, rising just 5.70m before fa

marta [7]

Answer:

Explanation:

Given

height from which ball is dropped $h_1=9.5\ m$

and it rises to a height of $h_2=5.7\ m$

A person caught the ball in mid-way i.e. at a height of $h_0=1.2\ m$ from pavement

time taken to reach the bottom is given by

$y=ut+\frac{1}{2}at^2$

where y=displacement

u=initial velocity

a=acceleration

t=time

here initial velocity is zero i.e.

$h_1=0+\frac{1}{2}gt_1^2$

$t_1=\sqrt{\frac{2h_1}{g}}$

Similarly to reach of height $h_2$ time taken is $t_2$

$t_2=\sqrt{\frac{2h_2}{g}}$

The ball is caught in mid-way i.e. ball travel a distance of $h_3=5.7-1.2=4.5\ m$

time taken is $t_3=\sqrt{\frac{2h_3}{g}}$

total time taken $t=t_1+t_2+t_3$

$t=\sqrt{\frac{2h_1}{g}}+\sqrt{\frac{2h_2}{g}}+\sqrt{\frac{2h_3}{g}}$

$t=\sqrt{\frac{2\cdot 9.5}{9.8}}+\sqrt{\frac{2\cdot 5.7}{9.8}}+\sqrt{\frac{2\cdot 4.5}{9.8}}$

$t=3.429\ s$

total time =3.429 s

4 0

3 years ago

4. With respect to a mass of 1kg. (a) How much change in elevation must it undergo to change its potential energy by 1 kJ? (g=9.

mojhsa [17]

Answer:

102 m upwards.

Explanation:

Just from a qualitative analysis we can tell the mass it needs to go upwards. How much we determine with the fact that the increase will be - in absolute value - equal to the work gravity does on it to go down that same distance.

Fixed that work being 1 kJ, we get

$\vec F \cdot \vec{ \Delta h} = |F| |\Delta h| cos 0 = ( 9.8 N )\Delta h \cdot 1 = 10^3 J\\\Delta h = \frac {10^3}{9,8} m \approx 102 m$

4 0

1 year ago

Martha is late for class, so she races her 75 kg body up a 4.0 meter flight of stairs in only 2.3 seconds. What is Martha’s powe

Tema [17]

Martha's power is 1,300 W and is denoted as option A.

<h3>What is Power?</h3>

This is defined as the amount of energy used or transferred per unit time and the S.I unit is Watts.

Power = Work/ time

Work = force × distance

= F = ma = 75kg× 10m/s² = 750N

Work = 750N × 4m = 3000J

Power = work/time = 3000J/ 2.3s ≈ 1300W

Read more about Power here brainly.com/question/1634438

#SPJ1

3 0

1 year ago