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borishaifa [10]
2 years ago
6

A piece of steel expands 10 cm when ur is heated from 20 to 50 degrees Celsius. How much would it expand if it was heated from 2

0 to 60 degrees Celsius
Physics
1 answer:
faltersainse [42]2 years ago
8 0

The steel would expand by 4. 8 * 10^-3 cm

<h3>How to determine the linear expansion</h3>

The change in length  ΔL is proportional to length  L. It is dependent on the temperature, substance, and length.

Using the formula:

ΔL= α LΔT

where  ΔL  is the change in length  L = 10cm

ΔT  is the change in temperature = 60° - 20° = 40° C

 α  is the coefficient of linear expansion = 1.2 x 10^-5 °C

Substitute into the formula

ΔL = 1.2 * 10^-5 * 10 * 40

ΔL = 4.8 * 10 ^-3 cm

Therefore, the steel would expand by 4. 8 * 10^-3 cm

Learn more about linear expansion here:

brainly.com/question/14325928

#SPJ1

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A wave has a frequency of 450 Hz and a wave length of 00.52 m what is the speed of the wave?
nata0808 [166]
Wave speed is always (frequency) x (wavelength)

                       Speed  =  (450 /sec) x (0.52 m)

                                     =      234 m/sec .
4 0
3 years ago
Read 2 more answers
you leave your house and travel for 6 minutes at an average velocity of 5 m/s. How far have you traveled?​
Anvisha [2.4K]

Answer:

1800 m

Explanation:

This question requires you to calculate the distance traveled after 6 minutes at a speed of 5m/s

Given speed as 5m/s and time is 6 minutes then;

Change minutes to seconds by multiplying the minutes by 60 seconds

1 minute=60 seconds

6 minutes= 6*60 = 360 seconds

Apply the formula for speed where;

S=D/T where S is speed, D is distance and T is time

To get D , rewrite formula as S*T

where S= 5m/s and T=360s

D= 5*360 =1800m

3 0
4 years ago
A car is moving in the positive direction along a straight highway and accelerates at a constant rate while going from point A t
rusak2 [61]

Answer:

The time where the avergae speed equals the instaneous speed is T/2

Explanation:

The velocity of the car is:

v(t) = v0 + at

Where v0 is the initial speed and a is the constant acceleration.

Let's find the average speed. This is given integrating the velocity from 0 to T and dividing by T:

v_{ave} = \frac{1}{T}\int\limits^T_0 {v(t)} \, dt

v_ave = v0+a(T/2)

We can esaily note that when <u><em>t=T/2</em></u><u><em> </em></u>

v(T/2)=v_ave

Now we want to know where the car should be, the osition of the car is:

x(t) = x_A + v_0 t + \frac{1}{2}at^2

Where x_A is the position of point A. Therefore, the car will be at:

<u><em>x(T/2) = x_A + v_0 (T/2) + (1/8)aT^2</em></u>

8 0
3 years ago
Find electric field at point p which is a distance l away from the both +q and -q
denis-greek [22]

Answer:

\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

Explanation:

As given point p is equidistant from both the charges

It must be in the middle of both the charges

Assuming all 3 points lie on the same line

Electric Field due a charge q at a point ,distance r away

=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{r^{2} }

Where

  • q is the charge
  • r is the distance
  • E is the permittivity of medium

Let electric field due to charge q be F1 and -q be F2

I is the distance of P from q and also from charge -q

⇒

F1=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }

F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

⇒

F1+F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

8 0
4 years ago
A total charge Q is distributed uniformly over a large flat insulating surface of area A . If the electric field magnitude is eq
Andrew [12]

Answer:

a. 1000N/C

Explanation:

Data mentioned in the question

Electrical field magnitude = 1000 NC

Perpendicular distance = 0.1 m

Perpendicular distance = 0.2 m

Based on the above information, the electric field is

As we know that

E = \frac{\sigma}{2\times E_o}

where,

\sigma = surface charge density

E = distance from nearby point to sheet i.e be independent

The distance at 0.1 and 0.2, the electric field would remain the same

So,

Based on the above explanation, the first option is correct

3 0
3 years ago
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