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2 years ago

Help!!! I need it today Thank you in advance

Physics

1 answer:

svlad2 [7]2 years ago

5 0

Answer:

F = - k (x-xo) a graph of the weight or applied force against the elongation obtaining a line already proves Hooke's law.

Explanation:

The student wants to prove hooke's law which has the form

F = - k (x-xo)

To do this we hang the spring in a vertical position and mark the equilibrium position on a tape measure, to simplify the calculations we can make this point zero by placing our reference system in this position.

Now for a series of known masses let's get them one by one and measure the spring elongation, building a table of weight vs elongation,

we must be careful when hanging the weights so as not to create oscillations in the spring

we look for the mass of each weight

W = mg

m = W / g

and we write them in a new column, we make a graph of the weight or applied force against the elongation and it should give a straight line; the slope of this line is sought, which is the spring constant.

The fact of obtaining a line already proves Hooke's law.

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What is the length of an aluminum rod at 65Â°C if its length at 15Â°C is 1.2 meters? A. 0.00180 meter B. 1.201386 meters C. 1.2

Deffense [45]

Answer:

Option B is the correct answer.

Explanation:

Thermal expansion

$\Delta L=L\alpha \Delta T$

L = 1.2 meter

ΔT = 65 - 15 = 50°C

Thermal Expansion Coefficient for aluminum, α = 24 x 10⁻⁶/°C

We have change in length

$\Delta L=L\alpha \Delta T=1.2\times 24\times 10^{-6}\times 50=1.44\times 10^{-3}m$

New length = 1.2 + 1.44 x 10⁻³ = 1.2014 m

Option B is the correct answer.

3 0

3 years ago

Read 2 more answers

Which statement BEST explains the model?

I am Lyosha [343]

Answer:

Explanation:

Ur welcome just took on test

8 0

3 years ago

A puck of mass 0.70 kg approaches a second, identical puck that is stationary on frictionless ice. The initial speed of the movi

natali 33 [55]

Answer:

$v_1 = \ 5.196 \frac{m}{s}$

$v_2 = 3 \frac{m}{s}$

Explanation:

For this problem, we just need to remember conservation of momentum, as there are no external forces in the horizontal direction:

$\vec{p}_i = \vec{p}_f$

where the suffix i means initial, and the suffix f means final.

The initial momentum will be:

$\vec{p}_i = m_1 \ \vec{v}_{1_i} + m_2 \ \vec{v}_{2_i}$

as the second puck is initially at rest:

$\vec{v}_{2_i} = 0$

Using the unit vector $\vec{i}$ pointing in the original line of motion:

$\vec{v}_{1_i} = 6.0 \frac{m}{s} \hat{i}$

$\vec{p}_i = 0.70 \ kg \ 6.0 \frac{m}{s} \ \hat{i} + 0.70 \ kg \ 0$

$\vec{p}_i = 4.2 \ \frac{kg \ m}{s} \ \hat{i}$

So:

$\vec{p}_i = 4.2 \ \frac{kg \ m}{s} \ \hat{i} = \vec{p}_f$

$\vec{p}_f = 4.2 \ \frac{kg \ m}{s} \ \hat{i}$

Knowing the magnitude and directions relative to the x axis, we can find Cartesian representation of the vectors using the formula

$\ \vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )$

So, our velocity vectors will be:

$\vec{v}_{1_f} = v_1 \ ( \ cos(30 \°) \ , \ sin (30 \°) \ )$

$\vec{v}_{2_f} = v_2 \ ( \ cos(-60 \°) \ , \ sin (-60 \°) \ )$

We got

$\vec{p}_f = 0.7 \ kg \ \vec{v}_{1_f} + 0.7 \ kg \ \vec{v}_{2_f}$

$4.2 \ \frac{kg \ m}{s} \ \hat{i} = 0.7 \ kg \ v_1 \ ( \ cos(30 \°) \ , \ sin (30 \°) \ ) + 0.7 \ kg \ v_2 \ ( \ cos(-60 \°) \ , \ sin (-60 \°) \ )$

So, we got the equations:

$4.2 \ \frac{kg \ m}{s} = 0.7 \ kg \ v_1 \ cos(30 \°) + 0.7 \ kg \ v_2 \ cos(-60 \°)$

and

$0 = 0.7 \ kg \ v_1 \ sin(30 \°) + 0.7 \ kg \ v_2 \ sin(-60 \°)$ .

From the last one, we get:

$0 = 0.7 \ kg \ ( v_1 \ sin(30 \°) + \ v_2 \ sin(-60 \°) )$

$0 = v_1 \ sin(30 \°) + \ v_2 \ sin(-60 \°)$

$v_1 \ sin(30 \°) = - \ v_2 \ sin(-60 \°)$

$v_1 = \ v_2 \ \frac{sin(60 \°)}{ sin(30 \°) }$

and, for the first one:

$4.2 \ \frac{kg \ m}{s} = 0.7 \ kg \ ( v_1 \ cos(30 \°) + v_2 \ cos(60 \°) )$

$\frac{4.2 \ \frac{kg \ m}{s}}{ 0.7 \ kg} = v_1 \ cos(30 \°) + v_2 \ cos(60 \°)$

$6 \ \frac{m}{s} = (\ v_2 \ \frac{sin(60 \°)}{ sin(30 \°) } ) \ cos(30 \°) + v_2 \ cos(60 \°)$

$6 \ \frac{m}{s} = v_2 (\ \frac{sin(60 \°)}{ sin(30 \°) } ) \ cos(30 \°) + cos(60 \°)$

$6 \ \frac{m}{s} = v_2 * 2$

so:

$v_2 = 6 \ \frac{m}{s} / 2 = 3 \frac{m}{s}$

and

$v_1 = \ 3 \frac{m}{s} \ \frac{sin(60 \°)}{ sin(30 \°) }$

$v_1 = \ 5.196 \frac{m}{s}$

3 0

3 years ago

A 0.43 m long and 0.43 m wide loop is moved at a constant velocity of 0.15 m/s into a perpendicular constant magnetic field of 0

olya-2409 [2.1K]

Answer:

The magnitude of the induced voltage in the loop is 20 mV.

Explanation:

given;

length of loop, L = 0.43 m

width of loop,w = 0.43 m

velocity of moved loop, v = 0.15m/s

magnetic field strength,B = 0.31 T

To determine the magnitude of the induced voltage in the loop, we apply Faraday's law;

magnitude induced E.M.F = BLv

magnitude induced E.M.F = 0.31 x 0.43 x 0.15 = 0.02 V = 20 mV

Therefore, the magnitude of the induced voltage in the loop is 20 mV.

7 0

3 years ago

A light year is defined as the distance that light can travel in 1 year. What is the value of 1 light year in meter? Show your c

Ipatiy [6.2K]

Answer:

d=9.462×10^15 meters

Explanation:

<u>Relation between distance, temps and velocity:</u>

d=v*t

t=1year*(365days/1year)*/(24hours/1day)*(3600s/1h)=31536000s

So:

1 light year=d=3*10^8m/s*3.154*10^7s=9.462×10^15 meters

6 0

3 years ago