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Mars2501 [29]
2 years ago
11

Equal masses of granite, iron, copper, and lead are placed in sunlight. Based on specific heat, which material will warm up the

fastest?
Physics
2 answers:
alekssr [168]2 years ago
7 0

Based on specific heat, lead will be warm up the fastest among all other elements.

Explanation:

The specific heat is defined as the amount of heat required per unit mass of the object to increase the temperature of the object by 1 °C.

The object will get warm up when the heat transfer from the sunlight to object occurs leading to increase in the temperature of object.

Thus, the specific heat of an object will be directly proportional to the added heat.

It can also be defined as the object with less specific heat requires less heat energy to get warm up.

This means the object with less specific heat will get warm up in sunlight the fastest.

As all the objects like  lead, copper, granite and iron are taken in equal masses, so the influence of mass on the addition of heat will be neglected as it will be same for all object. Only the specific heat will be changing for all object.  

The specific heat of lead is very low compared to granite, iron and copper. So lead will get heat up earlier or fastest in sunlight.

bogdanovich [222]2 years ago
5 0
Answer is lead, lead will heat up the fastest as it has the highest specific heat amongst the choices. The higher the specific heat of the substance the easier it is for that substance temperature to change for every mole of substance with respect to a degree in kelvin.
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the steel bed of a suspension bridge is 200m long at 20 C. If the extremes of temperature to which it might be exposed are -30 C
raketka [301]

Answer:

The steel bed will contract by  0.13 m, and expand by 0.052 m

Explanation:

For contraction,

α = ΔL/(LΔθ)..................... Equation 1

Where α = Linear expansivity of  steel, ΔL = decrease in length/ Increase in length, L = Original length, Δθ = Change in temperature

make ΔL the subject of the equation

ΔL = α(LΔθ)................. Equation 2

Given: α = 13×10⁻⁶/C, L = 200 m, Δθ = -30-20 = -50 °C

Substitute into equation 2

ΔL = 13×10⁻⁶(200)(-50)

ΔL = -0.13 m

Similarly, For expansion,

Using equation 2

ΔL =  α(LΔθ)

Given: α = 13×10⁻⁶/C, L = 200 m, Δθ = 40-20 = 20  °C

Substitute into equation 2

ΔL =  13×10⁻⁶(200)(20)

ΔL = 0.052 m.

Hence the steel bed will contract by  0.13 m, and expand by 0.052 m

3 0
2 years ago
Rod AB is held in place by the cord AC. Knowing that the tension in the cord is 1350 N and that c 5 360 mm, determine the moment
riadik2000 [5.3K]

Answer:

291.598 N-m

291.6 N-m

Explanation:

Let's first take a  look at the free bodily diagrammatic representation.

The first diagram will aid us in answering  question (a), so as the second diagram will facilitate effective understanding when solving for question (b).

Let's first determine our angle θ from the diagram

To find angle θ ; we have :

tan θ  = \frac{360+240}{450}

tan θ  = \frac{600}{450}

tan θ  = 1.333

θ  = tan⁻¹ (1.333)

θ  = 53.13°

Now, to determine the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied at point A.

We have:

M__B}=(Fcos \theta *240)-(Fsin \theta *450)

where Force(F) = Force in the cord AC = 1350 N and θ  = 53.13° ; we have:

M__B}=(1350&cos 53.13^0 *240)-(1350sin 53.13^0 *450)

M__B}= 194400.463-485999.348

M__B}=-291598.885 N-mm\\

M__B}=-291.598 N-m

Since the negative sign illustrates just the clockwise movement ; then the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied at point A = 291.598 N-m

b) From the second diagram, taking the moment at point B (M__B}),

we have:

M__B}=-Fcos \theta *360 - Fsin \theta * 0

M__B}=-Fcos \theta *360 - 0

M__B}=-Fcos \theta *360

where Force(F) =  1350 N and θ  = 53.13° ; we have:

M__B}= -1350*cos53.13^0*360

M__B}= -291600 N-mm

M__B}= -291.6 N-m

Since the negative sign illustrates just the clockwise movement ; then the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied at point C = 291.6 N-m

6 0
2 years ago
An iron ball and an aluminum ball of mass 100 g each are heated to the same temperature and then cooled to a temperature of 20°C
Mekhanik [1.2K]
As the temperature changes and their masses are the same, heat lost by the balls is directly proportional to their specific heat values. The heat lost by the aluminum ball is higher implies aluminum has higher specific heat.
4 0
2 years ago
Karyotypes are done by matching up _____________________________ so that they are paired up. Question 11 options:
Rufina [12.5K]

Homologous Chromosomes

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Read 2 more answers
Students use a stretched elastic band to launch carts of known mass horizontally on a track. The elastic bands exert a force F,
rodikova [14]

Answer:

the correct answer is  E

A graph of the cart's maximum speed squared as a function of x^3

Explanation:

For this exercise let's use Newton's second law

        F = m a

force has the form

        F = k x²

and acceleration is related to velocity

        a = dv / dt

Let's use the chain rule or L'Hospital

        a = dv /dx   dx/dt

        a = dv /dx    v

let's substitute

     k x² = m v dv / dx

     k /m x² dx = v dv

we integrate

     k /m    x³ /3 = v² / 2

     v² = (2k /3m)   x³

This is the expression for the variation of the speed as a function of the position, to make a linear graph realism the changes of variable

      y = v²

     x´ = x³

       y = (2k/3m)  x´

if we graph y vs x 'we have a linear graph whose slope is

      m = 2k / 3m

By reviewing the different answers, the correct answer is  E

4 0
2 years ago
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