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3 years ago

13

Do you think energy can be “used up?” Why or why not?

1 answer:

My name is Ann [436]3 years ago

8 0

I dint think energy can be used up because energy cant be created nor destroyed

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a small table has a mass of 4kg, stands on four legs, each leg having an area of 0.001 m2. what is the pressure exerted by the t

Gemiola [76]

Answer:

P = 10 kPa

Explanation:

Given that,

The mass of a small table, m = 4 kg

The area of each leg = 0.001 m²

We need to find the pressure exerted by the table on the floor. Pressure is equal to the force per unit area. So

$P=\dfrac{mg}{4\times A}\\\\P=\dfrac{4\times 10}{4\times 0.001}\\P=10000\ Pa\\\\or\\\\P=10\ kPa$

So, the required pressure is 10 kPa.

3 0

2 years ago

An experiment is set up to test the angular resolution of an optical device when red light (wavelength ????r ) shines on an aper

Neko [114]

Explanation:

As per Rayleigh criterion, the angular resolution is given as follows:

$\theta=\frac{1.22 \lambda}{D}$

From this expression larger the size of aperture, smaller will be the value of angular resolution and hence, better will be the device i.e. precision for distinguishing two points at very high angular difference is higher.

4 0

3 years ago

What is the difference between a series circuit and a parallel circuit?

Angelina_Jolie [31]

Parallel has more than one circuit or form of energy

series has only one form of energy circuit

6 0

3 years ago

A 20000 kg railroad car is rolling at 1.00 m/s when a 1000 kg load of gravel is suddenly dropped in. part a what is the car's sp

Talja [164]

Using conservation of energy and momentum we get m1*v1=(m1+m2)*v2 so rearranging for v2 and plugging the given values in we get:
(200000kg*1.00m/s)/(21000kg)=.952m/s

8 0

3 years ago

39 g aluminum spoon (specific heat 0.904 J/g·°C) at 24°C is placed in 166 mL (166 g) of coffee at 83°C and the temperature of th

tatuchka [14]

<u>Answer:</u> The final temperature of the solution is $80.14^oC$

<u>Explanation:</u>

The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, $\text{heat}_{absorbed}=\text{heat}_{released}$

To calculate the amount of heat released or absorbed, we use the equation:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

Also,

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ ..........(1)

where,

q = heat absorbed or released

$m_1$ = mass of aluminium = 39 g

$m_2$ = mass of coffee = 166 g

$T_{final}$ = final temperature = ?

$T_1$ = temperature of aluminium = $24^oC$

$T_2$ = temperature of coffee = $83^oC$

$c_1$ = specific heat of aluminium = $0.904J/g^oC$

$c_2$ = specific heat of coffee= $4.1801J/g^oC$

Putting all the values in equation 1, we get:

$39\times 0.904\times (T_{final}-24)=-[166\times 4.1801\times (T_{final}-83)]$

$T_{final}=80.14^oC$

Hence, the final temperature of the solution is $80.14^oC$

4 0

3 years ago