The heat energy gained by the water in one minute is ?

2. How have human activities and natural processes influenced the gradual or sudden change in global temperature? (You need a mi

mote1985 [20]

Answer:

Human activities and natural processes have influenced the change in the global temperature by the following processes

1) Green house gas such as carbon dioxide, methane, ozone, nitrous oxide and fluorinated gases produced by the combustion of fossil fuels the use of industrial chemicals, the production of coal, and natural gas

2) Deforestation which reduces the natural process of conversion of carbon dioxide to oxygen, thereby, increasing the greenhouse gases in the atmosphere

3) The accumulation of the greenhouse gases in the atmosphere results in the trapping of heat in the atmosphere, causing the atmospheric temperature to rise

4) Changes in the amount of energy produced by the Sun can result in an increase or decrease in the atmospheric temperature

5) Volcanic activity that occurs at a sufficiently large scale can produce sulfur dioxide that blocks the rays of the Sun from reaching the Earth, resulting in a change of atmospheric temperature.

Explanation:

7 0

3 years ago

The two lenses in a microscope are separated by 19.5 cm, and the focal length of the eyepiece lens is 2.75 cm.

Naya [18.7K]

Answer:

Explanation:

a ) If the image of this object is viewed with the eyepiece adjusted for minimum eyestrain (image at the far point of the eye) , the image from object lens must have been formed at the focus of eye lens . So the objective image must have been formed at 19.5 - 2.75 = 16.75 cm from the object lens.

b ) Let the object distance be u

For object lens

v = 16.75 cm , f = .35 cm

1/v - 1/u = 1/f

1/16.75 - 1/u = 1/ .35

.0597 - 1/u = 2.857

1/u = - 2.7973

u = .3575 cm

c ) Angular magnification

= $\frac{v_o}{u_o} \times\frac{D}{f_e}$

v₀ and u₀ are image and object distance for object lens , D = 25 cm and f_e is focal length of eye lens

= (16.75 / .3575) x( 25 / 2.75)

= 46.85 x 9.09

= 426

7 0

3 years ago

How are heat, work and internal energy related

iris [78.8K]

The change in internal energy that accompanies the transfer of heat, q, or work, w, into or out of a system can be calculated using the following equation: Note the value of heat and work as they are transferred into or out of a system.

6 0

4 years ago

Technician A says that gaskets are used to fill a space or gap between two objects to prevent leakage from occurring. Technician

natta225 [31]

Answer:C

Explanation:the both are telling us more about a gasket

4 0

4 years ago

(a) Calculate the self-inductance of a solenoid that is

adell [148]

Answer:

a) 0.142mH

b) 14mV

Explanation:

the complete answer is:

(a) Calculate the self-inductance of a solenoid that is <ghtly wound with wire of diameter 0.10 cm, has a cross-sec<onal area of 0.90 cm2 , and is 40 cm long. (b) If the current through the solenoid decreases uniformly from 10 to 0 A in 0.10 s, what is the emf induced between the ends of the solenoid

a) the self inductance of a solenoid is given by:

$L=\frac{\mu_o N^2 A}{L}$

μo: magnetic permeability of vacuum = 4\pi*10^{-7}N/A^2

A: cross sectional area = 0.9cm^2=9*10^{-5}m

L: length of the solenoid = 40cm = 0.4m

The N turns of the wire is calculated by using the diameter of the wire:

N = (40cm)/(0.10cm)=400

By replacing in the formula you obtain:

$L=\frac{(4\pi *10^{-7}N/A^2)(400)^2(9*10^{-5}m^2))}{(0.4m)}=1.42*10^{-4}H$

the self inductance is 1.42*10^{-4}H = 0.142mH

b) to find the emf you can use:

$emf=L\frac{\Delta I}{\Delta t}=(1.42*10^{-4}H)\frac{10A-0A}{0.10s}=0.014V=14mV$

the emf induced is 14mV

4 0

3 years ago