Answer:
Hammer
Anvil
Stirrup
Explanation:
The three tiny bones in the ear drum are:
Hammer, this is also known as malleus and it is attached to the eardrum
Anvil, this is also called incus and it found with the chain of bones in the middle ear.
Stirrup, also known as stapes and it attached to the membrane covered opening that connects the middle ear with the inner ear.
Answer:
Statement 2 is wrong
Explanation:
To check the statements in this exercise, let's describe the main properties of electromagnetic waves. Let's describe the characteristics
* they are transverse waves
* formed by the oscillations of the electric and magnetic fields
* the speed of the wave is the speed of light
with these concepts let's review the final statements
1) True. Formed by the oscillation of the two fields
2) False. They are transverse waves
3) True. Can travel by vacuum as they are supported by oscillations of the electric and magnetic fields
4) True. They all have the same speed of light
Statement 2 is wrong
<span>Energy is calculated by molecule dividing energy by mole by Avogadro's number (6.022*10^23)
941kJ=9.41*10^5 J
so energy by molecule
E= 9.41*10^5/6.022*10^23=1.563*10^-18 J
Wavelength (w) given by E=hc/w
where, E = energy
h = planks constant (6.6262 x 10-34 J·s)
c = speed of light (3 x 10^8 m/s )
So,
w= hc/E
= (6.6262*10^-34)*(3*10^8) /1.563*10^-18
= 127.2 Nm
Longest wavelength of radiation =127.2 Nm</span>
The maximum mass of a load that can be lifted by the jack and the distance covered are:
m = 160.2 Kg
h = 25 cm
Given that a certain hydraulic jack is used for lifting load. a force of 250N is applied on a small piston with a diameter of 80cm while the diameter of the larger piston is 2.0m.
The parameters given are
= 250
= Area of the small piston = π
= 22/7 x 
= 0.5 
= ?
= Area of the large piston = π
= π x 1
= 3.14
To calculate the force on the large piston, we will use the below formula
/ = /
Substitute all the parameters into the equation
250/0.5 = /3.14
= 1570 N
To calculate the maximum mass of a load that can be lifted by the jack, let us apply Newton second law
F = mg
1570 = 9.8m
m = 1570/9.8
m = 160.2 Kg
.(take g=9.81ms^-2)
If the applied force moves through a distance of 25cm, the distance through which the load is lifted will be
/ 0.25 = / h
250/0.125 = 1570/3.14h
make h the subject of the formula
6280h = 1570
h = 1570/6280
h = 0.25 m
Therefore, the distance through which the load is lifted is 25 cm
Learn more here: brainly.com/question/13596980
