Answer:
θ = 4.78º
with respect to the vertical or 4.78 to the east - north
Explanation:
This is a velocity compound exercise since it is a vector quantity.
The plane takes a direction, the air blows to the west and the result must be to the north, let's use the Pythagorean theorem to find the speed
v_fly² = v_nort² + v_air²
v_nort² = v_fly² + - v_air²
Let's use trigonometry to find the direction of the plane
sin θ = v_air / v_fly
θ = sin⁻¹ (v_air / v_fly)
let's calculate
θ = sin⁻¹ (10/120)
θ = 4.78º
with respect to the vertical or 4.78 to the north-east
Answer: 75 ft
Explanation:
Breaking distance = Speed²/ 20
= 30²/20
= 45 feet
Stopping distance = Speed + braking distance
= 30 + 45
= 75 ft
Answer:
your question is incomplete as the options are not given. I guess following is the complete question.
Which of these atoms is most likely to share electrons with other atoms?
a) chlorine (7 valence electrons)
b) calcium (2 valence electrons)
c) argon (8 valence electrons)
d) carbon (4 valence electrons)
e) potassium (1 valence electron)
The correct option is d) carbon (4 valence electrons)
Explanation:
Carbon has four electrons in its valence shell. In order to complete the 8 electrons in its valence shell carbon has to make four covalent bonds by sharing its four electrons with the other atom. Carbon atom will neither gain the electrons nor it losses the electrons to follow the octet rule. So in the above mentioned options carbon is the atom that will share maximum electrons.
→ Chlorine has 7 electrons, it will gain 1 electron. It will not do the sharing.
→ Calcium has 2 electrons, it will lost these 2 electrons to complete its shell.
→ Argon has already a completed shell. It will not react with other atom.
→ Potassium has only 1 valence electron which it will lose to complete its shell.
A conductor contains electrons that are bound so weakly to their atoms
that they can be ripped away with a small voltage, and sent scurrying in
the direction of the higher potential.
Answer:
magnification = - 30
overall magnification = -240
Explanation:
given data
Focal length of microscope objective f = 0.150 cm
Object distance from microscope objective do = 0.155 cm
magnification by eyepiece = 8 ×
to find out
What magnification is produced and overall magnification
solution
we consider here Image distance from microscope objective is = di
so that
Magnification produced by objective will be = - 
so we find here di by given equation that is
..................1
di = 4.65 cm
so that magnification by object will be
magnification = -
magnification = - 
magnification = - 30
and
overall magnification will be
overall magnification = magnification by objective × magnification by eyepiece ........................2
overall magnification = -30 × 8
overall magnification = -240
