Answer:
Given, Apparent weight(W₂)=4.2N
Weight of liquid displaced (u)=2.5N
Let weight of body in air = W₁
Solution,
U=W₁-W₂
W₁=4.2=2.5=6.7N
∴Weight of body in air is 6.7N
To find the ratio of planetary speeds Va/Vb we need the orbital velocity formula:
V=√({G*M}/R), where G is the gravitational constant, M is the mass of the distant star and R is the distance of the planet from the star it is orbiting.
So Va/Vb=[√( {G*M}/Ra) ] / [√( {G*M}/Rb) ], in our case Ra = 7.8*Rb
Va/Vb=[ √( {G*M}/{7.8*Rb} ) ] / [√( {G*M}/Rb )], we put everything under one square root by the rule: (√a) / (√b) = √(a/b)
Va/Vb=√ [ { (G*M)/(7.8*Rb) } / { (G*M)/(Rb) } ], when we cancel out G, M and Rb we get:
Va/Vb=√(1/7.8)/(1/1)=√(1/7.8)=0.358 so the ratio of Va/Vb = 0.358.
11. protect the cell and keep its shape.
12. chloroplast
Answer:
3.6km South East
Explanation:
Displacement is the shortest distance between the starting point and the ending point and the direction it is displaced in. To calculate the displacement we can use the pythagoras theorem because the 3km East and the 2km south form the two shorter sides of a right angled triangle between the starting and ending points. So, the displacement is the length C of the triangle which we can calculate as follows:
Pythagoras Theorem:
a^2+b^2=c^2
(2)^2+(3)^2=c^(2)
4+9=c^2
Square root 13 = c
c=3.6km (1dp)
The total displacement is 3.6km and is in the approximate direction of South East (because he travelled east and south).
Hope this helped!
Answer:
Hi Carter,
The complete answer along with the explanation is shown below.
I hope it will clear your query
Pls rate me brainliest bro
Explanation:
The magnitude of the magnetic field on the axis of a circular loop, a distance z from the loop center, is given by Eq.:
B
= NμοiR² / 2(R²+Z²)³÷²
where
R is the radius of the loop
N is the number of turns
i is the current.
Both of the loops in the problem have the same radius, the same number of turns, and carry the same current. The currents are in the same sense, and the fields they produce are in the same direction in the region between them. We place the origin at the center of the left-hand loop and let x be the coordinate of a point on the axis between the loops. To calculate the field of the left-hand loop, we set z = x in the equation above. The chosen point on the axis is a distance s – x from the center of the right-hand loop. To calculate the field it produces, we put z = s – x in the equation above. The total field at the point is therefore
B
= NμοiR²/2 [1/ 2(R²+x²)³÷² + 1/ 2(R²+x²-2sx+s²)³÷²]
Its derivative with respect to x is
dB
/dx= - NμοiR²/2 [3x/ (R²+x²)⁵÷² + 3(x-s)/(R²+x²-2sx+s²)⁵÷²
]
When this is evaluated for x = s/2 (the midpoint between the loops) the result is
dB
/dx= - NμοiR²/2 [3(s/2)/ (R²+s²/4)⁵÷² - 3(s/2)/(R²+s²/4)⁵÷²
] =0
independent of the value of s.
