Some basic formulas involving triangles
\ a^2 = b^2 + c^2 - 2bc \textrm{ cos } \alphaa 2 =b 2+2 + c 2
−2bc cos α
\ b^2 = a^2 + c^2 - 2ac \textrm{ cos } \betab 2=
m_b^2 = \frac{1}{4}( 2a^2 + 2c^2 - b^2 )m b2 = 41(2a 2 + 2c 2-b 2)
b
Bisector formulas
\ \frac{a}{b} = \frac{m}{n} ba =nm
\ l^2 = ab - mnl 2=ab-mm
A = \frac{1}{2}a\cdot b = \frac{1}{2}c\cdot hA=
\ A = \sqrt{p(p - a)(p - b)(p - c)}A=
p(p−a)(p−b)(p−c)
\iits whatever A = prA=pr with r we denote the radius of the triangle inscribed circle
\ A = \frac{abc}{4R}A=
4R
abc
- R is the radius of the prescribed circle
\ A = \sqrt{p(p - a)(p - b)(p - c)}A=
p(p−a)(p−b)(p−c)
Answer:
A. 5.16 s.
B. 5.66 s.
Step-by-step explanation:
A.
For a simple harmonic motion,
T = 2pi (sqrt * (l/g))
Given:
L1 = 3 cm
T1 = 4 s
L2 = 5 cm
T2 = ?
4 = 2pi*sqrt(3/g)
g = 7.4
At, L2,
T2 = 2pi*sqrt(5/7.4)
= 5.16 s.
B.
M1 = M1
M2 = 2*M1
For a simple harmonic motion,
T = 2pi (sqrt * (m/k))
4 = 2pi (sqrt * (M1/k))
M1/k = 0.405
Inputting the above values,
T2 = 2pi (sqrt * (2*M1/k))
= 2pi (sqrt * (2 * 0.405))
= 5.66 s.
Answer:
each of them 2 balloons
hope this helps
Step-by-step explanation:
