The sides AB & AC of a traingle ACB are produced to P & O respectively if the bisector of angle PBC & angle QCB intersect at O then angle BOC??????????
Answers
Answer:
EXPLANATION.
A motorcycle is moving vwith a speed of
=> 60 km/hr = 60 X 5/18 = 50/3 m/s.
=> The mass of the motorcycle = 120 kg.
To calculate the kinetic energy.
\rm \to \boxed{\green{\: formula \: of \: kinetic \: energy \: = \frac{1}{2} m {v}^{2}} }→
formulaofkineticenergy=
2
1
mv
2
\rm \to \: \frac{1}{2} \times 120 \times \frac{50}{3} \times \frac{50}{3}→
2
1
×120×
3
50
×
3
50
=> 16666.6 joule.
Deprivation of kinetic energy.
\rm \to \: \Delta \: k \: = w→Δk=w
\rm \to \: \Delta \: k \: = \int \: f(r).dr→Δk=∫f(r).dr
\rm \to \: \Delta \: k \: = \int \: ma.dr→Δk=∫ma.dr
\rm \to \Delta \: k \: = m \int \frac{dv}{dt} . \: dr→Δk=m∫
dt
dv
.dr
\rm \to \: \Delta \: k \: = m \int \: \frac{dr}{dt} . \: dv→Δk=m∫
dt
dr
.dv
\rm \to \: \Delta \: k \: = m \int \: v \: . \: dv→Δk=m∫v.dv
\rm \to \: \Delta \: k \: = \frac{1}{2} mv {}^{2} - \frac{1}{2}m v_{0} {}^{2}→Δk=
2
1
mv
2
−
2
1
mv
0
2
\boxed{\green{\rm \to \: ke \: = \: \frac{1}{2}m {v}^{2} }}
→ke=
2
1
mv
2