You visited us 1 times! Enjoying our articles? Unlock Full Access!

Newton's second law

Formula for t...

Question

Formula for true force is :

F = m \frac{d^{2} x}{d t^{2}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

F = \frac{d (m v)}{d t}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

F = m a

No worries! We‘ve got your back. Try BYJU‘S free classes today!

F = m \frac{d v}{d t}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

All

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct options are
A $F = \frac{d (m v)}{d t}$
C $F = m \frac{d^{2} x}{d t^{2}}$
According to Newton's second law,
Force = rate of change of linear momentum.
And $F = m a = m \frac{d^{2} x}{d t^{2}}$
Option A,B

Suggest Corrections

Similar questions

\frac{d^{2} Θ}{d t^{2}} = \frac{c}{Θ}

\frac{d^{2} x}{d t^{2}} + \frac{R}{m} \frac{d x}{M t} + x w^{2} = \frac{F}{m} c o s θ

\frac{d^{2} x}{d t^{2}} - \frac{k}{m} x = 0

m \frac{d^{2} x}{d t^{2}} + R \frac{d n}{d t} + m x ω^{2} = 0

{lim}_{x \to 0} \frac{\int_{- x}^{x} f (t) d t}{\int_{0}^{2 x} f (t + 4) d t} =

f

f (x) > 0

x

f (0) = 48 f (4)

Let f(x) is a real valued function defined by $f (x) = x^{2} + x^{2} \int_{- 1}^{1} t f (t) d t + x^{3} \int_{- 1}^{1} f (t) d t$ , then

\int_{0}^{x} f (t) d t = x + \int_{x}^{1} t f (t) d t

f (1)

x = f^{''} (t) c o s t + f^{'} (t) s i n t

y = {- f}^{''} (t) s i n t + f^{'} (t) c o s t .

\int {[{(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2}]}^{\frac{1}{2}} d t

f (x), f^{'} (x), f^{''} (x), f^{'''} (x) > 0

Join BYJU'S Learning Program