Deduce The Weight Of This Frog

from the question we know that ,

$p=0.5 kg$ $m/s$ or we can change to $\Rightarrow$ $mv= 0.5 kg$ $m/s$

$\frac {1}{2} mv^{2}= 1 J$ $\Rightarrow$ kinetic energy = $\frac {1}{2} mv^{2}$

$mv^{2} = 2 J$

$mv \times v = 2J$

back to the top, cause $mv = 0.5 kg$ $m/s$ so we can write

$0.5 kg$ $m/s \times v =2J$ $\Rightarrow$ $v=$ $4$ $m/s$

substitute $v=$ $4$ $m/s$ to $\Rightarrow$ $mv = 0.5 kg$ $m/s$

$m \times 4$ $m/s = 0.5 kg$ $m/s$

$m = \boxed {0.125} kg$

Note : $mv = mass \times velocity$

$m/s$ $=$ meter per second or velocity

1 Joule Equals $1 \frac {kg * m^{2}}{s^{2}}$

Okay .

We know that :

$Kinetic energy = \dfrac{mv^2}{2} = 1 J$

And that :

$Momentum = mv = 0.5 kg \cdot m/s$

Therefore , by manipulating the formula for KE (i.e Kinetic energy) ,

$\dfrac{mv^2}{2} = 1$

$\Rightarrow \dfrac{1}{2} \cdot (mv) \cdot v = 1$

$\Rightarrow \dfrac{1}{2} \cdot (0.5) \cdot v = 1$

$\Rightarrow v = 4$

Putting $v = 4$ in the formula for momentum ,

We get $mv = m \cdot 4 = \dfrac{1}{2}$

$\Rightarrow m = \dfrac{1}{8} = \boxed{0.125}$

P = 0.5 kg m/s

KE = 1 J

KE = (1/2)mv^2

P = mv

substituting equation for momentum into equation for kinetic energy...

KE = (1/2)(P/m)^2

1 kg m^2/s^2 = (1/2)(mass)[(0.5 kg m/s^2)/mass]^2

solving for mass, you obtain:

mass = 0.125 kg ]ans.

Kinetic Energy = 0.5 * Mass * Velocity^2

Momentum = Mass * Velocity

Kinetic Energy = 0.5 * Momentum * Velocity

v = Kinetic Energy * 2 / Momentum

 = 1 * 2 / 0.5

 = 4

Momentum = Mass * Velocity

Mass = Momentum / Velocity

 = 0.5 / 4

 = 0.125

Let $p$ = momentum, $KE$ = kinetic energy, $v$ = velocity, and $m$ = mass.

Recall that $KE = \frac{1}{2}mv^{2}$ and $p = mv$ .

We now have a system of equations:

$1 J = \frac{1}{2}mv^{2}$ $(1)$

$0.5 kg \times m/s = mv$ $(2)$

Substituting $(2)$ into $(1)$ , we get $\frac{1}{4}v = 1$ $m/s$ .

Therefore, $v = 4 m/s$ $(3)$ .

Substituting $(3)$ into $(2)$ , we get $m = \frac{1}{8} kg = 0.125 kg$ .

Hence, the frog's mass is $\boxed{0.125 kg}$ .

p = mv |||||||||||| 1/2 = mv ||||||| v= 1/2m |||||||||||||||||||||||||||||||||||||| EK = 1/2 mv^2 |||||||||||||| 1 = 1/2 m (1/2m)^2 |||||||||||||||||||| 1 = 1/2 . m . 1/4m^2 |||||||||||||||||||||| 1 = 1/8 m |||||||||||||| 8m = 1 |||||||||||||||| m = 1/8 |||||||||| m = 0,125 kg

KE=P^2/2m 1=.25/2m m=0.125kg

we know that K.E=P P/2M......SO m=P P/2 K.E=.5 .5/1*2=.125

we know, K.E= p^2/2m given m is mass of body, p is momentum nd KE is kinetic energy of the body.. so changing the equation as m=p^2/2KE nd placing the values we get the result

E=p^2/(2m)

It is the direct application of the formula

KE = p^2/2m

Where p is momentum of body

1J = 0.5^2/2m

1J = 1/8m

=>m = 1/8 = 0.125 Kg

First , 1/2 m v^2 = 1 joule , arrange it and we found v^2=2/m Then, input the value of v into the momentum, then we got the answer 0.125

Since kinetic energy $= \frac{p^2}{2m}$ , $1 = \frac{0.5^2}{2m}$ .

Therefore, $2m = 0.25$ and $m = 0.125$ .

So, the mass of the frog is $0.125$ kg.

KE = 1/2 mv.v
Here p ( momentum ) = 0.5 kg. m/s And KE (kinetic energy) = 1 joule (J )

Putting values of KE and p, we get

1 = 1/2 (0.5 ) v

2 = ( 0.5 ) v

V= 4

Also, p = mv

0.5 = m (4)

M = 0.125

MOMENTUM=mv-----------------------1, KINETIC ENERGY=1/2mv^2-----------------2, DIVIDING 1 AND 2,WE GET, MOMENTUM/KINETIC ENERGY=2/v

ACCORDING TO GIVEN CONDITION, 0.5=2/V V=2/0.5

KINETIC ENERGY=1/2mv^2 1 =1/2 m[4]^2 2/16 =m i.e 0.125 =m