For any equation to be balanced, be it elemental equation or radioactive decay equation, the number of atoms on the left must balance the number of atoms of the right. The same is true for alpha decay as the number of protons on the left side must equal the number of protons on the right side?

3 0

3 years ago

20-kilogram canoe is floating downriver at 2 m/s. what’s the kinetic energy?

ziro4ka [17]

Answer:

40J

Explanation:

Kinetic energy = 1/2 mv^2

Where m = mass and

v = velocity

Given mass = 20kg

v = 2m/s

K.E = 1/2 x 20 x2^2

= 1/2 x 20 x 2 x 2

= 80/2

= 40J

6 0

2 years ago

Read 2 more answers

Some help please ,.......

serious [3.7K]

It's c bro, .2x5 equals 1, which accounts for the 1m/s accelerations.

4 0

3 years ago

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Exactly one turn of a flexible rope with mass m is wrapped around a uniform cylinder with mass M and radius R.

Dennis_Churaev [7]

Answer:

$\omega=\sqrt{\omega_0^2(\frac{M+m}{M})}$

Explanation:

The rotational kinetic energy when the cylinder is with the rope is:

$E_k=\frac{1}{2}I_c\omega_0^2+\frac{1}{2}I_r\omega_0^2$

where we used the fact that both rope and cylinder hast the same w. This E_k must conserve, that is, E_k must equal E_k when the rope leaves the cylinder. Hence, the final w is given by:

$E_{k1}=E_{k2}\\\\\frac{1}{2}I_c\omega_0^2+\frac{1}{2}I_r\omega_0^{2}=\frac{1}{2}I_c\omega^2\\\\\omega=\sqrt{\omega_0^2(\frac{I_c+I_r}{I_c})}$ (1)

For Ic and Ir we can assume that the rope is a ring of the same radius of the cylinder. Then, we have:

$I_c=\frac{1}{2}MR^2\\\\I_r=mR^2$

Finally, by replacing in (1):

$\omega=\sqrt{\omega_0^2(\frac{M+m}{M})}$

hope this helps!!

7 0

2 years ago