Calculate the linear momentum for each of the following cases. (a) A proton with mass 1.67 ✕ 10-27 kg moving with a velocity of 5.09 ✕ 106 m/s straight up.
___kg·m/s upward

(b) A 12.0 g bullet moving with a velocity of 312 m/s to the right
___kg·m/s to the right

(c) A 69 kg sprinter running with a velocity of 8.0 m/s southwest.
___kg·m/s southwest

(d) Earth (m = 5.98 ✕ 1024 kg) moving in its orbit with a velocity equal to 2.98 ✕ 104 m/s forward.
___kg·m/s forward

4.725 kgm/s to the right

468.6 kgm/s southwest

1.782\times 10^{49}\ kgm/s1.782×1049 kgm/s forward

Explanation:

When the mass of an object is multiplied by its velocity we get momentum

m = Mass

v = Velocity

Proton

\begin{gathered}p=mv\\\Rightarrow p=1.67\times 10^{-27}\times 5.04\times 10^6\\\Rightarrow p=8.4168\times 10^{-21}\ kgm/s\end{gathered}p=mv⇒p=1.67×10−27×5.04×106⇒p=8.4168×10−21 kgm/s

The momentum of the proton is 8.4168\times 10^{-21}\ kgm/s8.4168×10−21 kgm/s upward

Bullet

\begin{gathered}p=mv\\\Rightarrow p=0.015\times 315\\\Rightarrow p=4.725\ kgm/s\end{gathered}p=mv⇒p=0.015×315⇒p=4.725 kgm/s

The momentum of the bullet is 4.725 kgm/s to the right

Sprinter

\begin{gathered}p=mv\\\Rightarrow p=71\times 6.6\\\Rightarrow p=468.6\ kgm/s\end{gathered}p=mv⇒p=71×6.6⇒p=468.6 kgm/s

The momentum of the sprinter is 468.6 kgm/s southwest

Earth

\begin{gathered}p=mv\\\Rightarrow p=5.98\times 10^{24}\times 2.98\times 10^4\\\Rightarrow p=1.782\times 10^{29}\ kgm/s\end{gathered}p=mv⇒p=5.98×1024×2.98×104⇒p=1.782×1029 kgm/s

The momentum of the sprinter is 1.782\times 10^{29}\ kgm/s1.782×1029 kgm/s forward

cant without the picture! ?

find the sum of the two horizontal vectors and the sum of the two vertical vectors

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