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photoshop1234 [79]
4 years ago
9

A baseball on a T-ball stand has no momentum until it is hit with a bat. When Tyler swings the bat, it has a momentum of 12 kg m

/s. After the bat hits the ball, the ball has a momentum of 8 kg m/s. What is the momentum of the bat AFTER it hits the ball
Physics
2 answers:
Troyanec [42]4 years ago
0 0
Positive direction : bat swing

total momentum before = after
12(bat) + 0(ball) = 8(ball) + bat-after

bat-after is positive then direction is the same
marysya [2.9K]4 years ago
0 0

Answer:

4 kg m/s

Explanation:

According to the Law of Conservation of Momentum, the total momentum must be the same before and after a collision. The total momentum before the bat hit the ball was 12 kg m/s. After the collision, the ball has a momentum of 8 kg m/s. Therefore, the bat must have a momentum of 4 kg m/s.

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Answer:

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3 years ago
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1.What is the kinetic energy of an object with a velocity of 7m / s and a mass of 14kg
Vlada [557]

Answer:

Explanation:

The formula for KE is

KE=\frac{1}{2}mv^2 and we fill in to solve for KE:

KE=\frac{1}{2}(14)(7.0)^2 I used 2 significant digits in the velocity because 1 isn't accurate enough. Therefore, my answer will be rounded to 2 sig fig's:

KE = 340 J (rounded from 343). And we do the exact same thing in #2:

KE=\frac{1}{2}(5.0)(5.0)^2 (Used 2sig fig's for both the mass and the velocity) to get

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How much tension must a rope withstand if it is used to accelerate a 1700-kg car
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4 years ago
The electric field between two parallel plates is uniform, with magnitude 646 N/C. A proton is held stationary at the positive p
Ahat [919]

Answer:

The distance from the positive plate at which the two pass each other is 0.0023 cm.

Explanation:

We need to find the acceleration of each particle first. Let's use the electric force equation.

F=Eq

ma=Eq

<u>For the proton</u>

m_{p}a_{p}=Eq_{p}

a_{p}=\frac{Eq_{p}}{m_{p}}

a_{p}=\frac{646*1.6*10^{-19}}{1.67*10^{−27}}

a_{p}=6.19*10^{10}\: m/s^{2}

<u>For the electron</u>

m_{e}a_{e}=Eq_{e}

a_{e}=\frac{Eq_{e}}{m_{e}}

a_{e}=\frac{646*1.6*10^{-19}}{9.1*10^{−31}}  

a_{e}=1.14*10^{14}\: m/s^{2}

Now we know that the plate separation is 4.26 cm or 0.0426 m. The travel distance of the proton plus the travel distance of the electron is 0.0426 m.

x_{p}+x_{e}=0.0426

Both of them have an initial speed equal to zero. So we have:

\frac{1}{2}a_{p}t^{2}+\frac{1}{2}a_{e}t^{2}=0.0426

t^{2}(a_{p}+a_{e})=2*0.0426

t^{2}=\frac{2*0.0426}{a_{p}+a_{e}}

t=\sqrt{\frac{2*0.0426}{6.19*10^{10}+1.14*10^{14}}}

t=2.73*10^{-8}\: s    

With this time we can find the distance from the positive plate (x(p)).

x_{p}=\frac{1}{2}a_{p}t^{2}

x_{p}=\frac{1}{2}6.19*10^{10}*(2.73*10^{-8})^{2}

x_{p}=0.0023\: cm

Therefore, the distance from the positive plate at which the two pass each other is 0.0023 cm.

I hope it helps you!

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Answer:

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