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Salsk061 [2.6K]
4 years ago
5

Vector 1 points along the z axis and has magnitude V1 = 80. Vector 2 lies in the xz plane, has magnitude V2 = 50, and makes a -3

7° angle with the x axis (points below the x axis). What is the scalar product V1·V2?
Physics
1 answer:
Y_Kistochka [10]4 years ago
5 0

Answer:

\vec{V1} . \vec{V2}  = 2574.08    

Explanation:

given data

magnitude V1 = 80

magnitude V2 = 50

angle a =  -37°

solution

\vec{V1} = 80 k

\vec{V2} = 50 cos{37} i  - 50sin{37} k

so that here \vec{V1} . \vec{V2}    is

\vec{V1} . \vec{V2}  = 80 k . ( 50 cos{37} i  - 50sin{37} k )

\vec{V1} . \vec{V2}  = 80 k . (  38.270 i + 32.176 k )

\vec{V1} . \vec{V2}  = 2574.08    

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Answer:

P=mv

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Explanation:

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4 years ago
A 10-g bullet moving horizontally with a speed of 2.0 km/s strikes and passes through a 4.0-kg block moving with a speed of 4.2
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Answer:

K=512J

Explanation:

Since the surface is frictionless, momentum will be conserved. If the bullet of mass m_1 has an initial velocity v_{1i} and a final velocity v_{1f} and the block of mass m_2 has an initial velocity v_{2i} and a final velocity v_{2f} then the initial and final momentum of the system will be:

p_i=m_1v_{1i}+m_2v_{2i}

p_f=m_1v_{1f}+m_2v_{2f}

Since momentum is conserved, p_i=p_f, which means:

m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}

We know that the block is brought to rest by the collision, which means v_{2f}=0m/s and leaves us with:

m_1v_{1i}+m_2v_{2i}=m_1v_{1f}

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v_{1f}=\frac{m_1v_{1i}+m_2v_{2i}}{m_1}

Considering the direction the bullet moves initially as the positive one, and writing in S.I., this gives us:

v_{1f}=\frac{(0.01kg)(2000m/s)+(4kg)(-4.2m/s)}{0.01kg}=320m/s

So kinetic energy of the bullet as it emerges from the block will be:

K=\frac{mv^2}{2}=\frac{(0.01kg)(320m/s)^2}{2}=512J

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