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Irina18 [472]
2 years ago
12

A large blue marble of mass 3.5 g is moving to the right with a velocity of 15 cm/s. The large marble hits a small red marble of

mass 1.2 g that is moving to the right with a velocity of 3.5 cm/s. After the collision, the blue marble moves to the right with a velocity of 5.5 cm/s.
What is the magnitude of the final velocity of the red marble?

9.2 cm/s
14 cm/s
24 cm/s
31 cm/s
Physics
2 answers:
d1i1m1o1n [39]2 years ago
9 0

Answer:

31 cm/s

Explanation:

Let's solve the problem by using conservation of momentum:

m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

where:

m_1 = 3.5 g is the mass of the blue marble

m_2=1.2 g is the mass of the red marble

u_1 = 15 cm/s is the initial velocity of the blue marble

u_2 = 3.5 cm/s is the initial velocity of the red marble

v_1 = 5.5 cm/s is the final velocity of the blue marble

We can find the final velocity of the red marble by re-arranging the equation and solving for v_2:

v_2 = \frac{1}{m_2}(m_1 u_1 + m_2 u_2 - m_1 v_1)=

=\frac{1}{1.2}(3.5\cdot 15 + 1.2 \cdot 3.5 - 3.5 \cdot 5.5  )=31 cm/s

Elan Coil [88]2 years ago
7 0
According to the elastic conservation momentum
m1v1 + m2v2 = m1V1 +m2V2

v: velocity before collision, V after collision
<span>the magnitude of the final velocity of the red marble
</span>m1v1 + m2v2 = m1V1 +m2V2
V2 is the final velocity we must find
V2 = 1 / m2 ( m1v1 + m2v2  - m1V1)
= 1/1.2 (3.5x15 + 1.2x 3.5 - 3.5x 5.5)= 63.29 cm/s

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B. to the right

Explanation:

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<u>Force on the test charge due to +Q:</u>

F_1=k.\frac{Q.q}{r^2}

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F_2=k.\frac{2Q.q}{(2r)^2}

F_2=k.\frac{Q.q}{2r^2}

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A walkman uses four standard 1.5 V batteries. How much resistance is in the circuit if it uses a current of 0.02A? *
hodyreva [135]

Answer:

75ohms

Explanation:

V= IR

V = 1.5volts

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Answer:

V₁  = 5.6 m/s

V₂ = 7.2 m/s

V₃ = 8.8 m/s

Explanation:

Average velocity: Average velocity can be defined as the ratio of the total  displacement to the total time taken. The S.I unit of Average velocity is m/s.

For the first 2 s,

V₁ = Δd₁/t

Where V₁  = Average velocity for the first 2 s

Where Δd₁= distance, t = time

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V₁ = 5.6 m/s

For the second 2 s,

V₂ =Δd₂/t

Where V₂ = average velocity for the second 2 s.

Δd₂= 40-25.6 = 14.4 m, t= 2 s

V₂ = 14.4/2

V₂ = 7.2 m/s

For the last 2 seconds,

V₃ =Δd₃/t

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Find the electric field inside a hollow plastic ball of radius R that has charge Q uniformly distributed on its outer surface. G
muminat

Answer:

The electric field inside the hollow plastic ball is zero.

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$ E(4\pi r^2)= \dfrac{Q_{enc}}{{\varepsilon _0 }}}$

$ E(4\pi r^2)=0$\\\\\boxed{E = 0}

The electric field inside the hollow plastic ball is zero.

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