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2 years ago

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1. Two blocks travel along a level frictionless surface. Block A is initially moving to the right at 5.0 m/s, while block B is i

nitially at rest. After the collision, block A continues to the right at 1.0 m/s. The mass of block A is 2.0 kg and the mass of block B is 3.0 kg. The positive x-direction is to the right. a. Find the final velocity of block B. Show your work. b. Is the sign of the change in momentum of block A as a result of the collision positive, negative, or is it zero

1 answer:

ad-work [718]2 years ago

6 0

Answer: 2.67 m/s

Explanation:

Given

Mass of block A is $m_a=2\ kg$

mass of block B is $m_b=3\ kg$

The initial velocity of block A $u_a=5\ m/s$

the initial velocity of block B is $u_b=0$

After collision velocity of block A is $v_a=1\ m/s$

Conserving momentum

$m_au_a+m_bu_b=m_av_a+m_bv_b\\\\2\times 5+3\times0=2\times 1+3\times v_b\\\\v_b=\dfrac{8}{3}=2.67\ m/s$

The momentum of block A after the collision is $P_a=2\times 1=2\ kg.m/s$

Therefore, there is no change in sign.

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A Foucault pendulum consists of a brass sphere with a diameter of 31.0 cm suspended from a steel cable 11.0 m long (both measure

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Answer:

43.7 °C

Explanation:

$\alpha_b$ = Coefficient of linear expansion of brass = $18\times 10^{-6}\ ^{\circ}C$

$\alpha_s$ = Coefficient of linear expansion of steel = $11\times 10^{-6}\ ^{\circ}C$

$L_{0b}$ = Initial length of brass = 31 cm

$L_{0s}$ = Initial length of steel = 11 m

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$\Delta L=\Delta L_b+\Delta L_s\\\Rightarrow \Delta L=L_{0b}\alpha_b\Delta T+L_{0s}\alpha_b\Delta T\\\Rightarrow \Delta T=\frac{\Delta L}{L_{0b}\alpha_b+L_{0s}\alpha_b}\\\Rightarrow \Delta T=\frac{0.003}{0.31\times 18\times 10^{-6}+11\times 10^{-6}\times 11}\\\Rightarrow \Delta T=23.7\ ^{\circ}C$

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3 years ago

Telephone signals are often transmitted over long distances by microwaves. What is the frequency of microwave radiation with a w

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Answer:

1) f= 8.6 GHz

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Explanation:

1)

Since microwaves are electromagnetic waves, they move at the same speed as the light in vacuum, i.e. 3*10⁸ m/s.
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$v = \lambda * f (1)$

Replacing by the givens, and solving for f, we get:

$f =\frac{c}{\lambda} =\frac{3e8m/s}{0.035m} = 8.57e9 Hz (2)$

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$v =\frac{d}{t} (3)$

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Solving for t, we get:

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