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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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Oh no! The Hulk just fell off the Empire State Building! Calculate how long it took him to fall straight down from the top of th

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10) Two students want to use a 12-meter long rope to create standing waves. They first measure the speed at which a single wave

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Answer:

To create a second harmonic the rope must vibrate at the frequency of 3 Hz

Explanation:

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2 years ago

In a Hydrogen atom an electron rotates around a stationary proton in a circular orbit with an approximate radius of r =0.053nm.

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Answer:

(a): $F_e = 8.202\times 10^{-8}\ \rm N.$

(b): $F_g = 3.6125\times 10^{-47}\ \rm N.$

(c): $\dfrac{F_e}{F_g}=2.27\times 10^{39}.$

Explanation:

Given that an electron revolves around the hydrogen atom in a circular orbit of radius r = 0.053 nm = 0.053 $\times 10^{-9}$ m.

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two charged particles of charges $q_1$ and $q_2$ respectively is given by

$F_e = \dfrac{k|q_1||q_2|}{r^2}$

where,

$k$ = Coulomb's constant = $9\times 10^9\ \rm Nm^2/C^2.$
$r$ = distance of separation between the charges.

For the given system,

The Hydrogen atom consists of a single proton, therefore, the charge on the Hydrogen atom, $q_1 = +1.6\times 10^{-19}\ C.$

The charge on the electron, $q_2 = -1.6\times 10^{-19}\ C.$

These two are separated by the distance, $r = 0.053\times 10^{-9}\ m.$

Thus, the magnitude of the electrostatic force of attraction between the electron and the proton is given by

$F_e = \dfrac{(9\times 10^9)\times |+1.6\times 10^{-19}|\times |-1.6\times 10^{-19}|}{(0.053\times 10^{-9})^2}=8.202\times 10^{-8}\ \rm N.$

Part (b):

The gravitational force of attraction between two objects of masses $m_1$ and $m_1$ respectively is given by

$F_g = \dfrac{Gm_1m_2}{r^2}.$

where,

$G$ = Universal Gravitational constant = $6.67\times 10^{-11}\ \rm Nm^2/kg^2.$
$r$ = distance of separation between the masses.

For the given system,

The mass of proton, $m_1 = 1.67\times 10^{-27}\ kg.$

The mass of the electron, $m_2 = 9.11\times 10^{-31}\ kg.$

Distance between the two, $r = 0.053\times 10^{-9}\ m.$

Thus, the magnitude of the gravitational force of attraction between the electron and the proton is given by

$F_g = \dfrac{(6.67\times 10^{-11})\times (1.67\times 10^{-27})\times (9.11\times 10^{-31})}{(0.053\times 10^{-9})^2}=3.6125\times 10^{-47}\ \rm N.$

The ratio $\dfrac{F_e}{F_g}$ :

$\dfrac{F_e}{F_g}=\dfrac{8.202\times 10^{-8}}{3.6125\times 10^{-47}}=2.27\times 10^{39}.$

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2 years ago

Sedaia [141]

Answer:

V = 20.5 m/s

Explanation:

Given,

The mass of the cart, m = 6 Kg

The initial speed of the cart, u = 4 m/s

The acceleration of the cart, a = 0.5 m/s²

The time interval of the cart, t = 30 s

The final velocity of the cart is given by the first equation of motion

v = u + at

= 4 + (0.5 x 30)

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Hence the final velocity of cart at 30 seconds is, v = 19 m/s

The speed of the cart at the end of 3 seconds

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Hence, the final velocity of the cart at the end of this 3.0 second interval is, V = 20.5 m/s

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2 years ago