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NemiM [27]
3 years ago
5

102) Consider massive gliders that slide friction- free along a horizontal air track Glider A has a mass of 1 kg, aspeed of 1 m/

s, and collides with Glider B that has a mass of 5 kg and is at rest. If they stick upon collision, theirspeed after collision will beA) lm/s.B) 1/6 m/s.C) 1/5 m/s.D) 1/4 m/s.E) none of these.
Physics
1 answer:
Darya [45]3 years ago
3 0

Answer:

B. 1/6 m/s

Explanation:

Parameters given:

Mass of first glider, m = 1 kg

Initial speed of first glider, u = 1 m/s

Mass of second glider, M = 5 kg

Initial speed of second glider, U = 0 m/s.

Using the principle of conservation of momentum, we have that:

Total initial momentum = Total final momentum

=> m*u + M*U = m*v + M*V

Where v is the final velocity of the first glider and V is the final velocity of the second glider.

Since they stick upon collision, their final velocities are the same, hence, v = V

m*u + M*U = m*v + M*v

m*u + M*U = (m + M)*v

Also, since the second glider is initially at rest, U = 0

m*u = (m + M)*v

Therefore,

v = (m*u) / (m + M)

v = (1 * 1) / (1 + 5)

v = 1/6 m/s

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DanielleElmas [232]

Answer:

A

Explanation:

4 0
3 years ago
38.4 mol of krypton is in a rigid box of volume 64 cm^3 and is initially at temperature 512.88°C. The gas then undergoes isobari
kolbaska11 [484]

Answer:

Final volumen first process V_{2} = 98,44 cm^{3}

Final Pressure second process P_{3} = 1,317 * 10^{10} Pa

Explanation:

Using the Ideal Gases Law yoy have for pressure:

P_{1} = \frac{n_{1} R T_{1} }{V_{1} }

where:

P is the pressure, in Pa

n is the nuber of moles of gas

R is the universal gas constant: 8,314 J/mol K

T is the temperature in Kelvin

V is the volumen in cubic meters

Given that the amount of material is constant in the process:

n_{1} = n_{2} = n

In an isobaric process the pressure is constant so:

P_{1} = P_{2}

\frac{n R T_{1} }{V_{1} } = \frac{n R T_{2} }{V_{2} }

\frac{T_{1} }{V_{1} } = \frac{T_{2} }{V_{2} }

V_{2} = \frac{T_{2} V_{1} }{T_{1} }

Replacing : T_{1} =786 K, T_{2} =1209 K, V_{1} = 64 cm^{3}

V_{2} = 98,44 cm^{3}

Replacing on the ideal gases formula the pressure at this piont is:

P_{2} = 3,92 * 10^{9} Pa

For Temperature the ideal gases formula is:

T = \frac{P V }{n R }

For the second process you have that T_{2} = T_{3}  So:

\frac{P_{2} V_{2} }{n R } = \frac{P_{3} V_{3} }{n R }

P_{2} V_{2}  = P_{3} V_{3}

P_{3} = \frac{P_{2} V_{2}}{V_{3}}

P_{3} = 1,317 * 10^{10} Pa

7 0
3 years ago
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spayn [35]
The hardest part of the job is to find the right formula to use, and write it down. You've already done that ! The rest is just turning the crank until an answer falls out.

You wrote. E = m g h.
Beautiful.
Now divide each side by (g h), and you'll have the formula for mass:

m = E / (g h).

You know all the numbers on the right side. Just pluggum in, do the arithmetic, and you'll have the mass.
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3 years ago
Phân biệt các đặc điểm khác nhau giữa chất rắn, chất lỏng
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Answer:

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3 years ago
A body weighing 108N moves with speed of 5m/s in a horizontal
poizon [28]

Answer:

i) 5 m/s^{2}  ii) 54 N  iii) 54 N

Explanation:

i) a = \frac{v^{2}}{r} ⇒ a = 5² ÷ 5 = 5 m/s^{2}

ii) m = \frac{W}{g} ⇒ m = 108 ÷ 10 = 10.8 kg , F = ma ⇒ F = 10.8 × 5 = 54 N

iii) F1 = F2 = 54 N

4 0
3 years ago
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