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kykrilka [37]
3 years ago
11

Please help me find a) and b):

Physics
1 answer:
Triss [41]3 years ago
7 0

Answer:

A=3.33m/s

B=3.11m/s

Explanation:

Speed = total distance/time taken to cover distance

hence,

S=d/t

s=32.0/9.60

s=3.33m/s

acceleration due to gravity

v=u+at

that is (v=final velocity,u=initial velocity,a=acceleration and t= time)

from the formula

2.10=32.0+a*9.60

2.10=32.0+9.60a

CLT

9.60a=32.0-2.10

9.60a=29.9

a=29.9/9.60

a=3.11m/s

hope it correct!?

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Answer:

No

Explanation:

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3 years ago
A 200 g piece of iron is heated to 100°C. It is then dropped into water to bring its temperature down to 22°C. What is the amoun
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The piece of iron is brought from a temperature of 100C to a temperature of 22C. In this process, the heat released by this piece of iron is
Q=m C_s \Delta T
where m=200 g is the mass of the piece of iron, C_s = 0.444 J/(g C) is the iron's specific heat, and the temperature variation is \Delta T = 22^{\circ}-100^{\circ}=-78^{\circ}C. Using these values, the amount of heat released by the iron is
Q=(200 g)(0.441 J/(gC))(-78C)=-6926J=-6.9 kJ
With negative sign because it is amount of heat released. This heat is then trasferred to the water, so the amount of heat absorbed by the water is Q=6.9 kJ.
8 0
3 years ago
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What would you do if natural calamity or disaster strikes you now ?
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Answer:

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Explanation:

hope that helps. Make sure to be safe and sheltered

5 0
2 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}}

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7 0
3 years ago
Calculate the capacitance of a system that stores 9.4 x 10-10 C of charge at
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Answer:

A. 1.88\times 10^{-11}\,F.

Explanation:

By definition of Electric Capacitance, the capacitance of the system (C), in farads, is described by the following formula:

C = \frac{q}{V} (1)

Where:

q - Electric charge, in coulombs.

V - Voltage, in volts.

If we know that q = 9.4\times 10^{-10}\,C and V = 50\,V, then the capacitance of the system is:

C = \frac{9.4\times 10^{-10}\,C}{50\,V}

C = 1.88\times 10^{-11}\,F

The correct answer is A.

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