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prohojiy [21]
3 years ago
15

An irregularly shaped stone was lowered into a graduated cylinder holding a volume of water equal to 20L. The height of water ro

se to 30.2mL. If the mass of the stone was 25g what was its density?
Physics
1 answer:
vagabundo [1.1K]3 years ago
8 0

Answer:

2.45 g/mL

Explanation:

(assuming that the original volume was 20 mL and not L)

Since you know the amount of water in the cylinder before and after you placed the rock in, you can subtract them to get the volume of the rock itself. 30.2-20=10.2 mL.

To get the rock's density, use the equation d=\frac{m}{v}, where m is mass and v is volume.

d=\frac{25 g}{10.2mL}\\\\d=2.45 g/mL

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What is the current flowing through the circuit shown? (V= 120 V, R₁ = 20 02,
spayn [35]

Assumed that resistors are connected in series .

  • R_net=20+50+10=80ohm

  • Voltage=V=120V

Current=I

  • V/R
  • 120/80
  • 3/2
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3 0
2 years ago
Sheila weighs 60 kg and is riding a bike. Her momentum on the bike is 340 kg • m/s. The bike hits a rock, which stops it complet
Vikki [24]

Answer:

v₂ = 5.7 m/s

Explanation:

We will apply the law of conservation of momentum here:

Total\ Initial\ Momentum = m_{1}v_{1} + m_{2}v_{2}\\

where,

Total Initial Momentum = 340 kg.m/s

m₁ = mass of bike

v₁ = final speed of bike = 0 m/s

m₂ = mass of Sheila = 60 kg

v₂ = final speed of Sheila = ?

Therefore,

340\ kg.m/s = m_{1}(0\ m/s) + (60\ kg)v_{2}\\v_{2} = \frac{340\ kg.m/s}{60\ kg}\\\\

<u>v₂ = 5.7 m/s </u>

6 0
3 years ago
An object that starts at a position of 5 m and travels for 3 seconds at a velocity of -9 m/s ends up at what position?
Rina8888 [55]

Answer:

Sorry don't know the answer

5 0
3 years ago
Need help with number 1
Margarita [4]
200 + 175 + 125 + 100 = 600
3 0
3 years ago
Helium gas contained in a piston cylinder assembly undergoes an isentropic polytropic process from the given initial state with
jeyben [28]

Answer:

Heat transfer during the process = 0

Work done during the process = - 371.87 KJ

Explanation:

Initial pressure P_{1} = 0.02 bar

Initial temperature T_{1} = 200 K

Final pressure P_{2} = 0.14 bar

Gas constant for helium R = 2.077 \frac{KJ}{kg k}

This is an isentropic polytropic process so temperature - pressure relationship is given by the following formula,

\frac{T_{2} }{T_{1} } = [\frac{P_{2} }{P_{1} } ]^{\frac{\gamma - 1}{\gamma} }

Put all the values in above formula we get,

⇒ \frac{T_{2} }{200} = [\frac{0.14 }{0.02 } ]^{\frac{1.4 - 1}{1.4} }

⇒  \frac{T_{2} }{200} = 1.74

⇒ T_{2} = 348.72 K

This is the final temperature of helium.

For isentropic polytropic process heat transfer to the system is zero.

⇒ ΔQ = 0

Work done W = m × ( T_{1} - T_{2} ) × \frac{R}{\gamma - 1}

⇒ W = 1 × ( 200 - 348.72 ) × \frac{2.077}{1.4 - 1}

⇒ W = 371.87 KJ

This is the work done in this process. here negative sign shows that work is done on the gas in the compression of gas.

3 0
3 years ago
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