Answer:
The maximum length of the specimen before deformation is 240.64 mm
Explanation:
Strain = stress ÷ elastic modulus
stress = load ÷ area
load = 2130 N
diameter = 3.4 mm = 3.4×10^-3 m
area = πd^2/4 = 3.142 × (3.4×10^-3)^2/4 = 9.08038×10^-6 m^2
stress = 2130 N ÷ 9.08038×10^-6 m^2 = 2.35×10^8 N/m^2
elastic modulus = 126 GPa = 126×10^9 Pa
Strain = 2.35×10^8 ÷ 126×10^9 = 0.00187
Length = extension ÷ strain = 0.45 mm ÷ 0.00187 = 240.64 mm
Answer:
in a liquid, particles will flow or glide over one another, but stay toward the bottom of the container.
Explanation:
the attractive forces between particles are strong enough to hold a specific volume but not strong enough to keep the molecules sliding over each other.
CH2O3
C=19.36\12 = 1.61
H=3.25\1.008 =3.22
O=77.39\16=4.83
then dived the moles with the smallest mole
C=1.61\1.61=1
H=3.22\1.61=2
O=4.83\1.61=3
E.F
CH2O3
Answer:
We need 420 cal of heat
Explanation:
Step 1: Data given
Mass of the aluminium = 200.0 grams
Temperature rises with 10.0 °C
Specific heat of aluminium = 0.21 cal/g°C
Step 2: Calculate the amount of heat required
Q =m * c* ΔT
⇒with Q = the amount of heat required= TO BE DETERMINED
⇒with m = the mass of aluminium = 200.0 grams
⇒with c = the specific heat of aluminium = 0.21 cal/g°C
⇒with ΔT = the change of temperature = 10.0°C
Q = 200.0 grams * 0.21 cal/g°C * 10.0 °C
Q = 420 cal
We need 420 cal of heat (option 2 is correct)
Answer : The final volume at STP is, 1000 L
Explanation :
According to the Boyle's, law, the pressure of the gas is inversely proportional to the volume of gas at constant temperature and moles of gas.
or,
where,
= initial pressure = 1520 mmHg = 2 atm (1 atm = 760 mmHg)
= final pressure at STP = 1 atm
= initial volume = 500.0 L
= final volume at STP = ?
Now put all the given values in the above formula, we get:
Therefore, the final volume at STP is, 1000 L
