D. The table pushes up on the vase with the same amount of force as gravity pulling it down.
Answer:
344.68 m
Explanation:
The computation of the far does the car travel before stopping is
Data provided in the question
Force = F = 8,868 N
mass = m = 2,500 kg
So,
accleration = a is
a = -3.54 m/s^2
The initial speed = u = 49.4 m / s
final speed = v = 0
Based on the above information
Now applying the following formula
v^ 2- u^ 2= 2aS
Therefore
= 344.68 m
Answer:
(a) -16.7 N s; (b) -167 N
Explanation:
Given: m = 0.530 kg; vi = 18.0 m/s; vf = 13.5 m/s; t = 0.100 s
Find: (a) Impulse, (b) Force
(a) Impulse = Momentum Change = m•Delta v = m•(vf - vi)= (0.530 kg)•( -13.5 m/s - 18.0 m/s)
Impulse = -16.7 kg•m/s = -16.7 N•s
where the "-" indicates that the impulse was opposite the original direction of motion.
(Note that a kg•m/s is equivalent to a N•s)
(b) The impulse is the product of force and time. So if impulse is known and time is known, force can be easily determined.
Impulse = F•t
F = Impulse/t = (-16.7 N s) / (0.100 s) = -167 N
where the "-" indicates that the impulse was opposite the original direction of motion.
Answer:
Explanation:
a. The wire's length is 10m long and has a mass 100g and a tension of 250N.
Frequency is given by the equation:
#where t=250N*10=2500N,
#substitute for actual values for the lowest frequency.
#n=1, lowest frequency
Hence, the lowest frequency for standing waves is 7.9057Hz
b.The wire's length is 10m long and has a mass 100g and a tension of 250N.
Frequency is given by the equation:
#where t=250N*10=2500N,
#The second lowest frequency happens at :
Hence, the second lowest frequency is 15.8114Hz
c.Given that the wire's length is 10m long and has a mass 100g and a tension of 250N.
Frequency is given by the equation:
#where t=250N*10=2500N,
The third lowest frequency happens at
Hence, the third lowest frequency is 23.7171Hz
Answer:
The vapor pressure is 748.77 torr
Explanation:
Using Clausius-Clapeyron equation:
where;
T₁ is the initial temperature = 85.0°F = 302.5 K
T₂ is the final temperature = 100 °C = 373 K
P₂ is the final pressure = 760 torr
P₁ is the initial pressure = vapor pressure = ?
R is gas constant = 8.314 J/K.mol
ΔHvap is the heat of vaporization of water = 40.7 kJ/mol
4.895(0.00331 - 0.00268) = 0.01489
= 1.015
P₁ = (760 torr)/(1.015) = 748.77 torr
Therefore, the vapor pressure is 748.77 torr