Answer:
If she ___(be) rich, she _____ (buy) a new sports car.
Group of answer choices
is/buys
be/buy
were/would buy
were/bought
b. 460.8 m/s
Explanation:
The relationship between the speed of the wave along the string, the length of the string and the frequency of the note is

where v is the speed of the wave, L is the length of the string and f is the frequency. Re-arranging the equation and substituting the data of the problem (L=0.90 m and f=256 Hz), we can find v:

c. 18,000 m
Explanation:
The relationship between speed of the wave, distance travelled and time taken is

where
v = 6,000 m/s is the speed of the wave
d = ? is the distance travelled
t = 3 s is the time taken
Re-arranging the formula and substituting the numbers into it, we find:

Answer:
1. all of them
2. cork and wax
3. iron, lead, and aluminum
4. none of them
Explanation:
1.Which material will displace a volume of water? all of them
When an object is introduced into a container with a volume of water, a volume of liquid equal to the volume of the object is displaced
2.Which material will displace a volume of water less than its own volume?
cork and wax
because the density of the object is less than that of the displaced liquid
3.Which material will displace a volume of water equal to its own volume?
iron, lead, and aluminum
because Arquimedes's principle: any body plunged inside a fluid in this case water experiences an ascending force called push, equivalent to the weight of the fluid removed by the body
4.Which material will displace a volume of water greater than its own volume?
None of them
Answer:
a) 0.018 kg
b) 262 kPa
Explanation:
The volume of the concentric cylinders would be:
V = π/4 * h * (D^2 - d^2)
V = π/4 * 13 * (52^2 - 33^2) = 16500 cm^3 = 0.0165 m^3
The state equation of gases:
p * V = m * R * T
Rearranging:
m = (p * V) / (R * T)
R is 287 J/(kg * K) for air
25 C = 298 K
m0 = 202000 * 0.0165 / (287 * 298) = 0.039 kg
After pumping more air the volume remains about the same, but temperature and pressure change.
30 C = 303 K
m1 = 303000 * 0.0165 / (287 * 303) = 0.057 kg
The mass that was added is
m1 - m0 = 0.057 - 0.039 = 0.018 kg
If that air is cooled to 0 C
0 C is 273 K
p = m * R * T / V
p = 0.057 * 278 * 273 / 0.0165 = 262000 Pa = 262 kPa