Answer:
Explanation:
The equation that relates heat Q with the temperature change of a substance of mass <em>m </em>and specific heat <em>c </em>is
.
We want to calculate the final temperature <em>T, </em>so we have:
Which for our values means (in this case we do not need to convert the mass to Kg since <em>c</em> is given in g also and they cancel out, but we add to our temperature in
to have it in
as it must be):
If we see the forces on the astronauts then there is only one force on each astronaut while they are in air.
This force is due to gravity of earth.
Now while the astronauts are in air and doing some activities then the net force on them is counterbalanced on them by centrifugal force.
so we can say the rate of fall of astronauts due to gravity is at same rate as the orbiting rate of the space shuttle.
this is given by
so all astronauts will experience the situation of free fall.
Answer:
a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
Explanation:
Given the data in the question;
as the equation of standing wave on a string is fixed at both ends
y = 2AsinKx cosωt
but k = 2π/λ and ω = 2πf
λ = 4 × 0.150 = 0.6 m
and f = v/λ = 260 / 0.6 = 433.33 Hz
ω = 2πf = 2π × 433.33 = 2722.69
given that A = 2.20 mm = 2.2×10⁻³
so = A × ω
= 2.2×10⁻³ × 2722.69 m/s
= 5.9899 m/s
therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b)
A' = 2AsinKx
= 2.20sin( 2π/0.6 ( 0.075) rad )
= 2.20 sin( 0.7853 rad ) mm
= 2.20 × 0.706825 mm
A' = 1.555 mm = 1.555×10⁻³
so
= A' × ω
= 1.555×10⁻³ × 2722.69
= 4.2338 m/s
Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
Answer:
a. 0.000002 m
b. 0.00000182 m
Explanation:
36 cm = 0.36 m
15 cm = 0.15 m
a) We can start by calculating the air-water pressure of the bucket submerged 20m below the water surface:
Suppose air is ideal gas, then if the temperature stays the same, the product of its pressure and volume stays the same
Where P1 = 1.105 Pa is the atmospheric pressure, V_1 is the air volume in the bucket on the suface:
As the pressure increases, the air inside the bucket shrinks. But the crossection area stays constant, so only h, the height of air, decreases:
b) If the temperatures changes, we can still reuse the ideal gas equation above: