Answer:
The change in height of the mercury is approximately 2.981 cm
Explanation:
Recall that the formula for thermal expansion in volume is:
from which we solved for the change in volume 
due to a given change in temperature 
We can estimate the initial volume of the mercury in the spherical bulb of diameter 0.24 cm ( radius R = 0.12 cm) using the formula for the volume of a sphere:
Therefore, the change in volume with a change in temperature of 36°C becomes:
Now, we can use this difference in volume, to estimate the height of the cylinder of mercury with diameter 0.0045 cm (radius r= 0.00225 cm):
Complete question :
NASA is concerned about the ability of a future lunar outpost to store the supplies necessary to support the astronauts. The supply storage area of the lunar outpost, where gravity is 1.63 m/s2, can only support 1 x 10^5 N. What is the maximum WEIGHT of supplies, as measured on EARTH, NASA should plan on sending to the lunar outpost
Answer:
601,220N
Explanation:
Given that:
Gravity at lunar outpost = 1.63m/s²
Acceleration due to gravity on earth = 9.8m/s²
Supported weight = 1 * 10^5 N
Maximum weight of supplies as measured on earth;
(Ratio of the gravities) * weight of supplies
(9.8m/s² / 1.63m/s²) * (1 * 10^5 N)
6.0122 * (1 * 10^5)
6.0122 * 10^5 N
= 601,220 N
340 ms
I got it right and I hope you do as well
-- "Declination zero" means the object is in the sky at some point directly over the Earth's equator.
-- If it's the sun and it appears to be over the equator, then that tells us that the Earth's axis is not tilted toward or away from it.
-- That in turn tells us that the Earth is at one of the two equinoxes in its orbit, either the Spring one or the Autumn one. <em> (D)</em>
-- (The first days of Summer and Winter coincide with solstices, not equinoxes.)
Explanation:
Elongation of the wire is:
ΔL = F L₀ / (E A)
where F is the force,
L₀ is the initial length,
E is Young's modulus,
and A is the cross sectional area.
ΔL = T (0.5 m) / ((2.0×10¹¹ Pa) (0.02 cm²) (1 m / 100 cm)²)
ΔL = T (1.25×10⁻⁶ m/N)
T = (80,000 N/m) ΔL
Draw a free body diagram of the mass at the bottom of the circle. There are two forces: tension force T pulling up and weight force mg pulling down.
Sum of forces in the centripetal direction:
∑F = ma
T − mg = mv²/r
T − mg = mω²r
T − (15 kg) (9.8 m/s²) = (15 kg) (2 rev/s × 2π rad/rev)² (0.5 m + ΔL)
T − 147 N = (2368.7 N/m) (0.5 m + ΔL)
Substitute:
(80,000 N/m) ΔL − 147 N = (2368.7 N/m) (0.5 m + ΔL)
(80,000 N/m) ΔL − 147 N = 1184.35 N + (2368.7 N/m) ΔL
(797631.3 N/m) ΔL = 1331.35 N
ΔL = 0.00167 m
ΔL = 1.67 mm
