Answer;
The temperature change for the second pan will be lower compared to the temperature change of the first pan
Explanation;
-The quantity of heat is given by multiplying mass by specific heat and by temperature change.
That is; Q = mcΔT
This means; the quantity of heat depends on the mass, specific heat capacity of a substance and also the change in temperature.
-Maintaining the same quantity of heat, with another pan of the same mass and greater specific heat capacity would mean that the change in temperature would be much less lower.
A. Use the thin lens equation to determine the image distance from the lens. Is the correct answer because I have no idea why
Answer;
=15855.40 kg/m^3
Explanation;
Volume (V) of the cylinder = pi x r^2 x h
V = 3.14 x (44/2 x 10^-3)^2 x 41.5 x 10^-3
V = 6.307 x 10^-5 m^3
By density = m/V
mass = 1 kg
density = 1/(6.307 x 10^-5) = 15855.40 kg/m^3
The distance you free-fall from rest is D = (1/2) (g) (T²) <== memorize this
Height of the platform = (1/2) (9.8 m/s²) (2.4 sec)²
Height = (4.9 m/s²) (5.76 s²)
Height = (4.9/5.76) meters
Height = 28.2 meters (a VERY high platform ... about 93 ft off the water !)
Without air-resistance, your horizontal speed doesn't change. It's constant. Traveling 3.1 m/s for 2.4 sec, you cover (3.1 m/s x 2/4 s) = 7.4 m horizontally.
Given: Mass of earth Me = 5.98 x 10²⁴ Kg
Radius of earth r = 6.37 x 10⁶ m
G = 6.67 x 10⁻¹¹ N.m²/Kg²
Required: Smallest possible period T = ?
Formula: F = ma; F = GMeMsat/r² Centripetal acceleration ac = V²/r
but V = 2πr/T
equate T from all equation.
F = ma
GMeMsat/r² = Msat4π²/rT²
GMe = 4π²r³/T²
T² = 4π²r³/GMe
T² = 39.48(6.37 x 10⁶ m)³/6.67 x 10⁻¹¹ N.m²/Kg²)(5.98 x 10²⁴ Kg)
T² = 1.02 x 10²² m³/3.99 x 10¹⁴ m³/s²
T² = 25,563,909.77 s²
T = 5,056.08 seconds or around 1.4 Hour
