Answer:
Conduction: Some metalloids, such as silicon and germanium, can act as electrical conductors under the right conditions, thus they are called semi-conductors. Luster: Silicon for example appears lustrous, but is not malleable or ductile (it is brittle - a characteristic of some nonmetals).
Explanation:
Answer:
a. All the laboratory equipment given are very basic equipment used in all the laboratories. Name of each equipment is as follows:
- V refers to the tape measure.
- W refers to dropper.
- X refers to an inoculating loop.
- Y refers to a stopwatch.
- Z refers to the microscope.
b. Use of each laboratory equipment identified is:
- Tape measures (V) is used to measure the length of objects or distance in a laboratory.
- Dropper (W) is used to measure unit of drop required to dispensed as one drop or several drops in any experiment.
- Inoculating loop (X) is used by microbiologists to cultivate microbes on plates and retrieving an inoculum from a culture of microorganisms.
- Stopwatch (Y) is used to measure the time of any experiment.
- Microscope (Z) is used to magnify an object to look at it in detail.
Answer:
Work done, W = 1786.17J
Explanation:
The question says "A 75.0-kg painter climbs a 2.75-m ladder that is leaning against a vertical wall. The ladder makes an angle of 30.0 ° with the wall. How much work (in Joules) does gravity do on the painter? "
Mass of a painter, m = 75 kg
He climbs 2.75-m ladder that is leaning against a vertical wall.
The ladder makes an angle of 30 degrees with the wall.
We need to find the work done by the gravity on the painter.
The angle between the weight of the painter and the displacement is :
θ = 180 - 30
= 150°
The work done by the gravity is given by :
Hence, the required work done is 1786.17 J.
Initial volume of mercury is
V = 0.1 cm³
The temperature rise is 35 - 5 = 30 ⁰C = 30 ⁰K.
Because the coefficient of volume expansion is 1.8x10⁻⁴ 1/K, the change in volume of the mercury is
ΔV = (1.8x10⁻⁴ 1/K)*(30 ⁰K)(0.1 cm³) = 5.4x10⁻⁴ cm³
The cross sectional area of the tube is
A = 0.012 mm² = (0.012x10⁻² cm²).
Therefore the rise of mercury in the tube is
h = ΔV/A
= (5.4x10⁻⁴ cm³)/(0.012x10⁻² cm²)
= 4.5 cm
Answer: 4.5 cm
To solve this problem, let us recall that the formula for
gases assuming ideal behaviour is given as:
rms = sqrt (3 R T / M)
where
R = gas constant = 8.314 Pa m^3 / mol K
T = temperature
M = molar mass
Now we get the ratios of rms of Argon (1) to hydrogen (2):
rms1 / rms2 = sqrt (3 R T1 / M1) / sqrt (3 R T2 / M2)
or
rms1 / rms2 = sqrt ((T1 / M1) / (T2 / M2))
rms1 / rms2 = sqrt (T1 M2 / T2 M1)
Since T1 = 4 T2
rms1 / rms2 = sqrt (4 T2 M2 / T2 M1)
rms1 / rms2 = sqrt (4 M2 / M1)
and M2 = 2 while M1 = 40
rms1 / rms2 = sqrt (4 * 2 / 40)
rms1 / rms2 = 0.447
Therefore the ratio of rms is:
<span>rms_Argon / rms_Hydrogen = 0.45</span>
