Answer:
30ms
Explanation:
you need to multiple the 10ms by 3s which gives you 30ms
Answer:
4 %
2 ) 3.42 %
Explanation:
Sensitivity requirement of 4 milligram means it is not sensitive below 4 milligram or can not measure below 4 milligram .
Given , 4 milligram is the maximum error possible .
Measured weight = 100 milligram
So percentage maximum potential error
= (4 / 100) x 100
4 %
2 )
As per measurement
weight of 6 milliliters of water
= 48.540 - 42.745 = 5.795 gram
6 milliliters of water should measure 6 grams
Deviation = 6 - 5.795 = - 0.205 gram.
Percentage of error =(.205 / 6 )x 100
= 3.42 %
Answer:
B. 6.6%
Explanation:
The percentage error of a measurement can be calculated using the formula;
Percent error = (experimental value - accepted value / accepted value) × 100
In this question, the calibrated 250.0 gram mass is the accepted value while the weighed mass of 266.5 g is the experimental or measured value.
Hence, the percentage error can be calculated thus;
Percent error = (266.5-250.0/250.0) × 100
Percent error = 16.5/250 × 100
Percent error = 0.066 × 100
Percent error = 6.6%
Answer:
Lens
Explanation:
Lenses fit this description.
Answer:
A. 2.82 eV
B. 439nm
C. 59.5 angstroms
Explanation:
A. To calculate the energy of the photon emitted you use the following formula:
(1)
n1: final state = 5
n2: initial state = 2
Where the energy is electron volts. You replace the values of n1 and n2 in the equation (1):
B. The energy of the emitted photon is given by the following formula:
(2)
h: Planck's constant = 6.62*10^{-34} kgm^2/s
c: speed of light = 3*10^8 m/s
λ: wavelength of the photon
You first convert the energy from eV to J:
Next, you use the equation (2) and solve for λ:
C. The radius of the orbit is given by:
(3)
where ao is the Bohr's radius = 2.380 Angstroms
You use the equation (3) with n=5:
hence, the radius of the atom in its 5-th state is 59.5 anstrongs
