You can estimate one more digit past the smallest division on the measuring device. If you look at a 10mL graduated cylinder, for example, the smallest graduation is tenth of a milliliter (0.1mL). That means when you read the volume, you can estimate to the hundredths place (0.01mL).
PV=nRT
(720/760)(0.200)=(0.800/x)(0.08206)(323.15)
(0.1894736842)=(0.800/x)(0.08206)(323.15)
.0071451809=(0.800/x)
x=MM=111.9635758 g/mol
Answer:
test 5 seemed to be the hardest for me to perceive in account i only saw three f's when there was indeed 6 it was very difficult to find the f's even going very slowly.
Explanation:
correct on edge
Answer:
v = 5.75 x 10⁶ m/s
Explanation:
The radius (r) of the circular orbit taken by a charged particle is related to its speed perpendicular to a magnetic field of strength B, and is given by
r =
--------------(i)
Where,
q = charge of the particle
m = mass of the particle
Making v subject of the formula in equation (i) above gives
v =
-------------------(ii)
Given;
r = 20cm = 0.2m
B = 0.3T
v = unknown
q = charge of proton = 1.6 x 10⁻¹⁹ C
m = mass of the proton = 1.67 x 10⁻²⁷kg
Substitute the values of m, q, B and r into equation (ii) above to get;
v = 
Solving for v gives:
v = 5.75 x 10⁶ m/s
Therefore, the velocity of the proton is 5.75 x 10⁶ m/s
It’s 10 joules. W=FD, W=5•2=10