Answer:
93.125 × 10^(19)
Explanation:
We are told the asteroid has acquired a net negative charge of 149 C.
Thus;
Q = -149 C
charge on electron has a value of:
e = -1.6 × 10^(-19) C
Now, for us to determine the excess electrons on the asteroid, we will just divide the net charge in excess on the asteroid by the charge of a single electron.
Thus;
n = Q/e
n = -149/(-1.6 × 10^(-19))
n = 93.125 × 10^(19)
Thus, it has 93.125 × 10^(19) more electrons than protons
Answer:
2 per s
Explanation:
divid 40 and 20 it gives you = 2
Answer:
6957.04N
Explanation:
Using
vf2=vi2+2ad
But vf = 0 .
So convert 50km/hr to m/s, and you need to convert 61 cmto m
(50km/hr)*(1hr/3600s)*(1000m/km) = 13.9m/s
61cm * (1m/100cm) = .61m
So n
0 = (13.9m/s)^2 + 2a(.61m)
a = 158.11m/s^2
So
using F = ma
F = 44kg(158.11m/s^2) = 6957.04N
Answer:
3.4 mT
Explanation:
L = 0.53 m
i = 7.5 A
Theta = 19 degree
F = 4.4 × 10^-3 N
Let B be the strength of magnetic field.
Force on a current carrying conductor placed in a magnetic field.
F = i × L × B × Sin theta
4.4 × 10^-3 = 7.5 × 0.53 × B × Sin 19
B = 3.4 × 10^-3 Tesla
B = 3.4 mT
