Answer:
b. The electric field points away from the source charge, if the source charge is positive
d. The electric field points toward the source charge, if the source charge is negative.
Explanation:
A positive source charge would create an electric field that would exert a repulsive effect upon a positive test charge. Thus, the electric field vector would always be directed away from positively charged objects. On the other hand, a positive test charge would be attracted to a negative source charge. Therefore, electric field vectors are always directed towards negatively charged object.
Also electric field strength depends only on test charge
The correct options include b and d
The electric field points away from the source charge, if the source charge is positive.
Also, the electric field points toward the source charge, if the source charge is negative.
Answer:
10.88 km
Explanation:
We shall represent displacement in terms of i , j unit vectors in the direction of east and north .
4.5 km due west
D₁ = - 4.5 i
6.7 km at an angle of 27° south of west
D₂ = - 6.7 cos27 i - 6.7 sin27j
= - 6.7 x .89 i - 6.7 x .45 j
= - 5.96i - 3 j
Total displacement
= D₁ + D₂
= - 4.5 i - 5.96i - 3 j
= -10.46 i - 3j
Magnitude = √ ( 10.46² + 3²)
= √ ( 109.41 + 9)
= √ 118.41
= 10.88 km .
1 watt = 1 joule/sec
2,000 watts = 2,000 joules/sec
(2,000 joule/sec) x (120 sec)
= (2,000 x 120) (joule-sec/sec)
= 240,000 joules .
1 angstrom = 10^-8 cm
6.5 x 10-4 cm = 65 000 x 10-8 cm = 65 000 angstroms
answer 65 000 angstroms
Answer:
10 days
Explanation:
The half-life of a radioactive sample is the time taken for half of the sample to decay.
In the diagram, the half-life corresponds to the time after which the % of cobalt-57 has halved. We can observe the following:
At t=10 days, the % of Co remaining is approximately 45%
At t=20 days, the % of Co remaining is approximately 22%
This means that the sample of cobalt-57 has halved in 10 days, so the half-life of cobalt-57 is 10 days.
