Answer:
Transition has to cross between solid and liquid in gray zone.No indoor organized public events and social gatherings are allowed, except with members of the same household.
Answer:
B. counterclockwise
Explanation:
We can solve the problem by using the right-hand rule:
- put your thumb finger of the right hand in the same direction of the current in the wire (upward)
- wrap the other fingers around the thumb
- the direction of the other fingers will give the direction of the magnetic field lines
By doing these steps, we see that the other fingers form concentric circles in a counterclockwise direction (seen from above), so this is the direction of the magnetic field lines.
Hello!
Which nuclei is NOT radioactive?
A) Am-241 B) Mg-24 C) Pu-241 D) U-238
Solving:
It is noteworthy that chemical elements located on the periodic table in the lanthanide and actinide groups are radioactive.
Am-241 (americium) belongs to the group of actinides and is a heavy and radioactive metal.
Mg-24 (magnesium) is an essential element for the body, mainly for the nervous system, in addition to synthesizing proteins and serves for hormonal control, belongs to the group of alkaline earth metals and is a non-radioactive nucleus.
Pu-241 (plutonium) is an element that is isotope of fission by plutonium, belongs to the group of actinides and is a heavy and radioactive metal.
U-238 (uranium) is an element that is isotope of non-fission uranium, belongs to the group of actinides and is a heavy and radioactive metal.
Answer:
B) Mg-24
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Answer:
(a) 8Ω (b) Ratio = Parra/P8 ohm = 1
Explanation:
Solution
Recall that,
An high-fidelity amplifier has one output for a speaker of resistance of = 8 Ω
Now,
(a) How can two 8-Ω speakers be arranged, when one = 4-Ω speaker, and one =12-Ω speaker
The Upper arm is : 8 ohm, 8 ohm
The Lower arm is : 12 ohm, 4 ohm
The Requirement is = (16 x 16)/(16 + 16) = 8 ohm
(b) compare your arrangement power output of with the power output of a single 8-Ω speaker
The Ratio = Parra/P8 ohm = 1
Answer:
16.4287
Explanation:
The force and displacement are related by Hooke's law:
F = kΔx
The period of oscillation of a spring/mass system is:
T = 2π√(m/k)
First, find the value of k:
F = kΔx
78 N = k (98 m)
k = 0.796 N/m
Next, find the mass of the unknown weight.
F = kΔx
m (9.8 m/s²) = (0.796 N/m) (67 m)
m = 5.44 kg
Finally, find the period.
T = 2π√(m/k)
T = 2π√(5.44 kg / 0.796 N/m)
T = 16.4287 s