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Solnce55 [7]
3 years ago
14

2. CHARLES MANSON WAS DIAGNOSED WITH

Physics
1 answer:
PilotLPTM [1.2K]3 years ago
3 0

Answer:

The answer is C. anti social

Explanation:

Hope that helps!

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Which type of circuit would be best to use for lights used for decorations?
Dahasolnce [82]

Answer:

led would be the best for decorations

Explanation:

6 0
2 years ago
Which statement best describes how work and power are different? a. To find work we need to know force and distance; to find pow
weqwewe [10]
A. To find work we need to know F and S; to find power we need to know F and V
6 0
3 years ago
A coin is resting on the bottom of an empty container. The container is then filled to the brim three times, each time with a di
Virty [35]

Answer:

Refractive index of liquid C > Refractive index of liquid B > Refractive index of liquid A

Explanation:

Let the depth of each section is h.

That means the real depth for each section is h.

Apparent depth is liquid A is 7 cm.

Apparent depth in liquid B is 6 cm.

Apparent depth in liquid C is 5 cm.

by the formula of the refractive index

n = real depth / apparent depth

where, n is the refractive index of the liquid.

For liquid A:

n_{A}=\frac{h}{7}    .... (1)

For liquid B:

n_{B}=\frac{h}{6}   ..... (2)

For liquid C:

n_{C}=\frac{h}{5}     ..... (3)

By comparing all the three equations

nc > nB > nA

Refractive index of liquid C > Refractive index of liquid B > Refractive index of liquid A

5 0
3 years ago
When an ideal gas does positive work in its surroundings, which of the ga's quantities increases?
Burka [1]
I think it's B hope it helps
4 0
3 years ago
Two devices with capacitances of 25 μf and 5.0 μf are each charged with separate 120 v power supplies. calculate the total energ
mash [69]
The energy stored in a capacitor is given by
U= \frac{1}{2} CV^2
where C is the value of the capacitance while V is the voltage difference applied to the capacitor.

Let's calculate the energy of the first capacitor:
U_1 =  \frac{1}{2} (25\cdot 10^{-6}F)(120 V)=1.5 \cdot 10^{-3}J

And now the energy of the second capacitor:
U_2 =  \frac{1}{2} (5 \cdot 10^{-6}F)(120 V)=3 \cdot 10^{-4}J

So, the total energy stored in the two capacitors is
U=U_1 +U_2 = 1.8 \cdot 10^{-3}J
3 0
3 years ago
Read 2 more answers
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