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Mumz [18]
4 years ago
13

Psychoactive drugs interact with which alter a persons mood,perception and behavior

Physics
1 answer:
lora16 [44]4 years ago
8 0
The central nervous system.
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A 0.20 kg mass (m1) hangs vertically from a spring and an elongation of the spring of 9.50 cm (r1) is recorded. With a mass (m2)
kupik [55]

Answer:

k=320N/m

Explanation:

Step one:

given data

Let the initial/equilibrum position be x

mass m1= 0.2kg

F1= 0.2*10= 2N

elongation e= 9.5cm= 0.095m

mass m2=1kg

F2=1*10= 10N

elongation e= 12cm= 0.12m

Step two:

From Hooke's law, which states that provided the elastic limits of a material is not exceeded the extention e is proportional to applied Force F

F=ke

2=k(0.095-a)

2=0.095k-ka----------1

10=k(0.12-a)

10=0.12k-ka----------2

solving equation 1 and 2 simultaneously

 

   10=0.12k-ka----------2

-   2=0.095k-ka----------1

   8=0.025k-0

divide both side by 0.025

k=8/0.025

k=320N/m

5 0
3 years ago
In a solution of soda, the gaseous carbon dioxide is the solvent, and the liquid cola is the solute.
satela [25.4K]
I think its true but im not sure
4 0
3 years ago
The deflection of a 160-cm-long beam in bending will be _____ times the deflection of an equivalent 120-cm-long beam.
katovenus [111]

Answer:

Number of time bend length = 1.33 (Approx.)

Explanation:

Given:

Length of beam = 160 centimeter

Length of beam after deflection = 120 centimeter

Find:

Number of time bend length

Computation:

Number of time bend length = Length of beam / Length of beam after deflection

Number of time bend length = 160 / 120

Number of time bend length = 1.33333

Number of time bend length = 1.33 (Approx.)

8 0
3 years ago
A capacitor with initial charge q0 is discharged through a resistor. What multiple of the time constant τ gives the time the cap
fiasKO [112]

Answer: a) 0.13*τ ; b) 2.08*τ

Explanation: In order to explain the discharg of a capacitor through a resistor, we have to consider the following:

Q(t)=Qo* exp(-t/τ) for a lose of 1/8-th of its charge

in this case, Q(t)=7/8*Qo=7/8*exp(-t/τ)

ln(7/8)*τ=-t

then, t= -ln(7/8)*τ =0.13

For a lose of 7/8 th  of its charge , we have

Q(t)=1/7*Qo*exp(-t/τ)

t=-ln(1/8)*τ=2.08

5 0
4 years ago
A 10 gauge copper wire carries a current of 23 A. Assuming one free electron per copper atom, calculate the magnitude of the dri
Reptile [31]

Question:

A 10 gauge copper wire carries a current of 15 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm².)

Answer:

3.22 x 10⁻⁴ m/s

Explanation:

The drift velocity (v) of the electrons in a wire (copper wire in this case) carrying current (I) is given by;

v = \frac{I}{nqA}

Where;

n = number of free electrons per cubic meter

q =  electron charge

A =  cross-sectional area of the wire

<em>First let's calculate the number of free electrons per cubic meter (n)</em>

Known constants:

density of copper, ρ = 8.95 x 10³kg/m³

molar mass of copper, M = 63.5 x 10⁻³kg/mol

Avogadro's number, Nₐ = 6.02 x 10²³ particles/mol

But;

The number of copper atoms, N, per cubic meter is given by;

N = (Nₐ x ρ / M)          -------------(ii)

<em>Substitute the values of Nₐ, ρ and M into equation (ii) as follows;</em>

N = (6.02 x 10²³ x 8.95 x 10³) / 63.5 x 10⁻³

N = 8.49 x 10²⁸ atom/m³

Since there is one free electron per copper atom, the number of free electrons per cubic meter is simply;

n = 8.49 x 10²⁸ electrons/m³

<em>Now let's calculate the drift electron</em>

Known values from question:

A = 5.261 mm² = 5.261 x 10⁻⁶m²

I = 23A

q = 1.6 x 10⁻¹⁹C

<em>Substitute these values into equation (i) as follows;</em>

v = \frac{I}{nqA}

v = \frac{23}{8.49*10^{28} * 1.6 *10^{-19} * 5.261*10^{-6}}

v = 3.22 x 10⁻⁴ m/s

Therefore, the drift electron is 3.22 x 10⁻⁴ m/s

6 0
4 years ago
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