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Lelu [443]
3 years ago
5

__________ is a synthetic drug with effects very similar to those of cocaine. A. Ecstasy B. Dramamine C. Hydrocodone D. None of

the above
Physics
2 answers:
ehidna [41]3 years ago
4 0
Answer is C. Hydrocodone, also known as dihyrocodeinone, is a semi-synthetic of opioid synthesized from codeine, found in opium poppy. Used to relieve moderate to severe pain.
zhannawk [14.2K]3 years ago
3 0
C. hydrocodone is what i think.
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Responding to an alarm, a 765-N firefighter slides down a pole to the ground floor, 3.5 m below. The firefighter starts at rest
Afina-wow [57]

Answer:

Explanation:

Let acceleration of fall be a .

v² = u² + 2as

v = 3.8 m /s

u = 0

s = 3.5 m

3.8² = 0 + 2 x a x 3.5

a = 2.06 m /s²

Since this acceleration is less than g , an upward force is acting on the firefighter in the form of friction . Let this force be F . Let mass of the firefighter be m .

m = 765 / 9.8

= 78.06 kg

mg - F = ma

765 - F = 78.06 x 2.06

765 - F = 160.8

F = 604.2 N .

4 0
2 years ago
In a simple circuit a 6-volt dry cell pushes charge through a single lamp which has a resistance of 3 Ω. According to Ohms law t
RSB [31]

Answer:

The answers and workings is in the Explanation section

Explanation:

<em>In a simple circuit a 6-volt dry cell pushes charge through a single lamp which has a resistance of 3 Ω. According to Ohms law the current through the circuit is _____________ Amps. </em>

According to Ohm’s law V =I*R  

Where V = Voltage, I = Current and R = Resistance

I = V/R =6/3 =2 Amps of current

Answer = 2 Amps

<em> If a second identical lamp is connected in series, the 6-volt battery must push a charge through a total resistance of ________ Ω</em>.  <em>The current in the circuit is then __________ Amps. </em>

Since the second lamp is connected in series with the first, the total resistance will R₂ = 3 + 3 = 6Ω

2 resistance in series  R₂ =  6Ω

The current in the circuit with the two lamps connected in series is I₂ =V/R₂ =6/6 = 1 Amps

The current is 1 Amps

Answer =  6Ω  and 1 Amps

<em>If a third identical lamp is connected in series, the total resistance is now _________Ω.  The current through all three lamps in series is now _________ Amps.  </em>

Since the third lamp is connected in series with the first and second, the total resistance will R₃ = 3 + 3 + 3 = 9Ω

total resistance of the 3 lamps R₃ =  9Ω

The current in the circuit with the three lamps connected in series is

I =V/R₃ =6/9 = 0.67 Amps

The current through the 3 lamps I₃ = 0.67 Amps

Answer =  9Ω  and 0.67 Amps

<em>The current through each individual lamp is __________ Amps.  </em>

Since all 3 lamps are connected in series, the same current will flow through each of the  3 lamps, and that current is I₃  

The current through each individual lamp is 0.67 Amp

Answer = 0.67 Amp

<em>What is the power when a voltage of 120 volts drives a current of 3 amps through a device?  </em>

The formula for power P = I*V =120*3 = 360 Watts

power P  = 360 Watts

Answer = 360 Watts

<em>What is the current when a 90-W light bulb is connected to 120 V? </em>

From P =I*V, make I the subject of the formula, I = P/V =90/120 = 0.75

Current= 0.75 Amps

Answer =  0.75 Amps

<em>How much current does a 75-W light bulb draw when connected to 120 V?</em>  

Current I =P/V = 75/120 = 0.625 Amps.

Answer = 0.625 Amps

<em>If part of an electric circuit dissipates energy at 6 W when it draws a current of 3 A, what voltage is impressed across it? </em>

Voltage V =P/I =6/3 =2 Volts

Answer = 2 Volts

<em> If a 60 W light bulb at 120 V is left on in your house to prevent burglary, and the power company charges 10 cents per kilowatt-hour, how much will it cost to leave the bulb on for 30 days? Show your work. </em>

24 hours make 1 day, so the number of hours the bulb was left on = 24 *30 = 720 hours

The power rating of the bulb is 60W = 60/1000 = 0.06 KiloWatt

Total power consumed in kilowatt-hour = 0.06 * 720 = 43.2 kilowatt-hour

Cost for 30 days = 0.1*43.2 = $4.32 ( note that 10 Cents = $0.1)

Answer =  $4.32

4 0
3 years ago
I need help on 1. B and show work
Nata [24]
Acceleration = (end speed minus beginning speed) divided by (time for the change).
4 0
3 years ago
A ball of mass 0.1kg is thrown vertically upwards with an initial velocityof 80 m/s. calculate the pontential energy (i) half wa
gregori [183]

Answer:

Stated below:

Explanation:

Let's calculate the maximum height

.

H=v^2/2g=320mH=v  

2

/2g=320m

PE=mgh=0.1*10*320=320 J

at halfway up PE will be half of max = 320/2=160 J

KE will be equal to PEmax=320 J.

Hope I helped! ☺

6 0
3 years ago
A small rock is thrown straight up from the edge of the roof of an 8.00 m tall building with an initial speed vo . The speed of
Serjik [45]

Answer:

20.47 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² = a

v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{24^2-0^2}{2\times 9.81}\\\Rightarrow s=29.3577\ m

Total height of the fall is 29.3577 m

Height the ball reached above the building is 29.3577-8=21.3577\ m

s=ut+\frac{1}{2}at^2\\\Rightarrow 21.3577=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{21.3577\times 2}{9.81}}\\\Rightarrow t=2.0866\ s

Time taken to reach the point from where the ball was thrown is 2.0866 s

This will also be the time it takes the ball to reach the maximum height

s=ut+\frac{1}{2}at^2\\\Rightarrow u=\dfrac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\dfrac{21.3577-\frac{1}{2}\times -9.81\times 2.0866^2}{2.0866}\\\Rightarrow u=20.47\ m/s

The initial velocity with which the rock was thrown was 20.47 m/s

7 0
3 years ago
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