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3 years ago

14

True or False Criminals structure crime. For instance, crack cocaine street dealers prefer the middle of a long block as the bes

t location for dealing.

1 answer:

LenaWriter [7]3 years ago

8 0

True. Though typically crack deals take place in a car, or in an alley. The middle of a long block is kind of exposed. It could happen anywhere in Chicago though.
But yeah, criminals structure crime.

You might be interested in

Acceleration = change of velocity divided by time interval = Δv/Δt.

MariettaO [177]

Answer:

a=2.378 m/s^2

Explanation:

a=Δv/Δt------eq(1)

Δv=Vf-Vi=120 km/h-0 km/h=120 km/h

or Δv=33.3 m/sec

or time=t=14s

putting values in eq(1)

a=33.3/14

a=2.378 m/s^2

6 0

3 years ago

A) An automobile light has a 1.0-A current when it is connected to a 12-V battery. Determine the resistance of the light.

kirill [66]

Answer:

The resistance in first case is 12 Ω, power delivered is 12 W, and potential difference is 0.01 V

Explanation:

Given:

(A)

Current $I = 1$ A

Voltage $V = 12$ V

For finding the resistance,

$V = IR$

$R = \frac{V}{I}$

$R = \frac{12}{1}$

$R =$ 12Ω

(B)

For finding power delivered,

$P = I^{2} R$

$P = (1) ^{2} \times 12$

$P = 12$ Watt

(C)

For finding the potential difference,

$V = IR$

$V = 5 \times 10^{-3} \times 2$

$V = 10 \times 10^{-3}$

$V = 0.01$ V

Therefore, the resistance in first case is 12 Ω, power delivered is 12 W, and potential difference is 0.01 V

4 0

3 years ago

How much heat is needed to raise the temperature of 50.0 g of water by 25.0°C

love history [14]

Answer:

Explanation:

In order to be able to solve this problem, you will need to know the value of water's specific heat, which is listed as

c

=

4.18

J

g

∘

C

Now, let's assume that you don't know the equation that allows you to plug in your values and find how much heat would be needed to heat that much water by that many degrees Celsius.

Take a look at the specific heat of water. As you know, a substance's specific heat tells you how much heat is needed in order to increase the temperature of

1 g

of that substance by

1

∘

C

.

In water's case, you need to provide

4.18 J

of heat per gram of water to increase its temperature by

1

∘

C

.

What if you wanted to increase the temperature of

1 g

of water by

2

∘

C

? You'd need to provide it with

increase by 1

∘

C



4.18 J

+

increase by 1

∘

C



4.18 J

=

increase by 2

∘

C



2

×

4.18 J

To increase the temperature of

1 g

of water by

n

∘

C

, you'd need to supply it with

increase by 1

∘

C



4.18 J

+

increase by 1

∘

C



4.18 J

+

...

=

increase by n

∘

C



n

×

4.18 J

Now let's say that you wanted to cause a

1

∘

C

increase in a

2-g

sample of water. You'd need to provide it with

for 1 g of water



4.18 J

+

for 1 g of water



4.18 J

=

for 2 g of water



2

×

4.18 J

To cause a

1

∘

C

increase in the temperature of

m

grams of water, you'd need to supply it with

for 1 g of water



4.18 J

+

for 1 g of water



4.18 J

+

,,,

=

for m g of water



m

×

4.18 J

This means that in order to increase the temperature of

m

grams of water by

n

∘

C

, you need to provide it with

heat

=

m

×

n

×

specific heat

This will account for increasing the temperature of the first gram of the sample by

n

∘

C

, of the the second gram by

n

∘

C

, of the third gram by

n

∘

C

, and so on until you reach

m

grams of water.

And there you have it. The equation that describes all this will thus be

q

=

m

⋅

c

⋅

Δ

T

, where

q

- heat absorbed

m

- the mass of the sample

c

- the specific heat of the substance

Δ

T

- the change in temperature, defined as final temperature minus initial temperature

In your case, you will have

q

=

100.0

g

⋅

4.18

J

g

∘

C

⋅

(

50.0

−

25.0

)

∘

C

q

=

10,450 J

Rounded to three sig figs and expressed in kilojoules, t

Explanation:

3 0

3 years ago

Read 2 more answers

How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe

Rus_ich [418]

Explanation:

(a) For an isothermal process, work done is represented as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

Putting the given values into the above formula as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

= $- 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})$

= $-12280.82 \times ln (0.09)$

= $-12280.82 \times -2.41$

= 29596.78 J

or, = 29.596 kJ (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

W = $\frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}$

= $\frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}$

= $\frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}$

= 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c) We know that for an isothermal process,

$P_{1}V_{1} = P_{2}V_{2}$

or, $P_{2} = \frac{P_{1}V_{1}}{V_{2}}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})$

= 11 atm

Hence, the required pressure is 11 atm.

(d) For adiabatic process,

$P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}$

or, $P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}$

= 28.7 atm

Therefore, required pressure is 28.7 atm.

6 0

3 years ago

A body with initial velocity 8.0 m/s moves along a straight line with constant acceleration and travels

Aleksandr [31]

Answer:

<em>(a) The average velocity is 16 m/s</em>

<em>(b) The acceleration is 0.4 m/s^2</em>

<em>(c) The final velocity is 24 m/s</em>

Explanation:

<u>Constant Acceleration Motion</u>

It's a type of motion in which the velocity (or the speed) of an object changes by an equal amount in every equal period of time.

Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, final speed is calculated as follows:

$v_f=v_o+at\qquad\qquad [1]$

The distance traveled by the object is given by:

$\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad\qquad [2]$

(a) The average velocity is defined as the total distance traveled divided by the time taken to travel that distance.

We know the distance is x=640 m and the time taken t= 40 s, thus:

$\displaystyle \bar v=\frac{x}{t}=\frac{640}{40}=16$

The average velocity is 16 m/s

Using the equation [1] we can solve for a:

$\displaystyle a=\frac{v_f-v_o}{t}$

(c) From [2] we can solve for a:

$\displaystyle a= 2\frac{x-v_ot}{t^2}$

Since vo=8 m/s, x=640 m, t=40 s:

$\displaystyle a= 2\frac{640-8\cdot 40}{40^2}=0.4$

The acceleration is 0.4 m/s^2

(b) The final velocity is calculated by [1]:

$v_f=8+0.4\cdot 40$

$v_f=8+16=24$

The final velocity is 24 m/s

3 0

3 years ago