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3 years ago

Investigators should collect virtually everything from a crime scene. True or false?

Physics

2 answers:

Marianna [84]3 years ago

8 0

Answer:

false

Explanation:

its a trick question. not all evidence should be collected from the crime scene due to not all the evidence be useful.

Stels [109]3 years ago

4 0

I think it would be TRUE but I’m not 100% sure.... please let me know if I’m right or wrong

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A 1000 kg satellite and a 2000 kg satellite follow exactly the same orbit around the earth. What is the ratio F1/F2 of the gravi

Tamiku [17]

Answer:

the ratio F1/F2 = 1/2

the ratio a1/a2 = 1

Explanation:

The force that both satellites experience is:

F1 = G M_e m1 / r² and

F2 = G M_e m2 / r²

where

m1 is the mass of satellite 1
m2 is the mass of satellite 2
r is the orbital radius
M_e is the mass of Earth

Therefore,

F1/F2 = [G M_e m1 / r²] / [G M_e m2 / r²]

F1/F2 = [G M_e m1 / r²] × [r² / G M_e m2]

F1/F2 = m1/m2

F1/F2 = 1000/2000

F1/F2 = 1/2

The other force that the two satellites experience is the centripetal force. Therefore,

F1c = m1 v² / r and

F2c = m2 v² / r

where

m1 is the mass of satellite 1
m2 is the mass of satellite 2
v is the orbital velocity
r is the orbital velocity

Thus,

a1 = v² / r ⇒ v² = r a1 and

a2 = v² / r ⇒ v² = r a2

Therefore,

F1c = m1 a1 r / r = m1 a1

F2c = m2 a2 r / r = m2 a2

In order for the satellites to stay in orbit, the gravitational force must equal the centripetal force. Thus,

F1 = F1c

G M_e m1 / r² = m1 a1

a1 = G M_e / r²

also

a2 = G M_e / r²

Thus,

a1/a2 = [G M_e / r²] / [G M_e / r²]

a1/a2 = 1

4 0

3 years ago

True or false: When an object becomes polarized, it acquires a charge and becomes a charged object.

PSYCHO15rus [73]

Answer:

i think its true

Explanation:

8 0

3 years ago

How does speed effect momentum?

drek231 [11]

Momentum = mass * speed
Speed is proportional to momentum; i.e. having twice the speed would result in double the momentum, a lower speed = smaller momentum, etc

7 0

3 years ago

In a power plant, pipes transporting superheated vapor are very common. Superheated vapor flows at a rate of 0.3 kg/s inside a p

grigory [225]

Answer: $h=160.84 W/m^2-K$

Explanation:

Given

mass flow rate=0.3 kg/s

diameter of pipe=5 cm

length of pipe=10 m

Inside temperature=22

Pipe surface =100

Temperature drop=30

specific heat of vapor(c)=2190 J/kg.k

heat supplied $Q=mc\Delta T=0.3\times 2190\times (30)$

Heat due to convection =hA(100-30)

$A=\pi d\cdot L$

$A=\pi 0.05\times 10=1.571 m^2$

$Q_{convection}=h\times 1.571\times (100-22)=122.538 h$

$Q=Q_{convection}$

19,710=122.538 h

$h=160.84 W/m^2-K$

5 0

3 years ago

Monochromatic light is incident on a pair of slits that are separated by 0.220 mm. The screen is 2.60 m away from the slits. (As

Naddik [55]

Answer:

$\lambda = 1.667 nm$

$\theta = 0.8681^o$

Explanation:

From the question we are told that

The distance of separation is $d = 0.220 \ mm = 0.00022 \ m$

The is distance of the screen from the slit is $D = 2.60 \ m$

The distance between the central bright fringe and either of the adjacent bright $y = 1.97 cm = 1.97 *10^{-2}\ m$

Generally the condition for constructive interference is

$d sin \tha(\theta ) = n \lambda$

From the question we are told that small-angle approximation is valid here.

So $sin (\theta ) = \theta$

=> $d \theta = n \lambda$

=> $\theta = \frac{n * \lambda }{d }$

Here n is the order of maxima and the value is n = 1 because we are considering the central bright fringe and either of the adjacent bright fringes

Generally the distance between the central bright fringe and either of the adjacent bright is mathematically represented as

$y = D * sin (\theta )$

From the question we are told that small-angle approximation is valid here.

$y = D * \theta$

=> $\theta = \frac{ y}{D}$

$\frac{n * \lambda }{d } = \frac{y}{D}$

$\lambda =\frac{d * y }{n * D}$

substituting values

$\lambda = \frac{0.00022 * 1.97*10^{-2} }{1 * 2.60 }$

$\lambda = 1.667 *10^{-6}$

$\lambda = 1.667 nm$

In the b part of the question we are considering the next set of bright fringe so n= 2

Hence

$dsin (\theta ) = n \lambda$

$\theta = sin^{-1}[\frac{ n * \lambda }{d} ]$

$\theta = sin^{-1}[\frac{ 2 * 1667 *10^{-9}}{ 0.00022} ]$

$\theta = 0.8681^o$

7 0

4 years ago