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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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What is the gauge pressure in Pascals inside a honey droplet of a 0.1 cm diameter? Assume that air is surrounding this droplet a

Vikki [24]

Answer:

The gauge pressure in Pascals inside a honey droplet is 416 Pa

Explanation:

Given;

diameter of the honey droplet, D = 0.1 cm

radius of the honey droplet, R = 0.05 cm = 0.0005 m

surface tension of honey, γ = 0.052 N/m

Apply Laplace's law for a spherical membrane with two surfaces

Gauge pressure = P₁ - P₀ = 2 (2γ / r)

Where;

P₀ is the atmospheric pressure

Gauge pressure = 4γ / r

Gauge pressure = 4 (0.052) / (0.0005)

Gauge pressure = 416 Pa

Therefore, the gauge pressure in Pascals inside a honey droplet is 416 Pa

5 0

3 years ago

Vector A has a magnitude of 63 units and points west, while vector B has the same magnitude and points due south. Find the magni

ozzi

Given :

Vector A has a magnitude of 63 units and points west, while vector B has the same magnitude and points due south.

To Find :

The magnitude and direction of

a) A + B .

b) A - B.

Solution :

Let , direction in north is given by +j and east is given by +i .

So , $A=-63i$ and $B=63j$

Now , A + B is given by :

$A+B=-63i+63j$

$| A+B | = 63\sqrt{2}$

Direction of A+B is 45° north of west .

Also , for A-B :

$A-B=-63i-63j$

$|A-B|=63\sqrt{2}$

Direction of A-B is 45° south of west .

( When two vector of same magnitude which are perpendicular to each other are added or subtracted the resultant is always 45° from each of them)

Hence , this is the required solution .

4 0

3 years ago

An electron in a television tube is accelerated uniformly from rest to a speed of 8.4\times 10^7~\text{m/s}8.4×10 7 m/s over

stich3 [128]

Answer:

P=3.42×10^-6 J/s

Explanation:

From the kinematics of motion with constant acceleration we know that :

vf^2=vi^2+2*a(xf-xi)

Where :

• vf , vi, are the the final and the initial velocity of the electron

• a is the acceleration of the electron

• xf , xi are the final and the initial position of the electron .

Strategy for solving the problem : at first from the given information we calculate the acceleration of the electron.

Givens: vf = 8.4 x 10^7 m/s , vi, = 0 m/s , xf = 0.025 m and xi = 0 m

vf^2 =vi^2+2*a(xf-xi)

vf^2-vi^2=2*a(xf-xi)

2*a(xf-xi)= vf^2-vi^2

a = (vf^2-vi^2)/2(xf-xi)

Pluging known information to get :

a = (vf^2-vi^2)/2(xf-xi)

= 1.411 × 10^17

From the acceleration and the previous Eq. we can calculate the final velocity of the electron but a new position xf = 0.01 m

so,

vf^2 =vi^2+2*a(xf-xi)

vf^2 =5.312× 10^7

From the following Eq. we can calculate the time elapsed in this motion .

xf =xi+vi*t+1/2*a*t

t=√2(xf-xi)/a

t=3.765×10^-10 s

now we can use the power P Eq.

P=W/Δt => ΔK/Δt

Where: the work done W change the kinetic energy K of the electron ,

ΔK=Kf-Ki=>1/2*m*vf^2-1/2*m*vi^2

P=1/2*m*vf^2-1/2*m*vi^2/Δt

P=3.42×10^-6 J/s

6 0

3 years ago

If a person weighs 818 N on earth and 5320 N on the surface of a nearby planet, what is the acceleration due to gravity on that

alexandr402 [8]

Answer:

$g'=63.74\ m/s^2$

Explanation:

It is given that,

Weight of the person on Earth, W = 818 N

Weight of a person is given by the following formula as :

$W=mg$

g is the acceleration due to gravity on earth

$m=\dfrac{W}{g}$

$m=\dfrac{818\ N}{9.8\ m/s^2}$

m = 83.46 kg

The mass of an object is same everywhere. It does not depend on the location.

Let W' is the weight of the person on the surface of a nearby planet, W' = 5320 N

g' is the acceleration due to gravity on that planet. So,

$g'=\dfrac{W'}{m}$

$g'=\dfrac{5320\ N}{83.46\ kg}$

$g'=63.74\ m/s^2$

So, the acceleration due to gravity on that planet is $63.74\ m/s^2$ . Hence, this is the required solution.

6 0

3 years ago

A rock thrown with speed 8.50 m/s and launch angle 30.0 ∘ (above the horizontal) travels a horizontal distance of d = 19.0 m bef

frez [133]

Draw a diagram to illustrate the problem as shown below.

The vertical component of the launch velocity is
v = (8.5 m/s)*sin30° = 4.25 m/s
The horizontal component of the launch velocity is
8.5*cos30° = 7.361 m/s

Assume that aerodynamic resistance may be ignored.
Because the horizontal distance traveled is 19 m, the time of travel is
t = 19/7.361 = 2.581 s

The downward vertical travel is modeled by
h = (-4.25 m/s)*(2.581 s) + 0.5*(9.8 m/s²)*(2.581 s)²
= 21.675 m

Answer: The height is 21.7 m (nearest tenth)

4 0

3 years ago