Jerry's boat capsized last Saturday, he had been fishing with a friend. His friend made it to shore and went
for help, but it was too late. The boat was pulled out of the river and taken to the police station, Jerry's body
was found floating in the missouri river the next day. The police assumed after speaking with the friend, that
Jerry drowned, but a quick look at the body by the ME and the manner of death had to be listed as pending
investigation. The ME found petechial hemorrhaging and there was no water in his lungs.
What do you think actually happened? Provide evidence and reasoning for your claim
HELPPP

Answer:

(B)  1.23 x 10⁴ J

Explanation:

Given;

radius of the sphere, r = 7.0 m

diameter of the sphere, d = 2r = 14.0 m

mass of the person sitting on the sphere, m = 90.0 kg

The gravitational potential energy of the person is given by;

P.E = mgh

where;

g is acceleration due to gravity = 9.8 m/s²

h is the height above the ground level = d = 14.0 m

P.E = mgh

P.E = (90)(9.8)(14)

P.E = 12348 J

P.E = 1.2348 x 10⁴ J

Therefore, the gravitational potential energy of the person is 1.2348 x 10⁴ J

Answer:

W = 3934 N ,     m = 401.43 kg

Explanation:

This problem can be solved using the rotational equilibrium relation, where we place the reference system at the pivot point and assume that the counterclockwise turns are positive.

     ∑τ = 0

     F 7 - W 1 = 0

    W = F 7/1

     W = 562 7

     W = 3934 N

W = mg

     m = W / g

     m = 3934 / 9.8

     m = 401.43 kg

Answer:

first thing first they really have to concern about this,

and second one is people should leave those hazardous places

Answer:

-Emergency evacuation roadways could be designed.

-An Evacuation plan could be organized.

-Warning Sirens could be designed.

Explanation:

Those 3 are the answer.

Answer:

pretty sure its kinetic

Explanation:

Answer:

Kinetic

Explanation:

The stored chemical potential energy of a battery converts to electrical kinetic energy to transport electricity to a light bulb, which radiates thermal kinetic energy.

Answer:

total sum of all numbers/ number of items in the set

Explanation:

Answer:

45.9m

Explanation:

Given parameters:

Final velocity  = 30m/s

Initial velocity  = 0m/s

Unknown:

Height of fall  = ?

Solution:

  The motion equation to solve this problem is given below;

            V² = U² + 2gH

            V²  = 0 + (2 x 9.8 x H)

           30²  = 19.6H

              H = [tex]\frac{900}{19.6}[/tex]   = 45.9m

Answer:

20.8 ft³

Explanation:

The following data were obtained from the question:

Initial pressure (P1) = 80 psi

Initial volume (V1) = 13 ft³

Final pressure (P2) = 50 psi

Final volume (V2) =?

The new volume of the gas can be obtained by using the Boyle's law equation as shown below:

P1V1 = P2V2

80 × 13 = 50 × V2

1040 = 50 × V2

Divide both side by 50

V2 = 1040 / 50

V2 = 20.8 ft³

Thus, the volume of the gas at a pressure of 50 psi is 20.8 ft³

Answer:

Explanation:

a ) The angle required = angle of repose = θ

Tanθ = .81

θ = 39⁰

b ) when angle of incline θ = 44

Net force on the block = mg sinθ - μ mg cosθ where μ is coefficient of kinetic friction

acceleration = gsinθ - μ g cosθ

= 9.8 ( sin44 - μ cos44 )

= 9.8 ( .695 - .69 x .72 )

= 9.8 ( .695 - .497 )

= 1.94 m /s²

Answer:

i think its d

Explanation:

i took this quiz and got 100 percent

Answer:

Y = 78.13 x 10⁹ Pa = 78.13 GPa

Explanation:

First we will find the centripetal force acting on the wire as follows:

F = mv²/r

where,

F = Force = ?

m = mass of rock = 0.34 kg

v = speed = 19 m/s

r = length of wire

Therefore,

F = (0.34)(19)²/r

F = 122.74/r  

now, we find cross-sectional area of wire:

A = πd²/4

where,

A = Area = ?

d = diameter of wire = 1 mm = 0.001 m

Therefore,

A = π(0.001)²/4

A = 7.85 x 10⁻⁷ m²

Now, we calculate the stress on wire:

Stress = F/A

Stress = (122.74/r)/(7.85 x 10⁻⁷)

Stress = 1.56 x 10⁸/r

Now, we calculate strain:

Strain = Δr/r

where,

Δr = stretch in length = 2 mm = 0.002 m

Therefore,

Strain = 0.002/r

now, for Young's modulus (Y):

Y = Stress/Strain

Y = (1.56 x 10⁸/r)/(0.002/r)

Y = 78.13 x 10⁹ Pa = 78.13 GPa

Answer:

Explanation:

the formula for calculating the linear velocity is expressed as;

v = wr

w is the angular velocity

r is the radius

r = d/2 = 18/2

r = 9 m

Given that 1rev/min = 0.10472rad/s

4.9 rev/min = x

x = 4.9 * 0.10472

x = 0.513128 rad/s

Substitute into the given formula;

v = 9 * 0.513128

V = 4.61m/s

Hence the velocity of the child in m/s is 4.6m/s

Answer:

T' = o.707T

Explanation:

[tex]T = 1/2\pi \sqrt{k/m} \\\\m'=2m then\\T' = 1/2\pi \sqrt{k/m'}\\ = 1/2\pi \sqrt{k/2m}\\T' = \frac{(1/2\pi \sqrt{k/m}\\)}{1.4} T' = \frac{T}{1.4}\\T' = 0.707T[/tex]

Answer: Calendar

Explanation: I'm pretty sure it is calendar.

Answer:

Yes

Explanation:

Answer:

Explanation:

Remember that all solids, liquids, and gases are made up of atoms, molecules, and / or ions. ( Also remember that the answer "maybe" most likely won't be the answer in most quizzes. )

Which answers your question, gases are made up of particles.

Hope this helps! <3

Explanation:

We need to find the index of refraction of a material for which the wavelength of light is 0.671 times its value in a vacuum.

The wavelength in any medium is given by :

[tex]\lambda_m=\dfrac{\lambda}{n}[/tex]

Where, [tex]\lambda[/tex] is wavelength in vacuum and n is refractive index of the medium.

[tex]n=\dfrac{\lambda}{\lambda_m}\\\\n=\dfrac{\lambda}{0.671\lambda}\\\\=\dfrac{1}{0.671}\\\\=1.49[/tex]

So, the medium can be PMMA (acrylic, plexiglas, lucite, perspex) which have a refractive index of 1.49.

The index of refraction of the material for which the wavelength of light is 0.671 times, the value of it in a vacuum is 1.49.

The index of refraction of a material is the ratio of wavelength of the wave in vacuum to the wavelength of the wave in the given medium.

The index of refraction is inversely proportional to the wavelength. It can be find out using the following formula.

[tex]n=\dfrac{\lambda}{\lambda_m}[/tex]

Here, (λ) is the wavelength in vacuum and (λm) is the wavelength in a medium.

The index of refraction of a material for which the wavelength of light is 0.671 times its value in a vacuum (0.671×λ).

Plug in the values in the above formula,

[tex]n=\dfrac{\lambda}{0.671\times\lambda}\\n=1.49[/tex]

Thus, the index of refraction of the material for which the wavelength of light is 0.671 times, the value of it in a vacuum is 1.49.

Learn more about the index of refraction here;

https://brainly.com/question/10729741

Answer:

The correct option is  D

Explanation:

From the question we are told that

  The maximum electric field strength is  [tex]E = 1.0 *10^{11} \ V/m[/tex]

   The  area is  [tex]A = 5.0 \ mm^2 = 5.0 *10^{-6} \ m^2[/tex]

Generally the force the laser applies is mathematically represented as

       [tex]F = \epsilon_o * E ^2 * A[/tex]

Here  [tex]\epsilon_o = 8.85*10^{-12} C/(V \cdot m)[/tex]

      [tex]F = 8.85*10^{-12} * (1.0 *10^{11}) ^2 * 5.00*10^{-6 }[/tex]

=>   [tex]F = 4.43 *10^{5} \ N[/tex]

Answer:

61 degrees, I just did the test.

Explanation:

Answer: 61 degrees

Explanation:

I just did the question and got it right

Answer:

a

 [tex]y_1 = 0.004589 \ m[/tex]

b

 [tex]y_2 =0.009179 \ m[/tex]

Explanation:

From the question we are told that

   The wavelength of the red light is  [tex]\lambda _r = 633 \ nm = 633 *10^{-9} \ m[/tex]

    The width of the slit is  [tex]d = 0.40 mm = 0.40 *10^{-3} \ m[/tex]

    The distance of the screen from the point of diffraction is [tex]D = 2.9 \ m[/tex]

Generally the width of the central bright fringe is mathematically represented as

       [tex]y_1 = \frac{\lambda * D}{d}[/tex]

=>    [tex]y_1 = \frac{633 *10^{-9} * 2.90 }{0.40 *10^{-3}}[/tex]

=>    [tex]y_1 = 0.004589 \ m[/tex]

Generally the width of the first bright fringe on either side of the central one is mathematically represented as

     [tex]y_2 = 2 * y_1[/tex]

=>   [tex]y_2 = 2 * 0.004589[/tex]

=>   [tex]y_2 =0.009179 \ m[/tex]

Explanation:

u = 0m/s

v = 30m/s

a = g = 10m/s²

s = H = ?

using the formula,

v² = u² + 2as

30² = 0² + 2(10)(H)

900 = 20H

H = 900/20

H = 45m

Answer:

the answer is

45m - height

Answer:

2 m/s

Explanation:

The total time = 1 hour

The vertical displacement = 1 - 1

Vertical displacement = 0

Horizontal displacement = 4 - 2

Horizontal displacement = 2

Total displacement = sqrt (2^2 - 0^2)

Displacement - 2

Average velocity is displacement/time

= 2x1

=  2 m/s

The average velocity is 2 metres per second.

Answer:

what i and confused

Explanation:

Answer:

nerd

Explanation:

I H4+3 U

Answer:

If a player is fouled while shooting a three-point shot and makes it anyway, he is awarded one free throw. Thus, he could score four points on the play. Inbounds- If fouled while not shooting, the ball is given to the team the foul was committed upon.

Answer:

This question appear incomplete

Explanation:

This question appear incomplete. However, when capacitors are connected in series, the total capacitance is usually less than the capacitance of individual capacitor. This is because the total capacitance can be calculated as 1 ÷ 1/C₁ + 1/C₂ +1/C₃...

The formula above does not refer to the charge; this is because capacitors connected in series have the same charge (regardless of the capacitance). Also, the largest potential difference occur in the capacitor with the smallest/lowest capacitance.

Answer:

Speed of the spacecraft right before the collision: [tex]\displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}[/tex].

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

Let [tex]m[/tex] denote the mass of this spacecraft. At a distance of [tex]R[/tex] from the center of the earth (with mass [tex]M_\text{e}[/tex]), the gravitational potential energy ([tex]\mathrm{GPE}[/tex]) of this spacecraft would be:

[tex]\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}[/tex].

Initially, [tex]R[/tex] (the denominator of this fraction) is infinitely large. Therefore, the initial value of [tex]\mathrm{GPE}[/tex] will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy ([tex]\rm KE[/tex]) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance [tex]R[/tex] between the spacecraft and the center of the earth would be approximately equal to [tex]R_\text{e}[/tex], the radius of the earth.

The [tex]\mathrm{GPE}[/tex] of the spacecraft at that moment would be:

[tex]\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}[/tex].

Subtract this value from zero to find the loss in the [tex]\rm GPE[/tex] of this spacecraft:

[tex]\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}[/tex]

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the [tex]\rm GPE[/tex] of this spacecraft would be equal to the size of the gain in its [tex]\rm KE[/tex].

Therefore, right before collision, the [tex]\rm KE[/tex] of this spacecraft would be:

[tex]\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}[/tex].

On the other hand, let [tex]v[/tex] denote the speed of this spacecraft. The following equation that relates [tex]v\![/tex] and [tex]m[/tex] to [tex]\rm KE[/tex]:

[tex]\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2[/tex].

Rearrange this equation to find an equation for [tex]v[/tex]:

[tex]\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}[/tex].

It is already found that right before the collision, [tex]\displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}[/tex]. Make use of this equation to find [tex]v[/tex] at that moment:

[tex]\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}[/tex].

Answer:

The distance traveled by the particle before it stops is 41.06 m.

Explanation:

We can find the distance traveled by the particle using the following equation:

[tex] v_{f}^{2} = v_{0}^{2} + 2ad [/tex]

Where:

[tex] v_{f}[/tex]: is the final velocity = 0 (when it stops)

[tex] v_{0}[/tex]: is the initial velocity = 30 m/s

a: is the acceleration = -4t m/s²

t: is the time

d: is the distance

First, we need to calculate the time:

[tex] v_{f} = v_{0} + at [/tex]

[tex] 0 = 30 m/s + (-4t m/s^{2})t [/tex]

[tex]0 = 30 m/s - 4t^{2} m/s^{3}[/tex]    

[tex]t = 2.74 s[/tex]

Now, the acceleration is:

[tex] a = -4t = -10.96 m/s^{2} [/tex]                    

Hence, the distance is:          

[tex] d = \frac{v_{f}^{2} - v_{0}^{2}}{2a} = \frac{-(30 m/s)^{2}}{2*(-10.96 m/s^{2})} = 41.06 m [/tex]      

Therefore, the distance traveled by the particle before it stops is 41.06 m.

I hope it helps you!