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Wewaii [24]
3 years ago
10

You hear a thunder 3 seconds after you see lightning. How far away is the storm?

Physics
1 answer:
kondor19780726 [428]3 years ago
3 0
3 seconds (330 m) / second =
990 meter
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What % of an object’s mass is above the water line if the object’s density is 0.82g/ml?
Sladkaya [172]

To develop this problem we will apply the Archimedes model. As well as the definitions of Weight based on mass and acceleration. The first in turn will be considered under the relationship of Density and Volume. From the values given we have to:

\rho_w =1 g/mL \rightarrow \text{Water Density}

\rho_o = 0.82g/mL \rightarrow \text{Object density}

Since it is in equilibrium, the weight of the object will have a reaction from the water, which will cause the sum of forces between the two objects to be zero, therefore

\sum F= 0

F_w-F_o = 0

F_w = F_o

m_w g = m_og

\rho_w V_w g = \rho_o V_o g

The value of gravity is canceled because it is a constant

\frac{V_w}{V_o} = \frac{\rho_o}{\rho_w}

\frac{V_w}{V_o} = \frac{0.82}{1}

\frac{V_w}{V_o} = 0.82

The portion of the object that is submerged corresponds to 82%, while the portion that is visible, above the water level will be 18%

4 0
3 years ago
A 1 036-kg satellite orbits the Earth at a constant altitude of 98-km. (a) How much energy must be added to the system to move t
Veronika [31]

Answer:

a) The Energy added should be 484.438 MJ

b) The  Kinetic Energy change is -484.438 MJ

c) The Potential Energy change is 968.907 MJ

Explanation:

Let 'm' be the mass of the satellite , 'M'(6×10^{24} be the mass of earth , 'R'(6400 Km) be the radius of the earth , 'h' be the altitude of the satellite and 'G' (6.67×10^{-11} N/m) be the universal constant of gravitation.

We know that the orbital velocity(v) for a satellite -

v=\sqrt{\frac{Gmm}{R+h} }         [(R+h) is the distance of the satellite   from the center of the earth ]

Total Energy(E) = Kinetic Energy(KE) + Potential Energy(PE)

For initial conditions ,

h = h_{i} = 98 km = 98000 m

∴Initial Energy (E_{i})  = \frac{1}{2}mv^{2} + \frac{-GMm}{(R+h_{i} )}

Substituting v=\sqrt{\frac{GMm}{R+h_{i} } } in the above equation and simplifying we get,

E_{i} = \frac{-GMm}{2(R+h_{i}) }

Similarly for final condition,

h=h_{f} = 198km = 198000 m

∴Final Energy(E_{f}) = \frac{-GMm}{2(R+h_{f}) }

a) The energy that should be added should be the difference in the energy of initial and final states -

∴ ΔE = E_{f} - E_{i}

        = \frac{GMm}{2}(\frac{1}{R+h_{i} } - \frac{1}{R+h_{f} })

Substituting ,

M = 6 × 10^{24} kg

m = 1036 kg

G = 6.67 × 10^{-11}

R = 6400000 m

h_{i} = 98000 m

h_{f} = 198000 m

We get ,

ΔE = 484.438 MJ

b) Change in Kinetic Energy (ΔKE) = \frac{1}{2}m[v_{f} ^{2} - v_{i} ^{2}]

                                                          = \frac{GMm}{2}[\frac{1} {R+h_{f} } - \frac{1} {R+h_{i} }]

                                                          = -ΔE                                                            

                                                          = - 484.438 MJ

c)  Change in Potential Energy (ΔPE) = GMm[\frac{1}{R+h_{i} } - \frac{1}{R+h_{f} }]

                                                             = 2ΔE

                                                             = 968.907 MJ

3 0
3 years ago
PLEASE HELP ASAP
DaniilM [7]

Your answers:

Q1. I don't have information on her situation, but an EMR is a medical record that is stored digitally in a secure computer database within an organization or healthcare system. This system allows electronic storage, retrieval, and modification of medical records, and is an alternative to traditional, paper medical records.

Q2. Their food proportions, do they eat every meal, do they eat their vegetables, fruits grains, protein, etc. Check for healthy and unhealthy habits of eating and exercise.  Additional notes on medical problems relating to food concerns you may have to look further into based on her answers.

Q3. The info isn't provided,  but what I do know is that detection requires awareness of risk factors for, and symptoms and signs of, anorexia nervosa (e.g., participation in activities valuing thinness, family history of an eating disorder, amenorrhea, lanugo hair) and bulimia nervosa (e.g., unsuccessful attempts at weight loss, history of childhood sexual abuse, family history of depression, erosion of tooth enamel from vomiting, parotid gland swelling, and gastroesophageal reflux). Providers must also remain alert for disordered eating in female athletes (the female athlete triad) and disordered eating in diabetics. Treatment requires a multidisciplinary team including a primary care practitioner, nutritionist, and mental health professional. The role of the primary care practitioner is to help determine the need for hospitalization and to manage medical complications (e.g., arrhythmias, refeeding syndrome, osteoporosis, and electrolyte abnormalities such as hypokalemia).

I'm sorry I cannot answer all your questions because there's a typing limit. :)

Thank you for your patience, and I hope I helped!

Would appreciate Brainliest ;)

4 0
3 years ago
Based on molecular orbital theory, the only molecule in the list below that has unpaired electrons is ________.
xenn [34]
The answer is Radicals.
6 0
2 years ago
Residential building codes typically require the use of 12-gauge copper wire (diameter 0.205 cm) for wiring receptacles. Such ci
Alecsey [184]

Given Information:  

Current = I = 20 A

Diameter = d = 0.205 cm = 0.00205 m

Length of wire = L = 1 m

Required Information:  

Energy produced = P = ?

Answer:  

P = 2.03 J/s

Explanation:  

We know that power required in a wire is

P = I²R

and R = ρL/A

Where ρ is the resistivity of the copper wire 1.68x10⁻⁸ Ω.m

L is the length of the wire and A is the area of the cross-section and is given by

A = πr²

A = π(d/2)²

A = π(0.00205/2)²

A = 3.3x10⁻⁶ m²

R = ρL/A

R = 1.68x10⁻⁸*(1)/3.3x10⁻⁶

R = 5.09x10⁻³ Ω

P = I²R

P = (20)²*5.09x10⁻³

P = 2.03 Watts or P = 2.03 J/s

Therefore, 2.03 J/s of energy is produced in 1.00 m of 12-gauge copper wire carrying a current of 20 A

8 0
3 years ago
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