1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
erastova [34]
2 years ago
9

PLEASE HELP ME PLS PLS Imagine an object held at some height above the ground. It is released and falls toward the ground. Ignor

ing air resistance or friction, which of the following must be true?
a
Before it falls, all of its energy must be potential energy
b
At the end of its fall, all of its energy must be converted to kinetic energy
c
During its fall, the combination of its kinetic and potential energy must equal the amount of potential energy with which it started
d
Total energy = Kinetic Energy + Potential Energy
e
All of the above
Physics
2 answers:
Stels [109]2 years ago
8 0

Option e is true. The total energy is the sum of all the energies present in the system. The potential energy in a system is due to its position in the system.

<h3>What is the law of conservation of energy?</h3>

According to the Law of conservation of energy. Although energy cannot be generated or destroyed, it may be transferred from one form to another.

The following statements are true;

a)All of its energy must be potential energy before it falls.

b)At the conclusion of its fall, all of its energy must be transformed to kinetic energy.

c)During its fall, the sum of its kinetic and potential energy must match the initial quantity of potential energy.

d)Total energy = Kinetic Energy + Potential Energy.

Hence, option e is correct.

To learn more about the law of conservation of energy, refer ;

brainly.com/question/2137260

#SPJ2

iragen [17]2 years ago
8 0
I think it’s E not due
You might be interested in
On the way to the moon, the Apollo astronauts reach a point where the Moon’s gravitational pull is stronger than that of Earth’s
Drupady [299]

Answer:

rm = 38280860.6[m]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2}  } \\

F_{m} =G*\frac{m_{m}*m_{a}  }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]

When we match these equations the masses cancel out as the universal gravitational constant

G*\frac{m_{e} *m_{a} }{r_{e}^{2}  } = G*\frac{m_{m} *m_{a} }{r_{m}^{2}  }\\\frac{m_{e} }{r_{e}^{2}  } = \frac{m_{m} }{r_{m}^{2}  }

To solve this equation we have to replace the first equation of related with the distances.

\frac{m_{e} }{r_{e}^{2}  } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m}  )^{2}  } = \frac{7.36*10^{22}  }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17}  \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c }  }{2*a}\\  where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) }  }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]

We work with positive value

rm = 38280860.6[m] = 38280.86[km]

6 0
3 years ago
I forgot to write also put drawings on my theory questions
11Alexandr11 [23.1K]
Hi love hope you had an amazing day! you’re beautiful!!
8 0
3 years ago
If 14:15=42:x find the value of x​
Yuliya22 [10]

Answer:

x=45

Explanation:

by taking 42 and dividing it by 14 you get 1/3, because 14 is 1/3 of 42 you can then see that the ratio is multiplied by 3. si then you can just multiply 15 by 3 to get 45

3 0
3 years ago
Read 2 more answers
Answer please need help
Basile [38]

the answer is C. It allows citizens to submit anonymous tips to the police.

8 0
3 years ago
Malcolm and Ravi raced each other. The average of their maximum speeds was 260 km/h If doubled, Malcolm's maximum speed would be
AfilCa [17]

Answer

Given,

Average speed of Malcolm and Ravi = 260 km/h

Let speed of the Malcolm be X and speed of the Ravi Y.

From the given statement

\dfrac{X+Y}{2}=260

X + Y = 520 ....(i)

2X - Y = 80  ....(ii)

Adding both the equations

3 X = 600

 X = 200 km/h

Putting value in equation (i)

Y = 520 - 200

Y = 320 Km/h

Speed of Malcolm = 200 Km/h

Speed of Ravi = 320 Km/h

8 0
3 years ago
Other questions:
  • Which of the following is a luminous object? the Moon Jupiter a comet the Sun
    7·2 answers
  • What is the mathematical relationship between voltage, resistance, and current?
    10·1 answer
  • If you push a 1000 kg box so in 2 seconds its velocity changes from 5m/s to 10 m/s how much force do you have to apply?
    7·1 answer
  • Match each measurement tool with what it measures.
    14·1 answer
  • Nora tied a string around a tennis ball, and then she swung it in a circle in front of her to demonstrate a planet orbiting the
    11·2 answers
  • Compare and contrast aerobic physical activity with anaerobic physical activity.
    12·2 answers
  • Which of the following are in the correct order from smallest to largest?
    9·1 answer
  • An object has a mass of 12 kg. On Planet A, the object weighs 117.6 N. The force of gravity on Planet A is
    12·2 answers
  • A projectile is launched with an initial speed of 40 mys from the foor of a tunnel whose height is 30 m. What angle of elevation
    10·1 answer
  • a 2.5 kg mass is hung from the 0 cm mark on a 1 kg meter stick. a mass of 0.5 kg is hung from the 100 cm mark of the meter stick
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!