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SVEN [57.7K]
3 years ago
12

I need actual help, not just the answers. I need explanations and formulas please.

Physics
1 answer:
yaroslaw [1]3 years ago
3 0
Using newton's law of gravity on a 1kg mass near the star. G is newton's grav constant. g is acceleration of grav at starGMstar/r^2 = gstar
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Suspend a heavy weight by two pieces of rope. Tension is greatest when the ropes
bija089 [108]
Assuming that the angle is the same for both ropes, then D. is the answer.  You have to consider also if the ropes are close together or far apart and if the force to move the object is in line with the ropes or perpendicular to them.
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Which of the following is a transformer used for?
mamaluj [8]

A transformer is used to increase or decrease a voltage.  BUT ... it has to be an AC voltage, or the transformer doesn't work.

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3 years ago
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Explain how a projectile might be modified to decrease the air resistance impacting its trajectory.
Vitek1552 [10]

Answer:

Explanation:

Explain how a projectile might be modified to decrease the air resistance impacting its trajectory.

5 0
2 years ago
If a student flicks a stationary 0.1 kg ball with 5N of force for 0.1 seconds. What is its final
Alisiya [41]

5m/s

Explanation:

Given parameters:

Mass of ball = 0.1kg

Force on the ball = 5N

time taken = 0.1s

Unknown:

final speed of the ball = ?

Solution:

According to newton's second law "the net force on a body is the product of its mass and acceleration".

  Force = mass x acceleration      equation 1

Acceleration =

  V is the final velocity

  U is the initial velocity

  T is the time taken

 U = O since it is a stationary body;

      a = \frac{V}{T}

Input "a" into equation 1

  F = m x \frac{V}{T}

 5 = 0.1 x \frac{V}{0.1}

 V = 5m/s

learn more:

Newton's laws brainly.com/question/11411375

#learnwithBrainly

3 0
3 years ago
Energy is conventionally measured in Calories as well as in joules. One Calorie in nutrition is one kilocalorie, defined as 1 kc
Sergeeva-Olga [200]

Answer:

a) The student must run flight of stairs to lose 1.00 kg of fat 709.5 times.

b) Average power

P(w)= 1062.07 [w]

P(hp)=1.42 [hp]

c) This activity is highly unpractical, because the high amount of repetitions he has to due in order to lose, just 1 Kg of fat.

Explanation:

First, lets consider the required amount of work to move the mass of the student. (considering running stairs just as a vertical movement)

Work:

W= F*d= m*g*d

Where m is the mass of the student, g is gravity (9.8 m/s) and d is the total distance going up the stairs (0.15m *85steps= 12.75m )

W= F*d= m*g*d=85* 9.8*12.75=10620.75 [J]

Converting from Joules to Kcals:

\frac{10620.75}{4186} =2.537 Kcal

Now lets take into account the efficiency of the human body (20%)

2.537 ---> 20%

 x       ---> 100%

x=\frac{2.537*100}{20} =12.685

So the student is consuming 12.685 KCals each time he runs up the stairs.

Now,

1 g --> 9 Kcals

1000 g --> 9000KCals

Burning 1 g of fat, requieres 9 KCals, 1000g burns 9000KCals. So in order to burn a 1Kg of fat:

\frac{9000Kcals}{12.685Kcals} =709.5 times

He must run up the stairs 709.5 times, to burn 1 Kg of fat.

********************

For b) just converting units, taking into account the time lapse. (53103.75 is the 100% of the energy in joules, from converting 12.685Kcals to joules)

Power=\frac{Joules}{Seconds} =\frac{53103.75}{50} =1062.075 [W]\\

P(hp)=\frac{P(w)}{745.7} =\frac{1062.075}{745.7} =1.42[hp]

*****

4 0
3 years ago
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