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Kay [80]
3 years ago
7

Question 3 (1 point)

Physics
1 answer:
FrozenT [24]3 years ago
3 0

The force of friction is equal to the pushing force, and the acceleration is zero

Explanation:

In this problem, we are pushing a piano along the floor, in the horizontal direction.

There are 2 forces acting in the horizontal direction on the piano:

  • The applied force, F_a, acting forward
  • The force of friction, F_f, acting backward

Therefore, the net force in the horizontal direction is

\sum F = F_a - F_f

According to Newton's second law, the net force is equal to the product between the piano's mass (m) and its acceleration:

\sum F = ma

Combining the two equations,

F_a - F_f = ma

However, we are also told that the piano moves at constant speed, therefore the acceleration is zero:

a=0

And so,

F_a - F_f = 0\\F_f = F_a

which means that the force of friction is equal to the applied force.

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

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Energy is conventionally measured in Calories as well as in joules. One Calorie in nutrition is one kilocalorie, defined as 1 kc
IgorC [24]

Answer:

The answers to the questions are;

a. The number of times the student run the flight of stairs to lose 1.00 kg of fat is 829.23 times.

b. The average power output, in watts and horsepower, as he runs up the stairs is 158.026 watts.

c. The act of climbing the stairs is not a practical way to lose weight has to lose 1 kg of fat, the student needs to workout for about 26.49 hrs or 1.104 days.

Explanation:

To solve the question, we write out the known variables as follows

1 g of fat = 9.00kcal

Number of steps the student climbs = 95 steps

Height of each step = 0.150 m

Time it takes for the student to reach the top of the stairs = 57.5 s.

Efficiency of human muscles = 20 %

Mass of student, m = 65 kg

a. From the question, the energy expended by the student in climbing the stairs is the "work done" by the student.

The "work done" is the height climbed resulting in the gaining of gravitational potential energy P. E..

That is work done, W, =  P. E. = m·g·h

Where:

h = The total height climbed by the student

g = Acceleration due to gravity = 9.81 m/s²

Therefore;

h = Height of each step × Number of steps the student climbs =

  = 0.150 m/(step) × 95 steps = 14.25 m

Therefore, P. E. = 65 kg × 9.81 m/s² × 14.25 m = 9086.5125 kg·m²/s²

                          = 9086.5125 J

We remember that the efficiency of the muscle is 20 %

The formula for efficiency is

Efficiency = \frac{Ene rgy Out put}{Energ y In put} \times 100 %

The work produced by the muscle =  Energy Output = 9086.5125 J

Energy input is given by

\frac{Out put} {Effici ency} = 9086.5125 J/ (0.2) = 45432.5625 J

= 45.432 kJ

From the question, 1 g of fat = 9.00 kcal and

1 kcal = 4186 J

Therefore 1 g of fat can release 9.00 kcal × 4186 J = 37674 J

Therefore 1 kg of fat = 1000 g = 1000 × 37674 J = 37674 kJ

To consume the energy in 1 kg of fat the student therefore will run up the foight of stairs \frac{37674 kJ}{45.432 kJ} times to make up the 37674 kJ energy contained in 1 kg of fat

That is  \frac{37674 kJ}{45.432 kJ} =  829.23 times

b. Power is the rate of doing work

That is Power output = \frac{ WorkO utput }{Time} = \frac{9086.5125 J}{57.5 s} = 158.026 watts

c. No as the activity student will have to spend a total time of

829.23 × 57.5 s = 47680.67 s climbing up the stairs alone  and

47680.67 s = ‪13.24 Hours climbing up of which if the time to climb down is the same s climbing up, then we ave total time = 2× ‪13.24 Hours  

= 26.49 hrs = 1.104 days exercising which is not humanly possible.

3 0
3 years ago
During a takeoff run, an aircraft starts from rest and has a lift-off speed of 120 km/h. a. What minimum constant acceleration d
svp [43]

Answer:

1.984 m/s^2

Explanation:

initial velocity of air craft, u = 0 m/s

final speed of the aircraft, v = 120 km/h

Convert the speed into m/s from km/h

So, v = 120 km/h = 33.33 m/s

distance, s = 280 m  

Let a be the acceleration of the aircraft.

Use third equation of motion

v^{2}=u^{2}+2\times a\times s

33.33^{2}=0^{2}+2\times a\times 280

a = 1.984 m/s^2

Thus, the acceleration of the aircraft is 1.984 m/s^2.

7 0
4 years ago
An airplane flies eastward and always accelerates at a constant rate. At one position along its path it has a velocity of 34.3 m
Tomtit [17]

Explanation:

We'll need two equations.

v² = v₀² + 2a(x - x₀)

where v is the final velocity, v₀ is the initial velocity, a is the acceleration, x is the final position, and x₀ is the initial position.

x = x₀ + ½ (v + v₀)t

where t is time.

Given:

v = 47.5 m/s

v₀ = 34.3 m/s

x - x₀ = 40100 m

Find: a and t

(47.5)² = (34.3)² + 2a(40100)

a = 0.0135 m/s²

40100 = ½ (47.5 + 34.3)t

t = 980 s

7 0
4 years ago
If a light bulb is missing or broken in a parallel circuit, will the other light bulb right ?
Nataly [62]

No because the path the electricity needs to follow is broken. In parallel circuit,electricity has more that one path to follow.
7 0
3 years ago
A stretched rubber band has ___________ energy. a. elastic kinetic energy b. gravitational potential energy c. elastic potential
densk [106]

Answer: Option (c) is the correct answer.

Explanation:

An elastic object is defined as the object that is able to retain its shape when a force is applied on it.

For example, when we pull a rubber band then it stretches and when we withdraw the force applied on it then it retain its shape.

As we know that potential energy is the energy obtained by an object due to its position.

So, when we stretch a rubber band then it will have elastic potential energy as position of the rubber band is changing and since, it will retain it shape hence it has elastic potential energy.

Thus, we can conclude that a stretched rubber band has elastic potential energy.

3 0
4 years ago
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