Only the component of force perpendicular to the direction motion does work. true or false

Why does the fed rarely change the reserve requirement

Brut [27]

The answer is:

It can be disruptive to the whole banking system.

8 0

3 years ago

Read 2 more answers

Two balls have their centers 2.0 m apart. One ball has a mass of m1 = 7.9 kg. The other has a mass of m2 = 6.1 kg. What is the g

maks197457 [2]

Answer:

3.036×10⁻¹⁰ N

Explanation:

From newton's law of universal gravitation,

F = Gm1m2/r² .............................. Equation 1

Where F = Gravitational force between the balls, m1 = mass of the first ball, m2 = mass of the second ball, r = distance between their centers.

G = gravitational constant

Given: m1 = 7.9 kg, m2 = 6.1 kg, r = 2.0 m, G = 6.67×10⁻¹¹ Nm²/C²

Substituting into equation 1

F = 6.67×10⁻¹¹×7.9×6.1/2²

F = 321.427×10⁻¹¹/4

F = 30.36×10⁻¹¹

F = 3.036×10⁻¹⁰ N

Hence the force between the balls = 3.036×10⁻¹⁰ N

8 0

4 years ago

A small plastic sphere of mass 2.0 g is charged to −5.0 nC and held 2.5 cm above a small glass bead of mass 0.015 g resting on a

LenKa [72]

Answer:uhhh

Explanation:

Uhhhh

5 0

4 years ago

Consider an airplane with a wing area of 240 ft2 , a span of 44 ft, an Oswald efficiency of 0.75, and a zero-lift drag coefficie

Darya [45]

Answer:

Drag force=1/2Cd*Area*density*V^2

A=240 ft^2

cd=0.75

Drage force=675 lbs

6 0

4 years ago

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