An object traveling around another object in space is in

According to the concept of length contraction, what happens to the length of an object as it approaches the speed of light and

lord [1]

When an object moves its length contracts in the direction of motion. The faster it moves the shorter it gets in the direction of motion.
The object in this question moves and then stops moving. So it's length first contracts and then expands to its original length when the motion stops.
The speed doesn't have to be anywhere near the speed of light. When the object moves its length contracts no matter how fast or slow it's moving.

8 0

4 years ago

A 410-kg piano is being unloaded from a truck by rolling it down a ramp inclined at 25°. There is negligible friction and the ra

SpyIntel [72]

Answer:BOONK GANG

Explanation:

AHAHAHA I USE CALC LIKE IN CLASS AHAHA

5 0

4 years ago

Un coche de carreras de Fórmula 1 (véase Figura 2.30) acelera desde el reposo a razón de 18 m.s-2 . Suma que se mueve en línea r

Lynna [10]

Answer:

I will answer in English.

Ok, we know that the acceleration is a = 18m/s^2, and we have that the initial velocity and position are both zero. (because it starts at rest)

then we have:

a(t) = 18m/s^2

for the velocity, we integrate over time (because the initial velocity is equal to zero we do not have any integration constant)

v(t) = (18m/s^2)*t

for the position we integrate again over time, and again, we do not have any integration constant

p(t) = (1/2)(18m/s^2)*t^2 = (9m/s^2)*t^2

a) The speed at t= 3s can be found by replacing t = 3s in the velocity equation.

v(3s) = (18m/s^2)*3s = 54m/s

b) the distance traveled by this time can be found by replacing t = 3s in the position equation.

p(3s) =(9m/s^2)*(3s)^2 = 81 m

c) first, we need to find what is the time when the position is equal to 200m.

p(t) = 200m = (9m/s^2)*t^2

√(200/9) s = t = 4.7s

Now we replace that time in the velocity equation and we get:

v (4.7s) = (18m/s^2)*4.7s = 84.6m/s

d) ok, to do this we know that.

1 hour has: 60*60 = 3600 seconds.

then we have the transformation k = 1h/3600s

1 km has 1000 meters.

then we have the transformation c = 1km/1000m

so we have that:

84.6m/s = 84.6m/s*(c/k) = 84.6*(3600/1000)km/h = 304.56 km/h

6 0

3 years ago

PLEASE HELP WITH THIS QUESTION WILL GIVE BRAINLIST TO BEST ANSWER

lys-0071 [83]

The net force is 12 N to the left.

4 0

3 years ago

Read 2 more answers

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