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3 years ago

Why do you have to use a referance point to determine whether or not an object is in motion

Physics

2 answers:

PSYCHO15rus [73]3 years ago

5 0

The answer is because the reference points allows you to determine whether or not the position changes.

SOVA2 [1]3 years ago

4 0

Because there's no such thing as "really" moving.
ALL motion is always relative to something.

Here's an example:
You're sitting in a comfy cushy seat, reading a book and listening
to your .mp3 player, and you're getting drowsy. It's so warm and
comfortable, your eyes are getting so heavy, finally the book slips
out of your hand, falls into your lap, and you are fast asleep.

-- Relative to you, the book is not moving at all.
-- Relative to the seat, you are not moving at all.
-- Relative to the wall and the window, the seat is not moving at all.
-- But your seat is in a passenger airliner. Relative to people on the
    ground, you are moving past them at almost 500 miles per hour !
-- Relative to the center of the Earth, the people on the ground are moving
   in a circle at more than 700 miles per hour.
-- Relative to the center of the Sun, the Earth and everything on it are moving
   in a circle at about 66,700 miles per hour !

How fast are they REALLY moving ?
There's no such thing.
It all depends on what reference you're using.

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A street light is on top of a 9 foot pole. Joe, who is 3 feet tall, walks away from the pole at a rate of 4 feet per second. At

Gekata [30.6K]

Answer:2 ft/s

Explanation:

Given

Length of Pole is 9 ft

Length of Joe is 3 ft

Joe walks away from Pole at the rate 4 ft/s

Let Joe is x m away from Pole so its shadow length is x'

From Similar triangle concept

$\frac{x'}{x+x'}=\frac{3}{9}$

3x'=x+x'

x=2x'

and it is given $\frac{\mathrm{d} x}{\mathrm{d} t}=4 ft/s$

Differentiating

$\frac{\mathrm{d} x}{\mathrm{d} t}=2\frac{\mathrm{d} x'}{\mathrm{d} t}$

$4=2\times \frac{\mathrm{d} x'}{\mathrm{d} t}$

$\frac{\mathrm{d} x'}{\mathrm{d} t}=2 ft/s$

6 0

3 years ago

Pleaseeeee help me with b, c, and d. There are no angles.

taurus [48]

Answer:

a. 150 J

b. 150 J

c. 0 J

d. 0 J

Explanation:

The given parameters are;

The horizontal force with which the man pulls the canister, F = 50 N

The distance he moves the vacuum cleaner, d = 3.0 m

a. Work done, W = Force applied, F × Distance moved by the force, d

Therefore, for the work done by the 50 N force on the canister, we have;

W = 50 N × 3.0 m = 150 N·m = 150 J

b. Given that he pulls the canister at a constant speed, we have;

The acceleration of the canister, a = 0 m/s²

Therefore, the net force on the canister, $F_{NET}$ = F - $F_{Friction}$ = m × a

Where;

m = The mass of the canister

a = The acceleration of the canister

F = The applied force = 50 N

$F_{Friction}$ = The force of friction

∴ $F_{NET}$ = m × a = m × 0 m/s² = 0 N

Therefore;

$F_{NET}$ = F - $F_{Friction}$ = 0 N

From which we have;

F = $F_{Friction}$ = 50 N (The applied force, F is equal to the force of friction,

The work done by friction = The force of friction × The distance in which the force of friction acts

∴ The work done by friction = $F_{Friction}$ × d - 50 N × 3.0 m = 150 J

The work done by friction = 150 J

c. The normal force, N acts perpendicular to the force of friction

The distance the canister moves in the perpendicular direction, $d_p$ = 0 m

∴ The work done by the normal direction = N × $d_p$ = N × 0 m = 0 J

The work done by the normal direction = 0 J

d. The vacuum weight, W, acts on the same line as the normal force but in the opposite direction to the normal force, N

Therefore, the weight, W, acts perpendicular to the line of motion of canister

The distance the canister moves in the direction of the weight, $d_{wieght}$ = 0 m

Therefore, the work done by the weight = W × $d_{wieght}$ = W × 0 m = 0 J

The work done by the weight = 0 J

7 0

3 years ago

a slender rod of mass m and length l is released from rest in a horizontal position. what is the rod's angular velocity when it

Makovka662 [10]

Answer:

M g H / 2 = M g L / 2 initial potential energy of rod

I ω^2 / 2 = 1/3 M L^2 * ω^2 / 2 kinetic energy attained by rod

M g L / 2 = 1/3 M L^2 * ω^2 / 2

g = 3 L ω^2

ω = (g / (3 L))^1/2

8 0

2 years ago

What is the main difference between macroscopic kinetic energy in a motion and microscopic kinetic energy of individual molecule

labwork [276]

Answer:

Explained

Explanation:

1.The macroscopic form of kinetic energy is organized whereas the microscopic kinetic energy disorganized form of energy.

2.Observing macroscopic kinetic energy is easier than the microscopic energy.

3. All macroscopic kinetic energy can be converted to useful work whereas microscopic kinetic energy can not be.

4. Macroscopic kinetic energy is possessed by moving objects whereas microscopic objects possessed by the virtue of motion, rotation, vibrations of atoms in substance.

Wind energy is macroscopic energy as it can be harnessed to produce electricity and perform some useful work. And discussed above useful work can only be performed using macroscopic kinetic energy.

8 0

3 years ago

Holly puts a box into the trunk of her car. Later, she drives around an unbanked curve that has a radius of 48 m. The speed of t

Troyanec [42]

Answer:

The minimum coefficient of static friction for the box on the floor of the trunk is 0.54.

Explanation:

Given that,

Radius of the unbanked curve, r = 48 m

The speed of the car on the curve, v = 16 m/s

We need to find the minimum coefficient of static friction for the box on the floor of the trunk. It can be calculated by balancing the centripetal force and the force of gravity as :

$\mu mg=\dfrac{mv^2}{r}$

$\mu=\dfrac{v^2}{rg}$

$\mu=\dfrac{(16)^2}{48\times 9.8}$

$\mu=0.54$

So, the minimum coefficient of static friction for the box on the floor of the trunk is 0.54. Hence, this is the required solution.

5 0

4 years ago