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3 years ago

Why do scientists ask questions?

2 answers:

5 0

Scientists ask questions so they can solve the world. So they can solve their problems and others.

Hope this is what u need!

Sliva [168]3 years ago

3 0

I think you have it backwards.

Everybody is curious. Everybody asks questions ... especially
when they're between, say, 5 and 17 years old.

Some people don't get any answers, and they're OK with that.

Some people get answers from their teachers or their friends,
they learn to repeat the the words they heard and use those
for answers, and they never worry too much about whether the
answer is true, or where else it can take them.

Some people grow up with the burning desire to know, to understand,
to ask questions that were never asked before, and to search and find
answers that were never found before.

Those are the people who become the scientists.

You might be interested in

Two bumper cars in an amusement park ride collide elastically as one approaches the other directly fromthe rear. Car A has a mas

zubka84 [21]

Answer:

a) The velocity of car B after the collision is 4.45 m/s.

The velocity of car A after the collision is 3.65 m/s.

b) The change of momentum of car A is - 370.45 kg · m/s

The change of momentum of car B is 370.45 kg · m/s

Explanation:

Hi there!

Since the cars collide elastically, the momentum and kinetic energy of the system do not change after the collision.

The momentum of the system is calculated adding the momenta of each car:

initial momentum = final momentum

mA · vA + mB · vB = mA · vA´ + mB · vB´

Where:

mA = mass of car A

vA = initial velocity of car A

mB = mass of car B

vB = initial velocity of car B

vA´= final velocity of car A

vB´ = final velocity of car B

Let´s replace with the data we have and solve the equation for vA´:

mA · vA + mB · vB = mA · vA´ + mB · vB´

435 kg · 4.50 m/s + 495 kg · 3.70 m/s = 435 kg · vA´ + 495 kg · vB´

3789 kg · m/s = 435 kg · vA´ + 495 kg · vB´

3789 kg · m/s - 495 kg · vB´ = 435 kg · vA´

(3789 kg · m/s - 495 kg · vB´)/435 kg = vA´

Let´s write this expression without units for a bit more clarity:

vA´= (3789 - 495 vB´)/435

The kinetic energy of the system is also conserved, then, the initial kinetic energy is equal to the final kinetic energy:

initial kinetic energy of the system = final kinetic energy of the system

1/2 · mA · vA² + 1/2 · mB · vB² = 1/2 · mA · (vA´)² + 1/2 · mB · (vB´)²

Replacing with the data:

initial kinetic energy = 1/2 · 435 kg · (4.50 m/s)² + 1/2 · 495 kg · (3.70)²

initial kinetic energy = 7792.65 kg · m²/s²

7792.65 kg · m²/s² = 1/2 · 435 kg · (vA´)² + 1/2 · 495 kg · (vB´)²

multiply by 2 both sides of the equation:

15585.3 kg · m²/s² = 435 kg · (vA´)² + 495 kg · (vB´)²

Let´s replace vA´ = (3789 - 495 vB´)/435

I will omit units for clarity in the calculation:

15585.3 = 435 · (vA´)² + 495 · (vB´)²

15585.3 = 435 · (3789 - 495 vB´)²/ 435² + 495 (vB´)²

15585.3 = (3789² - 3751110 vB´ + 245025 vB²) / 435 + 495 (vB´)²

multiply both sides of the equation by 435:

6779605.5 = 3789² - 3751110 vB´ + 245025 vB² + 215325 vB´²

0 = -6779605.5 + 3789² - 3751110 vB´ + 460350 vB´²

0 = 7576915.5 - 3751110 vB´ + 460350 vB´²

Solving the quadratic equation:

vB´ = 4.45 m/s

vB´ = 3.70 m/s (the initial velocity)

a) The velocity of car B after the collision is 4.45 m/s

The velocity of car A will be teh following:

vA´= (3789 - 495 vB´)/435

vA´= (3789 - 495 (4.45 m/s))/435

vA´ = 3.65 m/s

The velocity of car A after the collision is 3.65 m/s

b) The change of momentum of each car is calculated as the difference between its final momentum and its initial momentum:

ΔpA = final momentum of car A - initial momentum of car A

ΔpA = mA · vA´ - mA · vA

ΔpA = mA (vA´ - vA)

ΔpA = 435 kg (3.648387097 m/s - 4.50 m/s) (I have used the value of vA´ without rounding).

ΔpA = - 370.45 kg · m/s

The change of momentum of car A is - 370.45 kg · m/s

ΔpB = mB (vB´ - vB)

ΔpB = 495 kg (4.448387097 m/s - 3.70 m/s) (I have used the value of vB´ without rounding).

ΔpB = 370.45 kg · m/s

The change of momentum of car B is 370.45 kg · m/s

I have used the values of the final velocities without rounding so we can notice that the change of momentum of both cars is equal but of opposite sign.

7 0

3 years ago

How can an object have high speed, but a low (or zero) velocity.

Ne4ueva [31]

Something that goes round in circles could have a high speed but zero average velocity.

6 0

3 years ago

A block of mass 0.404 0.404 kg is hung from a vertical spring and allowed to reach equilibrium at rest. As a result, the spring

VARVARA [1.3K]

To solve this problem it is necessary to apply the concepts related to the Force from Hook's law as well as the definition of the period provided by the same definition.

We know that the Force can be defined as

$F = xk \rightarrow mg = kx \Rightarrow k = \frac{mg}{x}$

Where

k = Spring constant

x = Displacement

g = Gravity

m = mass

At the same time the period of a spring mass system is defined as

$T = 2\pi \sqrt{\frac{m}{k}}$

Where

m = Mass

k = Spring constant

Our values are given as,

m = 0.404kg

x = 0.666m

Replacing to find the value of the Spring constant we have that

$k = \frac{mg}{x}$

$k = \frac{(0.404)(9.8)}{0.666}$

$k = 5.944N/m$

Now using the formula of the period we know that

$T = 2\pi \sqrt{\frac{m}{k}}$

$T = 2\pi \sqrt{\frac{0.404}{5.944}}$

$T = 1.638s$

Finally, if the oscillation was 0.359m

The maximum height will be determined by the total length of that oscillation being equivalent to

$h=2a$

$h = 2*0.359$

$h = 0.718m$

4 0

3 years ago

In each of the four situations below an object is experiencing (nearly) uniform circular motion. State what force is providing t

Stells [14]

Answer:

Explanation:

Given that a centripetal force is a form of force that gives rise or causes a body to move in a curved path.

Hence;

1. When a car is being driven around a track, it is the FORCE OF FRICTION that is acting upon the turned wheels of the vehicle, which transforms into the centripetal force required for circular motion.

2. When a ball being is swung on the end of a string, TENSION FORCE acts upon the ball, which transforms the centripetal force required for circular motion.

3. When the moon is orbiting the earth, it is the FORCE OF GRAVITY acting upon the moon, which transforms the centripetal force required for circular motion.

4. A rotating wheel on the other hand has NO centripetal force because centripetal force is pull towards the center of a motion. However the speed of the object is tangent to the circle, while the direction of the force is also perpendicular to the direction of the rotating wheel.

4 0

3 years ago

What is a load force

vladimir2022 [97]

Answer:

A push or pull exerted on an object

8 0

4 years ago