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3 years ago

Who first studied the relationship between buoyant force and the weight of a displaced fluid?

1 answer:

6 0

I'm not sure he was the 'first', but the modern "law" of that situation is now known as "<em>Archimedes</em>' Principle".

Archie himself was a Greek mathematician, physicist, engineer, inventor, and astronomer who lived from 288 to 212 BCE.

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How can people reconcile technology with environmental preservation?

MatroZZZ [7]

1. Developing renewable energy technology

- Efficient energy storage and smarter grids .
- renewable and rechargeable batteries and fuel cell

2. Saving endangered wildlife

-smart collars for endangered species and reducing human - animal conflict
-Gene sequencing for detecting and researching on deadly animal diseases.

3. Adopting a smarter lifestyle

- smart homes that promote energy saving and green - living .
- electric cars which are three times more conventional vehicles .

Hope this helps :)

5 0

3 years ago

A 3.0-kg block starts at rest at the top of a 37° incline, which is 5.0 m long. Its speed when it reaches the bottom is 2.0 m/s.

Mama L [17]

Answer: $f_{r}$ = 16.49N

Explanation: The object is placed on an inclined plane at an angle of 37° thus making it weight have two component,

$W_{x}$ = horizontal component of the weight = mgsinФ

$W_{y}$ = vertical component of weight = mgcosФ

Due to the way the object is positioned, the horizontal component of force will accelerate the object thus acting as an applied force.

by using newton's law of motion, we have that

mgsinФ - $f_{r}$ = ma

where m = mass of object=5kg

a = acceleration= unknown

Ф = angle of inclination = 37°

g = acceleration due to gravity = 9.8 $m/s^{2}$

$f_{r}$ = frictional force = unknown

we need to first get the acceleration before the frictional force which is gotten by using the equation below

$v^{2} = u^{2} + 2aS$

where v = final velocity = 2m/s

u = initial velocity = 0m/s (because the object started from rest)

a= unknown

S= distance covered = length of plane = 5m

$2^{2} = 0^{2} + 2*a*5\\\\4= 10 *a\\\\a = \frac{4}{10} \\a = 0.4m/s^{2}$

we slot in a into the equation below to get frictional force

mgsinФ - $f_{r}$ = ma

3 * 9.8 * sin 37 - $f_{r}$ = 3* 0.4

17.9633 - $f_{r}$ = 1.2

$f_{r}$ = 17.9633 - 1.2

$f_{r}$ = 16.49N

4 0

3 years ago

A girl throws a ball of mass 0.8 kg against a wall. The ball strikes the wall horizontally with a speed of 11 m/s, and it bounce

Karolina [17]

Answer:

F = 352 N

Explanation:

we know that:

F*t = ΔP

so:

F*t = M $V_f$ -M $V_i$

where F is the force excerted by the wall, t is the time, M the mass of the ball, $V_f$ the final velocity of the ball and $V_i$ the initial velocity.

Replacing values, we get:

F(0.05s) = (0.8 kg)(11m/s)-(0.8 kg)(-11m/s)

solving for F:

F = 352 N

3 0

3 years ago

Help please this is important!

bixtya [17]

Answer:

Since the ball becomes positively charged, it will repel as like charges repel.

7 0

3 years ago

A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?

KATRIN_1 [288]

Complete Question

Planet D has a semi-major axis = 60 AU and an orbital period of 18.164 days. A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?

Answer:

The value is $T_R = 11.8 \ days$