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2 years ago

Scientific fact can never change 1)True 2)False

Physics

2 answers:

AlexFokin [52]2 years ago

8 0

Answer:

True.

Explanation:

A scientific fact is a fact because it has been proven correct, and therefore cannot change.

Ksivusya [100]2 years ago

4 0

Answer: 2)False

Explanation:

The scientific fact may include the statement which is based on direct observation of a scientific phenomena, scientific event or scientific process occurring in the natural world. The scientific fact is the output of what scientists believe and understand. The knowledge of scientific fact can change with the passage of time due to implementation of new methodologies, technologies and innovations.

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You throw a ball downward from a window at a speed of 2.5 m/s. how fast will it be moving when it hits the sidewalk 2.1 m below?

Otrada [13]

List out all the variables that you do know;

acceleration=-9.8 ms⁻¹ (this remains constant on Earth)
Final velocity=?
Displacement (s)= -2.1 m
Initial Velocity(u)=2.5 ms⁻¹

v²=u²+2as
v²=(2.5)²+2(-9.8)(-2.1)
v²=47.41
v=√47.41
v=6.88549 ≈ 6.9 ms⁻¹

Hope I helped :)

4 0

3 years ago

In a closed system, the loss of momentum of one object ________ the gain in momentum of another object.

Virty [35]

In a closed system, the loss of momentum of one object is same as________ the gain in momentum of another object

according to law of conservation of momentum, total momentum before and after collision in a closed system in absence of any net external force, remains conserved . that is

total momentum before collision = total momentum after collision

P₁ + P₂ = P'₁ + P'₂

where P₁ and P₂ are momentum before collision for object 1 and object 2 respectively.

P'₁ - P₁ = - (P'₂ - P₂)

so clearly gain in momentum of one object is same as the loss of momentum of other object

8 0

2 years ago

A balloonist happens to drop his metal ballpoint pen from his balloon as he is taking notes on his flight. Because his pen has a

Lerok [7]

So this is dealing with the conservation of energy. So you set kinetic energy equal to potential energy, so it looks like this:

1/2mv^2=mgh. The m's cancel out, so it is 1/2v^2=gh.

To find out what the height h is, divide g on both sides, so...

h=0.5v^2/g. v=22m/s, g=9.81m/s^2, so h=(0.5)(22^2)/(9.81)=24.67m

5 0

3 years ago

A 12-pack of Omni-Cola (mass 4.30 kg) is initially at rest ona horizontal floor. It is then pushed in a straight line for1.20 m

erastovalidia [21]

Answer:

a) W = 46.8 J and b) v = 3.84 m/s

Explanation:

The energy work theorem states that the work done on the system is equal to the variation of the kinetic energy

W = ΔK = $k_{f}$ -K₀

a) work is the scalar product of force by distance

W = F . d

Bold indicates vectors. In this case the dog applies a force in the direction of the displacement, so the angle between the force and the displacement is zero, therefore, the scalar product is reduced to the ordinary product.

W = F d cos θ

W = 39.0 1.20 cos 0

W = 46.8 J

b) zero initial kinetic language because the package is stopped

W - $W_{fr}$ = $k_{f}$ -K₀

W - fr d= ½ m v² - 0

W - μ N d = ½ m v

on the horizontal surface using Newton's second law

N-W = 0

N = W = mg

W - μ mg d = ½ m v

v² = (W -μ mg d) 2/m

v = √(W -μ mg d) 2/m

v = √[(46.8 - 0.30 4.30 9.8 1.20) 2/4.3 ]

v = √(31.63 2/4.3)

v = 3.84 m/s

8 0

2 years ago

An adiabatic nozzle has an inlet area of 1 m^2 and an outlet area of 0.25 m^2. Water enters the nozzle at a rate of 5 m^3/s and

svlad2 [7]

Answer:

v=20m/S

p=-37.5kPa

Explanation:

Hello! This exercise should be resolved in the next two steps

1. Using the continuity equation that indicates that the flow entering the nozzle must be the same as the output, remember that the flow equation consists in multiplying the area by the speed

Q=VA

for he exitt

Q=flow=5m^3/s

A=area=0.25m^2

V=Speed

solving for V

$V=\frac{Q}{A} \\V=\frac{5}{0.25} =20m/s$

velocity at the exit=20m/s

for entry

$V=\frac{5}{1} =5m/s$

To find the pressure we use the Bernoulli equation that states that the flow energy is conserved.

$\frac{P1}{\alpha } +\frac{v1^2}{2g} =\frac{P2} {\alpha } +\frac{v2^2}{2g}$

where

P=presure

α=9.810KN/m^3 specific weight for water

V=speed

g=gravity

solving for P1

$(\frac{p1}{\alpha } +\frac{V1^2-V2^2}{2g})\alpha =p2\\(\frac{150}{9.81 } +\frac{5^2-20^2}{2(9.81)})9.81 =p2\\P2=-37.5kPa$

the pressure at exit is -37.5kPa

7 0

3 years ago