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3 years ago

Good science does not depend on interactions within the scientific community true or false

Physics

1 answer:

vlabodo [156]3 years ago

6 0

Answer:

I would say that is false. Science can only be perfect after at least some sort of scientific communication and interaction.

You might be interested in

Which of the following phenomena are due to the electric interaction? (Select all that apply.) surface tension in water friction

Eddi Din [679]

Answer:

Surface tension in water

Friction between tires and pavement

Dissolution of salt in water

Explanation:

Surface tension in water: It is due to the electrostatic force of attraction (cohesive force) between water molecules.

Friction between tires and pavement: It is due to the attractive force between tires and pavement.

Dissolution of salt in water: The ions of $Na ^ +$ and $Cl ^ -$ separate due to the strong attraction of water molecules.

5 0

3 years ago

Please help, and show steps. Thank you very much!

Vikentia [17]

V = 8 * 10^2 km/h = 800km/h
S= 1,8* 10^3 km = 1800km
t = ?
v = S/t
t = S/v
t = 1800km/ 800km/h
t ≈ 2,25h (135min)

6 0

3 years ago

An amplifier has a 50 watt output and a 5 watt input. what is the gain in decibels for this amplifier, rounded to the nearest de

Marrrta [24]

Gain in decibels is given by;

Gain db = 10*log (Po/Pi), where Po = Power output, Pin = Power input

Substituting;

Gain in db = 10 * log (50/5) = 10 db

6 0

3 years ago

Example 3 :

kherson [118]

The friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 × 10^8 respectively. Also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 × 10^8 respectively.

<h3>How to determine the friction factor</h3>

Using the formula

μ = viscosity = 0. 06 Pas

d = diameter = 120mm = 0. 12m

V = velocity = 1m/s and 3m/s

ρ = density = 0.9

a. Velocity = 1m/s

friction factor = 0. 52 × $\frac{0. 06}{0. 12* 1* 0. 9}$

friction factor = 0. 52 × $\frac{0. 06}{0. 108}$

friction factor = 0. 52 × 0. 55

friction factor $= 0. 289$

b. When V = 3mls

Friction factor = 0. 52 × $\frac{0. 06}{0. 12 * 3* 0. 9}$

Friction factor = 0. 52 × $\frac{0. 06}{0. 324}$

Friction factor = 0. 52 × 0. 185

Friction factor $= 0.096$

Loss When V = 1m/s

Head loss/ length = friction factor × 1/ 2g × velocity^2/ diameter

Head loss = 0. 289 × $\frac{1}{2*6. 6743 * 10^-11}$ × $\frac{1^2}{0. 120}$ × $\frac{1}{100}$

Head loss = 1. 80 × 10^8

Head loss When V = 3m/s

Head loss = $0. 096$ × $\frac{1}{1. 334 *10^-10}$ × $\frac{3^2}{0. 120}$ × $\frac{1}{100}$

Head loss = 5. 3× 10^8

Thus, the friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 ×10^8 respectively also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 ×10^8 respectively.

Learn more about friction here:

brainly.com/question/24338873

#SPJ1

4 0

1 year ago

A conveyer belt moves a 40 kg at a velocity of 2/ms what is the kinetic energy of the box while it is on the conveyor belt

CaHeK987 [17]

KE = (1/2)·(mass)·(speed)²

It doesn't matter whether the object is in a car, on a boat,
falling, on a conveyor belt, or being carried by ants.

KE = (1/2)·(40 kg)·(2 m/s)²

KE = (20 kg)·(4 m²/s²)

KE = 80 kg-m²/s²

KE = 80 Joules

5 0

3 years ago