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3 years ago

Mapa conceptual y diferenciar los sistemas de unidades de medida: Internacional e inglés

Physics

2 answers:

Zigmanuir [339]3 years ago

8 0

Answer:

dude i have no clue

Explanation:

my bad

Keith_Richards [23]3 years ago

5 0

Pene enorme pene enorme pene enorme

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When jogging outside you accidently bump into a curb. Your feet stop but your body continues to move forward and you end up on t

oee [108]

Inertia I think because I've heard it around school and in science

6 0

3 years ago

Two identical masses are connected to two different flywheels that are initially stationary. Flywheel A is larger and has more m

inysia [295]

Answer:

a) True. There is dependence on the radius and moment of inertia, no data is given to calculate the moment of inertia

c) True. Information is missing to perform the calculation

Explanation:

Let's consider solving this exercise before seeing the final statements.

We use Newton's second law Rotational

τ = I α

T r = I α

T gR = I α

Alf = T R / I (1)

T = α I / R

Now let's use Newton's second law in the mass that descends

W- T = m a

a = (m g -T) / m

The two accelerations need related

a = R α

α = a / R

a = (m g - α I / R) / m

R α = g - α I /m R

α (R + I / mR) = g

α = g / R (1 + I / mR²)

We can see that the angular acceleration depends on the radius and the moments of inertia of the steering wheels, the mass is constant

Let's review the claims

a) True. There is dependence on the radius and moment of inertia, no data is given to calculate the moment of inertia

b) False. Missing data for calculation

c) True. Information is missing to perform the calculation

d) False. There is a dependency if the radius and moment of inertia increases angular acceleration decreases

4 0

3 years ago

Water flows through a horizontal pipe of varying cross-section. In the first section, the cross-sectional area is 10 cm2 and flo

Stels [109]

Answer:

(a) the flow speed of the second section is 11 m/s

(b) the pressure of the second section is 6.33 x 10⁴ Pa

Explanation:

Given;

flow rate in the first section, Q₁ = 2750 cm³/sec

area of the first cross section, A₁ = 10 cm²

pressure in the first cross section, P₁ = 1.2 x 10⁵ Pa

area of the second section, A₂ = 2.5 cm²

(a) the flow speed of the second section (V₂)

Apply continuity equation;

Q₁ = Q₂

Q₁ = A₂V₂

V₂ = Q₁ / A₂

V₂ = (2750) / (2.5)

V₂ = 1100 cm/s = 11 m/s

(b) the pressure of the second section (P₂)

Apply Bernoulli's equation;

P₁ + ¹/₂ρV₁² = P₂ + ¹/₂ρV₂²

where;

ρ is density of water = 1000 kg/m³

V₁ is the speed of water in the first section;

Q₁ = A₁V₁

V₁ = Q₁ / A₁

V₁ = (2750) / (10)

V₁ = 275 cm/s = 2.75 m/s

P₂ = P₁ + ¹/₂ρV₁² - ¹/₂ρV₂²

P₂ = P₁ + ¹/₂ρ(V₁² - V₂²)

P₂ = 1.2 x 10⁵ Pa + ¹/₂ x 1000 (2.75² - 11²)

P₂ = 1.2 x 10⁵ Pa + 500(-113.438)

P₂ = 1.2 x 10⁵ Pa - 0.567 x 10⁵ Pa

P₂ = 0.633 x 10⁵ Pa

P₂ = 6.33 x 10⁴ Pa

8 0

2 years ago

33] You have long hair, you should:

Sveta_85 [38]

Answer:

Explanation:

hair can be a safety hazard

4 0

3 years ago

Read 2 more answers

A sinusoidal transverse wave of amplitude ym = 8.4 cm and wavelength = 5.3 cm travels on a stretched cord. Find the ratio of the

Scilla [17]

Answer:

The ratio is 9.95

Solution:

As per the question:

Amplitude, $y_{m} = 8.4\ cm$

Wavelength, $\lambda = 5.3\ cm$

Now,

To calculate the ratio of the maximum particle speed to the speed of the wave:

For the maximum speed of the particle:

$v_{m} = y_{m}\times \omega$

where

$\omega = 2\pi f$ = angular speed of the particle

Thus

$v_{m} = 2\pi fy_{m}$

Now,

The wave speed is given by:

$v = f\lambda$

Now,

The ratio is given by:

$\frac{v_{m}}{v} = \frac{2\pi fy_{m}}{f\lambda}$

$\frac{v_{m}}{v} = \frac{2\pi \times 8.4}{5.3} = 9.95$

8 0

3 years ago