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SpyIntel [72]
3 years ago
8

HELP ASAP!! WILL MARK BRAINLIEST!! WRITE IN YOUR OWN WORDS!! What is something that you would like to see a physicist develop in

the future?
Physics
1 answer:
Schach [20]3 years ago
8 0

Answer:

Phones as sunglasses with a mic. I put on my glasses and I say what's the weather today, The sunglasses will tell me the weather and can be charged just like phones

Explanation:

You might be interested in
Need help with dot product
slavikrds [6]

\textbf{A}\cdot\textbf{B} = 11.5

Explanation:

The dot product between two vectors \textbf{A} and \textbf{B} is defined as

\textbf{A}\cdot\textbf{B} = AB\cos{\theta}

where A and B are the magnitudes of the vectors \textbf{A} and \textbf{B}, respectively and \theta is the angle between the two. Since A = 3, B = 5 and \theta = 40°, the dot product \textbf{A}\cdot\textbf{B} is

\textbf{A}\cdot\textbf{B} = (3)(5)(0.766) = 11.5

5 0
3 years ago
4.- Una vagoneta de 1000 kg de peso parte del reposo en el punto 1 y desciende, sin rozamiento, por la vía indicada en la figura
Akimi4 [234]

Answer:

A) 49,050 N

B) 16 m

Explanation:

Question:

El dibujo de la pregunta se obtiene de un documento titulado "TRABAJO DIVERSO Y ENERGÍA" que se encuentra en línea y se presenta aquí.

La masa dada del vagón, m = 1,000 kg

La altura del punto en el que descansa el vagón, punto 1, h₁ = 12 m

A) El radio en el punto 2, el punto más bajo, R = 6 m

La fuerza, 'N', que la vía ejerce sobre el vagón en el punto 1 viene dada por la siguiente relación;

N = El peso del vagón + La fuerza de movimiento del vagón

∴ N = m × g + m × a

Dónde;

g = La aceleración debida a la gravedad ≈ 9,81 m / s²

a = La aceleración del vagón

Observamos que para el movimiento circular, la fuerza de movimiento del vagón, m × a = La fuerza centrípeta que actúa sobre el vagón = m × v² / R

∴ m × a = m × v² / R

Dónde;

v² = La velocidad del vagón en el punto 2 = 2 · g · h₁

Por lo tanto;

N = m × g + m × a = m × g + m × v² / R = m × g + m × 2 · g · h₁ / R

∴ N = 1000 × 9,81 + 1000 × 2 × 9,81 × 12/6 = 49,050

La fuerza que ejerce el vagón en el punto 2, N = 49,050 N

B) En el punto 3, tenemos;

N = m · g - m · a₃

La fuerza centrípeta en el punto 3, m · a₃ = m · v₃² / R₃

∴ La altura en el punto 3, h₃ = 4 m

El cuadrado de la velocidad en el punto 3, v₃² = 2 · g · (h₁ - h₃)

Para que el vagón esté seguro en el punto 3, la fuerza de la vía sobre el vagón, N = 0 para que el vagón permanezca en la vía actuando

Por lo tanto;

N = m · g - m · a₃ = 0

m · g = m · a₃ = m · v₃² / R₃ = m · (2 ​​· g · (h₁ - h₃)) / R₃

∴ R₃ = (2 · g · (h₁ - h₃)) / g = (2 · (h₁ - h₃)) = 2 × (12 - 4) = 16

El radio de curvatura en el punto 3 para que el punto sea seguro es R₃ = 16 m.

5 0
3 years ago
What is the difference between uniform and non-uniform magnetic field lines?
timurjin [86]
Uniform magnetic lines as the name suggests are composed of lines that point to the same direction and of the same magnitude. This means all of these lines exert the same force. Non-uniform relates to different directions of the lines and a varied magnitude.
8 0
3 years ago
The Earth's biosphere is consists of
Firdavs [7]

Answer:

The part of the earth and its atmosphere in which living organisms exist or that is capable of supporting life.

Explanation:

It’s in the definition of biosphere. “Bio” means “life,” so “biosphere” is that portion of the Earth capable of supporting life.

8 0
3 years ago
Consider a semi-spherical bowl machined out of wood (A drawing of a semi-spherical bowl can be found in UNIT- 4 Introduction, so
Mrrafil [7]

Answer:

The total surface are of the bowl is given by: 0.0532*pi m² (approximately 0.166533 m²)

Explanation:

The total surface area of the semi-spherical bowl can be decomposed in three different sections: 1) an outer semi-sphere of radius 12 cm, 2) an inner semi-sphere of radius 10 cm, and 3) the edge, which is a 2-dimensional ring with internal radius of 10 cm and external radius of 12 cm. We will compute the areas independently and then sum them all.

a) Outer semi-sphere:

A1 = 2*pi*r² = 2*pi*(12 cm)² = 288*pi cm² = 904.78 cm²

b) Inner semi-sphere:

A2 = 2*pi*(10 cm)² = 200*pi cm² = 628.32 cm²

c) Edge (Ring):

A3 = pi*(r1² - r2²) = pi*((12 cm)²-(10 cm)²) = pi*(144-100) cm² = 44*pi cm² = 138.23 cm²

Therefore, the total surface area of the bowl is given by:

A = A1 + A2 + A3 = 288*pi cm² + 200*pi cm² + 44*pi cm² = 532*pi cm² (approximately 1665.33 cm²)

Changing units to m², as required in the problem, we get:

A = 532*pi cm² * (1 m² / 10, 000 cm²) = 0.0532*pi m² (approximately 0.166533 m²)

5 0
3 years ago
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