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3 years ago

La aceleración de un proyectil que describe una trayectoria parabólica cerca de la superficie terrestre es constante o variable

Physics

1 answer:

Ann [662]3 years ago

5 0

Answer:

Es una constante porque sigue funcionando y debido a la gravedad, tú tendrías que ser tú quien la detenga si está bajando en lugar de hacia arriba.

Explanation:

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A budding magician holds a 2.24-mW laser pointer, wondering whether he could use it to keep an object floating in the air with t

Sonbull [250]

Answer:

W = 1.49 10⁻¹¹ kg

Explanation:

For this problem, let's use Newton's equation of equilibrium

F - W = 0

F = W (1)

Strength can be found from the definition of pressure

P = F / A

F = P A

The radiation pressure for a reflective surface is

P = 2 I / c)

We substitute in equation 1

2 I / c A = W

The intensity is defined by the ratio of the power between the area

I = P / A

P = I A

We substitute

2 P / c = W

W = 2 2.24 10-3 / 3 108

W = 1.49 10⁻¹¹ kg

7 0

4 years ago

What are two processes that result in rocks being broken down into smaller pieces ?

meriva

Mechanical and Chemical. (Weathering and erosion)

5 0

3 years ago

Read 2 more answers

A mass executes SHM at the end of a light spring. (a) What fraction of the total energy of the system is potential and what frac

PolarNik [594]

Answer:

Explained

Explanation:

A) The total energy of the system is defined by the energy at maximum amplitude, which we'll call A. At that point, the energy of the system is

E = 1/2×m×A^2;

since energy is conserved, this is also the total amount of energy that the system ever has.

So at x=1/2A,

the potential energy of the system is 1/8×m×A^2

which is one-fourth of the system's total energy. Therefore, the remaining three-fourths is kinetic.

B) (i) Doubling the maximum amplitude will quadruple the total energy:

$E= \frac{1}{2}m(2A)^2$

(ii) Doubling the maximum amplitude will double the maximum velocity

$\frac{1}{2}m(2A)^2= \frac{1}{2}mV^2$

(iii) Doubling the maximum amplitude will double the maximum acceleration: m×a = -k(2A)

(iv) Doubling the maximum amplitude leaves the period unchanged:

$T= 2\pi\sqrt{\frac{m}{k} }$

(neither m nor k has changed).

6 0

3 years ago

A force is acting on a 3-kg object during 5 seconds. If the object initially has a speed of 1.5m/s and after applying the force

uranmaximum [27]

Answer:

1.5 N

Explanation:

You've left us to guess what the question is. I will Assume it is what's the force?

Givens

m = 3 kg

vi = 1.5 m/s

vf = 4 m/s

t = 5 seconds

Formula

F = m * (vf - vi)/t

Solution

F = 3 * (4 - 1.5) / 5

F = 1.5 N

3 0

3 years ago

Calculate the volume of this regular solid.

fenix001 [56]

Answer:

$2122.64 cm^3$

Explanation:

Some data in the problem are missing.

Missing values:

Radius of the cylinder: 13 cm

Height of the cylinder: 4 cm

The volume of a cylinder is given by

$V=\pi r^2 h$

where

r is the radius of the base of the cylinder

h is the height of the cylinder

In this problem, we have:

r = 13 cm (radius)

h = 4 cm (height)

Therefore, the volume of the cylinder is:

$V=\pi (13)^2(4)=2122.64 cm^3$

6 0

3 years ago