What are three examples of constructive forces

In the electromagnetic spectrum, _______ is the _______ frequency color of visible light. A. green; highest B. red; lowest C. vi

MaRussiya [10]

Red is the lowest because it has the shortest wavelengths

5 0

3 years ago

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A rotating space station is said to create “artificial gravity”—a loosely-defined term used for an acceleration that would be cr

Grace [21]

Answer: 0.313 rad/s

Explanation:

The equation that relates the velocity $V$ and the angular velocity $\omega$ in the uniform circular motion is:

$V=\omega.r$ (1)

Where $r=d/2=100m$ is the radius of the space station (with a diaeter of 200m) that describes the uniform circular motion.

Isolating $\omega$ from (1):

$\omega=\frac{V}{r}$ (2)

On the other hand, we are told the “artificial gravity” produced by the cetripetal acceleration $a_{c}$ is $9.8m/s^{2}$ , and is given by the following equation:

$a_{c}=\frac{V^{2}}{r}$ (3)

Isolating $V$ :

$V=\sqrt{a_{c}.r}$ (4)

$V=31.3049m/s$ (5)

Substitutinng (5) in (2):

$\omega=\frac{31.3049m/s}{100m}$ (6)

$\omega=0.313rad/s$ This is the angular velocity that would produce an “artificial gravity” of 9 $9.8m/s^{2}$ .

6 0

3 years ago

In part one of this experiment, a 0.20 kg mass hangs vertically from a spring and an elongation below the support point of the s

statuscvo [17]

To solve this problem it is necessary to apply the concepts related to Hooke's Law as well as Newton's second law.

By definition we know that Newton's second law is defined as

$F = ma$

m = mass

a = Acceleration

By Hooke's law force is described as

$F = k\Delta x$

Here,

k = Gravitational constant

x = Displacement

To develop this problem it is necessary to consider the two cases that give us concerning the elongation of the body.

The force to keep in balance must be preserved, so the force by the weight stipulated in Newton's second law and the force by Hooke's elongation are equal, so

$k\Delta x = mg$

So for state 1 we have that with 0.2kg there is an elongation of 9.5cm

$k (9.5-l)=0.2*g$

$k (9.5-l)=0.2*9.8$

For state 2 we have that with 1Kg there is an elongation of 12cm

$k (12-l)= 1*g$

$k (12-l)= 1*9.8$

We have two equations with two unknowns therefore solving for both,

$k = 3.136N/cm$

$l = 8.877cm$

In this way converting the units,

$k = 3.136N/cm(\frac{100cm}{1m})$

$k = 313.6N/m$

Therefore the spring constant is 313.6N/m

3 0

3 years ago

What is initial velocity

Lapatulllka [165]

Initial Velocity is the velocity at time interval t = 0 and it is represented by u. It is the velocity at which the motion starts. They are four initial velocity formulas: (1) If time, acceleration and final velocity are provided, the initial velocity is articulated as. u = v – at.

3 0

3 years ago

When a man stands on a bathroom scale here on Earth, it reads 640 N . Assume each planet to be a perfect sphere with the followi

noname [10]

Answer:

242.19702 N

578.46718 N

681.02785 N

Explanation:

M = Mass of the corresponding planet

r = Radius of the corresponding planet

g = Acceleration due to gravity = 9.81 m/s²

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

Mass of person

$m=\frac{W}{g}\\\Rightarrow m=\frac{640}{9.81}=65.23955\ kg$

Mass is the property of an object, it is constant irrespective of the forces acting on it so the mass of the person on each planet would be the same.

Gravitational force on Mars

$F=\frac{GMm}{r^2}\\\Rightarrow F=\frac{6.67\times 10^{-11}\times 6.419\times 10^{23}\times 65.23955}{(3.396\times 10^{6})^2}\\\Rightarrow F=242.19702\ N$

Magnitude of the gravitational force Mars would exert on the man if he stood on its surface is 242.19702 N

Gravitational force on Venus

$F=\frac{GMm}{r^2}\\\Rightarrow F=\frac{6.67\times 10^{-11}\times 4.869\times 10^{24}\times 65.23955}{(6.052\times 10^{6})^2}\\\Rightarrow F=578.46718\ N$

Magnitude of the gravitational force Venus would exert on the man if he stood on its surface is 578.46718 N

Gravitational force on Saturn

$F=\frac{GMm}{r^2}\\\Rightarrow F=\frac{6.67\times 10^{-11}\times 5.685\times 10^{26}\times 65.23955}{(6.027\times 10^{7})^2}\\\Rightarrow F=681.02785\ N$

Magnitude of the gravitational force Saturn would exert on the man if he stood on its surface is 681.02785 N

6 0

4 years ago