A ball was dropped and had a mass of .2 kg and was falling with a force of 2 N, what was its acceleration?

Garrett Hardin's "Tragedy of the Commons" describes the tendency of people to overuse publicly-owned natural resources until the

Aliun [14]

Answer:

is it bad if i keep thinking about p ussy

Explanation:

8 0

3 years ago

How long does it take (in minutes) for light to reach venus from the sun, a distance of 1.152 × 108 km?

7nadin3 [17]

Using the precise speed of light in a vacuum ( $299,792,458 \ \frac{m}{s}$ ), and your given distance of $1.152 * 10^{8} km$ , we can convert and cancel units to find the answer. The distance in m, using $\frac{1000 \ m}{1 \ km}$ , is $1.152 * 10^{11} m$ . Next, for the speed of light, we convert from s to min, using $\frac{1 \ min}{60 \ s}$ , so we divide the speed of light by 60. Finally, dividing the distance between the Sun and Venus by the speed of light in km per min, we find that it is 6.405 min.

7 0

3 years ago

A wave has a wavelength of 3.0 m, a frequency of 25.0 Hz, and an amplitude of 14.0 cm. The wave travels in the positive x-direct

Mrac [35]

Answer:

The total number of oscillations made by the wave during the time of travel is 1.4 Oscillations. Strictly speaking, the number of complete oscillations is 1.

Explanation:

The required quantity is the number of complete oscillations made by the traveling wave. The amplitude time and frequency are not needed to calculate the number of oscillations as it is the ratio of the distance traveled to the wavelength( minimum distance that must be traveled to complete one oscillation) of the wave. So the total number of oscillations is 1.4 while the number of complete oscillations is 1 (strictly speaking). The detailed solution to this question can be found in the attachment below. Thank you!

4 0

3 years ago

A 5.5kg radio is pushed across the table. If the acceleration is 5m/s to the right, find the net force exerted on the radio.

Sholpan [36]

Answer:

Net force exerted on the radio is 27.5 Newton.

Given:

Mass = 5.5 kg

Acceleration = 5 $\frac{m}{s^{2} }$

To find:

Force exerted on the radio = ?

Formula used:

F = ma

Where F = net force

m = mass

a = acceleration

Solution:

According to Newton's second law of motion,

F = ma

Where F = net force

m = mass

a = acceleration

F = 5.5 × 5

F = 27.5 Newton

Hence, Net force exerted on the radio is 27.5 Newton.

4 0

3 years ago

When a board with a box on it is slowly tilted to a larger and larger angle, common experience shows that the box will at some p

Shtirlitz [24]

Answer:

the reason for the acceleration month that the coefficient of kinetic friction is less than the coefficient of satic frictionExplanation:

This exercise uses Newton's second law with the condition that the acceleration is zero, by the time the body begins to slide. At this point the balance of forces is

fr- w || = 0

The expression for friction force is that it is proportional to the coefficient of friction by normal.

fr = μ N

When the system is immobile, the coefficient of friction is called static coefficient and has a value, this is due to the union between the surface, when the movement begins some joints are broken giving rise to coefficient of kinetic friction less than static.

In consequence a lower friction force, which is why the system comes out of balance and begins to accelerate.

μ kinetic <μ static

In all this movement the normal with changed that the angle of the table remains fixed.

Consequently, the reason for the acceleration month that the coefficient of kinetic friction is less than the coefficient of satic friction

3 0

3 years ago