PHYSICS COLLEGE

A 2,000 kg car travels with a tangentialvelocity of 12 m/s around a circular
track with a radius of 30 meters. What
is the car's rate of centripetal
acceleration?

Answers

Answer 1

Answer:

The car's rate of centripetal acceleration in the circular path is 4.8 m/s².

The given parameters;

mass of the car, m = 2,000 kg
velocity of the car, v = 12 m/s
radius of the circular path, r = 30 m

The centripetal acceleration of the car is calculated as follows;

$a_c = (v^2)/(r)$

where;

v is the tangential speed of the car
r is the radius of the circular path

Substitute the given parameters and solve for the centripetal acceleration;

$a_c = (12^2)/(30) \n\na_c = 4.8 \ m/s^2$

Thus, the car's rate of centripetal acceleration is 4.8 m/s².

Learn more here:brainly.com/question/11700262

Answer 2

Answer: a= v²/R
a = 12²/30 =4.8 m/s²

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What is the wavelength of the photons emitted by hydrogen atoms when they undergo n =5 to n =3 transitions?

Answers

Answer:

$\lambda=1282nm$

Explanation:

The wavelength of the photons emitted due to an atomic electron transition in a hydrogen atom, is given by the Rydberg formula:

$(1)/(\lambda)=R_H((1)/(n_1^2)-(1)/(n_2^2)})$

Here $R_H$ is the Rydberg constant for hydrogen and $n_1,n_2$ are the lower and higher quantum number for the energy levels of the atomic electron transition, respectively. Replacing the given values and solving for $\lambda$

$(1)/(\lambda)=1.097*10^7m^(-1)((1)/(3^2)-(1)/(5^2)})\n(1)/(\lambda)=7.81*10^5m^(-1)\n\lambda=(1)/(7.81*10^5m^(-1))\n\lambda=1.282*10^(-6)m\n\lambda=1.282*10^(-6)m*(1nm)/(10^(-9)m)\n\lambda=1282nm$

Ultraviolet light is typically divided into three categories. UV-A, with wavelengths between 400 nm and 320 nm, has been linked with malignant melanomas. UV-B radiation, which is the primary cause of sunburn and other skin cancers, has wavelengths between 320 nm and 280 nm. Finally, the region known as UV-C extends to wavelengths of 100 nm. (a) Find the range of frequencies for UV-B radiation. (b) In which of these three categories does radiation with a frequency of 7.9 * 1014 Hz belong

Answers

Answer:

a) The UV-B has frequencies between $9.375x10^(14)Hz$ and $1.071x10^(15)Hz$

b) The radiation with a frequency of $7.9x10^(14)Hz$ belong to the UV-A category.

Explanation:

(a) Find the range of frequencies for UV-B radiation.

Ultraviolet light belongs to the electromagnetic spectrum, which distributes radiation along it in order of different frequencies or wavelengths.

Higher frequencies:

Gamma ray

X ray

Ultraviolet rays

Visible region

Lower frequencies:

Infrared

Microwave

Radio waves

That radiation is formed by electromagnetic waves, which are transverse waves formed by an electric field and a magnetic field perpendicular to it. Any of those radiations will have a speed of $3x10^{8]m/s$ in vacuum.

The velocity of a wave can be determined by means of the following equation:

$c = \nu \cdot \lambda$ (1)

Where c is the speed of light, $\nu$ is the frequency and $\lambda$ is the wavelength.

Then, from equation 1 the frequency can be isolated.

$\nu = (c)/(\lambda)$ (2)

Before using equation 2 to determine the range of UV-B it is necessary to express $\lambda$ in units of meters in order to match with the units from c.

$\lambda = 320nm . (1m)/(1x10^(9)nm)$ ⇒ $3.2x10^(-7)m$

$\lambda = 280nm . (1m)/(1x10^(9)nm)$ ⇒ $2.8x10^(-7)m$

$\nu = (3x10^(8)m/s)/(3.2x10^(-7)m)$

$\nu = 9.375x10^(14)s^(-1)$

$\nu = 9.375x10^(14)Hz$

$\nu = (3x10^(8)m/s)/(2.8x10^(-7)m)$

$\nu = 1.071x10^(15)Hz$

Hence, the UV-B has frequencies between $9.375x10^(14)Hz$ and $1.071x10^(15)Hz$

(b) In which of these three categories does radiation with a frequency of $7.9x10^(14)Hz$ belong.

The same approach followed in part A will be used to answer part B.

Case for UV-A:

$\lambda = 400nm . (1m)/(1x10^(9)nm)$ ⇒ $4x10^(-7)m$

$\nu = (3x10^(8)m/s)/(4x10^(-7)m)$

$\nu = 7.5x10^(14)s^(-1)$

$\nu = 7.5x10^(14)Hz$

Hence, the UV-A has frequencies between $7.5x10^(14)Hz$ and $9.375x10^(14)Hz$ .

Therefore, the radiation with a frequency of $7.9x10^(14)Hz$ belongs to UV-A category.

Suppose a fast-pitch softball player does a windmill pitch, moving her hand through a circular arc with her arm straight. She releases the ball at a speed of 25.5 m/s (about 57.0 mph ). Just before the ball leaves her hand, the ball's radial acceleration is 1060 m/s2 . What is the length of her arm from the pivot point at her shoulder

Answers

Answer:

61.3 cm

Explanation:

Radial acceleration of the object in circular motion is given by formula

$a = (v^2)/(R)\n$

Given:

$a = 1060 m/s^2\nv = 25.5 m/s$

Plugging in the values in the formula

$1060 = (25.5^2)/(R)\nR = 0.613 m$

so length of his arm is 61.3 cm

Dogs can hear higher-pitched whistles that humans do. How do you
think the sound frequencies that dogs can
hear compare to the frequencies that humans
can hear?

Answers

Dogs can hear sounds at higher frequencies than humans. The range of sound frequencies that dogs can hear is approximately 40 Hz to 60,000 Hz, while the range for humans is 20 Hz to 20,000 Hz. This means that dogs can hear ultrasonic sounds that are beyond the range of human hearing.

What is sound about?

In terms of physics, sound is a vibration that travels through a transmission medium like a gas, liquid, or solid as an acoustic wave.

Sound is the reception of these waves and the brain's perception of them in terms of human physiology and psychology. Dogs have the ability to hear ultrasonic sounds that are audible only to them.

Learn more about sound on:

brainly.com/question/16093793

#SPJ1

A 5-kg moving at 6 m/s collided with a 1-kg ball at rest. The ball bounce off each other and the second ball moves in the same direction as the first ball at 10 m/sec. What is the velocity of the first ball after the collision?

Answers

Given :

A 5-kg moving at 6 m/s collided with a 1-kg ball at rest.

The ball bounce off each other and the second ball moves in the same direction as the first ball at 10 m/sec.

To Find :

The velocity of the first ball after the collision.

Solution :

We know, by conservation of momentum :

$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$

Putting all given values with directions ( one side +ve and other side -ve ).

$5* 6 + 1* 0 =5 * v_1 + 1* 10\n\n5v_1=10-30\n\nv = -4 \ m/s$

Therefore, the velocity of first ball after the collision is 4 m/s after in opposite direction.

Hence, this is the required solution.

Lab: Weather Patterns

Answers

A weather pattern is defined as a period of time when the weather remains consistent. In the lab, a lot of observation about weather is obtained

What is the definition of a weather pattern?

A weather pattern is defined as a period of time when the weatherremains consistent. Weather changes are crucial to humanexistence.

because they influence our everyday activities and provide moisture for crops.

The rain does not always end within the day, and gloomy days might last just as long as sunny days. Tornadoes and hurricanes, for example, may inflict tremendous damage.

In the lab the following observation about weather is obtained;

1. We will find the graphs and statistics that indicate signs of climate change and engage with an interactivegraphic.

2. You'll also look at and debate maps of global temperature and precipitation patterns that are changing.

3. This lab will teach you about Earth's biomes and the close relationship that exists between them and the climates that serve to define them.

To learn more about the weatherpattern refer to the link;

brainly.com/question/2497685

Final answer:

The question pertains to meteorology, climatology, and atmospheric science. These are disciplines that study weather and climate, respectively, and their effects on the planet. Atmospheric Science is a broad field that includes both and employs physics principles.

Explanation:

The question refers to the subjects of meteorology, climatology, and atmospheric science. Meteorology is the study of the atmosphere, atmospheric phenomena, and atmospheric impacts on the Earth's weather. It involves the prediction of weather in the short term based on thousands of measurements of variables such as air pressure and temperature.

Climatology, on the other hand, is the study of climate, which involves analyzing averaged weather conditions over longer time periods using atmospheric data. Unlike meteorologists, climatologists focus on patterns and effects that occur over longer timescales of decades, centuries, and millennia.

Atmospheric Science is a broad field that encompasses both meteorology and climatology, as well as other disciplines that study the atmosphere. This discipline is typically based heavily on physics and involves the study of weather and climate patterns, predictions of developments in weather and climatic events, and the analysis of the effects of these events on the planet and its inhabitants.