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3 years ago

14

by how much does the speed of a vehicle moving ina straight line change each second when it is accelerating at 2km/h/s? At 4 km/

h/s ? At 10 km /h/s?

1 answer:

Svetlanka [38]3 years ago

8 0

Answer:

At 2km/h*s it is changing speed every 2 seconds.

At 4km/h*s it is changing speed every 4 seconds.

At 10km/h*s it is changing speed every 10 seconds.

You might be interested in

If a net force acting on a moving object causes its speed to increase, what

Klio2033 [76]

Answer:

It increases.

Explanation:

If a force acts in the same direction as the object's motion, then the force speeds the object up. Either way, a force will change the velocity of an object. And if the velocity of the object is changed, then the momentum of the object is changed.

7 0

3 years ago

Mr. Ben drove from Town A to Town B. For the first 3 h, he traveled at an average speed of

Vanyuwa [196]

Answer:

It’s 7 hours

Explanation:

You have to use the formula your teacher has given to you plug in the numbers then solve be sure to use a calculator made for physics it helps a lot :)

8 0

3 years ago

If you want quick access to large amount of data, you would use a

RSB [31]

You would use d. table

6 0

3 years ago

he fan blades on a jet engine make one thousand revolutions in a time of 54.9 ms. What is the angular frequency of the blades?

Gnesinka [82]

So, the angular frequency of the blades approximately <u>36.43π rad/s</u>.

<h3>Introduction</h3>

Hi ! Here I will discuss about the angular frequency or what is also often called the angular velocity because it has the same unit dimensions. <u>Angular frequency occurs, when an object vibrates (either moving harmoniously / oscillating or moving in a circle)</u>. Angular frequency can be roughly interpreted as the magnitude of the change in angle (in units of rad) per unit time. So, based on this understanding, the angular frequency can be calculated using the equation :

$\boxed{\sf{\bold{\omega = \frac{\theta}{t}}}}$

With the following condition :

$\sf{\omega}$ = angular frequency (rad/s)
$\sf{\theta}$ = change of angle value (rad)
t = interval of the time (s)

<h3>Problem Solving</h3>

We know that :

$\sf{\theta}$ = change of angle value = 1,000 revolution = 1,000 × 2π rad = 2,000π rad/s >> Remember 1 rev = 2π rad/s.
t = interval of the time = 54.9 s.

What was asked :

$\sf{\omega}$ = angular frequency = ... rad/s

Step by step :

$\sf{\omega = \frac{\theta}{t}}$

$\sf{\omega = \frac{2,000 \pi}{54.9}}$

$\boxed{\sf{\omega \approx 36.43 \pi \: rad/s}}$

<h3>Conclusion :</h3>

So, the angular frequency of the blades approximately 36.43π rad/s.

8 0

2 years ago

"One of the main projects being carried out by the Hubble Space Telescope is to measure the distances of galaxies located in gro

pickupchik [31]

Answer:

finding Cepheid variable and measuring their periods.

Explanation:

This method is called finding Cepheid variable and measuring their periods.

Cepheid variable is actually a type of star that has a radial pulsation having a varying brightness and diameter. This change in brightness is very well defined having a period and amplitude.

A potent clear link between the luminosity and pulsation period of a Cepheid variable developed Cepheids as an important determinants of cosmic criteria for scaling galactic and extra galactic distances. Henrietta Swan Leavitt revealed this robust feature of conventional Cepheid in 1908 after observing thousands of variable stars in the Magellanic Clouds. This in fact turn, by making comparisons its established luminosity to its measured brightness, allows one to evaluate the distance to the star.

5 0

3 years ago