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3 years ago

Describe a situation where you might have high velocity but low acceleration

1 answer:

6 0

In a level cruise on a passenger airliner, flying in a straight line at a steady 550 miles per hour.
Acceleration is zero.

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A convex lens has a focal length of 16.5 cm. Where on the lens axis should an object be placed in order to get a virtual, enlarg

alexandr402 [8]

Answer:

Object should be placed at a distance, u = 7.8 cm

Given:

focal length of convex lens, F = 16.5 cm

magnification, m = 1.90

Solution:

Magnification of lens, m = - $\frac{v}{u}$

where

u = object distance

v = image distance

Now,

1.90 = $\frac{v}{u}$

v = - 1.90u

To calculate the object distance, u by lens maker formula given by:

$\frac{1}{F} = \frac{1}{u}+ \frac{1}{v}$

$\frac{1}{16.5} = \frac{1}{u}+ \frac{1}{- 1.90u}$

$\frac{1}{16.5} = \frac{1.90 - 1}{1.90u}$

$\frac{1}{16.5} = \frac{ 0.90}{1.90u}$

u = 7.8 cm

Object should be placed at a distance of 7.8 cm on the axis of the lens to get virtual and enlarged image.

6 0

4 years ago

A 0.750 kg block is attached to a spring with spring constant 13.0 N/m . While the block is sitting at rest, a student hits it w

trapecia [35]

To solve this problem we will apply the concepts related to energy conservation. From this conservation we will find the magnitude of the amplitude. Later for the second part, we will need to find the period, from which it will be possible to obtain the speed of the body.

A) Conservation of Energy,

$KE = PE$

$\frac{1}{2} mv ^2 = \frac{1}{2} k A^2$

Here,

m = Mass

v = Velocity

k = Spring constant

A = Amplitude

Rearranging to find the Amplitude we have,

$A = \sqrt{\frac{mv^2}{k}}$

Replacing,

$A = \sqrt{\frac{(0.750)(31*10^{-2})^2}{13}}$

$A = 0.0744m$

(B) For this part we will begin by applying the concept of Period, this in order to find the speed defined in the mass-spring systems.

The Period is defined as

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

Replacing,

$T = 2\pi \sqrt{\frac{0.750}{13}}$

$T= 1.509s$

Now the velocity is described as,

$v = \frac{2\pi}{T} * \sqrt{A^2-x^2}$

$v = \frac{2\pi}{T} * \sqrt{A^2-0.75A^2}$

We have all the values, then replacing,

$v = \frac{2\pi}{1.509}\sqrt{(0.0744)^2-(0.750(0.0744))^2}$

$v = 0.2049m/s$

7 0

3 years ago

What happens to light waves from a star as the star moves away from Earth?

AURORKA [14]

<h2>Answer: Light waves have a redshift due to the Doppler effect </h2>

The astronomer Edwin Powell Hubble observed several celestial bodies, and when obtaining the spectra of distant galaxies he observed the spectral lines were displaced towards the red (red shift), whereas the nearby galaxies showed a spectrum displaced to the blue.

From there, Hubble deduced that the farther the galaxy is, the more redshifted it is in its spectrum. <u>The same happens with the stars and this phenomenom is known as the Doppler effect. </u>

This phenomenon refers to the change in a wave perceived frequency (or wavelength=color) when the emitter of the waves, and the receiver (or observer in the case of light) move relative to each other. For example, as a star moves away from the Earth, its espectrum turns towards the red.

8 0

3 years ago

To heat 1g of water by 1 C requires<br> A) 1 calorie<br> b)1 Carlorie<br> c) 1 Joule<br> d) 1 watt

blsea [12.9K]

I think the answer would be 1 watt but i'm not sure

8 0

3 years ago

Create a model in the space below that demonstrates how action potentials ensue. A sample model can be found under the Unit Pack

Pepsi [2]

Answer:

K+NA+30

Explanation:

8 0

3 years ago