Acceleration is displacement divided by time interval<br> True<br> Or False

What is the pressure 100m below the surface of the sea , if the density of sea water is 1150kg/m3

Olin [163]

The pressure at 100 meters below the surface of sea water with a density of 1150kg is 145.96 psi.

4 0

3 years ago

Which situation best describes the act of reducing?

pishuonlain [190]

In my opinion, i think the first one is the best one since we're cutting down on paper wasted in mails when its better to do it online.

"paying bills online instead of sending paper through mail"

3 0

3 years ago

Two cars are heading towards one another. Car A is moving with an acceleration of aA = 4 m/s2. Car B is moving with an accelerat

Paladinen [302]

Answer:

Car B reaches car A in 19.7 s.

Explanation:

Hi there!

The equation of the position of an object moving in a straight line at constant acceleration is as follows:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the object at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration

When both cars meet, their positions are the same. At the meeting point:

position of car A = position of car B

xA = xB

x0A + v0A · t + 1/2 · aA · t² = x0B + v0B · t + 1/2 · aB · t²

Let´s place the origin of the frame of reference at the point where A is located. In that case x0A = 0 and x0B = 2900 m. Since both cars are initially at rest, v0A and v0B = 0. So, the equation gets reduced to this:

1/2 · aA · t² = x0B + 1/2 · aB · t²

If we replace with the data we have and solve for t:

1/2 · 4 m/s² · t² = 2900 m - 1/2 · 11 m/s² · t²

2 m/s² · t² = 2900 m - 5.5 m/s² · t²

5.5 m/s² · t² + 2 m/s² · t² = 2900 m

7.5 m/s² · t² = 2900 m

t² = 2900 m / 7.5 m/s²

t = 19.7 s

Car B reaches car A in 19.7 s.

4 0

3 years ago

Ask Questions Action and reaction force pairs are all around you, but they aren't always obvious. Write down

NARA [144]

Answer:

Writing with a pencil. The pencil pushes on the paper. The paper pushes on the pencil.

Explanation:

Newton's third law.

5 0

2 years ago

An example of when total internal reflection occurs is when all the light passing from a region of higher index of refraction to

Amiraneli [1.4K]

Answer:

is reflected back into the region of higher index

Explanation:

Total internal reflection is a phenomenon that occurs when all the light passing from a region of higher index of refraction to a region of lower index is reflected back into the region of higher index.

According to Snell's law, refraction of ligth is described by the equation

$n_1 sin \theta_1 = n_2 sin \theta_2$

where

n1 is the refractive index of the first medium

n2 is the refractive index of the second medium

$\theta_1$ is the angle of incidence (in the first medium)

$\theta_2$ is the angle of refraction (in the second medium)

Let's now consider a situation in which

$n_1 > n_2$

so light is moving from a medium with higher index to a medium with lower index. We can re-write the equation as

$sin \theta_2 = \frac{n_1}{n_2}sin \theta_1$

Where $\frac{n_1}{n_2}$ is a number greater than 1. This means that above a certain value of the angle of incidence $\theta_1$ , the term on the right can become greater than 1. So this would mean

$sin \theta_2 > 1$

But this is not possible (the sine cannot be larger than 1), so no refraction occurs in this case, and all the light is reflected back into the initial medium (total internal reflection). The value of the angle of incidence above which this phenomen occurs is called critical angle, and it is given by

$\theta_c =sin^{-1}(\frac{n_2}{n_1})$

8 0

3 years ago

Acceleration is displacement divided by time interval True Or False