_______is a speed in a certain direction.

In most cases, what happens to a liquid when it cools?

erik [133]

Answer:

Option (A) and (F)

Explanation:

As the liquid cools down, it means the temperature decreases the density of the liquid increases in most o the cases.

Now the volume is inversely proportional to the density of substance so density increases.

7 0

4 years ago

You stand on a merry-go-round which is spinning at f = 0:25 revolutions per second. You are R = 200 cm from the center. (a) Find

ivanzaharov [21]

Answer:

(a) ω = 1.57 rad/s

(b) ac = 4.92 m/s²

(c) μs = 0.5

Explanation:

(a)

The angular speed of the merry go-round can be found as follows:

ω = 2πf

where,

ω = angular speed = ?

f = frequency = 0.25 rev/s

Therefore,

ω = (2π)(0.25 rev/s)

ω = 1.57 rad/s

(b)

The centripetal acceleration can be found as:

ac = v²/R

but,

v = Rω

Therefore,

ac = (Rω)²/R

ac = Rω²

therefore,

ac = (2 m)(1.57 rad/s)²

ac = 4.92 m/s²

(c)

In order to avoid slipping the centripetal force must not exceed the frictional force between shoes and floor:

Centripetal Force = Frictional Force

m*ac = μs*R = μs*W

m*ac = μs*mg

ac = μs*g

μs = ac/g

μs = (4.92 m/s²)/(9.8 m/s²)

μs = 0.5

7 0

3 years ago

A main goal of most environmental scientists is to achieve _________________________.

sergiy2304 [10]

Answer:

The answer is biodiversity

Explanation:

4 0

3 years ago

Calculat the acceleration of a person at latitude 40degreesowing to the rotation of the earth.

beks73 [17]

Answer:

acceleration of person = 9.77 m/s²

Explanation:

given data

latitude = 40 degree

to find out

Calculate the acceleration of a person

solution

we know that here 40 degree = 0.698 rad

so

acceleration of person = g - ω²R ...............1

and 1 rotation complete in 24 hours = 360 degree

here g is 9.81

so we know Earth angular speed ω = 7.27 × $10^{-5}$ rad/s and R is earth radius that is 6.37 × $10^{6}$ m

so

put here value in equation 1 we get

acceleration of person = g - ω²R

acceleration of person = 9.81 - (7.27 × $10^{-5}$ )² × 6.37 × $10^{6}$

acceleration of person = 9.77 m/s²

6 0

4 years ago

Ignoring the electron spin what is the largest possible energy difference, if the magnetic field is 2.02 tesla?

Nonamiya [84]

Complete Question:

When specially prepared Hydrogen atoms with their electrons in the 6f state are placed into a strong uniform magnetic field, the degenerate energy levels split into several levels. This is the so called normal Zeeman effect.

Ignoring the electron spin what is the largest possible energy difference, if the magnetic field is 2.02 Tesla?

Answer:

ΔE = 1.224 * 10⁻²² J

Explanation:

In the 6f state, the orbital quantum number, L = 3

The magnetic quantum number, $m_{L} = -3, -2, -1, 0, 1, 2, 3$

The change in energy due to Zeeman effect is given by:

$\triangle E = m_{L} \mu_{B} B$

Magnetic field B = 2.02 T

Bohr magnetron, $\mu_{B} = 9.274 * 10^{-24} J/T$

$\triangle E = 6 * 9.274 * 10^{-24} * 2.2\\$

ΔE = 1.224 * 10⁻²² J

5 0

3 years ago