Find the radius of a cylindrical fire hose that is 200 ft long and has a volume of 39.25 ft*.

At the top of a roller coaster you have 90j of potential energy and 10 j of kinetic energy at the bottom of the roller coaster y

Svetlanka [38]

The energy has to be conserved thus the final energy level will always be equal to the initial state .. the initial state has 90j of potential energy and 10j of kinetic energy the total energy is equal to = 100j .

now for the second condition the kinetic energy is 90j and the total energy is 100j thus 100j-90j = 10j of potential energy

5 0

2 years ago

20 cubic inches of a gas with an absolute pressure of 5 psi is compressed until it’s press reaches 10 psi. What is the new volum

jek_recluse [69]

Let x be the new volume

10x = 20(5)
10x = 100
x = 10

Have a beautiful day!

5 0

2 years ago

Describe the role of the teacher and the learner in a natural science and technology classroom

dem82 [27]

Teacher -> teach the topic
Learner -> learn the topic
Is that what you’re asking?

5 0

2 years ago

In one of the classic nuclear physics experiments at the beginning of the 20th century, an alpha particle was accelerated toward

Vladimir79 [104]

Answer:

The answer is " $1.01 \times 10^{-13}$ "

Explanation:

Using the law of conservation for energy. Equating the kinetic energy to the potential energy.

$KE=U=\frac{kqq'}{r}\\\\$

Calculating the closest distance:

$\to r=\frac{kqq'}{KE}\\\\$

$=\frac{k(2e)(79e)}{KE}\\\\=\frac{k(2)(79)e^2}{KE}\\\\=\frac{9.0\times 10^9 \ N \cdot \frac{m^2}{c}(2)(79)(1.6 \times10^{-19} \ C)^2}{(2.25\ meV) (\frac{1.6 \times 10^{-13} \ J}{1 \ MeV})}\\\\$

$=\frac{9.0\times 10^9 \times 2\times 79\times 1.6 \times10^{-19}\times 1.6 \times10^{-19} }{(2.25 \times 1.6 \times 10^{-13}) }\\\\=\frac{3,640.32\times 10^{-29}}{3.6 \times 10^{-13} }\\\\=\frac{3,640.32}{3.6} \times 10^{-16}\\\\=1011.2 \times 10^{-16}\\\\=1.01 \times 10^{-13}$

5 0

2 years ago

When a magnetic field is first turned on, the flux through a 20-turn loop varies with time according to Φm=5.0t2−2.0t, where Φm

goldfiish [28.3K]

Answer:

A) ( - 200t + 40 ) volts

B) b) anticlockwise , c) anticlockwise , d) clockwise , e) clockwise

Explanation:

Given data:

magnetic flux (Φm) = 5.0t^2 − 2.0t

number of turns = 20

<u>a) determine induced emf </u>

E = - N $\frac{d\beta }{dt}$

= - N ( 10t - 2 ) = - 20 ( 10t - 2 )

= - 200t + 40 volts

<u>b) Determine direction of induced current </u>

i) at t = 0

E = - 0 + 40 ( anticlockwise direction )

ii) at t = 0.10

E = -20 + 40 = 20 ( anticlockwise direction )

iii) at t = 1

E = - 200 + 40 = - 160 ( clockwise direction)

iv) at t = 2

E = -400 + 40 = - 360 ( clockwise direction )

8 0

2 years ago