Ask question

Ask question

All categories

3 years ago

15

5N of force is applied to move a large nail a distance of 10 cm from an electromagnet on a frictionless table. The nail is then

released. What is the best description of the change from potential to kinetic energy as the nail moves back toward the electromagnet?
A) All of the energy is kinetic as soon as the nail is released.
B) The rate of change from potential to kinetic energy is constant.
C) The rate of change from potential to kinetic energy is exponential.
D) The kinetic energy is 25 J when the nail is 5 cm from the electromagnet.

1 answer:

DedPeter [7]3 years ago

7 0

Answer:

C) The rate of change from potential to kinetic energy is exponential.

Explanation:

As we know that total mechanical energy must be conserved here

As we know that there is no friction force on this system of electromagnet and nail

So here we can say that that

Magnetic potential energy of the nail + electromagnet system will convert into kinetic energy of the nail as it is released.

So we will have

Initial potential energy = work done to move it away by 10 cm

$U = 5(0.10)$

$U = 0.5 J$

now as the nail is released then this potential energy will start to convert into kinetic energy

So here correct answer must be

C) The rate of change from potential to kinetic energy is exponential.

You might be interested in

Which answer choice provides the best set of labels for Wave A and Wave B?

hodyreva [135]

A has less energy and lower frequency, while B has greater energy and higher frequency.

3 0

3 years ago

Read 2 more answers

APAKAH VOLUME 5 KG ES SAMA DENGAN VOLUME 5KG AIR APA YG MEMBEDAKAN KEDUA ZAT TERSEBUT

horrorfan [7]

Answer:c

Explanation:ganglandd

4 0

3 years ago

A cord is used to vertically lower an initially stationary block of mass M = 3.6 kg at a constant downward acceleration of g/7.

dalvyx [7]

Answer:

(a) $W_c=127.008 J$

(b) $W_g=148.176 J$

(c) K.E. = 21.168 J

(d) $v=3.4293m.s^{-1}$

Explanation:

Given:

mass of a block, M = 3.6 kg

initial velocity of the block, $u=0 m.s^{-1}$

constant downward acceleration, $a_d= \frac{g}{7}$

$\Rightarrow$ That a constant upward acceleration of $\frac{6g}{7}$ is applied in the presence of gravity.

∴ $a=- \frac{6g}{7}$

height through which the block falls, d = 4.2 m

(a)

Force by the cord on the block,

$F_c= M\times a$

$F_c=3.6\times (-6)\times\frac{9.8}{7}$

$F_c=-30.24 N$

∴Work by the cord on the block,

$W_c= F_c\times d$

$W_c= -30.24\times 4.2$

We take -ve sign because the direction of force and the displacement are opposite to each other.

$W_c=-127.008 J$

(b)

Force on the block due to gravity:

$F_g= M.g$

∵the gravity is naturally a constant and we cannot change it

$F_g=3.6\times 9.8$

$F_g=35.28 N$

∴Work by the gravity on the block,

$W_g=F_g\times d$

$W_g=35.28\times 4.2$

$W_g=148.176 J$

(c)

Kinetic energy of the block will be equal to the net work done i.e. sum of the two works.

mathematically:

$K.E.= W_g+W_c$

$K.E.=148.176-127.008$

K.E. = 21.168 J

(d)

From the equation of motion:

$v^2=u^2+2a_d\times d$

putting the respective values:

$v=\sqrt{0^2+2\times \frac{9.8}{7}\times 4.2 }$

$v=3.4293m.s^{-1}$ is the speed when the block has fallen 4.2 meters.

6 0

3 years ago

A particle is confined to the x-axis between x = 0 and x = 1 nm. The potential energy U = 0 inside this region and U is infinite

DiKsa [7]

Answer:

Explanation:

According to heisenberg uncertainty Principle

Δx Δp ≥ h / 4π , where Δx is uncertainty in position , Δp is uncertainty in momentum .

Given

Δx = 1 nm

Δp ≥ h /1nm x 4π

≥ 6.6 x 10⁻³⁴ / 10⁻⁹ x 4 π

≥ . 5254 x ⁻²⁵

h / λ ≥ . 5254 x ⁻²⁵

6.6 x 10⁻³⁴ /. 5254 x ⁻²⁵ ≥ λ

12.56 x 10⁻⁹ ≥ λ

longest wave length = 12.56 n m

6 0

3 years ago

PLEASE HELP The graph shows the amplitude of a passing wave over time in seconds (s). What is the approximate frequency of the w

d1i1m1o1n [39]

Answer:

$F = 0.3\ Hz$

Explanation:

Given

See attachment for the graph

Required

Determine the frequency

Frequency (F) is calculated as:

$F = \frac{1}{T}$

Where

T = Time to complete a period

From the attachment, the wave complete a cycle or period in 3 seconds..

So:

$F = \frac{1}{3s}$

$F = 0.333\ Hz$

$F = 0.3\ Hz$ --- Approximated

7 0

2 years ago

Read 2 more answers