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3 years ago

Please answer the following question

Physics

2 answers:

Elden [556K]3 years ago

6 0

Answer:

Where is the resistance value?

Question is incomplete

Liula [17]3 years ago

6 0

Can you show the rest of the problems?

You might be interested in

The nucleus of a hydrogen atom is a single proton, which has a radius of about 1.2 × 10-15 m. The single electron in a hydrogen

Nutka1998 [239]

Answer: 0.86 × 10^14

Explanation:

Given the following :

Radius of proton = 1.2 × 10-15 m

Radius of hydrogen atom = 5.3 × 10-11 m

Density of proton could be calculated thus:

Mass of proton = 1.67 × 10^-27 kg

Using the formula :

(4/3) × pi × r^3

(4/3) × 3.142 × (1.2 × 10^-15)^3 = 7.24 × 10^-45

Density = mass / volume

Density = (1.67 × 10^-27) / ( 7.24 × 10^-45)

= 0.2306 × 10^18

Density of hydrogen atom:

Mass of hydrogen atom= 1.67 × 10^-27 kg

Using the formula :

(4/3) × pi × r^3

(4/3) × 3.142 × (5.3 × 10^-11)^3 = 6.24 × 10^-31

Density = mass / volume

Density = (1.67 × 10^-27) / ( 6.24 × 10^-31)

= 0.2676 × 10^4

Ratio is thus:

Density of proton / density of hydrogen atom

0.2306 × 10^18 / 0.2676 × 10^4 = 0.8617 × 10^14

6 0

3 years ago

How do you find the direction of displacement?

andreyandreev [35.5K]

You draw a straight line from the start point to the end point. It doesn't matter what route was actually followed for the trip.

3 0

3 years ago

An electron and a proton have the same kinetic energy upon entering a region of constant magnetic field and their velocity vecto

kupik [55]

Answer: rp/re= me/mp= 544 * 10^-6.

Explanation: To calculate this problem we have to consider the circular movement by the electron and proton inside a magnetic field.

Then the dynamic equation for the circular movement is given by:

Fcentripetal= m*ω^2.r

q*v*B=m*ω^2.r

we write this for each particle then we have the following:

q*v*B=me* ω^2*re

q*v*B=mp* ω^2*rp

rp/re=me/mp=9.1*10^-31/1.67*10^-27=544*10^-6

4 0

3 years ago

(d) Suppose you use a spring to launch a payload horizontally from the asteroid so that the payload ends up far from the asteroi

nydimaria [60]

Answer:

ks= 133.2 N/m

Explanation:

Assuming that we can neglect the gravitational potential energy of the mass, and that no other forces acting on the payload, total mechanical energy must be conserved.
This energy, at any time, is part elastic potential energy (stored in the spring) and part kinetic energy.
When the spring is initially compressed, the payload is at rest, so all energy is elastic potential.
Once the spring has returned to its natural state, all this elastic potential energy must have been turned into kinetic energy.
If the payload is launched horizontally, and no gravity is present,this means that its final speed will be horizontal only also, according to Newton's First Law.
So, we can write the following equation:

$\Delta U + \Delta K = 0 (1)$

where ΔU = -1/2*k*(Δx)² (2)
and ΔK = 1/2*m*v² (3)
Replacing in (2) and (3) by the givens, and simplifying, we can find the stiffness ks as follows:

$k_{s} =\frac{m*v^{2}}{\Delta x^{2}} = \frac{29 kg*(3m/s)^{2}}{(1.4m)^{2}} = 133.2 N/M (4)$

5 0

2 years ago

A 5000-lb wrecking ball hangs from a 50-ft cable of density 10 lb/ft attached to a crane. Calculate the work done if the crane l

aniked [119]

Answer:

total work is = 52450 J

Explanation:

given data

mass = 5000-lb

density = 10 lb/ft

height = 50 ft

solution

as we will treat here cable and ball are separate

and

here work need to lift cable is

w = (10Δy )(9.8 y ) j

and

now summing all segment of cable

so passing limit Δy to 0

so total work need

= $\int\limits^{10}_0 {98y} \, dy$

= $[49 y^2]^{50}_0$

= 2450J

so lifting 5000 lb wrcking 50 m required additional 5000 + 2450

so total work is = 52450 J

3 0

3 years ago