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3 years ago

The lead nucleus has a diameter of 14.2 fm. What is the electric field strength at the surface of a lead nucleus

Physics

1 answer:

Snowcat [4.5K]3 years ago

6 0

Answer:

E = kQ / r^2 = 2.34*10^21 V/m

where k = 8.99*10^9 m/F

Q = 82 * 1.6*10^-19 C

r = 7.1*10^-15 m

Explanation:

14.2 FM = 2.34×10^21 V/m

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Two pebbles are dropped into a pool of water as shown. If a new wave with a

Kazeer [188]

Answer:

Constructive interference

Explanation:

3 0

3 years ago

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Ship A is 32 miles north of ship B and is sailing due south at 16 mph. Ship B is sailing due east at 12 mph. At what rate is the

Zielflug [23.3K]

Answer:

$\dfrac{dz}{dt} =-5.6\ mile/h$

Explanation:

distance between ship A and B = 32 mile

Ship A velocity in south, dx/dt = -16 mph

Ship B is sailing toward east with speed, dy/st = 12 mph

time = 1 hour

rate of change of distance between them = ?

x be the distance travel after t time

X = 32 + x

Let distance between them be z

now, using Pythagoras theorem to calculate distance between ships after 1 hours

z² = x² + y²

z² = (32 + x)² + 12²

z² = (32 - 16)² + 12²

z = √400

z = 20 miles

now, calculation of rate of change of distnace

z² = (32 + x)² + y²

differentiating both side w.r.t. time

$2 z \dfrac{dz}{dt} = 2(32+x)\dfrac{dx}{dt} + 2 y\dfrac{dy}{dt}$

$z \dfrac{dz}{dt} =(32-16)\dfrac{dx}{dt} +y\dfrac{dy}{dt}$

$20\times \dfrac{dz}{dt} =16\times (-16) +12\times 12$

$\dfrac{dz}{dt} =\dfrac{-112}{20}$

$\dfrac{dz}{dt} =-5.6\ mile/h$

hence, the rate is the distance between them changing at the end of 1 hour is equal to $\dfrac{dz}{dt} =-5.6\ mile/h$

7 0

3 years ago

Ethan dives 10 meters into the Atlantic Ocean. How much pressure does he experience?

ycow [4]

The answer is A. 1 atm.

Solution:

Pressure in terms of height can be calculated using the formula:

P = pgh

where: p = density (kg/m^3); g = gravitational constant (m/s^2); h = height (m)

Using SI units: P = 1000(kg/m^3)*9.81(m/s^2)*10(m) = 98,100 Pascals

Convert Pascals to atm:

98,100 Pa (1 atm/101325 Pa) = 0.968 atm = rounded up to 1 atm

6 0

3 years ago

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One of the strongest emission lines observed from distant galaxies comes from hydrogen and has a wavelength of 122 nm (in the ul

NeTakaya

Answer:

$v=2.4*10^8m/s$

Explanation:

From the question we are told that

Wavelength of emission $\lambda=122nm$

Observation distance $d=366nm$

Generally the s equation is given as

$f'=f\sqrt{\frac{(1-\frac{v}{c} )}{1+\frac{v}{c} }$

where

F is inversely proportional to T

$d=\lambda\sqrt{\frac{(1-\frac{v}{c} )}{1+\frac{v}{c} }$

$\frac{v}{c} =\frac{(1-\frac{\lambda}{d})}{(1+\frac{\lambda}{d}}$

$\frac{v}{c}=\frac{1-(\frac{122}{366} )^2}{1+(\frac{122}{366})^2}$

$\frac{v}{c}=\frac{0.8888888889}{1.11111111}$

$\frac{v}{c}=0.80$

$v=0.80*3*10^6$

$v=2.4*10^8m/s$

7 0

3 years ago

A child's water pistol shoots water through a 1.0-mm-diameter hole. If the pistol is fired horizontally 70 cm above the ground,

lesya692 [45]

Answer:

2.52 ml/s

Explanation:

Unit conversions:

1 mm = 0.001m

70 cm = 0.7 m

Let g = 10m/s2. If the pistol is fired horizontally at first, it did not have an vertical velocity, only horizontal velocity. So g is the only thing that affects the vertical motion of water.

We can calculate the time it takes for the squirts to hit the ground in the following equation of motion:

$h = gt^2/2$

where h = 0.7 m is the vertical distance, and t is the time it takes, which is what we are solving for:

$t^2 = 2h/g = 2*0.7/10 = 0.14$

$t = \sqrt{0.14} = 0.374 s$

So the squirts takes 0.374s to hit the ground, and within that time it travels a distance of 1.m horizontally. Neglect air resistance, we can calculate the horizontal velocity:

$v = s/t$