answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
xz_007 [3.2K]
2 years ago
4

What is the approximate radius R of 208 82Pb? Express your answer in meters to two significant figures. R = 7.10x10-15 mAssuming

that each nucleus is roughly spherical and that its mass is roughly equal to A (in atomic mass units u), what is the density of a nucleus with nucleon number A?Express your answer in terms of A, R0, .
Physics
1 answer:
lesya692 [45]2 years ago
4 0

Answer:

Using

Radius = RoA^ 1/3

And we are given

To= 1.42 fm A= 208

So R = 1.4E-15 x 208^1/3

Radius= 8.4fm

But we know that density =mass/volume

So D = 3/4 x A/ πR³

You might be interested in
Sean, after being so happy for two full days that he reported he "never needed much sleep," now is stating he is so sad that he
Kaylis [27]
D or A i'm not that sure
5 0
2 years ago
Read 2 more answers
An object moving at a constant velocity travels 274 m in 23 s. what is its velocity?
svetlana [45]
V= 274 meters / 23 sec

V= 11.91 meters per sec
6 0
2 years ago
Read 2 more answers
Charge q1 is distance s from the negative plate of a parallel-plate capacitor. Charge q2=q1/3 is distance 2s from the negative p
Svetlanka [38]

Answer:

The ratio (U₁/U₂) = 6

Explanation:

U, the potential energy is given as

U = kqQ/r

k = Coulomb's constant

q = charge we're concerned about

Q = charge of the negative plate of the capacitor

r = distance of q from the negative plate of the capacitor.

For charge q₁

U₁ = kq₁Q/s

U₂ = kq₂Q/2s

But q₂ = q₁/3

U₂ becomes U₂ = kq₁Q/6s

U₁ = kq₁Q/s

U₂ = kq₁Q/6s

(U₁/U₂) = 6

5 0
2 years ago
A thin ring of radius 73 cm carries a positive charge of 610 nC uniformly distributed over it. A point charge q is placed at the
kow [346]

Answer:

q = - 93.334 nC

Explanation:

GIVEN DATA:

Radius of ring  73 cm

charge on ring 610 nC

ELECTRIC FIELD p FROM CENTRE IS AT 70 CM

E  =  2000 N/C

Electric field due tor ring is guiven as

E = \frac{KQx}{[x^2+ R^2]^{3/2}}

E = \frac{9\time 10^9 \times 610\times 10^[-9} 0.70}{(0.70^2 + 0.73^2)^{3/2}}

E1 = 3714.672 N/C

electric field due to point charge q

E  =\frac[kq}{x^2}

E = \frac{9\times 10^9 \times q}{0.70^2}

E2 = 1.837\times 10^{10}\times q

now the eelctric charge at point P is

E = E1 + E22000 =  3714.672 + 1.837\times 10[10} \times q

solving for q

q = - 93.334 nC

7 0
2 years ago
An infinitely long cylinder of radius R has linear charge density λ. The potential on the surface of the cylinder is V0, and the
tamaranim1 [39]

Answer:

V = V_0 - (lamda)/(2pi(epsilon_0))*ln(R/r)

Explanation:

Attached is the full solution

5 0
2 years ago
Other questions:
  • Identify the arrows that show input force
    14·2 answers
  • The deepest point of the pacific ocean is 11,033 m, in the mariana trench. what is the gauge pressure in the water at that point
    6·1 answer
  • A transverse wave on a string has an amplitude A. A tiny spot on the string is colored red. As one cycle of the wave passes by,
    7·1 answer
  • A positive point charge Q1 = 2.5 x 10-5 C is fixed at the origin of coordinates, and a negative point charge Q2 = -5.0 x 10-6 C
    7·1 answer
  • Carbon dioxide (CO2) gas within a piston–cylinder assembly undergoes a process from a state where p1 = 5 lbf/in.2 , V1 = 2.5 ft3
    9·1 answer
  • A uniformly dense solid disk with a mass of 4 kg and a radius of 2 m is free to rotate around an axis that passes through the ce
    6·2 answers
  • Car A rounds a curve of 150‐m radius at a constant speed of 54 km/h. At the instant represented, car B is moving at 81 km/h but
    11·2 answers
  • The surface is tilted to an angle of 37 degrees from the horizontal, as shown above in Figure 3. The blocks are each given a pus
    7·1 answer
  • A pendulum makes 50 complete swings in 2 min 40 s.<br> What is the time period for 1 complete swing?
    12·2 answers
  • Four rods that obey Hooke's law are each put under tension. (a) A rod 50.0 cm50.0 cm long with cross-sectional area 1.00 mm21.00
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!