In which of the following scenarios is the total momentum of the system conserved?

A sphere of radius R has total charge Q. The volume charge density (C/m3) within the sphere is rho(r)=C/r2, where C is a constan

san4es73 [151]

Answer: C = Q/4πR

Explanation:

Volume(V) of a sphere = 4πr^3

Charge within a small volume 'dV' is given by:

dq = ρ(r)dV

ρ(r) = C/r^2

Volume(V) of a sphere = 4/3(πr^3)

dV/dr = (4/3)×3πr^2

dV = 4πr^2dr

Therefore,

dq = ρ(r)dV ; dq =ρ(r)4πr^2dr

dq = C/r^2[4πr^2dr]

dq = 4Cπdr

FOR TOTAL CHANGE 'Q', we integrate dq

∫dq = ∫4Cπdr at r = R and r = 0

∫4Cπdr = 4Cπr

Q = 4Cπ(R - 0)

Q = 4CπR - 0

Q = 4CπR

C = Q/4πR

The value of C in terms of Q and R is [Q/4πR]

7 0

3 years ago

Write three nuclear equations to represent the nuclear decay sequence that begins with the alpha decay of u-235 followed by a be

vazorg [7]

Answer:

We know that in the decay process the sum of molecular number as well as molecular weight should be constant.

The following three reaction are as follows

1 .

Alpha decay of parent nuclide

$_{92}^{235}\textrm{U} \rightarrow _{90}^{231}\textrm{Th}+_{2}^{4}\textrm{alpha }$

The molecular number of alpha particle is 2 and molecular weight is 4 g/mol.

2.

Beta decay of daughter nuclide

$_{90}^{231}\textrm{Th}\rightarrow _{91}^{231}\textrm{Pa}+_{-1}^{0}\textrm{beta}+v$

v is the neutrino emission,The charge on the beta particle is zero.

3.

Alpha decay

$_{91}^{231}\textrm{Pa}\rightarrow_{89}^{227}\textrm{Pa}+_{2}^{4}\textrm{alpha}$

5 0

4 years ago

Read 2 more answers

An isolated conducting sphere has a 15 cm radius. One wire carries a current of 1.0000046 A into it while another wire carries a

lbvjy [14]

Answer: It would take 3.23ms

Explanation: Please see the attachments below

7 0

3 years ago

Suppose that an object is moving along a vertical line. Its vertical position is given by the equation L(t) = 2t3 + t2-5t + 1, w

Tatiana [17]

Answer:

The average velocity is

$266\frac{m}{s},274\frac{m}{s}$ and $117\frac{m}{s}$ respectively.

Explanation:

Let's start writing the vertical position equation :

$L(t)=2t^{3}+t^{2}-5t+1$

Where distance is measured in meters and time in seconds.

The average velocity is equal to the position variation divided by the time variation.

$V_{avg}=\frac{Displacement}{Time}$ = Δx / Δt = $\frac{x2-x1}{t2-t1}$

For the first time interval :

t1 = 5 s → t2 = 8 s

The time variation is :

$t2-t1=8s-5s=3s$

For the position variation we use the vertical position equation :

$x2=L(8s)=2.(8)^{3}+8^{2}-5.8+1=1049m$

$x1=L(5s)=2.(5)^{3}+5^{2}-5.5+1=251m$

Δx = x2 - x1 = 1049 m - 251 m = 798 m

The average velocity for this interval is

$\frac{798m}{3s}=266\frac{m}{s}$

For the second time interval :

t1 = 4 s → t2 = 9 s

$x2=L(9s)=2.(9)^{3}+9^{2}-5.9+1=1495m$

$x1=L(4s)=2.(4)^{3}+4^{2}-5.4+1=125m$

Δx = x2 - x1 = 1495 m - 125 m = 1370 m

And the time variation is t2 - t1 = 9 s - 4 s = 5 s

The average velocity for this interval is :

$\frac{1370m}{5s}=274\frac{m}{s}$

Finally for the third time interval :

t1 = 1 s → t2 = 7 s

The time variation is t2 - t1 = 7 s - 1 s = 6 s

Then

$x2=L(7s)=2.(7)^{3}+7^{2}-5.7+1=701m$

$x1=L(1s)=2.(1)^{3}+1^{2}-5.1+1=-1m$

The position variation is x2 - x1 = 701 m - (-1 m) = 702 m

The average velocity is

$\frac{702m}{6s}=117\frac{m}{s}$

5 0

3 years ago

The dragster has a mass of 1.3 Mg and a center of mass at G. A parachute is attached at C provides a horizontal braking force of

adell [148]

Answer:

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²

Explanation:

The additional information to the question is embedded in the diagram attached below:

The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m

Balancing the equilibrium about point A;

F(1.1) - mg (1.25) = $ma_a (0.35)$

$1.8v^2(1.1)$ - 1200(9.8)(1.25) = 1200a(0.35)

$1.8v^2(1.1)$ - 14700 = 420 a ------- equation (1)

$F_x = ma_x \\ \\ = 1.8v^2 = 1200 \ a$ --------- equation (2)

Replacing equation 2 into equation 1 ; we have :

${1.1 * 1200 \ a} - 14700 = 420 a$

1320 a - 14700 = 420 a

1320 a - 420 a =14700

900 a = 14700

a = 14700/900

a = 16.33 m/s²

The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²

5 0

3 years ago