Which is the smallest unit of life?

Sphere 1 with radius R_1 has positive charge q, Sphere 2 with radius 4.50 R_1 is far from sphere 1 and initially uncharged. The

tia_tia [17]

Answer:

Explanation:

capacitance of sphere 2 will be 4.5 times sphere 1

a ) when spheres are in contact they will have same potential finally . So

V_1 / V_2 = 1

b )

Charge will be distributed in the ratio of their capacity

charge on sphere1 = q x 1 / ( 1 + 4.5 )

= q / 5.5

fraction = 1 / 5.5

c ) charge on sphere 2

= q x 4.5 / 5.5

fraction = 4.5 / 5.5

d ) surface charge density of sphere 1

= q /( 5.5 x A ) where A is surface area

surface charge density of sphere 2

= q x 4.5 /( 5.5 x 4.5² A ) where A is surface area

= q /( 5.5 x 4.5 A )

q_1/q_2 = 4.5

6 0

3 years ago

A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and

Svetlanka [38]

Answer:

a = 1.152s

b = 0.817 m

c = 7.29m/s

Explanation: let the following

From the first equation of linear motion

V = u+at..........1

parameters be represented as :

t = Time taken

v = Final velocity

a = Acceleration due to gravity = 9.8m/s²

u = Initial velocity = 4 m/s

s = Displacement

V = 0

Substitute the values into equation 1

0 = 4-9.8(t)

-4 = -9.8t

t = 4/9.8

t = 0.408s

From : s = ut+1/2at^2.........2

S = 4×0.408+0.5(-9.8)×0.408^2

S= 1.632-4.9(0.166)

S = 1.632-0.815

S = 0.817m

Her highest height above the board is 0.817 m

Total height she would fall is 0.817+1.90 = 2.717 m

From equation 2

s = ut+1/2at^2

2.717 m = 0t+0.5(9.8)t^2

2.717 m = 0+4.9t^2

2.717 m = 4.9t^2

2.717/4.9 = t^2

0.554 =t^2

t =√0.554

t = 0.744s

Hence, her feet were in the air for 0.744+0.408seconds

= 1.152s

Also recall from equation 1

V= u+at

V = 0+9.8(0.744)

V = 7.29m/s

Hence, the velocity when she hits the water is 7.29m/s

Finally,

a = 1.152s

b = 0.817 m

c = 7.29m/s

4 0

3 years ago

An electron (q=-1.602×10-19C) is placed .03m away from spherical object with a net charge of -7.2 C.

vovangra [49]

Answer:

Explanation:

electric field at the location of electron

= 9 x 10⁹ x 7.2 / .03²

= 72 x 10¹² N/C

force on electron = electric field x charge on electron

= 72 x 10¹² x 1.6 x 10⁻¹⁹

= 115.2 x 10⁻⁷ N .

C )

work done = charge on electron x potential difference at two points

potential at .03 m

= 9 x 10⁹ x 7.2 / .03

= 2.16 x 10¹² V

potential at .001 m

= 9 x 10⁹ x 7.2 / .001

= 64.8 x 10¹² V

potential difference = (64.8 - 2.16 )x 10¹² V

= 62.64 x 10¹² V .

work done = 62.64 x 10¹² x 1.6 x 10⁻¹⁹

= 100.224 x 10⁻⁷ J .

D )

There will be no change in the magnitude of force on positron except that the direction of force will be reversed . In case of electron , there will be repulsion and in case of positron , there will be attraction .

Work done in case of electron will be positive and work done in case of positron will be negative .

electric field due to charge will be same in both the cases .

8 0

3 years ago

A dipole of moment 0.5 e·nm is placed in a uniform electric field with a magnitude of 8 times 104 N/C. What is the magnitude of

Over [174]

Answer:

The net torque is zero

Explanation:

Let's assume that the dipole is compose of two equal but oposite charges e, and it cam be represented by a rod with one end having a charge e and the other end with a charge of -e. Notice that the dipole is parallel to the electric field thus the force felt by both of the charges will be parallel to the electric field. This means that there will be no components of the forces that are perpendicular to the rod which is a requirement for it to have a torque.

8 0

3 years ago

Dante is leading a parade across the main street in front of city hall. Starting at city hall, he marches the parade 4 blocks ea

Ipatiy [6.2K]

Answer:

The correct option is A)

Displacement: 6.71 m, Direction: 63.4 degrees north of east

Explanation:

Given that Dante is leading a parade across the main street in front of city hall.

Let, Initial location of parade is 0i+0j

One block of city is one units on the XY- graph

Statement 1: Parade marches the parade 4 blocks east, then 3 blocks south

New location of parade is 4i-3j

Statement 2: The parade marches 1 block west and 9 blocks north and finally stops.

Final location of parade is (4i-3j)+(-1i+9j)=3i+6j

Displacement is given by

Displacement = (Final destination)-(Initial destination)

Displacement = (3i+6j)-(0i+0j)=3i+6j

Thus,

Magnitude of displacement = $\sqrt{3^{2}+6^{2}}$

= 6.71 m

Direction of displacement = $tan^{-1}(\frac{Y}{X} )$

= $tan^{-1}(\frac{6}{3} )$

= 63.43 NE

Therefore, the correct option is A) Displacement: 6.71 m, Direction: 63.4 degrees north of east

5 0

3 years ago