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Ostrovityanka [42]
3 years ago
10

What is the magnitude of the electric field at a distance 60 cm from the center of the sphere? The radius of the sphere 30 cm, t

he charge on the sphere is 1.56 × 10−5 C and the permittivity of a vacuum is 8.8542 × 10−12 C 2 /N · m2 . Answer in units of N/C.

Physics
2 answers:
Serga [27]3 years ago
5 0

Explanation:

Below is an attachment containing the solution.

Nastasia [14]3 years ago
3 0

Answer

3.9*10^{5}N/C

Explanation:

from the expression for determing the magnetic field strength

E=\frac{q}{4\pi e_{0}d^2 }\\

since the charge is given as

q=1.56*10^{-5}c\\

and the distance is

d=60cm=0.6m

We can calculate the constant k

K=\frac{1}{4\pi e_{0}}\\ K=\frac{1}{4\pi *8.8542*10^{-12}}\\k=8.98*10^{9}\\

if we substitute values, we arrive at

E=\frac{8.98*10^9 *1.56*10^{-5}}{0.6^2} \\E=389133.33\\E=3.9*10^{5}N/C

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Immediately after being struck by a hammer, the nail (mass of 50 g) has a velocity of 50 m/s. The total frictional force is 62.5
Marina86 [1]

Answer:

nail will stop after traveling 1000 m    

Explanation:

We have given mass m = 50 gram = 0.05 kg

Frictional force which is used to stop the mass F = 62.5 kN

Initial velocity is given u = 50 m/sec

From newton's law force is equal to F = ma, here m is mass and a is acceleration

So a=\frac{F}{m}=\frac{62.5\times 10^3}{0.05}=1.25\times 10^6m/sec^2

As finally nail stops so final velocity v = 0 m/sec

From third equation of motion v^2=u^2+2as

So 0^2=50^2-2\times 1.25\times 10^6\times s

s = 1000 m

So nail will stop after traveling 1000 m

8 0
3 years ago
A compound wall consists of parallel layers of two different materials, 10 cm of brick and 2 cm of wood. If the difference of te
RSB [31]

The temperature difference between the wooden wall is 16⁰C, and the heat current per square metre of the wall is 7,325 W/m².

<h3>Temperature difference between the wooden wall</h3>

The temperature difference between the wooden wall is calculated as follows;

  • Let the brick wall =  wall A
  • Let the wooden wall = wall B
  • Let the area of the walls = A

\frac{dQ_A}{dt } = \frac{dQ_B}{dt} \\\\\frac{K_A \Delta T_A \times  A}{L_A} = \frac{K_B \Delta T _B \times A}{L_B}\\\\\frac{K_A \Delta T_A }{L_A} = \frac{K_B \Delta T _B}{L_B}\\\\\frac{0.5 \times 20}{0.1} = \frac{0.125 \times \Delta T _B }{0.02} \\\\100 = 6.25\Delta T _B\\\\\Delta T _B = \frac{100}{6.25} \\\\\Delta T _B = 16 \ ^oC

<h3 /><h3>Heat flowing in the walls</h3>

Q = KL(ΔT)

Q = 0.5 x 0.1 x (20 + 273)

Q = 14.65 W

<h3>Heat current per square meter of the Walls</h3>

QA = W/A

QA = (14.65)/A

  • Let the area of the wall = 10 cm x 2 cm = 0.1 m x 0.02 m = 0.002 m².

QA = 14.65/0.002

QA = 7,325 W/m²

Learn more about thermal conductivity here: brainly.com/question/11213835

4 0
3 years ago
If a negatively charged rod is held near a neutral metal ball, the ball is attracted to the rod. This happens:_______
qaws [65]

Answer:

<em>c. because the rod polarizes the metal.</em>

Explanation:

Bringing the negatively charged rod close to the neutral metal ball causes the neutral metal ball to be polarized with induced positive charge on it. The polarizing of the formally neutral metal ball is due to the negative charge on the metal rod (bodies induce a charge opposite of their own charge on a nearby neutral body). The ball and rod then attract themselves because bodies with opposite charges attract each other, unlike bodies with same charges that repel each other.

7 0
3 years ago
How can a model be created to predict how forces affect objects at a distance?
olchik [2.2K]

Answer:

Gravitational force increases as the masses of the objects increase and decreases as the distance between the objects increases. Balanced forces acting on an object cause no change in the motion of the object. When unbalanced forces act on an object, the sum of the forces is not equal to zero.

Explanation:

put it in your own words

8 0
3 years ago
a loaded sack of total mass is 1000 gramme falls down from the floor of a lorry 200cm high, calculate the workdone by the gravit
Helen [10]

Answer:

W = 20 J

Explanation:

Given that,

The mass of a loaded sack, m = 1000 g = 1 kg

It falls down from the floor of a lorry 200 cm high, h = 2 m

We need to find the work done by the gravity. The work done by an object under the action of gravity is given by :

W = mgh

Substitute all the values,

W = 1 × 10 × 2

= 20 J

Hence, the required work done by gravity is equal to 20 J.

8 0
3 years ago
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