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3 years ago

Electro-magnets are created by...

Physics

1 answer:

mojhsa [17]3 years ago

6 0

Wrapping in insulated wire around a medal with ferromagnetic properties and applying an electric current

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In a college homecoming competition, eighteen students lift a sports car. While holding the car off the ground, each student exe

Nata [24]

Answer:

Explanation:

Given

Each student exert a force of $F=400 N$

Let mass of car be m

there are 18 students who lifts the car

Total force by 18 students $F=18\times 400=7200 N$

therefore weight of car $W=7200$

mass of car $m=\frac{W}{g}$

$m=\frac{7200}{9.8}=734.69 kg$

(b) $7200 N \approx 1618.624\ Pound-force$

$734.69 kg\approx 1619.71 Pounds$

6 0

3 years ago

a car travels from station A to B at 30kmph during its return trip A it travels 30 km pH for the first half distance and it 70 k

8_murik_8 [283]

Explanation:

If you cannot visualize it, just assume that the distance from station A to B is 420km. Each half is 210km.

When the car travels from A to B, it takes 420/30 = 14 hours.

When the car travels from B to the halfway point, it takes 210/30 = 7 hours.

When the car travels from the halfway point to A, it takes 210/70 = 3 hours.

Total time taken = 14 + 7 + 3 = 24 hours.

Total distance = 420km * 2 = 840km.

Hence, the average speed of the car is 840/24 = 35km/h.

3 0

3 years ago

A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its sur

muminat

Answer:

a) E = 0

b) $E = \dfrac{k_e \cdot q}{ r^2 }$

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) $\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}$

From which we have;

$E \cdot A = \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}} = \dfrac{0}{\varepsilon _{0}} = 0$

E = 0/A = 0

E = 0

b) $\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}$

$E \cdot A = \dfrac{+q }{\varepsilon _{0}}$

$E = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}$

By Gauss theorem, we have;

$E\oint dS = \dfrac{q}{\varepsilon _{0}}$

Therefore, we get;

$E \cdot (4 \cdot \pi \cdot r^2) = \dfrac{q}{\varepsilon _{0}}$

The electrical field outside the spherical shell

$E = \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }= \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }$

$k_e= \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }$

Therefore, we have;

$E = \dfrac{k_e \cdot q}{ r^2 }$

5 0

3 years ago

What is the momentum of and 200 kg car traveling south at 22 m/sec?

motikmotik

The momentum of the car is 4.4x10^3 kg•m/sec

5 0

3 years ago

1. What does the endocrine system have in common with the muscular system?

Lostsunrise [7]

Answer:

B Both are directly related to movement.

7 0

3 years ago