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3 years ago

9

The principles of magnetism apply everywhere on earth. What does this tell us about God and His character?

1 answer:

Bas_tet [7]3 years ago

3 0

Answer:

God is omnipresent.

Explanation:

This means God is everywhere and He works where ever we are in the world

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Which of the following lists the composition of the Earth’s outer core?

iris [78.8K]

Well,

The outer core of the Earth is mostly composed of iron and nickel.

The correct option is C.

5 0

3 years ago

A 49.0 kg wheel, essentially a thin hoop with radius 0.730 m, is rotating at 114 rev/min. It must be brought to a stop in 22.0 s

belka [17]

Explanation:

Mass of the wheel, m = 49 kg

Radius of the hoop, r = 0.73 m

Initial angular speed of the wheel, $\omega_i=114\ rev/min = 11.93\ rad/s$

Final angular speed of the wheel, $\omega_f=0$

Time, t = 22 s

(a) If I is the moment of inertia of the hoop. It is equal to,

$I=mr^2$

$I=49\times (0.73)^2$

$I=26.11\ kg-m^2$

We know that the work done is equal to change in kinetic energy.

$W=\Delta E$

$W=\dfrac{1}{2}I(\omega_f^2-\omega_i^2)$

$W=-\dfrac{1}{2}\times 26.11\times (11.93^2)$

W = -1858.05 Joules

(b) Let P is the average power. It is given by :

$P=\dfrac{W}{t}$

$P=\dfrac{1858.05\ J}{22\ s}$

P =84.45 watts

Hence, this is the required solution.

4 0

3 years ago

6. The average atomic mass of hydrogen is amu. 7. If a molecule is made of similar kind of atoms, then it is called atomic molec

Roman55 [17]

Answer/
I have no idea i’m so sorry

Explanation/
Have a good day

8 0

3 years ago

Which type of mixture is air explain

igor_vitrenko [27]

Answer:Three elements make up over 99.9 percent of the composition of dry air: these are nitrogen, oxygen, and argon.

Explanation:

7 0

3 years ago

Read 2 more answers

An airplane is flying through a thundercloud at a height of 2000 m (This is a very dangerous thing to do because of updrafts, tu

Doss [256]

Answer:

$400000\ \text{N/C}$

Explanation:

$q_1$ = Charge at 3000 m = 40 C

$q_2$ = Charge at 1000 m = -40 C

$r_1$ = 3000 m

$r_2$ = 1000 m

k = Coulomb constant = $9\times10^9\ \text{Nm}^2/\text{C}^2$

Electric field due to the charge at 3000 m

$E_1=\dfrac{k|q_1|}{r_1^2}\\\Rightarrow E_1=\dfrac{9\times 10^9\times 40}{3000^2}\\\Rightarrow E_1=40000\ \text{N/C}$

Electric field due to the charge at 1000 m

$E_2=\dfrac{k|q_2|}{r_2^2}\\\Rightarrow E_2=\dfrac{9\times 10^9\times 40}{1000^2}\\\Rightarrow E_2=360000\ \text{N/C}$

Electric field at the aircraft is $E_1+E_2=40000+360000=400000\ \text{N/C}$ .

7 0

3 years ago