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3 years ago

Does Earth´s magnetic field move?

Physics

1 answer:

AlekseyPX3 years ago

6 0

Answer:

Yes it does.

Explanation:

"The North Magnetic Pole moves over time due to magnetic changes in Earth's core. " - Wikipedia.

It does move around as the magnetic north does.

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A tug-of-war game is played by five c/hildren: three on one team and two on the other. How much force will the two child team ha

dem82 [27]

Answer:

no, 3 porces is more tha 2 so the power between the 3 should be more than 2

Explanation:

4 0

3 years ago

A small charged sphere is attached to a thread and placed in an electric field. The other end of the thread is anchored so that

VMariaS [17]

Answer:

$E = 9.66\times 10^{-6} N/C$

direction is Horizontal

Explanation:

As we know that the string is horizontal here

so the tension force in the string is due to electrostatic force on it

now we will have

$F = qE$

so here the force is tension force on it

$F = 6.57 \times 10^{-2} N$

$Q = 6.80 \times 10^3 C$

now we have

$6.57 \times 10^{-2} = (6.80 \times 10^3)E$

$E = 9.66\times 10^{-6} N/C$

direction is Horizontal

4 0

3 years ago

Adhira loves to ride her bike around the neighborhood. She starts riding 1.2 miles at 30° S of E. Then, she rides another 2.0 mi

zepelin [54]

Answer:

D = 1.8677 miles , θ = 24.28º at South of West

Explanation:

This is an exercise in adding vectors, the easiest way to solve them is to decompose the vectors and add each component algebraically. Let's use trigonometry

first displacement. d = 1.2 miles to 30º south of East

cos ( 360-30) = cos (-30) = x₁ / d

sin (-30) = y₁ / d

x₁ = d cos (-30)

y₁ = d sin (-30)

x₁ = 1.2 cos (-30) = 1,039 miles

y₁ = 1.2 sin (-30) = -0.6 miles

second shift. d = 2.0 miles to 20º West of South

cos (270-20) = x₂ / d

cos (250) = y₂ / d

x₂ = 2.0 cos 250 = -0.684 miles

y₂ = 2.0 sin250 = -1.879 miles

Third displacement. d = 1.6 miles to 30º South of West

cos (180 + 30) = x₃ / d

sin (210) = y₃ / d

x₃ = 1.6 cos 210 = -1.3856 miles

y₃ = 1.6 sin 210 = -0.8 miles

Fourth displacement. d = 2.6 miles to 15º West of North

cos (90 + 15) = x₄ / d

sin (105) = y₄ / d

x₄ = 2.6 cos 105 = -0.6729 miles

y₄ = 2.6 sin 105 = 2,511 miles

having all the components we add

x-axis (West-East direction)

X = x₁ + x₂ + x₃ + x₄

X = 1.039 -0.684 - 1.3846 - 0.6729

X = -1.7025 miles

Y = y₁ + y₂ + y₃ + y₄

Y = -0.6 -1.879 -0.8 +2.511

Y = -0.768

The modulus of this displacement is we use the Pythagorean theorem

D = √ (X² + Y²)

D = √ (1.7025² + 0.768²)

D = 1.8677 miles

let's use trigonometry to find the direction

tan θ = Y / X

θ = tan⁻¹ Y / x

θ = tan⁻¹ (0.768 / 1.7025)

θ = 24.28º

as the two components are negative this angle is in the third quadrant

therefore in cardinal direction form is

θ = 24.28º at South of West

4 0

3 years ago

For a caffeinated drink with a caffeine mass percent of 0.65% and a density of 1.00 g/mL, how many mL of the drink would be requ

slava [35]

Explanation:

First we will convert the given mass from lb to kg as follows.

157 lb = $157 lb \times \frac{1 kg}{2.2046 lb}$

= 71.215 kg

Now, mass of caffeine required for a person of that mass at the LD50 is as follows.

$180 \frac{mg}{kg} \times 71.215 kg$

= 12818.7 mg

Convert the % of (w/w) into % (w/v) as follows.

0.65% (w/w) = $\frac{0.65 g}{100 g}$

= $\frac{0.65 g}{(\frac{100 g}{1.0 g/ml})}$

= $\frac{0.65 g}{100 ml}$

Therefore, calculate the volume which contains the amount of caffeine as follows.

12818.7 mg = 12.8187 g = $\frac{12.8187 g}{\frac{0.65 g}{100 ml}}$

= 1972 ml

Thus, we can conclude that 1972 ml of the drink would be required to reach an LD50 of 180 mg/kg body mass if the person weighed 157 lb.

5 0

3 years ago

Con base en el tema ley de ohm, resistividad y resistencia resuelva los siguientes

Bumek [7]

Answer:Explanation:gfgfgfgfgfgfgfgfgfg

3 0

3 years ago