All categories

3 years ago

Where is earth's magnetic north pole located?

Physics

2 answers:

scoundrel [369]3 years ago

7 0

Answer:ellesmere island, Canada

andre [41]3 years ago

4 0

Answer: it is located over Ellesmere Island, Canada

Explanation

it is now drifting away from North America and toward Siberia.

You might be interested in

You’ve made the hypothesis that the stepper the slope , the faster a ball will be rolling when it reaches the bottom .

BigorU [14]

Answer:

B) how steep the slope is

Explanation:

Because you have to know how is the influence of the steep of the slope in the time that a ball reaches the bottom. The steep of the slope is the variable that you would have to change in an experiment.

I hope this is useful for you

regards

4 0

3 years ago

The latent heat of fusion of water at 0 °C is 6.025 kJ mol'' and the molar heat

Scilla [17]

Answer:

$\Delta H_{tot} = 2258.025\,kJ$

Explanation:

The amount of heat released from water is equal to the sum of latent and sensible heats. Let suppose that water is initially at a temperature of $25^{\circ}C$ . Then:

$\Delta H_{tot} = \Delta H_{s, w} + \Delta H_{f,w} + \Delta H_{s,i}$

$\Delta H_{tot} = n\cdot (c_{w}\cdot \Delta T_{w} + L_{f} + c_{i}\cdot \Delta T_{i})$

Finally, the amount of heat released from water is now computed by replacing variables:

$\Delta H_{tot} = (1\,mol)\cdot \left[\left(75.3\,\frac{kJ}{mol\cdot K} \right)\cdot (25^{\circ}C-0^{\circ}C)+ 6.025\,\frac{kJ}{mol} + \left(37.7\,\frac{kJ}{mol\cdot K} \right)\cdot (0 + 10^{\circ}C)\right]$ $\Delta H_{tot} = 2258.025\,kJ$

4 0

3 years ago

What type of composition does an igneous rock have that contains mostly plagioclase feldspar and about 30 percent dark silicate

mojhsa [17]

The type of composition that igneous rocks have that contains mostly plagioclase feldspar and about 30 percent dark silicate minerals would be andesitic. I am hoping that this answer has satisfied your query about and it will be able to help you.

5 0

3 years ago

Read 2 more answers

A ladder rests against a vertical wall at a point 12 feet from the floor. The angle formed by the ladder and the floor is 63°. C

GenaCL600 [577]

Answer:

length of the ladder is 13.47 feet

base of wall to latter distance 6.10 feet

angle between ladder and the wall is 26.95°

Explanation:

given data

height h = 12 feet

angle 63°

to find out

length of the ladder ( L) and length of wall to ladder ( A) and angle between ladder and the wall

solution

we consider here angle between base of wall and floor is right angle

we apply here trigonometry rule that is

sin63 = h/L

put here value

L = 12 / sin63

L = 13.47

so length of the ladder is 13.47 feet

and

we can say

tan 63 = h / A

put here value

A = 12 / tan63

A = 6.10

so base of wall to latter distance 6.10 feet

and

we say here

tanθ = 6.10 / 12

θ = 26.95°

so angle between ladder and the wall is 26.95°

8 0

3 years ago

The cornea behaves as a thin lens of focal lengthapproximately 1.80 {\rm cm}, although this varies a bit. The material of whichi

Keith_Richards [23]

Answer:

Explanation:

a )

from lens makers formula

$\frac{1}{f} =(\mu-1)(\frac{1}{r_1} -\frac{1}{r_2})$

f is focal length , r₁ is radius of curvature of one face and r₂ is radius of curvature of second face

putting the values

$\frac{1}{1.8} =(1.38-1)(\frac{1}{.5} -\frac{1}{r_2})$

1.462 = 2 - 1 / r₂

1 / r₂ = .538

r₂ = 1.86 cm .

= 18.6 mm .

b )

object distance u = 25 cm

focal length of convex lens f = 1.8 cm

image distance v = ?

lens formula

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$\frac{1}{v} - \frac{1}{-25} = \frac{1}{1.8}$

$\frac{1}{v} = \frac{1}{1.8} -\frac{1}{25}$

.5555 - .04

= .515

v = 1.94 cm

c )

magnification = v / u

= 1.94 / 25

= .0776

size of image = .0776 x size of object

= .0776 x 10 mm

= .776 mm

It will be a real image and it will be inverted.

5 0

3 years ago