Do magnets have to touch each other for a magnetic field to be present?

Present day glaciers are found primarily in _______________.

Mazyrski [523]

<h3><u>Answers;</u></h3>

Antarctica and Greenland

Present day glaciers are found primarily in <em><u>Antarctica and Greenland</u></em>.

<h3><u>Explanation;</u></h3>

<em><u>The two major ice sheets that exists today are found primarily in Antarctica and Greenland. Ice sheets are large masses of glacial ice that are also known as continental glaciers.</u></em>
Most ice in Antarctica and Greenland spill out into the ocean from a few spots. The Antarctica and Greenland ice sheets combined comprise more than 99 percent of freshwater ice found on Earth.

8 0

3 years ago

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State one way to decrease the moment of a given force about a given axis of rotation.

wlad13 [49]

Answer:

The moment of a given force about a given axis of rotation can be decreased by decreasing the perpendicular distance of force from the axis of rotation.

8 0

3 years ago

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Find the volume of a sphere of radius 10 mm.

vredina [299]

Answer:

Explanation: This is done using the equation:

$\frac{4}{3} π R^{3}$

Because the Radius is a know value. We have the following.

$\frac{4}{3} π (10mm)^{3}$

Which is:

4188.7902 mm

5 0

3 years ago

An object with a mass M = 250 g is at rest on a plane that makes an angle θ = 30 o above the horizontal. The coefficient of kine

liubo4ka [24]

Answer:

v = 79.2 m/s

Solution:

As per the question:

Mass of the object, m = 250 g = 0.250 kg

Angle, $\theta = 30^{\circ}$

Coefficient of kinetic friction, $\mu_{k} = 0.100$

Mass attached to the string, m = 0.200 kg

Distance, d = 30 cm = 0.03 m

Now,

The tension in the string is given by:

$Mgsin\theta + \mu_{k}Mgcos\theta + Ma = T$ (1)

Also

T = m(g + a)

Thus eqn (1) can be written as:

$Mgsin\theta + \mu_{k}Mgcos\theta + Ma = m(g - a)$

$Mgsin\theta + \mu_{k}Mgcos\theta + Ma = mg + ma$

$mg - Mgsin\theta - \mu_{k}Mgcos\theta = (M - m)a$

$a = \frac{0.2\times 9.8 - 0.250\times 9.8\times sin30^{\circ} - 0.1\times 0.250\times 9.8\times cos30^{\circ}}{0.250 - 0.200}$

$a = 10.45\ m/s^{2}$

Now, the speed is given by the third eqn of motion with initial velocity being zero:

$v^{2} = u^{2} + 2ad$

where

u = initial velocity = 0

Thus

$v = \sqrt{2ad}$

$v = \sqrt{2\times 10.45\times 0.03} = 0.792\ m/s$

3 0

3 years ago

A flat coil of wire has an inductance of 40. 0 mh and a resistance of 6. 00 ω. It is connected to a 21. 2-v battery at the insta

MissTica

For a flat coil of wire has an inductance of 40. 0 mh and a resistance of 6. 00 ω, the rate of energy being delivered is mathematically given as

P= 53 W

<h3>What rate is energy being delivered by the battery?</h3>

Generally, the equation for the Battery power is mathematically given as

P = I (dt)V

Therefore

P= 2.50 A * 21.2V

P= 53 W

In conclusion, rate of energy being delivered

P= 53 W