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Yakvenalex [24]

2 years ago

9

Joe follows the procedure below to make a compass.

2 answers:

Anna [14]2 years ago

8 0

The answer would be C.

koban [17]2 years ago

7 0

Answer:

C: to magnetize the needle

Explanation:

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A driver averaged 64 mph and took 3½ hours to travel from st. Louis to chicago. Based on this, what is the distance between st.

sammy [17]

Average speed of the driver is given as

$v = 64 mph$

if he moved for total time t = 3.5 hours

so the distance between the tow is given as

$d = v* t$

$d = 64 * 3.5$

$d = 224 miles$

so the distance between St. Louis and Chicago is 224 miles

6 0

3 years ago

A machinist with normal vision has a near point at 25 cm. The machinist wears eyeglasses in order to do close work. The power of

Alik [6]

Answer:

17.4 cm

Explanation:

Power of lens = +1.75 diopters

Focal length of lens

$f=\frac{100}{1.75}=\frac{400}{7}$

This is a convex lens as focal the diopter given is positive which makes the focal length positive. Image distance will be negative.

v = -25

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\\Rightarrow \frac{1}{-25}-\frac{1}{u}=\frac{1}{\frac{400}{7}}\\\Rightarrow -\frac{1}{u}=\frac{7}{400}+\frac{1}{25}\\\Rightarrow \frac{1}{u}=-\frac{7}{400}-\frac{1}{25}\\\Rightarrow \frac{1}{u}=-0.0575\\\Rightarrow u=-17.4\ cm$

∴ The new near point is 17.4 cm

7 0

3 years ago

If 50 km thick crust having an average density of 3.0 g/cm3 has a surface elevation of 2.5 km above sea level, what would you pr

RUDIKE [14]

Answer:

To calculate the predicted surface elevation of a 50km thick crust above a surface of 2.5km we are given a density of 3 gram per centimeter cube.

The displacement of the material will be calculated by subtracting the surface elevation of 2.5 km from the 50 km thick crust. Therefore 50-25= 47.5 km.

Thus let the density of the material be Pm

50*3= 47.5*Pm

Therefore: Pm= (50*3)/47.5= 3.16gram per centimeter cube

Thus with an average density of 2.8gram per centimeter cube

50*2.8= (50-x)*3.16

(50-x)= (50*2.8)/3.16

50-x=44.3

x=50-44.3= 5.7

Explanation:

To calculate the predicted surface elevation of a 50km thick crust above a surface of 2.5km we are given a density of 3 gram per centimeter cube.

The displacement of the material will be calculated by subtracting the surface elevation of 2.5 km from the 50 km thick crust. Therefore 50-25= 47.5 km.

Thus let the density of the material be Pm

50*3= 47.5*Pm

Therefore: Pm= (50*3)/47.5= 3.16gram per centimeter cube

Thus with an average density of 2.8gram per centimeter cube

50*2.8= (50-x)*3.16

(50-x)= (50*2.8)/3.16

50-x=44.3

x=50-44.3= 5.7

5 0

2 years ago

When the displacement of a mass on a spring in simple harmonic motion is A/2 from the equilibrium position, what fraction of the

KonstantinChe [14]

Answer:

The ratio is KE : TM = 0.75

Explanation:

from the question we are told that

The displacement of a mass on a spring in simple harmonic motion is A/2 from the equilibrium position

Generally the total mechanical energy of the mass is mathematically represented as

$TM = \frac{1}{2} * k * A^2$

Here k is the spring constant , A is the total displacement of the the mass from maximum compression to maximum extension of the spring

Generally this total mechanical energy is mathematically represented as

$TM = KE + PE$

=> $KE = TM - PE$

Here the potential energy of the mass is mathematically represented as

$PE = \frac{1}{ 2} * k * [ x ]^2$

Here x is the displacement of the mass from maximum compression or extension of the spring to equilibrium position and the value is

$x = \frac{A}{2}$

So

$PE = \frac{1}{ 2} * k * [ \frac{A}{2} ]^2$

So

$KE = \frac{1}{2} * k * A^2 - \frac{1}{2} * k * [\frac{A}{2} ]^2$

=> $KE = \frac{1}{2} * k * A^2 - \frac{1}{8} * k * A ^2$

=> $KE = 0.375 * k * A^2$

So the ratio of $KE : TM$ is mathematically represented as

$\frac{KE}{TM} = \frac{0.375 k A^2 }{0.5 k A^2}$

=> $\frac{KE}{TM} = 0.75$

3 0

2 years ago

The structural diversity of carbon-based molecules is determined by which properties?

Leokris [45]

Explanation:

The structural diversity of carbon-based molecules is determined by following properties:

1. the ability of those bonds to rotate freely,

2.the ability of carbon to form four covalent bonds,

3.the orientation of those bonds in the form of a tetrahedron.

5 0

2 years ago