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Compute the density in g/cm3 of a piece of metal that has a mass of 0.530 kg and a volume of 79 cm3

irinina [24]

Answer:

Density =6.71g/cm³

Explanation:

Density =mass /volume

0.530kg = 530g

Density =530/79= 6.7089

7 0

3 years ago

A 65-cm segment of conducting wire carries a current of 0.35 A. The wire is placed in a uniform magnetic field that has a magnit

dimaraw [331]

Answer:

e) 67°

Explanation:

the force on the wire can be calculated using the expression below

F = BILsinФ

But we are looking for the angle between the wire segment and the magnetic field, then we can make Ф the subject of the formula from the above expresion, then we have,

Ф =sin⁻¹ (F/BIL)

The parameters is defined as

I =current that is been carried by the wire= 0.35 A

Ф = angle between the wire segment and the magnetic field, which is the unknown?

L = length of the wire=65 cm

B = magnetic field = 1.24

F= force on the wire = 0.26 N,

Ф =sin⁻¹ (F/BIL)

Ф =sin⁻¹ X .....................eqn(#)

Where X= (F/BIL)

We can calculate for X= (F/BIL), from eqn(#) by substituting value of Force, Lenght and

magnetic field

X=(F/BIL)= 0.26/(1.24×0.35×0.65)

= 0.26/0.2821

=0.922

Then substitute X into eqn (Ф =sin⁻¹ X)

Then

Ф =sin⁻¹ (0.922)

Ф=66.42°

Ф=67° approximately

Therefore, the angle between the wire segment and the magnetic field is 67°

7 0

3 years ago

7. A force stretches a wire by 1 mm. a. A second wire of the same material has the same cross section and twice the length. How

Lera25 [3.4K]

Answer:

(a) The second wire will be stretched by 2 mm

(b) The third wire will be stretched by 0.25 mm

Explanation:

Tensile stress on every engineering material is given as the ratio of applied force to unit area of the material.

σ = F / A

Tensile strain on every engineering material is given as the ratio of extension of the material to the original length

δ = e / L

The ratio of tensile stress to tensile strain is known as Young's modulus of the material.

$Y = \frac{FL}{Ae}$

Part A

cross sectional area and applied force are the same as the original but the length is doubled

$\frac{FL_1}{A_1e_1} = \frac{FL_o}{A_oe_o} \\\\\frac{L_1}{e_1} = \frac{L_o}{e_o}\\\\e_1 = \frac{L_1e_o}{L_o} \\\\But, L_1 =2L_o\\\\e_1 = \frac{2L_oe_o}{L_o}\\e_1 = 2e_o$

The second wire will be stretched by 2 mm

Part B

a third wire with the same length but twice the diameter of the first

$\frac{FL}{A_1e_1} = \frac{FL}{A_oe_o} \\\\\frac{1}{A_1e_1} = \frac{1}{A_oe_o}\\\\\frac{4}{\pi d_1^2e_1} = \frac{4}{\pi d^2_oe_o}\\\\\frac{1}{d_1^2e_1} = \frac{1}{d^2_oe_o}\\\\d_1^2e_1 = d^2_oe_o\\\\e_1 = \frac{d^2_oe_o}{d_1^2} \\\\e_1 =(\frac{d_o}{d_1})^2e_o\\\\But, d_1 = 2d_o\\\\e_1 =(\frac{d_o}{2d_o})^2e_o\\\\e_1 =(\frac{1}{2})^2e_o\\\\e_1 =(\frac{1}{4})e_o$

e₁ = ¹/₄ x 1 mm = 0.25 mm

The third wire will be stretched by 0.25 mm

3 0

4 years ago

HELP ME ANSWER THIS PLZ I WILL GIVE BRAINLIEST

disa [49]

Answer:

In mitosis, the products are identical to the parent cell, while in meiosis the products are different from the parent cell.

Explanation:

Mitosis replicates DNA at the beginning of the process, so it's always the same, but Meiosis uses a special process called crossing over which causes two chromosomes to swap information, which means that the DNA has changed.

5 0

3 years ago

The graph shows projected changes in the populations of the world.

malfutka [58]

Answer: C. Africa

Explanation:

The data given on the graph shows that:

Asia will grow from around 1,300 million to 5,000 million in 2050 which is an increase of:

= 5,000 /1,300 = 3.84 times

Europe will decrease over that period.

Africa will go from around 300 million to 2,000 million which is an increase of:

= 2,000 / 300

= 6.67 times

Mexico

, Central America, Caribbean Islands:

= 900 / 120

= 7.5 times

United States, Canada, and Greenland:

= 400/120

= 3.33

Oceania:

= 50 / 13

= 3.85

From the options give, Africa will see the greatest increase at 6.67 times its population in 1950.

3 0

3 years ago

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