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2 years ago

3. True or false. All objects that are made of metal are magnetic. Explain why or why not.

Physics

1 answer:

Eduardwww [97]2 years ago

7 0

$false$

Refer to attachment for your answer

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Name an element in the same period as rubidium

Artyom0805 [142]

Answer: Potassium(K)

Explanation:

its an alkali metal placed under sodium and its over rubidium, its also the first element of period 4

5 0

2 years ago

A force of 20 N is executed to raise a rock weighing 30 N. What is the actual mechanical advantage?

xxMikexx [17]

I am absolutely sure its 1.5

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3 years ago

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A car drives for 30 km with a speed of 30m/s. How much time does it take the car to travel this distance?

Artist 52 [7]

   Time = (distance) / (speed)

   = (30 km) / (30 m/s)

   = (30,000 m) / (30 m/s)

   = (30,000 / 30) sec

   =    1,000 seconds

   =    16 minutes 40 seconds

6 0

3 years ago

One of the most important properties of materials in many applications is strength. Two of the qualitative measures of the stren

katrin2010 [14]

To solve this exercise it is necessary to take into account the concepts related to Tensile Strength and Shear Strenght.

In Materials Mechanics, generally the bodies under certain loads are subject to both Tensile and shear strenghts.

By definition we know that the tensile strength is defined as

$\sigma = \frac{F}{A}$

Where,

$\sigma =$ Tensile strength

F = Tensile Force

A = Cross-sectional Area

In the other hand we have that the shear strength is defined as

$\sigma_y = \frac{F_y}{A}$

where,

$\sigma_y =$ Shear strength

$F_y =$ Shear Force

$A_0 =$ Parallel Area

PART A) Replacing with our values in the equation of tensile strenght, then

$311*10^6 = \frac{F}{(15*10^{-6})(30*10^{-2})}$

Resolving for F,

$F= 1399.5N$

PART B) We need here to apply the shear strength equation, then

$\sigma_y = \frac{F_y}{A}$

$210*10^6 = \frac{F_y}{15*10^{-6}30*10^{-2}}$

$F_y = 945N$

In such a way that the material is more resistant to tensile strength than shear force.

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3 years ago

Saturn has an orbital period of 29.46 years. In two or more complete sentences, explain how to calculate the average distance fr

d1i1m1o1n [39]

For astronomical objects, the time period can be calculated using:
T² = (4π²a³)/GM
where T is time in Earth years, a is distance in Astronomical units, M is solar mass (1 for the sun)
Thus,
T² = a³
a = ∛(29.46²)
a = 0.67 AU
1 AU = 1.496 × 10⁸ Km
0.67 * 1.496 × 10⁸ Km
= 1.43 × 10⁹ Km

3 0

3 years ago

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