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3 years ago

14

Where are alkaline earth metals found on the periodic table? Group 1 Group 2 Groups 3–12 Group 17'

2 answers:

anzhelika [568]3 years ago

8 0

Answer:

B- Group 2... It says my answer is too short so how ya'll doing today?

kogti [31]3 years ago

8 0

Answer:

Elements of this group are beryllium, magnesium, calcium, strontium, barium and radium. Elements of group 1 are called alkali metals and elements of group 17 are called halogens. Thus, we can conclude that out of the given options alkaline earth metals are found on the periodic table in group 2

Explanation:

Hope this helps:)

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What is work please give ans

Fed [463]

Work= force x displacement :)

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2 years ago

A girl runs once around a circular track with a radius of 100m at speed of 10m/s.

goblinko [34]

Answer:

0 m

Explanation:

Displacement is the shortest distance from one ppont to another. Here, the girl returns to the same spot after running. Hence, displacement is <u>0</u><u> </u><u>m</u>.

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2 years ago

Len [333]

Answer:

$d=5\ g/cm^3$

Explanation:

Given that,

Mass of the object, m = 100 grams

Volume of the object, V = 20 cm³

We need to find the density of the object. We know that, density is equal to mass per unit volume. So,

$d=\dfrac{m}{V}\\\\d=\dfrac{100\ g}{20\ cm^3}\\\\d=5\ g/cm^3$

So, the density of the object is equal to $5\ g/cm^3$ .

6 0

3 years ago

You pull with a force of 77 N on a piece of luggage of mass 23 kg, but it does

Vinvika [58]

Answer:

The force of static friction acting on the luggage is, Fₓ = 180.32 N

Explanation:

Given data,

The mass of the luggage, m = 23 kg

You pulled the luggage with a force of, F = 77 N

The coefficient of static friction of luggage and floor, μₓ = 0.8

The formula for static frictional force is,

Fₓ = μₓ · η

Where,

η - normal force acting on the luggage 'mg'

Substituting the values in the above equation,

Fₓ = 0.8 x 23 x 9.8

= 180.32 N

Hence, the minimum force require to pull the luggage is, Fₓ = 180.32 N

5 0

3 years ago

Read 2 more answers

A 2.0-kg laptop sits on the horizontal surface of the seat of a car moving at 8.0 m/s. The driver starts slowing down to stop. F

ivanzaharov [21]

Answer: $32.65\ m$

Explanation:

Given

mass of laptop m=2 kg

The velocity of car u=8 m/s

The coefficient of static friction is $\mu_s=0.4$

The coefficient of kinetic friction is $\mu_k=0.2$

As the car is moving, so the coefficient of kinetic friction comes into play

deceleration offered by friction $\mu_kg=0.2\times 9.8\ m/s^2$

Using the equation of motion $v^2-u^2=2as\\$

insert the values

$0^2-8^2=2(-0.2\times 9.8)s\\\\s=\dfrac{64}{1.96}\\\\s=32.65\ m$

4 0

3 years ago