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3 years ago

Why do elements within a group have similar chemical properties

1 answer:

4 0

Elements in the same group in the periodic table have similar chemical properties. This is because their atoms have the same number of electrons in the highest occupied energy level. Group 1 elements are reactive metals called the alkali metals.Group 0 elements are unreactive non-metals called the noble gases.

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Kevin decides to soup up his car by replacing the car's wheels with ones that have 1.4 times the diameter of the original wheels

True [87]

Answer:

No.

Explanation:

Given that Kevin decides to soup up his car by replacing the car's wheels with ones that have 1.4 times the diameter of the original wheels. Note that the speedometer in a car is calibrated based on the tire's diameter and on the distance the tire covers in each revolution. (a) Will the reading of the speedometer change ?

Considering the formula

V = wr

Where

V = linear speed

W = angular speed

r = radius of the wheel.

But W = 2πrf

Where the the 2 and pi are constant. The radius of the first wheel will be small but counter balance with the larger frequency.

While the radius of the second wheel may be large but it will be of a small frequency.

We can therefore conclude that the reading on the speedometer will not change. Because speedometer will read the linear speed V.

7 0

3 years ago

A billion years ago, Earth and its moon were just 200000 kilometers apart. Express this distance in meters.

Luda [366]

Answer:

The value is $x = 200000000 \ m$

Explanation:

From the question we are told that

The distance between the earth and the moon is $d = 200000 \ km$

Generally ,

$1 \ km = 1000 \ m$

$200000 \ km \to x \ m$

=> $x = \frac{200000 * 1000}{1 }$

=> $x = 200000000 \ m$

4 0

3 years ago

Refer to Concept Simulation 4.4 for background relating to this problem. The drawing shows a large cube (mass = 28.9 kg) being a

Usimov [2.4K]

Answer:

smallest magnitud is P=33.3 N

Explanation:

We are analyze the situation as an external force is applied and there is a friction force. We have a problem with Newton's second law.

F = ma

As the two blocks go together they must have the same acceleration, so we can calculate this for the entire system

P = (m1 + m2) a

a = P / (m1 + m2)

In this case there is no friction force because the small block does not touch the ground.

In order to calculate the friction force, we must analyze each system component separately.

The large block on the X axis has an applied P force and as it moves feels a force from the small block. In the Y axis has the weight (W1) and the reaction to normal (N1)

For the small block on the X axis, the force it feels is the thrust of the large block, note that this is an action and reaction force between the two blocks, it is the same definition we have of the normal one, so we can call this force (N)

Y axis it has the weight (W2) down, the force of friction (fr) that opposes the movement, so it is directed upwards. we write these equations

N = m2 a

fr -W2 = 0

fr = W2

The definition of friction force is

fr = μ N

Let's replace and calculate

μ (m2 a) = m2 g

μ (P / (m1 + m2)) = g

P = g /μ (m1 + m2)

Let's calculate the value of this force

P = 9.8 / 0.710 (28.9 +4.4)

P = 13.80 (33.3)

P = 33.3 N

This is the minimum friction force that prevents the block from sliding down

6 0

3 years ago

A 46.8-g golf ball is driven from the tee with an initial speed of 58.8 m/s and rises to a height of 24.7 m. (a) Neglect air res

Andre45 [30]

Answer:

a) the kinetic energy of the ball at its highest point is 69.58 J

b) its speed when it is 8.11 m below its highest point is 55.97 m/s

Explanation:

Given that;

mass of golf ball m = 46.8 g = 0.0468 kg

initial speed of the ball v₁ = 58.8 m/s

height h = 24.7 m

acceleration due to gravity = 9.8 m/s²

the kinetic energy of the ball at its highest point = ?

from the conservation of energy;

Kinetic energy at the highest point will be;

K.Ei + P.Ei = KEf + PEf

now the Initial potential energy of the ball P.Ei = 0 J

1/2mv² + 0 J = KEf + mgh

K.Ef = 1/2mv² - mgh

we substitute

K.Ef = [1/2 × 0.0468 × (58.8 )²] - [0.0468 × 9.8 × 24.7]

K.Ef = 80.904 - 11.3284

K.Ef = 69.58 J

Therefore, the kinetic energy of the ball at its highest point is 69.58 J

b) when the ball is 8.11 m below the highest point, speed = ?

so our raw height h' will be ( 24.7 m - 8.11 m) = 16.59 m

so our velocity will be v₂

also using the principle of energy conservation;

K.Ei + P.Ei = KEh + PEh

1/2mv² + 0 J = 1/2mv₂² + mgh'

1/2mv₂² = 1/2mv² - mgh'

multiply through by 2/m

v₂² = v² - 2gh'

v₂ = √( v² - 2gh' )

we substitute

v₂ = √( (58.8)² - 2×9.8×16.59 )

v₂ = √( 3457.44 - 325.164 )

v₂ = √( 3132.276 )

v₂ = 55.97 m/s

Therefore, its speed when it is 8.11 m below its highest point is 55.97 m/s

5 0

3 years ago

A battery supplies ________ to the circuit.

MariettaO [177]

Answer:

Electrical energy

Explanation:

Hope It helps you 

6 0

2 years ago

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