How we can best share the observation of the light through one hole the paper

Alan wrote the following examples of changes in substance.

Yuliya22 [10]

A campfire being lighted and plants converting carbon-dioxide and water into glucose and oxygen are both forms of chemical change.

Therefore, the answer is:

B. Both are examples of chemical change.

5 0

3 years ago

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The gravitational force experienced by Earth due to the Moon is ________ the gravitational force experienced by the Moon due to

Vsevolod [243]

The gravitational force experienced by Earth due to the Moon is equal to the gravitational force experienced by the Moon due to Earth.

Explanation:

The force that attracts any two objects/bodies with mass towards each other is defined as gravitational force. Generally the gravitational force is attractive, as it always pulls the masses together and never pushes them apart.

The gravitational force can be calculated effectively using the following formula: F=GMmr^2

where “G” is the gravitational constant.

Though gravity has the ability to pull the masses together, it is the weakest force in the nature.

The mass of the Earth and moon varies, but still the gravitational force felt by the Earth and Moon are alike.

5 0

3 years ago

What is R2 in the circuit? WILL GIVE BRAINLIEST !!!!

Nataly_w [17]

Answer:

1. Rₑq = 4 Ω

2. R₂ = 6 Ω

3. Vₜ = 12 V, V₁ = 12 V, V₂ = 12 V

4. Iₜ = 3 A, I₁ = 1 A, I₂ = 2 A

Explanation:

1. Determination of the equivalent resistance

Voltage (V) = 12 V

Current (I) = 3 A

Resistance (Rₑq) =?

V= IRₑq

12 = 3 × Rₑq

Divide both side by 3

Rₑq = 12 / 3

Rₑq = 4 Ω

Thus, the equivalent resistance (Rₑq) = 4 Ω

2. Determination of R₂.

Equivalent resistance (Rₑq) = 4 Ω

Resistance 1 (R₁) = 12 Ω

Resistance 2 (R₂)

Since the resistor are in parallel arrangement, the value of R₂ can be obtained as follow:

Rₑq = R₁ × R₂ / R₁ + R₂

4 = 12 × R₂ / 12 + R₂

Cross multiply

4(12 + R₂) = 12R₂

48 + 4R₂ = 12R₂

Collect like terms

48 = 12R₂ – 4R₂

48 = 8R₂

Divide both side by 8

R₂ = 48 / 8

R₂ = 6 Ω

3. Determination of the total voltage (Vₜ), V₁ and V₂.

From the question given above, the total voltage is 12 V

Since the resistors are arranged in parallel connection, the same voltage will go through them.

Thus,

Vₜ = V₁ = V₂ = 12 V

4. Determination of the total current (Iₜ), I₁ and I₂

From the question given above, the total current (Iₜ) is 3 A

Next, we shall determine I₁. Since the resistors are arranged in parallel connection, different current will pass through each resistor respective.

Vₜ = V₁ = 12 V

R₁ = 12 Ω

I₁ =?

V₁ = I₁R₁

12 = I₁ ×12

Divide both side by 12

I₁ = 12 / 12

I₁ = 1 A

Next, we shall determine I₂. This can be obtained as follow:

Iₜ = 3 A

I₁ = 1 A

I₂ =?

Iₜ = I₁ + I₂

3 = 1 + I₂

Collect like terms

I₂ = 3 – 1

I₂ = 2 A

5 0

3 years ago

Two wires are made from the same material. One wire has a resistance of 0.10 ω. The other wire is twice as long as the first wir

Anna71 [15]

Answer:

Explanation:

Given that,

First wire has a resistance of

R1 = 0.1 Ω

We are told that

Second wire is twice as long as the first wire.

Then,

Let the first wire has length

L1 = x

Then, second wire will have

L2 = 2x

Also, the radius of the second wire is half the radius of the first wire

Let the first wire has radius

r1 = y

Then, it area is

A1 = πr1² = πy²

Then, the second wire has radius

r2 = ½y

It area is also

A2 = πr2² = π(½y)² = ½πy²

Since the wire is made of the same material, then, they will have the same resistivity ρ

Then, we want to find the resistance of the second wire

Using

R = ρL/A

Where

R = resistance

ρ = sensitivity

L = length

A = area

Then, make resistivity subject of formula since it is a constant

RA = ρL

Then, ρ = RA/L

Then,

R1 • A1 / L1 = R2 • A2 / L2

Substituting each value

0.1 × πy² / x = R2 × ¼πy² / 2x

Cross multiply

0.1 × πy² × 2x = R2 ×¼π y² × x

Then,

R2 = 0.1 × πy² × 2x / ¼πy² × x

R2 = 0.1 × 2 / ¼

R2 = 0.8Ω

The resistance of the second wire is 0.8Ω

6 0

4 years ago

A thin uniform cylindrical turntable of radius 2.2 m and mass 35 kg rotates in a horizontal plane with an initial angular speed

fenix001 [56]

Explanation:

The given data is as follows.

M = 35 kg, radius (r) = 2.2 m,

m = 17 kg, = 11 rad/s

We assume that will be the final angular speed.

Now, according to the conservation of angular momentum.

$L_{1} = L_{2}$

or, $I_{1} \times \omega_{1} = I_{2} \times \omega_{2}$

Putting the given values into the above formula as follows.

$I_{1} \times \omega_{1} = I_{2} \times \omega_{2}$

or,

= $\frac{(0.5 \times 35 \times (2.2)^{2}) \times 11}{(0.5 \times 35 \times (2.2)^{2} + 17 \times (1.5)^{2})}$

= 7.58 rad/s

Thus, we can conclude that the angular speed of the clay and turntable is 7.58 rad/s.

4 0

4 years ago

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