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3 years ago

Dalton was one of this first scientist to experiment prove that

1 answer:

5 0

John Dalton. Modern Atomic Theory (John Dalton) Experiments with gases that first became possible at the turn of the nineteenth century led John Dalton in 1803 to propose a modern theory of the atom based on the following assumptions. 1. Matter is made up of atoms that are indivisible and indestructible.

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If a small rock is dropped from a height of 3.1 m how fast will it be moving when it reaches the ground 0.80 seconds later

garik1379 [7]

X=1/2 at^2
3.1=1/2 a *0.64
a=9.68
v=at
v=0.8*9.6875=7.75

5 0

3 years ago

True or false gravitational pull decreases with an increase of distance between two objects

ohaa [14]

Answer:

true! : )

(i underlined the place where the answer is the other information is just as important but if you do not want to read it you do not have to)

Explanation:

Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces. So as two objects are separated from each other, the force of gravitational attraction between them also decreases. the greater the mass, the greater the gravitational pull. <u>gravitational pull decreases with an increase in the distance between two objects.</u> Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces. So as two objects are separated from each other, the force of gravitational attraction between them also decreases.

5 0

3 years ago

(1) Predator-prey is the most important type of ecological relationship.

DIA [1.3K]

1) True 2) True 3)True I hope I helped

4 0

3 years ago

a particle moves at a constant speed in a circular path. the instantaneous velocity and instantaneous acceleration vectors are:

Lilit [14]

The instantaneous velocity always tangential to circular path and instantaneous acceleration always points towards the center of circle.

What is instantaneous?

The definition of instantaneous velocity is the rate of change of location during a time that is virtually zero .The replica was. The definition of instantaneous velocity is the speed of a moving item at a certain instant in time.

The rate of change of location during a relatively brief period of time is known as instantaneous velocity.

In the limit when the time (and thus the displacement) between the two places approaches zero, the instantaneous acceleration is the average acceleration between two points on the route.

so, both of them are perpendicular to each other.

To learn about instantaneous velocity

brainly.com/question/1222392

#SPJ4

8 0

1 year ago

How to solve it? Three capacitors with capacities of 600 pF, 300 pF, 200 pF are connected in series. The 60 V voltage is applied

adell [148]

Answer:

1. Voltage across 600 pF is 10 V.

2. Voltage across 300 pF is 20 V.

3. Voltage across 200 pF is 30 V.

Explanation:

We'll begin by calculating the total capacitance of capacitor. This can be obtained as follow:

Capicitance 1 (C₁) = 600 pF

Capicitance 2 (C₂) = 300 pF

Capicitance 3 (C₃) = 200 pF

Total capacitance (Cₜ) =?

1/Cₜ = 1/C₁ + 1/C₂ + 1/C₃

1/Cₜ = 1/600 + 1/300 + 1/200

1/Cₜ = 1 + 2 + 3 / 600

1/Cₜ = 6/600

1/Cₜ = 1/100

Cₜ = 100 pF

Next, we shall convert 100 pF to Farad (F). This can be obtained as follow:

1 pF = 1×10¯¹² F

Therefore,

100 pF = 100 pF × 1×10¯¹² F / 1 pF

100 pF = 1×10¯¹⁰ F

Thus, 100 pF is equivalent to 1×10¯¹⁰ F.

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

Voltage (V) = 60 V

Capicitance (C) = 1×10¯¹⁰ F

Charge (Q) =?

Q = CV

Q = 60 × 1×10¯¹⁰ F

Q = 6×10¯⁹ C

1. Determination of the voltage across 600 pF.

Capicitance 1 (C₁) = 600 pF = 6×10¯¹⁰ F

Charge (Q) = 6×10¯⁹ C

Voltage 1 (V₁) =?

Q = C₁V₁

6×10¯⁹ = 6×10¯¹⁰ × V₁

Divide both side by 6×10¯¹⁰

V₁ = 6×10¯⁹ / 6×10¯¹⁰

V₁ = 10 V

2. Determination of the voltage across 300 pF.

Capicitance 2 (C₂) = 300 pF = 3×10¯¹⁰ F

Charge (Q) = 6×10¯⁹ C

Voltage 2 (V₂) =?

Q = C₂V₂

6×10¯⁹ = 3×10¯¹⁰ × V₂

Divide both side by 3×10¯¹⁰

V₂ = 6×10¯⁹ / 3×10¯¹⁰

V₂ = 20 V

3. Determination of the voltage across 200 pF.

Capicitance 3 (C₃) = 200 pF = 2×10¯¹⁰ F

Charge (Q) = 6×10¯⁹ C

Voltage 3 (V₃) =?

Q = C₃V₃

6×10¯⁹ = 2×10¯¹⁰ × V₃

Divide both side by 2×10¯¹⁰

V₃ = 6×10¯⁹ / 2×10¯¹⁰

V₃ = 30 V

7 0

3 years ago