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3 years ago

By the end of the 19th century, more than half of the population believed in the presence of atoms.

Physics

1 answer:

suter [353]3 years ago

6 0

Answer:

hello your question is vague hence I will provide a general answer about what was known about atoms in the 19th Century

answer :

By the end of the 19th century more than half of the population believed in the presence of atoms and also in the 19th Century is was a known fact that atoms determine the chemical properties of an element

Explanation:

By the end of the 19th century more than half of the population believed in the presence of atoms and also by the 19th Century is was a known fact that atoms determine the chemical properties of an element.

John Dalton reintroduced the presence of atoms in 1800 with evidence which he used to develop atomic theory. with this theory people believed in the presence of atoms in nature and also that atoms determine the chemical properties of an element .

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A 25,000 kg traveling east collides with a 2,000 kg truck standing still on the tracks. After the collision the train and truck

Elis [28]

Answer:

24.084 m/s

Explanation:

From the law of conservation of linear momentum

Total momentum before collision equals to the total momentum after collision

Since momentum=mv where m is mass and v is velocity

$M_{truck}V_{truck}=V_{common}*(M_{truck} +M_{standing})$ where $M_{truck}$ is the mass of the truck, $V_{truck}$ is velocity of the truck, $V_{common}$ is the common velocity of moving and standing truck after collision and $M_{standing}$ is the mass of the standing truck

Making $V_{truck}$ the subject we obtain

$V_{truck}=\frac { V_{common}*(M_{truck} +M_{standing})}{M_{truck}}$

Substituting $M_{truck}$ as 25000 Kg, $V_{common}$ as 22.3 m/s, $M_{standing}$ as 2000 Kg we obtain

$V_{truck}=\frac { 22.3 m/s *(25000 Kg +2000 Kg)}{25000}= 24.084 m/s$

Therefore, assuming no friction and considering that after collision they still move eastwards hence common velocity and initial truck velocities are positive

The truck was moving at 24.084 m/s

3 0

3 years ago

Sasha sits on a horse on a carousel 3.5 m from the center of the circle. She makes a revolution once every 8.2 seconds. What is

Leokris [45]

Answer: 2.7 m/s

Explanation:

Given the following :

Period (T) = 8.2 seconds

Radius = 3.5 m

The tangential speed is given as:

V = Radius × ω

ω = angular speed = (2 × pi) / T

ω = (2 × 22/7) / 8.2

ω = 6.2857142 / 8.2

ω = 0.7665505

Therefore, tangential speed (V) equals;

r × ω

3.5 × 0.7665505 = 2.6829268 m/s

2.7 m/s

6 0

3 years ago

The first step in a star formation is ______?

Mariana [72]

First look at the star

8 0

3 years ago

Read 2 more answers

Block A of mass M is on a horizontal surface of negligible friction. An identical block B is attached to block A by a light stri

miv72 [106K]

Answer:

T’= 4/3 T

The new tension is 4/3 = 1.33 of the previous tension the answer e

Explanation:

For this problem let's use Newton's second law applied to each body

Body A

X axis

T = m_A a

Axis y

N- W_A = 0

Body B

Vertical axis

W_B - T = m_B a

In the reference system we have selected the direction to the right as positive, therefore the downward movement is also positive. The acceleration of the two bodies must be the same so that the rope cannot tension

We write the equations

T = m_A a

W_B –T = M_B a

We solve this system of equations

m_B g = (m_A + m_B) a

a = m_B / (m_A + m_B) g

In this initial case

m_A = M

m_B = M

a = M / (1 + 1) M g

a = ½ g

Let's find the tension

T = m_A a

T = M ½ g

T = ½ M g

Now we change the mass of the second block

m_B = 2M

a = 2M / (1 + 2) M g

a = 2/3 g

We seek tension for this case

T’= m_A a

T’= M 2/3 g

Let's look for the relationship between the tensions of the two cases

T’/ T = 2/3 M g / (½ M g)

T’/ T = 4/3

T’= 4/3 T

The new tension is 4/3 = 1.33 of the previous tension the answer e

4 0

3 years ago

Which is a compound A Kr B O 2 c cl2 d CuF2

Volgvan

I think the answer is CuF2

7 0

3 years ago