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3 years ago

How is momentum conserved is a Newton's cradle when one steel ball hits the other

Physics

1 answer:

Vikentia [17]3 years ago

4 0

Answer:

Newton's Cradle experiment perfectly demonstrates the law of conservation of momentum which states that in a closed system, momentum before the collision is equal to momentum after the collision of the system.

As the first ball swings in the air, it gains momentum. When it strikes the second ball, it loses momentum and second ball gains equal amount of momentum. The second ball transfers the momentum to third, then fourth and till the last. The last ball when gains the same momentum swings up in the air. This continues. This experiment is done in drag free condition. This means there is no loss of momentum or opposing forces present.

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Two forces that are not equal in size are

Bumek [7]

They are unbalanced forces ..... Hope this helps :3

4 0

3 years ago

What is the best use of an atomic model to explain the charge of the particles in thomson's beams

hoa [83]

The best use of an atomic model to explain the charge of the particles in Thomson's beams is:

An atom's smaller negative particles are at a distance from the central positive particles, so the negative particles are easier to remove.

Explanation:

In Thomson's model, an atom comprises of electrons that are surrounded by a group of positive particles to equal the electron's negative particles, like negatively charged “plums” that are surrounded by positively charged “pudding”.

Atoms are composed of a nucleus that consists of protons and neutrons . Electron was discovered by Sir J.J.Thomson. Atoms are neutral overall, therefore in Thomson’s ‘plum pudding model’:

atoms are spheres of positive charge
electrons are dotted around inside

Thomson's conclusions made him to propose the Rutherford model of the atom where the atom had a concentrated nucleus of positive charge and also large mass.

6 0

3 years ago

Here, mA = 3.65 kg and mB = 7.05 kg. The string connecting the two objects is of negligible mass and the pulley is frictionless.

mr_godi [17]

Answer:

The magintude of the acceleration for both objects is $3.11m/s^2$

Explanation:

Drawing a free body diagram on the two boxes we can analyze the system more easily.

we can take the acceleration going up as positive for reference purposes.

for mA let's suppose that is ascending so:

$T_A-m_A*g=m_A*a$

and for mB (descending):

$T_B-m_B*g=-m_B*a$

$T_A=T_B$

because the two boxes has the same acceleration because they are attached together:

$m_B*g-m_B*a-m_a*g=m_a*a\\(m_B-m_A)*g=(m_a+m_B)*a\\a=\frac{(7.05kg-3.65kg)*9.8m/s^2}{3.65kg+7.05kg}\\\\a=3.11m/s^2$

So the magintude of the acceleration for both objects is $3.11m/s^2$

5 0

3 years ago

The design speed of a multilane highway is 60 mi/hr. What is the minimum stopping sight distance that should be provided on the

kicyunya [14]

Answer:

Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.

Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.

Explanation:

Part a

When Road is Level

The stopping sight distance is given as

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}$

Here

SSD is the stopping sight distance which is to be calculated.
u is the speed which is given as 60 mi/hr
t is the perception-reaction time given as 2.5 sec.
a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
G is the grade of the road, which is this case is 0 as the road is level

Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0)}\\SSD=220.5 +342.86 ft\\SSD=563.36 ft$

So the minimum stopping sight distance is 563.36 ft.

Part b

When Road has a maximum grade of 4%

The stopping sight distance is given as

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}$

Here

SSD is the stopping sight distance which is to be calculated.
u is the speed which is given as 60 mi/hr
t is the perception-reaction time given as 2.5 sec.
a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade

For upgrade of 4%, Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35+0.04)}\\SSD=220.5 +307.69 ft\\SSD=528.19 ft$

So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.

For downgrade of 4%, Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0.04)}\\SSD=220.5 +387.09 ft\\SSD=607.59ft$

So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.

As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.

3 0

3 years ago

A blue line with 5 orange tick marks then one red tick mark then 4 orange tick marks. The number zero is above the red tick mark

Digiron [165]

Answer:

16 cm

Explanation:

Given that,

The object begins from 0 and moves 3cm towards left side followed by 7 cm towards the right and then, 6 cm towards the left side.

Let the x-axis to be the +ve and on the right side and -ve on the left

Thus, displacement would be:

= 0 -3 + 7 -6

= -2 cm

This implies that the object displaces 2cm towards the left.

While the total distance covered by the object equal to,

= 0cm + 3cm + 7cm + 6cm

= 16 cm

Thus, 16 cm is the total distance.

3 0

3 years ago