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3 years ago

→13 Points← List the four key principles for Newton’s 3rd law:

Physics

1 answer:

Paha777 [63]3 years ago

8 0

The third law states, all forces between two objects exist in equal magnitude and opposite direction.

for example, If you push against something it puses back against you. Your hand pushes on the table, and the table pushes back just as hard against your hand. If it didn't push back, your hand would go straight through the table.

third law of motion is also known as law of action and reaction.

hope this helps

mark this as a brainliest if it helped !:)

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1)Light of wavelength 588.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 55.5 cm from the slit

Talja [164]

Answer:

These are Diffraction Grating Questions.

Q1. To determine the width of the slit in micrometers (μm), we will need to use the expression for distance along the screen from the center maximum to the nth minimum on one side:

Given as

y = nDλ/w Eqn 1

where

w = width of slit

D = distance to screen

λ = wavelength of light

n = order number

Making x the subject of the formula gives,

w = nDλ/y

Given

y = 0.0149 m

D = 0.555 m

λ = 588 x 10-9 m

and n = 3

w = 6.6x10⁻⁵m

Hence, the width of the slit w, in micrometers (μm) = 66μm

Q2. To determine the linear distance Δx, between the ninth order maximum and the fifth order maximum on the screen

i.e we have to find the difference between distance along the screen (y₉-y₅) = Δx

Recall Eqn 1, y = nDλ/w

given, D = 27cm = 0.27m

λ = 632 x 10-9 m

w = 0.1mm = 1.0x10⁻⁴m

For the 9th order, n = 9,

y₉ = 9 x 0.27 x 632 x 10-9/ 1.0x10⁻⁴m = 0.015m

Similarly, for n = 5,

y₅ = 5x 0.27 x 632 x 10-9/ 1.0x10⁻⁴m = 0.0085m

Recall, Δx = (y₉-y₅) = 0.015 - 0.0085 = 0.0065m

Hence, the linear distance Δx between the ninth order maximum and the fifth order maximum on the screen = 6.5mm

8 0

3 years ago

A large mass collides with a stationary, smaller mass. How will the masses behave if the collision is inelastic?

iragen [17]

Logically both masses will collide and well make a reaction. first of all depending on the small mass it will either merge or unite with the big mass or it will bounce away from it . if this happen it will make a reaction that will affect both masses. Hope this helps if it is incorrect please let me know :)

3 0

3 years ago

An amusement park ride raises people high into the air, suspends them for a moment, and then drops them at a rate of free-fall a

blsea [12.9K]

Answer: apparent weighlessness.

Explanation:

1) Balance of forces on a person falling:

i) To answer this question we will deal with the assumption of non-drag force (abscence of air).

ii) When a person is dropped, and there is not air resistance, the only force acting on the person's body is the Earth's gravitational attraction (downward), which is the responsible for the gravitational acceleration (around 9.8 m/s²).

iii) Under that sceneraio, there is not normal force acting on the person (the normal force is the force that the floor or a chair exerts on a body to balance the gravitational force when the body is on it).

2) This is, the person does not feel a pressure upward, which is he/she does not feel the weight: freefalling is a situation of apparent weigthlessness.

3) True weightlessness is when the object is in a place where there exists not grativational acceleration: for example a point between two planes where the grativational forces are equal in magnitude but opposing in direction and so they cancel each other.

Therefore, you conclude that, assuming no air resistance, a person in this ride experiencing apparent weightlessness.

3 0

2 years ago

Read 2 more answers

A ceiling fan has three blades. The moment of inertia of a blade is 0.2kgm^2. The net torque exerted on fan blades is 8Nm. Find

olchik [2.2K]

Answer:

(A) the angular acceleration of the blades is 13.33 m/s.

Explanation:

Given;

moment of inertia of a blade, I = 0.2 kgm²

net torque exerted on fan blades, ∑τ = 8Nm

Torque is given as product of moment of inertia and angular acceleration;

τ = Iα

where;

α is the angular acceleration

Since there are three blades of the ceiling fan, the net torque is given as;

∑τ = (3I)α

∑τ = 3Iα

α = ∑τ / 3I

α = (8) / (3 x 0.2)

α = 13.33 m/s

Therefore, the angular acceleration of the blades is 13.33 m/s.

8 0

2 years ago

You have been assigned to investigate a traffic accident. The masses of car A and car B are 1300 kg and 1200 kg, respectively. C

jarptica [38.1K]

Answer:

The velocity of A before impact = 17.90 m/s

Explanation:

Coefficient of restitution = (speed of seperation)/(speed of approach)

= (v₁ - v₂)/(u₂ - u₁)

where v₁ = velocity of the car A after the impact = ?

v₂ = velocity of the car B after the impact = ?

u₂ = velocity of the car B before the impact = 0 m/s (it was initially at rest)

u₁ = velocity of car A before the impact = ?

First of, we can solve for v₂, the velocity of car B after the impact, from some of the information given in the question.

- Skid marks indicate car B slid 10 m after the impact

- The coefficient of kinetic friction the tires and road is 0.8.

According to the work energy theorem, the work done by frictional force in stopping the car B is equal to the change in kinetic energy of the car B. (All after collision)

W = ΔK.E

ΔK.E = (1/2)(1200)(v₂²) - 0 (final kinetic energy is 0 since the car comes to stop eventually)

ΔK.E = (600v₂²) J

W = F × d

where F = frictional force = μmg = 0.8×1300×9.8 = 10,192 N

d = distance the car skids over before stopping = 10 m

W = 10,192 × 10 = 101,920 J

W = ΔK.E

101,920 = 600v₂²

v₂² = (101920/600) = 169.867

v₂ = 13.03 m/s

But recall,

Coefficient of restitution = (v₁ - v₂)/(u₂ - u₁)

For the sake of convention, we take the direction of car A's initial velocity to be the positive direction.

u₁ = ?

u₂ = 0 m/s

v₁ = ?

v₂ = +13.03 m/s

Coefficient of restitution = 0.4

0.4 = (v₁ - 13.03)/(0 - u₁)

-0.4u₁ = v₁ - 13.03

v₁ = 13.03 - 0.4u₁

But this is a collision. In a collision, the linear momentum is usually conserved.

Momentum before collision = Momentum after collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

1300u₁ + (1200×0) = 1300v₁ + (1200×13.03)

1300u₁ + 0 = 1300v₁ + 15639.95

1300u₁ = 1300v₁ + 15639.95

But recall, from the coefficient of restitution relation,

v₁ = 13.03 - 0.4u₁

Substituting this into the momentum balance equation.

1300u₁ = 1300v₁ + 15639.95

1300u₁ = 1300(13.03 - 0.4u₁) + 15639.95

1300u₁ = 16943.28 - 520u₁ + 15639.95

1820u₁ = 32,583.23

u₁ = (32,583.23/1820)

u₁ = 17.90 m/s

Therefore, the velocity of A before impact = 17.90 m/s

Hope this Helps!!!

4 0

3 years ago