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Physics chapter 4:
Forces in action revision:
Newtons 2nd law:
The net force-mass & acceleration
(For a constant mass,
F = ma
AIN
Fner=2-1=1

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Physics chapter 4:
Forces in action revision:
Newtons 2nd law:
The net force-mass & acceleration
(For a constant mass,
F = ma
AIN
Fner=2-1=1

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Physics chapter 4:
Forces in action revision:
Newtons 2nd law:
The net force-mass & acceleration
(For a constant mass,
F = ma
AIN
Fner=2-1=1

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Physics chapter 4:
Forces in action revision:
Newtons 2nd law:
The net force-mass & acceleration
(For a constant mass,
F = ma
AIN
Fner=2-1=1

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Physics chapter 4:
Forces in action revision:
Newtons 2nd law:
The net force-mass & acceleration
(For a constant mass,
F = ma
AIN
Fner=2-1=1

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Physics chapter 4:
Forces in action revision:
Newtons 2nd law:
The net force-mass & acceleration
(For a constant mass,
F = ma
AIN
Fner=2-1=1

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Physics chapter 4:
Forces in action revision:
Newtons 2nd law:
The net force-mass & acceleration
(For a constant mass,
F = ma
AIN
Fner=2-1=1

Sign up

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Access to all documents

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Physics chapter 4:
Forces in action revision:
Newtons 2nd law:
The net force-mass & acceleration
(For a constant mass,
F = ma
AIN
Fner=2-1=1

Sign up

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Physics chapter 4: Forces in action revision: Newtons 2nd law: The net force-mass & acceleration (For a constant mass, F = ma AIN Fner=2-1=1N in upwards. direction. IN -2 1 newton = the force required to accelerate a mass of 1 kg by 1 ms² Rin -2 | N=1 kg x 1 ms ²² Forces: weight fner = IN m Img Upthrust: Friction: Normal Forces: 2 Friction (8) Fnet-ON M w = mg always acts vertically downwards. g=9.81m5²² ↑N = normal 12N W IN Force - 2 electrons cannot occupy the same space at the same time. &w=mg - wil always be equal + opposite to Tension: gravity assuming that the surface is level. CA rope pulling an object) mg Free body diagrams: IN 2N → draw to scale → use dot (•) for object. Fnet in norizontal direction = 2-1 = IN to the right. Example Q: A helicopter is lifting a mass of 100 kg at terminal velocity. 1) Find the tension in the cable: F=ma net torke =0 100 ×9.81=981 N T= mg 2) The Force is now increased and the load experiences an acceleration of I ms2 Find the new tension in the cable: Fnet = T-W₂₁ ma= =T-W T = ma+w T = 100 × 1 +100 × 9.81 T= 1081N acceleration is decreasing. 3. terminal Drag: The acceleration is not constant. Da v² If an object travels through a fluid, it will experience drag. factors that affect drag in air: → Surface Area →Speed velocity. acceleration is 9 = IF A=0 TEW: 1) mg >no resultant force 21 mg Fnet-mg-D = 9.81*100=981N 3) 60 √ D = mg Fnet = 0 Practical investigation of the motion in a fuuid: 11 Flil a...

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Alternative transcript:

tube with viscous Fluid. 21 mark equal consequtive (10cm) on the tube. 31 drop a bau bearing, and record time it takes to reach each interval. 4) take multiple readings and find the mean to reduce the uncertai- nty. bau bearing 110 cm use a magnet to retrieve img ball bearing. Forces on a slope: •N= normal reaction 90-e 90° to surface. V terminal velocity. masin t Parallel to Plane: Perpendicular to piane: mg cos o The center of gravity = The point at which the entire weight of an object seems to act. The centre of mass = The point where the entire mass of the object seems to act How to determine center of gravity: 1. Pin an object + hang a plumbline. 2. Rotate the object and repeat. 3. Again. 4. intersection of those lines = the center of gravity. Moments: The product of the force and the perpendicular distance from the pivot to the line of action of the Force. unit = Nm (Force & distance) pivot 7 m = F x Clockwise = Clockwise moment M=FX m=6×0.1 F2 Anticlockwise = 0.5 Nm 1. 3. SOH CAH TOA x Pivot 0.2m FT:5N 30 0.2m H wall 30 f 'al m= T=5N 0.2m 30 FOO Sin 30=0=x H 0.2 String = x X-0.2 Sin 30 x₂ 0.1m shelt Perpendicular d From pivot to line of action of force. 2. araw 90°From Slope to pivot T-5N 30° ↑ 0.2m we want this length = X sin Sin 30 = 0 H 0.2 x = 0.2 sin 30° x = 0.1m The principle of moments: An object will be in equilibrium if: 7 11 There is no net torce acting on the object. 21 Sum of anticlockwise moments = Sum of clockwise moments. FX = F₂ X₂ Xi Torave: Force couples and torames: A Force couple consists of 2 caual opposite Forces acting in the same. plane F₁ 30 x ₂ T=Fxd F₁ = F₂ = F Triangle of Forces: F kg m-3 ↓ F₂ N VW N 30 W Density and pressure: ↓ R=m j=d F₂ F₁ = F₂ in magnitude but opposite in direction. otherwise there win be acceleration. True only if the object is in equilibrium. -2 NM = Pa F J N 50% W Pressure in a Huid: >A M 1. Archimedes Principle: The upthrust acting on a body submerged in a liquid is equal to the weight of the liquid that has been displaced. Lmg P= F = mg A A rotation. d m = ev = pvg = pang png A A M = W water M = mass water g : M=pVg M = p Axg moments Practice Q's: F↓ upiniust density area height +gravity. X : P=png dz JFz m=Fxd anticlockwise M₂ F₂ xd₂ Clockwise : T=5N (30 0.2m M₁ = Fxd perp : = 5 × a perp 1 / 0.2 a perpendicular sin 30° = 0 = ∞ 5x 0.1=0.5 Nm ✓ 2. CW = ACW 0.2 sin 30x 2:0km 1. State the principle of moments: -Sum or CIOCKwise moments = surm or anticlockwise moments For objects. in equilibrium. 50 x 0.46 W x 0.14 50*0.46 = W 3 0.14 164 N w = mg m = @= 164 = 60.7 kg 99.81

forces in action

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Physics chapter 4:
Forces in action revision:
Newtons 2nd law:
The net force-mass & acceleration
(For a constant mass,
F = ma
AIN
Fner=2-1=1
Physics chapter 4:
Forces in action revision:
Newtons 2nd law:
The net force-mass & acceleration
(For a constant mass,
F = ma
AIN
Fner=2-1=1
Physics chapter 4:
Forces in action revision:
Newtons 2nd law:
The net force-mass & acceleration
(For a constant mass,
F = ma
AIN
Fner=2-1=1
Physics chapter 4:
Forces in action revision:
Newtons 2nd law:
The net force-mass & acceleration
(For a constant mass,
F = ma
AIN
Fner=2-1=1
Physics chapter 4:
Forces in action revision:
Newtons 2nd law:
The net force-mass & acceleration
(For a constant mass,
F = ma
AIN
Fner=2-1=1

motion

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Physics chapter 4: Forces in action revision: Newtons 2nd law: The net force-mass & acceleration (For a constant mass, F = ma AIN Fner=2-1=1N in upwards. direction. IN -2 1 newton = the force required to accelerate a mass of 1 kg by 1 ms² Rin -2 | N=1 kg x 1 ms ²² Forces: weight fner = IN m Img Upthrust: Friction: Normal Forces: 2 Friction (8) Fnet-ON M w = mg always acts vertically downwards. g=9.81m5²² ↑N = normal 12N W IN Force - 2 electrons cannot occupy the same space at the same time. &w=mg - wil always be equal + opposite to Tension: gravity assuming that the surface is level. CA rope pulling an object) mg Free body diagrams: IN 2N → draw to scale → use dot (•) for object. Fnet in norizontal direction = 2-1 = IN to the right. Example Q: A helicopter is lifting a mass of 100 kg at terminal velocity. 1) Find the tension in the cable: F=ma net torke =0 100 ×9.81=981 N T= mg 2) The Force is now increased and the load experiences an acceleration of I ms2 Find the new tension in the cable: Fnet = T-W₂₁ ma= =T-W T = ma+w T = 100 × 1 +100 × 9.81 T= 1081N acceleration is decreasing. 3. terminal Drag: The acceleration is not constant. Da v² If an object travels through a fluid, it will experience drag. factors that affect drag in air: → Surface Area →Speed velocity. acceleration is 9 = IF A=0 TEW: 1) mg >no resultant force 21 mg Fnet-mg-D = 9.81*100=981N 3) 60 √ D = mg Fnet = 0 Practical investigation of the motion in a fuuid: 11 Flil a...

Physics chapter 4: Forces in action revision: Newtons 2nd law: The net force-mass & acceleration (For a constant mass, F = ma AIN Fner=2-1=1N in upwards. direction. IN -2 1 newton = the force required to accelerate a mass of 1 kg by 1 ms² Rin -2 | N=1 kg x 1 ms ²² Forces: weight fner = IN m Img Upthrust: Friction: Normal Forces: 2 Friction (8) Fnet-ON M w = mg always acts vertically downwards. g=9.81m5²² ↑N = normal 12N W IN Force - 2 electrons cannot occupy the same space at the same time. &w=mg - wil always be equal + opposite to Tension: gravity assuming that the surface is level. CA rope pulling an object) mg Free body diagrams: IN 2N → draw to scale → use dot (•) for object. Fnet in norizontal direction = 2-1 = IN to the right. Example Q: A helicopter is lifting a mass of 100 kg at terminal velocity. 1) Find the tension in the cable: F=ma net torke =0 100 ×9.81=981 N T= mg 2) The Force is now increased and the load experiences an acceleration of I ms2 Find the new tension in the cable: Fnet = T-W₂₁ ma= =T-W T = ma+w T = 100 × 1 +100 × 9.81 T= 1081N acceleration is decreasing. 3. terminal Drag: The acceleration is not constant. Da v² If an object travels through a fluid, it will experience drag. factors that affect drag in air: → Surface Area →Speed velocity. acceleration is 9 = IF A=0 TEW: 1) mg >no resultant force 21 mg Fnet-mg-D = 9.81*100=981N 3) 60 √ D = mg Fnet = 0 Practical investigation of the motion in a fuuid: 11 Flil a...

Can't find what you're looking for? Explore other subjects.

Knowunity is the # 1 ranked education app in five European countries

Knowunity is the # 1 ranked education app in five European countries

Knowunity was a featured story by Apple and has consistently topped the app store charts within the education category in Germany, Italy, Poland, Switzerland and United Kingdom. Join Knowunity today and help millions of students around the world.

Ranked #1 Education App

Download in

Google Play

Download in

App Store

Still not sure? Look at what your fellow peers are saying...

iOS User

I love this app so much [...] I recommend Knowunity to everyone!!! I went from a C to an A with it :D

Stefan S, iOS User

The application is very simple and well designed. So far I have found what I was looking for :D

SuSSan, iOS User

Love this App ❤️, I use it basically all the time whenever I'm studying

Alternative transcript:

tube with viscous Fluid. 21 mark equal consequtive (10cm) on the tube. 31 drop a bau bearing, and record time it takes to reach each interval. 4) take multiple readings and find the mean to reduce the uncertai- nty. bau bearing 110 cm use a magnet to retrieve img ball bearing. Forces on a slope: •N= normal reaction 90-e 90° to surface. V terminal velocity. masin t Parallel to Plane: Perpendicular to piane: mg cos o The center of gravity = The point at which the entire weight of an object seems to act. The centre of mass = The point where the entire mass of the object seems to act How to determine center of gravity: 1. Pin an object + hang a plumbline. 2. Rotate the object and repeat. 3. Again. 4. intersection of those lines = the center of gravity. Moments: The product of the force and the perpendicular distance from the pivot to the line of action of the Force. unit = Nm (Force & distance) pivot 7 m = F x Clockwise = Clockwise moment M=FX m=6×0.1 F2 Anticlockwise = 0.5 Nm 1. 3. SOH CAH TOA x Pivot 0.2m FT:5N 30 0.2m H wall 30 f 'al m= T=5N 0.2m 30 FOO Sin 30=0=x H 0.2 String = x X-0.2 Sin 30 x₂ 0.1m shelt Perpendicular d From pivot to line of action of force. 2. araw 90°From Slope to pivot T-5N 30° ↑ 0.2m we want this length = X sin Sin 30 = 0 H 0.2 x = 0.2 sin 30° x = 0.1m The principle of moments: An object will be in equilibrium if: 7 11 There is no net torce acting on the object. 21 Sum of anticlockwise moments = Sum of clockwise moments. FX = F₂ X₂ Xi Torave: Force couples and torames: A Force couple consists of 2 caual opposite Forces acting in the same. plane F₁ 30 x ₂ T=Fxd F₁ = F₂ = F Triangle of Forces: F kg m-3 ↓ F₂ N VW N 30 W Density and pressure: ↓ R=m j=d F₂ F₁ = F₂ in magnitude but opposite in direction. otherwise there win be acceleration. True only if the object is in equilibrium. -2 NM = Pa F J N 50% W Pressure in a Huid: >A M 1. Archimedes Principle: The upthrust acting on a body submerged in a liquid is equal to the weight of the liquid that has been displaced. Lmg P= F = mg A A rotation. d m = ev = pvg = pang png A A M = W water M = mass water g : M=pVg M = p Axg moments Practice Q's: F↓ upiniust density area height +gravity. X : P=png dz JFz m=Fxd anticlockwise M₂ F₂ xd₂ Clockwise : T=5N (30 0.2m M₁ = Fxd perp : = 5 × a perp 1 / 0.2 a perpendicular sin 30° = 0 = ∞ 5x 0.1=0.5 Nm ✓ 2. CW = ACW 0.2 sin 30x 2:0km 1. State the principle of moments: -Sum or CIOCKwise moments = surm or anticlockwise moments For objects. in equilibrium. 50 x 0.46 W x 0.14 50*0.46 = W 3 0.14 164 N w = mg m = @= 164 = 60.7 kg 99.81