The difference between newtons first law of motion and newtons third law of motion

A 1.5-kilogram cart initially moves at 2.0 meters per second. It is brought to rest by a constant net force in 0.30 second. What

AnnZ [28]

Acceleration = (change in speed) / (time for the change)

Change in speed = (speed at the end) minus (speed at the beginning.

The cart's acceleration is

 (0 - 2 m/s) / (0.3 sec)

 = ( -2 / 0.3 ) (m/s²) = -(6 and 2/3) m/s² .

Newton's second law of motion says

 Force = (mass) x (acceleration) .

For this cart: Force = (1.5 kg) x ( - 6-2/3 m/s²)

 = ( - 1.5 x 20/3 ) (kg-m/s²)

 = - 10 newtons .

The force is negative because it acts opposite to the direction 
in which the cart is moving, it causes a negative acceleration, 
and it eventually stops the cart.

6 0

3 years ago

A skier with a mass of 63 kg starts from rest and skis down an icy (frictionless) slope that has a length of 50 m at an angleof

victus00 [196]

Answer:

Explanation:

When the skier reaches the bottom of the slope , height lost by it

h = 50 sin32 m

= 26.5 m

potential energy lost

= mgh

Gain of kinetic energy

= 1/2 mv²

mgh = 1/2 mv²

v = √ 2gh

= √ (2x9.8 x 26.5)

= 22.8 m /s

b )

Let μ be the coefficient of kinetic friction required.

friction force acting

= μmg

work done by friction in displacement of d (40 m ) on horizontal surface

- μmg x d

This negative work will be equal to positive kinetic energy of the skier on horizontal surface .

= μmg x d = (1/2) m v²

μ = v² / (2 gd)

= 519.4 / (2 x 9.8 x 140 )

= .19

8 0

4 years ago

An aluminum screw has a density of 2.7 g/mL and a volume of 15mL. What is the mass of the screw ?

vladimir1956 [14]

the answer is 40.5 because you have to multiply the density and volume of the object to get the mass.

6 0

3 years ago

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.8×106N , one an angle 11degrees west of north an

sp2606 [1]

Answer: 2,9 ×10¹⁰J.

Explanation:

1. The work is a measure of energy and is calculated as the product of the displacement times the parallel force to such displacement.

That means that the only components of the force that contribute to work are those that result parallel to the displacement.

2. Since it is given that the two tugboats "pull the tanker a distance 0.83km toward the north", that is the displacement, and you have to calculate the net force toward the north.

3. Tugboat #1.

a) Force magnitude: F₁ = 1.8×10⁶N

b) Angle: α = 11° West of North

c) North component of the force F₁: Fy₁ = F₁cos(α) = 1.8×10⁶N × cos(11°) = 1.77×10⁶N

4. Tugboat #2:

a) Force magnitude: F₂ = 1.8×10⁶N

b) Angle: = 11° East of North

c) North component of the force F₂: Fy₂ = F₂cos(β) = 1.8×10⁶N × cos(11°) = 1.77×10⁶N =

5. Total net force, Fn:

Fn = Fy₁ + Fy₂ = 1.77×10⁶N + 1.77×10⁶N = 3.54×10⁶N

6. Work, W:

Displacement, d = 0.83 km = 8,300 m

W = Fn×d = 3.54×10⁶N×8,300m = 29,000 ×10⁶J = 2,9 ×10¹⁰J

The answer is rounded to two significant figures because both data, Force and displacement, have two significant figures.

7 0

3 years ago

You have two pieces of wire that allow electrons to pass through them when they are connected to each other. If you insert a pie

hoa [83]

Out of the following choices given, if you insert a piece of rubber between the wires, the electrons do not flow. The rubber material is an insulator. The correct answer is D.

5 0

3 years ago

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