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3 years ago

What are two differences between gravity and air resistance

Physics

1 answer:

Natalka [10]3 years ago

4 0

1. Air resistance force usually upwards, but gravity doenwards.

2. Objects are affected by gravity the same, but air resistance can affect the speed of an object's descent.

Sorry if im wrong

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A 15.0-kg object sitting at rest is struck elastically in a head-on collision with a 10.5-kg object initially moving at 3.0 m/s.

DIA [1.3K]

Answer:

The final velocity of the 15.0-kg object after the collision is 2.47 m/s in forward direction.

Explanation:

Given;

mass of the object, m₁ = 15 kg

initial velocity of this object, u₁ = 0

mass of the second object, m₂ = 10.5 kg

initial velocity of this object, u₂ = 3.0 m/s

let the final velocity of the first object = v₁

also, let the final velocity of the second object = v₂

Apply the principle of conservation of linear momentum

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(15 x 0) + (10.5 x 3) = 15v₁ + 10.5v₂

31.5 = 15v₁ + 10.5v₂ ----- (1)

One directional velocity;

u₁ + v₁ = u₂ + v₂

0 + v₁ = 3 + v₂

v₂ = v₁ - 3 ------(2)

Substitute (2) into (1);

31.5 = 15v₁ + 10.5v₂

31.5 = 15v₁ + 10.5(v₁ - 3)

31.5 = 15v₁ + 10.5v₁ - 31.5

63 = 25.5v₁

v₁ = 63 / 25.5

v₁ = 2.47 m/s

Therefore, the final velocity of the 15.0-kg object after the collision is 2.47 m/s in forward direction.

7 0

3 years ago

An inner city revitalization zone is a rectangle that is twice as long as it is wide. The width of the region is growing at a ra

lukranit [14]

Answer:

$\frac{dA}{dt} = 28800 \ m^2/year$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Geometry

Area of a Rectangle: A = lw

Algebra I

Exponential Property: $w^n \cdot w^m = w^{n + m}$

Calculus

Derivatives

Differentiating with respect to time

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Explanation:

Step 1: Define

Area is A = lw

2w = l

w = 300 m

$\frac{dw}{dt} = 24 \ m/year$

Step 2: Rewrite Equation

Substitute in l: A = (2w)w
Multiply: A = 2w²

Step 3: Differentiate

Differentiate the new area formula with respect to time.

Differentiate [Basic Power Rule]: $\frac{dA}{dt} = 2 \cdot 2w^{2-1}\frac{dw}{dt}$
Simplify: $\frac{dA}{dt} = 4w\frac{dw}{dt}$

Step 4: Find Rate

Use defined variables

Substitute: $\frac{dA}{dt} = 4(300 \ m)(24 \ m/year)$
Multiply: $\frac{dA}{dt} = (1200 \ m)(24 \ m/year)$
Multiply: $\frac{dA}{dt} = 28800 \ m^2/year$

3 0

3 years ago

Read 2 more answers

A machine lifts a 35 kg object doing 6860 J worth of work. How much power is produced by the machine if it lifts the object in 4

timofeeve [1]

Answer:

Explanation:

We need the power equation for this which is

P = Work/time

We have everything we need to solve this (the mass of the object is extra information):

P = 6860/4

P = 1715W

5 0

3 years ago

On an icy road, a 1100 kg car moving at 55 km/h strikes a 480 kg truck moving in the same direction at 37 km/h . The pair is soo

statuscvo [17]

Answer:

Speed of the wreckage = 49.29 km/hr

Explanation:

This question is solved simply by using the conservation of momentum law.

The momentum of the three moving bodies are calculated below:

Momentum of Car 1 : 1100 * 55 = 60500 kg.km/hr

Momentum of Truck : 480 * 37 = 17760 kg.km/hr

Momentum of Car 2 : 1300 * 49 = 63700 kg.km/hr

Total mass of all three vehicles: 1100 + 480 + 1300 = 2880 kg

The final momentum equals the initial momentum if it is conserved. Thus we have the following equation:

Final Momentum = Initial Momentum

Final Velocity * Total mass = Momentum of all three vehicles combined

Final Velocity * 2880 = 60500 + 17760 + 63700

Final Velocity = 49.29 km/hr

7 0

4 years ago

Read 2 more answers

Which of the following quantities would be acceptable representations of weight? Check all that apply.

Oxana [17]

Only 12 lb and 1600 kN .

7 0

3 years ago