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2 years ago

.An airplane accelerates down a runway at

Physics

1 answer:

solniwko [45]2 years ago

6 0

Starting from rest, the plane travels a distance

x = 1/2 at²

with acceleration a after time t. In this instance, it travels

x = 1/2 (3.20 m/s²) (32.8 s)²

x ≈ 1720 m

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(BRAINLIEST W/ WORK SHOWN!!!) Help with Physics question?

VashaNatasha [74]

Answer:

$\boxed {3.43 x 10^{3}}$

Explanation:

We know that speed is defined as distance moved per unit time hence expressed as $v=\frac {d}{t}$ where v is speed in m/s, d is distance in m and t is time in seconds. Making d the subject of the above formula then

$d=vt$

Substituting 343 m/s for d and 10 s for t then

$d= 343\times10= 3430= 3.43 x 10^{3}$

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2 years ago

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2 years ago

A 600-g mass traveling at 8.0 m/s undergoes a head-on elastic collision with a 200-g mass traveling toward it also at 8.0 m/s. W

Pepsi [2]

Answer:

vf₂ = 16 m/s

Explanation:

Theory of collisions

Linear momentum is a vector magnitude (same direction of the velocity) and its magnitude is calculated like this:

P=m*v

where

P:Linear momentum

m: mass

v:velocity

There are 3 cases of collisions : elastic, inelastic and plastic.

For the three cases the total linear momentum quantity is conserved:

P₀ = Pf Formula (1)

P₀ : Initial linear momentum quantity

Pf : final linear momentum quantity

Data

m₁ = 600 g

m₂= 200 g

v₀₁ = 8 m/s :initial velocity of m₁

v₀₂ = - 8 m/s :initial velocity of m₂ ; (-) indicates that m₂ is moving in the opposite direction to m₁

Problem development

We appy the formula (1):

We assume that at the end of the collide both masses move in the initial direction of m₁ , then vf₁ and vf₁ have a sign (+)

P₀ = Pf

m₁*v₀₁ + m₂*v₀₂ = m₁*vf₁ + m₂*vf₂

(600) *(8)+ (200)*( -8) = (600) *vf₁ + (200)*vf₂

4800 - 1600 = (600) *vf₁ + (200)*vf₂

3200 = (600) *vf₁ + (200)*vf₂ : We divide both sides of the equation by 100

32= 6*vf₁ +2*vf₂ Equation (1)

For elastic collision, the elastic restitution coefficient (e) is equal to 1 :

$e= \frac{ v_{f2}- v_{f1} }{ v_{o1}- v_{o2} }$

$1= \frac{ v_{f2}- v_{f1} }{ 8- (-8) }$

16 = vf₂ - vf₁

vf₁ = vf₂ - 16 Equation (2)

We replace Equation (2) in the Equation (1)

32= 6*vf₁ +2*vf₂

32= 6*( vf₂ - 16) +2*vf₂

32= 6*vf₂ - 96 + 2*vf₂

32 + 96= 6*vf₂ + 2*vf₂

128 = 8 *vf₂

vf₂ = 128 / 8

vf₂ = 16 m/s

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