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2 years ago

a car advertisement States that a certain car can accelerate from rest to70km/h in 7seconds find the car's average acceleration

Physics

1 answer:

Aneli [31]2 years ago

8 0

Answer:

Acceleration is 2.78 m/s²

Explanation:

Since the car is from rest, initial velocity, u is zero.

from first newton's equation of motion:

${ \bf{v = u + at}} \\$

v is final velocity

u is initial velocity

a is acceleration

t is time

70 km/h is (70 × 1000 ÷ 3600) m/s = 19.44 m/s

${ \sf{19.44 = 0 + (a \times 7)}} \\ { \sf{7a = 19.44 }} \\ { \sf{a = 2.78 \: {ms}^{ - 2} }}$

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A projectile is fired into the air from the top of a 200-m cliff above a valley as shown below. Its initial velocity is 60 m/s a

anastassius [24]

a) y(max) = 337.76 m

b) t₁ = 5.30 s the time for y maximum

c)t₂ = 13.60 s time for y = 0 time when the fly finish

d) vₓ = 30 m/s vy = - 81.32 m/s

e)x = 408 m

Equations for projectile motion:

v₀ₓ = v₀ * cosα v₀ₓ = 60*(1/2) v₀ₓ = 30 m/s ( constant )

v₀y = v₀ * sinα v₀y = 60*(√3/2) v₀y = 30*√3 m/s

a) Maximum height:

The following equation describes the motion in y coordinates

y = y₀ + v₀y*t - (1/2)*g*t² (1)

To find h(max), we need to calculate t₁ ( time for h maximum)

we take derivative on both sides of the equation

dy/dt = v₀y - g*t

dy/dt = 0 v₀y - g*t₁ = 0 t₁ = v₀y/g

v₀y = 60*sin60° = 60*√3/2 = 30*√3

g = 9.8 m/s²

t₁ = 5.30 s the time for y maximum

And y maximum is obtained from the substitution of t₁ in equation (1)

y (max) = 200 + 30*√3 * (5.30) - (1/2)*9.8*(5.3)²

y (max) = 200 + 275.40 - 137.64

y(max) = 337.76 m

Total time of flying (t₂) is when coordinate y = 0

y = 0 = y₀ + v₀y*t₂ - (1/2)* g*t₂²

0 = 200 + 30*√3*t₂ - 4.9*t₂² 4.9 t₂² - 51.96*t₂ - 200 = 0

The above equation is a second-degree equation, solving for t₂

t = [51.96 ±√ (51.96)² + 4*4.9*200]/9.8

t = [51.96 ±√2700 + 3920]/9.8

t = [51.96 ± 81.36]/9.8

t = 51.96 - 81.36)/9.8 we dismiss this solution ( negative time)

t₂ = 13.60 s time for y = 0 time when the fly finish

The components of the velocity just before striking the ground are:

vₓ = v₀ *cos60° vₓ = 30 m/s as we said before v₀ₓ is constant

vy = v₀y - g *t vy = 30*√3 - 9.8 * (13.60)

vy = 51.96 - 133.28 vy = - 81.32 m/s

The sign minus means that vy change direction

Finally the horizontal distance is:

x = vₓ * t

x = 30 * 13.60 m

x = 408 m

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A major artery with a cross sectional area of 1.00cm^2 branches into 18 smaller arteries, each with an average cross sectional a

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Here we can say that rate of flow must be constant

so here we will have

$A_1v_1 = 18 A_2v_2$

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$A_1 = 1 cm^2$

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now from above equation

$1 cm^2 v_1 = 18(0.400 cm^2)v_2$

$\frac{v_2}{v_1} = \frac{1}{18\times 0.4}$

$\frac{v_2}{v_1} = 0.14$

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In a high orbit space vehicle or in the same space, said force decreases considerably or simply disappears, generating the sensation of weightlessness.

Remember that the Force of Gravity is given under the principle

$F_g = \frac{GMm}{r^2}$

Where,

G = Gravitational Universal constant

M = Mass of the planet

m = mass of the object

r = Distance from center of the planet

When the radius grows considerably the gravitational force begins to decrease.

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Mass represents the density of an object multiplied with the volume it occupies. As a result, an object's density is found by dividing its mass by its volume. So the answer is a.

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Answer:

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a car advertisement States that a certain car can accelerate from rest to70km/h in 7seconds find the car's average acceleration​

a car advertisement States that a certain car can accelerate from rest to70km/h in 7seconds find the car's average acceleration