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artcher [175]
3 years ago
14

You are sitting on a deck of your house surrounded by oak trees. You hear the sound of an acorn hitting the deck. You wonder if

an acorn will do much damage if instead of the deck it hits your head. Let the mass of the acorn be 3.5 g. Suppose it drops from a height of 14 m (relative to your head). Assume that the acorn sinks in 0.50 cm of your head skin before coming to a stop.
Physics
1 answer:
Black_prince [1.1K]3 years ago
8 0

Answer: 96N

Explanation:

To calculate the velocity of the impact On the persons head, we have

h = gt²/2

14 = 9.81t²/2

t² = 28/9.8

t² = 2.86

t = 1.69s

V = u + at

V = 0 + 9.81*1.69

V = 16.58m/s

a(average) = (v1² + v2²) /2Δy

a(average) = 16.58² + 0)/2 * 0.005

a(average) = 274.8964/0.01

a(average) = 27489.64m/s²

Using newton's second law of motion,

F(average) = m * a(average)

F(average) = 0.0035 * 27489.64

F(average) = 96.21N

Therefore the force needed by the acorn to do much damage starts from 96N

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how Long will it take a plane flying north from Miami to reach New York City 1800 km away if the average velocity of the plane i
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8 0
3 years ago
In part (a), suppose that the box weighs 128 pounds, that the angle of inclination of the plane is θ = 30°, that the coefficient
morpeh [17]

Answer:

v(t) = 21.3t

v(t) = 5.3t

v(t) = 48 -48 e ^{ \frac{t}{9}}

Explanation:

When no sliding friction and no air resistance occurs:

m\frac{dv}{dt} = mgsin \theta

where;

\frac{dv}{dt} = gsin \theta , 0 < \theta <  \frac{ \pi}{2}

Taking m = 3 ; the differential equation is:

3 \frac{dv}{dt}= 128*\frac{1}{2}

3 \frac{dv}{dt}= 64

\frac{dv}{dt}= 21.3

By Integration;

v(t) = 21.3 t + C

since v(0) = 0 ; Then C = 0

v(t) = 21.3t

ii)

When there is sliding friction but no air resistance ;

m \frac{dv}{dt}= mg sin \theta - \mu mg cos \theta

Taking m =3 ; the differential equation is;

3 \frac{dv}{dt}=128*\frac{1}{2} -\frac{\sqrt{3} }{4}*128*\frac{\sqrt{3} }{4}

\frac{dv}{dt}= 5.3

By integration; we have ;

v(t) = 5.3t

iii)

To find the differential equation for the velocity (t) of the box at time (t) with sliding friction and air resistance :

m \frac{dv}{dt}= mg sin \theta - \mu mg cos \theta - kv

The differential equation is :

= 3 \frac{dv}{dt}=128*\frac{1}{2} - \frac{ \sqrt{ 3}}{4}*128 *\frac{ \sqrt{ 3}}{2}-\frac{1}{3}v

= 3 \frac{dv}{dt}=16 -\frac{1}{3}v

By integration

v(t) = 48 + Ce ^{\frac{t}{9}

Since; V(0) = 0 ; Then C = -48

v(t) = 48 -48 e ^{ \frac{t}{9}}

7 0
3 years ago
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