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3 years ago

14

When a speeding truck hits a stationary car, the car is deformed and heat is generated. What can you say about the kinetic energ

y of the system after the collision?

1 answer:

guapka [62]3 years ago

4 0

I can say that the kinetic energy after the collision is less than it was before the collision. I can say this with confidence because you've said that some energy was used to deform the car, plus there was energy lost from the system in the form of heat.

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To analyze the motion of a body that is traveling along a curved path, to determine the body's acceleration, velocity, and posit

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To solve this problem we will apply the kinematic equations of linear motion and centripetal motion. For this purpose we will be guided by the definitions of centripetal acceleration to relate it to the tangential velocity. With these equations we will also relate the linear velocity for which we will find the points determined by the statement. Our values are given as

$R = 350ft$

$a_t = 1.1ft/s^2$

PART A )

$a_c = \frac{V^2}{R}$

$a_c = \frac{V^2}{350}$

Calculate the velocity of the motorcycle when the net acceleration of the motorcycle is $5.25ft/s^2$

$a = \sqrt{a_t^2+a_r^2}$

$5.25 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}$

$27.5625 = 1.21 + \frac{v^4}{122500}$

$v=42.3877ft/s$

Now calculate the angular velocity of the motorcycle

$v = r\omega$

$42.3877 = 350\omega$

$\omega = 0.1211rad/s$

Calculate the angular acceleration of the motorcycle

$a_t = r\alpha$

$1.1 = 350\alpha$

$\alpha = 3.1428*10^{-3}rad/s^2$

Calculate the time needed by the motorcycle to reach an acceleration of

$5.25ft/s^2$

$\omega = \alpha t$

$0.1211 = 3.1428*10^{-3}t$

$t = 38.53s$

PART B) Calculate the velocity of the motorcycle when the net acceleration of the motorcycle is $6.75ft/s^2$

$a = \sqrt{a_t^2+a_r^2}$

$6.75 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}$

$45.5625 = 1.21 + \frac{v^4}{122500}$

$v=48.2796ft/s$

PART C)

Calculate the radial acceleration of the motorcycle when the velocity of the motorcycle is $21.5ft/s$

$a_r = \frac{v^2}{R}$

$a_r = \frac{21.5^2}{350}$

$a_r =1.3207ft/s^2$

Calculate the net acceleration of the motorcycle when the velocity of the motorcycle is $21.5ft/s$

$a = \sqrt{a_t^2+a_r^2}$

$a = \sqrt{(1.1)^2+(1.3207)^2}$

$a = 1.7187ft/s^2$

PART D) Calculate the maximum constant speed of the motorcycle when the maximum acceleration of the motorcycle is $6.75ft/s^2$

$a = \sqrt{a_t^2+a_r^2}$

$6.75 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}$

$45.5625 = 1.21 + \frac{v^4}{122500}$

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3 years ago

What can lead to groundwater shortages

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Answer:

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3 years ago

A block of mass m = 150 kg rests against a spring with a spring constant of k = 880 N/m on an inclined plane which makes an angl

weeeeeb [17]

Answer:

b) k Δx - W cos θ - μ mg cos θ = m a , c) θ = 86.6º, d) Δx = 1.18 m

Explanation:

a) In the attachment we can see a diagram of the forces in this problem and the coordinate axes for its decomposition.

F is the force applied by the spring, while it is compressed, this force disappears when the block leaves the spring

b) Let's apply Newton's second law for when the spring is compressed

let's use trigonometry to break down the weight

sin θ = Wₓ / W

cos θ = W_y / W

Wₓ = W sin θ

W_y = W cos θ

Y axis

N - W_y = 0

N = W_y

N = W cos θ

X axis

F -Wₓ -fr = ma

the force applied by the spring is given by hooke's law

F = k Δx

friction force has the expression

fr = μ N

fr = μ W cos θ

we substitute

k Δx - W cos θ - μ mg cos θ = m a ( 1)

c) If the plane has no friction, what is the angle so that Δx = 0.1m

We write the equation 1, with fr = 0 and since the system is still a = 0

k Δx - W cos θ -0 = 0

cos θ = $\frac{k \Delta x}{ m g}$

cos θ = $\frac{880 \ 0.1}{ 150 \ 9.8}$

cos θ = 0.0598

θ = cos⁻¹ 0.0598

θ = 86.6º

d) In this part they give the angle θ = 45º and there is no friction, they ask the compression

the acceleration is zero, we substitute in 1

k Δx - W cos θ - 0 = 0

Δx = $\frac{mg \ cos \ \theta}{k}$

Δx = $\frac{ 150 \ 9.8 \ cos45}{880}$

Δx = 1.18 m

7 0

3 years ago