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The brakes on a truck fail as it approaches a car stopped at a red light. Use Newton’s first and second law of motion to explain

PolarNik [594]

newtons first law states that a object in motion will stay in motion unless acted on by an outside force and an object at rest will stay at rest for the same reason. the force of a moving object is equal to its mass times its acceleration.

3 0

3 years ago

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A model rocket rises with constant acceleration to a height of 4.2 m, at which point its speed is 27.0 m/s. How much time does i

geniusboy [140]

Answers:

a) $t=0.311 s$

b) $a=86.847 m/s^{2}$

c) $y=1.736 m$

d) $V=17.369 m/s$

Explanation:

For this situation we will use the following equations:

$y=y_{o}+V_{o}t+\frac{1}{2}at^{2}$ (1)

$V=V_{o} + at$ (2)

Where:

$y$ is the <u>height of the model rocket at a given time</u>

$y_{o}=0$ is the i<u>nitial height </u>of the model rocket

$V_{o}=0$ is the<u> initial velocity</u> of the model rocket since it started from rest

$V$ is the <u>velocity of the rocket at a given height and time</u>

$t$ is the <u>time</u> it takes to the model rocket to reach a certain height

$a$ is the <u>constant acceleration</u> due gravity and the rocket's thrust

<h2>a) Time it takes for the rocket to reach the height=4.2 m</h2>

The average velocity of a body moving at a constant acceleration is:

$V=\frac{V_{1}+V_{2}}{2}$ (3)

For this rocket is:

$V=\frac{27 m/s}{2}=13.5 m/s$ (4)

Time is determined by:

$t=\frac{y}{V}$ (5)

$t=\frac{4.2 m}{13.5 m/s}$ (6)

Hence:

$t=0.311 s$ (7)

<h2>b) Magnitude of the rocket's acceleration</h2>

Using equation (1), with initial height and velocity equal to zero:

$y=\frac{1}{2}at^{2}$ (8)

We will use $y=4.2 m$ :

$4.2 m=\frac{1}{2}a(0.311)^{2}$ (9)

Finding $a$ :

$a=86.847 m/s^{2}$ (10)

<h2>c) Height of the rocket 0.20 s after launch</h2>

Using again $y=\frac{1}{2}at^{2}$ but for $t=0.2 s$ :

$y=\frac{1}{2}(86.847 m/s^{2})(0.2 s)^{2}$ (11)

$y=1.736 m$ (12)

<h2>d) Speed of the rocket 0.20 s after launch</h2>

We will use equation (2) remembering the rocket startted from rest:

$V= at$ (13)

$V= (86.847 m/s^{2})(0.2 s)$ (14)

Finally:

$V=17.369 m/s$ (15)

5 0

3 years ago

Find the centroidal moment of inertia about the x axis for the T-shaped area below. The following steps MUST be taken to solve t

Alex Ar [27]

Answer:

Explanation:

Find attach the solution

4 0

3 years ago

Urgent!!!!! A student heated 235 g of water in a beaker until the water reached 100°C. The student removed the beaker from the h

Murrr4er [49]

It’s says that the temperature in the room is 23C so the air temperature in the air would be 23C

4 0

3 years ago

A friend pushes a sled across horizontal snow and when it gets up to speed the friend jumps on. After the friend jumps on, the s

Shalnov [3]

Answer:

v’ =( $\frac{1}{1+ \frac{M}{m} }$ ) v

we see that the greater the difference, the more the sled slows down.

friction force

Explanation:

When the man pushes the sled he does work and the sled acquires a speed and as long as it is supplied with an energy equal to the work of the chipping force with the snow, the speed is maintained.

When he jumps on the sled, a collision occurs and the initial moment

p₀ = mv

is increased by the increase in mass

m_f= (m + M_{man} ) v '

In this case there is no longer any external force applied and the only external force is friction, which causes the sled to stop, even when it is small, but the significant reduction in speed is due to the increase in masses.

p₀ = p_f

mv = (m + M_{man}) v '

v ’= $\frac{m}{m+M}$ v

v’ =( $\frac{1}{1+ \frac{M}{m} }$ ) v

Therefore, we see that the greater the difference, the more the sled slows down.

The only forces that act on the sled with the man are the friction that is responsible for the decrease in speed and weight with the normal

4 0

3 years ago