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2 years ago

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A 1 200-kg car traveling initially at vCi 5 25.0 m/s in an easterly direction crashes into the back of a 9 000-kg truck moving i

n the same direction at vTi 5 20.0 m/s (Fig. P9.22). The velocity of the car immediately after the collision is vCf 5 18.0 m/s to the east. (a) What is the velocity of the truck immediately after the colli

1 answer:

sukhopar [10]2 years ago

8 0

Answer:

The velocity of the truck after the collision is 20.93 m/s

Explanation:

It is given that,

Mass of car, m₁ = 1200 kg

Initial velocity of the car, $v_{Ci}=25\ m/s$

Mass of truck, m₂ = 9000 kg

Initial velocity of the truck, $v_{Ti}=20\ m/s$

After the collision, velocity of the car, $v_{Cf}=18\ m/s$

Let $v$ is the velocity of the truck immediately after the collision. The momentum of the system remains conversed.

$initial\ momentum=final\ momentum$

$1200\ kg\times 25\ m/s+9000\ kg\times 20\ m/s=1200\ kg\times 18+9000\ kg\times v$

$210000-21600=9000\ kg\times v$

$v=20.93\ m/s$

So, the velocity of the truck after the collision is 20.93 m/s. Hence, this is the required solution.

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Kim has a metal casting company which makes commemorative coins. She has 0.12 cubic meters of silver which she needs to make int

a_sh-v [17]

Answer:

The coin has a diameter of 2.67 cm

Explanation:

First, we need to find the volume of each coin by dividing the total volume of silver by the number of coins. We have also to do a conversion of units in terms of centimeters as follows:

$V=0.12 m^3\times (\frac{100cm}{1m})^3=12000\ cm^3\\V_c=\frac{120000\ cm^3}{1.07\times10^5}= 1.121 cm^3$

Then, we define the coin as a tiny cilinder to determine its diameter. In that order we use the cilinder's volumen equation as follows:

$V=\pi r^2h\\r = \sqrt \frac{V}{\pi h}= \sqrt\frac{1.121 cm^3}{\pi \times 0.2cm}=1.336 cm$

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2 years ago

A transformer changes the 10,000 v power line to 120 v. if the primary coil contains 750 turns, how many turns are on the second

inn [45]

For a transformer, the ratio between the number of turns of primary and secondary coil is the same as the ratio between the voltages on the two coils:
$\frac{N_p}{N_s}= \frac{V_p}{V_s}$
Where $N_p$ and $N_s$ are the number of turns in the primary and secondary coils, while $V_p$ and $V_s$ are the voltages on the two coils.

Using the data of the problem: $N_p=750$ , $V_p=10000 V$ and $V_s=120 V$ , we can find $N_s$ , the number of turns of the secondary coil:
$N_s=N_p \frac{V_s}{V_p}=750 \frac{120 V}{10000 V}=9$

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2 years ago

Sheila (m=56.8 kg) is in her saucer sled moving at 12.6 m/s at the bottom of the sledding hill near Bluebird Lake. She approache

FromTheMoon [43]

Answer:

y = 54.9 m

Explanation:

For this exercise we can use the relationship between the work of the friction force and mechanical energy.

Let's look for work

W = -fr d

The negative sign is because Lafourcade rubs always opposes the movement

On the inclined part, of Newton's second law

Y Axis

N - W cos θ = 0

The equation for the force of friction is

fr = μ N

fr = μ mg cos θ

We replace at work

W = - μ m g cos θ d

Mechanical energy in the lower part of the embankment

Em₀ = K = ½ m v²

The mechanical energy in the highest part, where it stopped

$Em_{f}$ = U = m g y

W = ΔEm = $Em_{f}$ - Em₀

- μ m g d cos θ = m g y - ½ m v²

Distance d and height (y) are related by trigonometry

sin θ = y / d

y = d sin θ

- μ m g d cos θ = m g d sin θ - ½ m v²

We calculate the distance traveled

d (g syn θ + μ g cos θ) = ½ v²

d = v²/2 g (sintea + myy cos tee)

d = 9.8 12.6 2/2 9.8 (sin16 + 0.128 cos 16)

d = 1555.85 /7.8145

d = 199.1 m

Let's use trigonometry to find the height

sin 16 = y / d

y = d sin 16

y = 199.1 sin 16

y = 54.9 m

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hoa [83]

The stone's altitude at time $t$ is given by

$y=73.5\,\mathrm m+\left(24.5\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

where $g=9.80\dfrac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity. The stone reaches the ground when $y=0$ :

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Rom4ik [11]

Answer:ring closure result in fewer molecules

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