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3 years ago

6

A 1200-kg station wagon is moving along a straight highway at Another car, with mass 1800 kg and speed has its center of mass 40

.0 m ahead of the center of mass of the station wagon (Fig. E8.54). (a) Find the position of the center of mass of the system consisting

1 answer:

taurus [48]3 years ago

5 0

Answer:

Center of mass lies 24 m in front of center of mass of second wagon.

Explanation:

Suppose A 1200 kg station wagon is moving along a straight highway at 12.0 m/s. Another car with mass 1800 kg and speed 20.0 m/s.

Given that,

Mass of first wagon = 1200 kg

Mass of second wagon = 180 kg

Distance = 40 m

We need to calculate the position of the center of mass of the system

Using formula of center mass

$x_{cm}=\dfrac{m_{1}x_{1}+m_{2}x_{2}}{m_{1}+m_{2}}$

$x_{cm}=\dfrac{1200\times0+1800\times40}{1200+1800}$

$x_{cm}=24\ m$

Hence, Center of mass lies 24 m in front of center of mass of second wagon.

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A tugboat tows a ship with a constant force of magnitude F1. The increase in the ship's speed during a 10 s interval is 5.0 km/h

Ratling [72]

Answer

given,

time = 10 s

ship's speed = 5 Km/h

F = m a

a is the acceleration and m is mass.

In the first case

F₁=m x a₁

where a₁ = difference in velocity / time

F₁ is constant acceleration is also a constant.

Δv₁ = 5 x 0.278

Δv₁ = 1.39 m/s

$a_1=\dfrac{1.39}{10}$

a₁ = 0.139 m/s²

F₂ =m x a₂

F₃ = F₂ + F₁

Δv₃ = 19 x 0.278

Δv₃ = 5.282 m/s

a₃=Δv₂ / t

$a_3=\dfrac{5.282}{10}$

a₃ = 0.5282 m²/s

m a₃=m a₁ + m a₂

a₃ = a₂ + a₁

0.5282 = a₂ + 0.139

a₂=0.3892 m²/s

F₂ = m x 0.3892...........(1)

F₁ = m x 0.139...............(2)

F₂/F₁

ratio = $\dfrac{0.3892}{0.139}$

ratio = 2.8

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4 years ago

Your mom is making lunch which forces are acting on your mom

N76 [4]

The forces acting on your mom while cooking is Air resistance and the force of friction

<u>Explanation:</u>

<u>1. Air resistance:</u>

In simple words, Air resistance can be stated as the type of friction between the air and the other materials.
In this scenario, there will be an air resistance and the air hits the mom while cooking via the doors or windows

<u>2. The force of friction:</u>

In simple words, friction can be stated as, the resistance that one surface or object encounters when moving over another.
While cooking the food mom would experience the friction since friction is the transfer of heat, and cooking is the process of receiving that heat.

7 0

3 years ago

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An ideal spring hangs from the ceiling. A 1.25-kg mass is hung from the spring. After all vibrations have died away, the spring

ch4aika [34]

The kinetic energy of the mass at the instant it passes back through its equilibrium position is about 1.20 J

$\texttt{ }$

<h3>Further explanation</h3>

Let's recall Elastic Potential Energy formula as follows:

$\boxed{E_p = \frac{1}{2}k x^2}$

where:

<em>Ep = elastic potential energy ( J )</em>

<em>k = spring constant ( N/m )</em>

<em>x = spring extension ( compression ) ( m )</em>

Let us now tackle the problem!

$\texttt{ }$

<u>Given:</u>

mass of object = m = 1.25 kg

initial extension = x = 0.0275 m

final extension = x' = 0.0735 - 0.0275 = 0.0460 m

<u>Asked:</u>

kinetic energy = Ek = ?

<u>Solution:</u>

<em>Firstly , we will calculate the spring constant by using </em><em>Hooke's Law</em><em> as follows:</em>

$F = k x$

$mg = k x$

$k = mg \div x$

$k = 1.25(9.8) \div 0.0275$

$k = 445 \frac{5}{11} \texttt{ N/m}$

$\texttt{ }$

<em>Next , we will use </em><em>Conservation of Energy</em><em> formula to solve this problem:</em>

$Ep_1 + Ek_1 = Ep_2 + Ek_2$

$\frac{1}{2}k (x')^2 + mgh + 0 = \frac{1}{2}k x^2 + Ek$

$Ek = \frac{1}{2}k (x')^2 + mgh - \frac{1}{2}k x^2$

$Ek = \frac{1}{2}k ( (x')^2 - x^2 ) + mgh$

$Ek = \frac{1}{2}(445 \frac{5}{11}) ( 0.0460^2 - 0.0275^2 ) + 1.25(9.8)(0.0735)$

$\boxed {Ek \approx 1.20 \texttt{ J}}$

$\texttt{ }$

<h3>Learn more</h3>

Kinetic Energy : brainly.com/question/692781

Acceleration : brainly.com/question/2283922

The Speed of Car : brainly.com/question/568302
Young Modulus : brainly.com/question/9202964
Simple Harmonic Motion : brainly.com/question/12069840

$\texttt{ }$

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Elasticity

8 0

3 years ago

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8. Il An 8.00 kg package in a mail-sorting room slides 2.00 m down a

Vitek1552 [10]

Answer:

See below

Explanation:

Normal force = m g cos 53 = 8 kg * 9.8 m/s^2 * cos 53 = 47.1823 N

no work is done by this force

Force friction = coeff friction * force normal = .4 * 47.1823 = 7.55 N

work of friction = 7.55 * 2 m = 15.1 j

Force Downplane = mg sin 53 = 62.61 N

work = 62.61 * 2 = 125.22 j

Net Force downplane = force downplane - force friction = 55.06 N

net Work = force * distance = 55.06 N * 2 M = 110.12 j

3 0

2 years ago

A person throws a ball straight up. He releases the ball at a height of 1.75 m above the ground and with a velocity of 12.0 m/s.

Lyrx [107]

Answer:

a) 1.22 s

b) 9.089 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-12}{-9.81}\\\Rightarrow t=1.22\ s$

Time taken by the ball to reach the highest point is 1.22 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=12\times 1.22+\frac{1}{2}\times -9.81\times 1.22^2\\\Rightarrow s=7.339\ m$

The maximum height the ball will reach above the ground is 1.75+7.339 = 9.089 m

8 0

3 years ago