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3 years ago

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Two lab carts are on a frictionless track, attached by a spring. The carts are pushed together by a student in a Physics lab, co

mpressing the spring fully. The student lets go of the carts and they move in opposite directions as the spring extends. Which of the following statements describes what happens to the car/spring system after the cars are released?
A. The kinetic energy remains constant.
B. The kinetic energy increases.
C. The momentum remains constant.
D. The momentum increases.

2 answers:

-Dominant- [34]3 years ago

8 0

Answer:

A) The kinetic energy remains constant

Feliz [49]3 years ago

8 0

A.the kinetic remains constant

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What will cause the volume of an ideal gas to triple in value?

trapecia [35]

Answer: The volume of an ideal gas will triple in value if the pressure is reduced to one-third of its initial value

Explanation:

We can determine this from the gas laws. Using Boyle's law, which states that "the pressure of a given mass of an ideal gas is inversely proportional to its volume at a constant temperature"

Mathematically, P ∝ (1/V)

Since P ∝ (1/V), we can then write that

P = k(1/V)

Where P is the pressure, V is the volume and k is the proportionality constant

PV = k

We can then write that

P1V1 = P2V2 = P3V3 = ...

Hence, P1V1 = P2V2

Where P1 is the initial pressure of the gas

P2 is the final pressure of the gas

V1 is the initial volume of the gas

and V2 is the final volume of the gas

From the question, we want to determine what will make the new volume be thrice the initial volume.

Hence,

P1 = P

V1 = V

P2= ??

V2 = 3V

Therefore,

P × V = P2 × (3V)

P2 = PV/3V

P2 = P/3 = 1/3(P)

This means the volume of an ideal gas will triple in value if the pressure is reduced to one-third of its initial value

6 0

3 years ago

Read 2 more answers

Help please, I don't get it

Sergio [31]

Answers:

a) $\hat F=(0.83,-0.55) N$

b) $\hat D=(-0.44,-0.89) m$

c) $\hat V=(-0.47,0.88) m/s$

Explanation:

A unit vector is a vector whose magnitude (length) is equal to 1. This kind of vector is identified as $\hat v$ and the way to calculate is as follows:

$\hat v=\frac{\vec v}{|v|}$

Where:

$\vec v=(x,y)$ is the vector

$|v|=\sqrt{x^{2}+y^{2}}$ is the magnitude of the vector

Having this information clarified, let's begin with the answers:

a) Force Vector

$\vec F=(9.0 \hat i - 6.0 \hat j) N$

Magnitude of $\vec F$ :

$|F|=\sqrt{(9.0 \hat i)^{2}+(-6.0 \hat j)^{2 }}N=10.81 N$

<u />

<u>Unit vector:</u>

$\hat F=\frac{\vec F}{|F|}$

$\hat F=\frac{(9.0 \hat i - 6.0 \hat j) N}{10.81 N}$

$\hat F=\frac{9.0}{10.81} N-\frac{6.0}{10.81}N$

$\hat F=(0.83,-0.55) N$

b) Displacement Vector

$\vec D=(-4.0 \hat i - 8.0 \hat j) m$

Magnitude of $\vec D$ :

$|D|=\sqrt{(-4.0 \hat i)^{2}+(-8.0 \hat j)^{2 }}m=8.94 m$

<u />

<u>Unit vector:</u>

$\hat D=\frac{\vec D}{|D|}$

$\hat D=\frac{(-4.0 \hat i - 8.0 \hat j) m}{8.94 m}$

$\hat D=\frac{-4.0}{8.94} Nm+\frac{-8.0}{8.94}m$

$\hat D=(-0.44,-0.89) m$

c) Velocity Vector

$\vec V=(-3.50 \hat i + 6.50 \hat j) m/s$

Magnitude of $\vec V$ :

$|V|=\sqrt{(-3.50 \hat i)^{2}+(6.50 \hat j)^{2}}m/s=7.38 m/s$

<u />

<u>Unit vector:</u>

$\hat V=\frac{\vec V}{|V|}$

$\hat V=\frac{(-3.50 \hat i +6.50 \hat j) m/s}{7.38 m/s}$

$\hat V=\frac{-3.50}{7.38} m/s+\frac{6.50}{7.38}m/s$

$\hat V=(-0.47,0.88) m/s$

8 0

4 years ago

A 500 g model rocket is on a cart that is rolling to the right at a speed of 3.0 m/s . The rocket engine, when it is fired, exer

ivanzaharov [21]

Answer:

The rocket has to be launched 8 m from the hoop

Explanation:

Let's analyze this problem, the rocket is on a car that moves horizontally, so the rocket also has the same speed as the car; The initial horizontal rocket speed is (v₀ₓ = 3.0 m/s).

On the other hand, when starting the engines we have a vertical force, which creates an acceleration in the vertical axis, let's use Newton's second law to find this vertical acceleration

F -W = m a

a = (F-mg) / m

a = F/m -g

a = 7.0/0.500 - 9.8

a = 4.2 m/s²

We see that we have a positive acceleration and that is what we are going to use in the parabolic motion equations

Let's look for the time it takes for the rocket to reach the height (y = 15m) of the hoop, when the rocket fires its initial vertical velocity is zero (I'm going = 0)

y = $v_{oy}$ t + ½ a t²

y = 0 + ½ a t²

t = √ 2y/a

t = √( 2 15 / 4.2)

t = 2.67 s

This time is also the one that takes in the horizontal movement, let's calculate how far it travels

x = v₀ₓ t

x = 3 2.67

x = 8 m

The rocket has to be launched 8 m from the hoop

8 0

3 years ago

A star is estimated to be 4.8 km away from the earth. When we see this star in the night sky how old is the image. Let the speed

andre [41]

Time = distance / speed

Time = (4,800 meters) / (3 x 10⁸ m/s)

<em>Time = 0.000016 second</em>

This number is not one of the choices on the list. My hunch is that you copied the distance wrong.

If the estimated distance to the star is actually 4.8 x 10¹⁵ km, instead of 4.8 km, then the answer would be close to 500 years <em>(B)</em>.

There's no way a star can be "4.8 km away from the Earth". You can <em>walk</em> that far in about an hour, and passenger jet airplanes fly <em>twice</em> as far as that away from the Earth !

5 0

3 years ago

What is t=3seconds in scrience

alekssr [168]

Well....t usually stands for time and the unit seconds proves that

8 0

3 years ago