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3 years ago

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A block mass m (0.25 kg) is pressed against (but is not attached to) an ideal spring of force constant k (100 N/m) and negligibl

e mass, compressing the spring a distance 0.1 m. After it is released, the box slides along a leveled surface for 2 m before fully stops. If we repeat this experiment but instead use a spring having force constant 2k,
a. the box will go up the incline twice as high as before.
b. just as it moves free off the spring, the kinetic energy of the box will be twice as great as before.
c. just as it moves free off the spring, the speed of the box will be
d. All the above choices are correct.
e. None of the above choices are correct.

1 answer:

VMariaS [17]3 years ago

5 0

Answer:

d. All the above choices are correct.

Explanation:

When a spring of spring constant k is compressed by distance x , the potential energy stored in it is equal to

E = 1/2 k x²

If spring constant is 2 k , potential energy stored

E = 1/2 2k x²

= k x²

which is twice the earlier potential energy.

In the first case , the energy of spring is imparted to box . The energy given to box is spent by frictional force due to which box comes to rest.

So energy of box acquired from spring = work done by frictional force.

So energy of box acquired from spring = F X d , F is frictional force , d is displacement .

In the second case ,

energy acquired by box becomes two times

Work done by frictional force will also become two times to put box at rest

So displacement will be two times ( because frictional force is constant )

so option a is correct .

option b is also correct .

Because kinetic energy of box will be twice as explained above .

So option d will be correct.

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Suppose each object emitted a burst of light right now.Rank the objects from left to right based on the amount of time itwould t

Wittaler [7]

Answer:

the order of arrival is from highest to lowest

star other side of Andromeda> star near side of andromeda> other side of milky way > center of the milky way> nevulosa orion> Pluto> Sum

Explanation:

The light that comes from stars and galaxies travels in a vacuum so its speed is constant and with a value of c = 3 108 m / s, so time will be directly proportional to the distance to the object

x = c t

the order of arrival is from highest to lowest

star other side of Andromeda> star near side of andromeda> other side of milky way > center of the milky way> nevulosa orion> Pluto> Sum

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3 years ago

A beam of light strikes a sheet of glass at an angle of 57.0° with the normal in air. You observe that red light makes an angle

Contact [7]

<h2>Answers: </h2>

1) 1.359, 1.403

2) $2.207(10)^{8}m/s$ , $2.138(10)^{8}m/s$

Explanation:

The described situation is known as Refraction.

Refraction is a phenomenon in which a wave (the light in this case) bends or changes it direction when passing through a medium with a refractive index different from the other medium.

In this context, the Refractive index $n$ is a number that describes how fast light propagates through a medium or material, and is defined as the relation between the speed of light in vacuum ( $c=3(10)^{8}m/s$ ) and the speed of light $v$ in the second medium:

$n=\frac{c}{v}$ (1)

On the other hand we have the Snell’s Law:

$n_{1}sin(\theta_{1})=n_{2}sin(\theta_{2})$ (2)

Where:

$n_{1}$ is the first medium refractive index . We are told is the air, hence $n_{1}\approx 1$

$n_{2}$ is the second medium refractive index

$\theta_{1}$ is the angle of the incident ray

$\theta_{2}$ is the angle of the refracted ray

Knowing this, let's begin with the answers:

<h2><u>1) Indexes of refraction for red and violet light</u></h2><h2 /><h2>1a) Red light</h2>

Using equation (2) according to Snell's Law and $\theta_{1}=57.0\º$ $\theta_{2}=38.1\º$ :

$(1)sin(57.0\º)=n_{2}sin(38.1\º)$

Finding $n_{2}$ :

$n_{2}=\frac{sin(57.0\º)}{sin(38.1\º)}$

$n_{2}=1.359$ (3)>>>Index of Refraction for red light

<h2>1b) Violet light</h2>

Again, using equation (2) according to Snell's Law and $\theta_{1}=57.0\º$ $\theta_{2}=36.7\º$ :

$(1)sin(57.0\º)=n_{2}sin(36.7\º)$

Finding $n_{2}$ :

$n_{2}=\frac{sin(57.0\º)}{sin(36.7\º)}$

$n_{2}=1.403$ (4) >>>Index of Refraction for violet light

<h2><u>2) Speeds of red and violet light</u></h2><h2 /><h2>1a) Red light</h2>

Here we are going to use equation (1):

$n_{red}=\frac{c}{v_{red}}$

$v_{red}=\frac{c}{n_{red}}$

Substituting (3) in this equation:

$v_{red}=\frac{3(10)^{8}m/s}{1.359}$

$v_{red}=2.207(10)^{8}m/s$ >>>>Speed of red light

<h2>1a) Violet light</h2>

Using again equation (1):

$n_{violet}=\frac{c}{v_{violet}}$

$v_{violet}=\frac{c}{n_{violet}}$

Substituting (4) in this equation:

$v_{violet}=\frac{3(10)^{8}m/s}{1.403}$

$v_{red}=2.138(10)^{8}m/s$ >>>>Speed of violet light

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3 years ago

A proton is released from rest at the origin in a uniform electric field in the positive x direction with magnitude 850 N/C. Wha

Gwar [14]

A proton is released from rest at the origin in a uniform electric field in the positive x direction with magnitude 850 N/C. The change in the electric potential energy of the proton-field system when the proton travels to x = 2.50m is -3.40 × 10⁻¹⁶ J (Option B)

<h3 /><h3>How is the change in electric potential energy of the proton-field system calculated?</h3>

Work done on the proton =Negative of the change in the electric potential energy of the proton field
In the given case, W = -qΔV
-W = qΔV
= qEcosθ
Therefore, work done on the proton = -e(8.50× $10^2$ N/C)(2.5m)(1)
= -3.40× $10^-^1^6$ J
Any change in the potential energy indicates the work done by the proton.
Therefore the positive sign shows that the potential energy increases when the proton does the work.
The negative sign shows that the potential energy decreases when the proton does the work.

To learn more about electric potential energy, refer

brainly.com/question/14306881

#SPJ4

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1 year ago

2- A car on a straight highway goes in the positive direction for 8 km and then backs up for 3.6 km. What are the distance and d

IRISSAK [1]

Answer:

11.6km

4.4km in the negative direction

Explanation:

Distance is the total length of path covered and traveled by a body.

So, for this car on a straight line;

Total distance = 8km + 3.6km = 11.6km

Displacement is the distance traveled along a path and the direction it takes.

It is a vector quantity with magnitude and directional attributes.

For this journey;

Displacement = 8km - 3.6km = 4.4km in the negative direction.

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What has to happen to an uncharged aluminum atom so that it will bond with oxygen?

iren [92.7K]

It has to lose electrons to become positively charged, so it can attract a negatively charged oxygen ion.

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