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Fernando is an engineer at an amusement park who is experimenting with changes to the setup for a magnetic roller coaster ride. In one ride, there are two identical roller coaster cars (red and yellow) that start on opposite sides of a large magnet located in the center of a station. He moves both cars, so each car ends up one space farther away from the large magnet.How did the potential energy of the two cars change? Did the potential energy of one car change more than the other? Why do you think so?

Answer:

none it's Hydroelectric energy

Answer:

it's 9m

Explanation:

Answer:

9m

Explanation:

edge2o2o

Answer:

well for me it's

Explanation:

L2-L1/L1(O2-O1)

O is the change in temperature

Answer:

Inelastic collision.

Explanation:

Because both the momentum of the man and that of the car, is conserved because they are both moving with the same velocity.

Also kinetic energy is not conserved because the car moved with different kinetic energy, as well as the man.

When they came into contact, the final kinetic energy is different from the initial kinetic energy.

Answer:

perfectly inelastic

Explanation:

Answer:

See list below

Explanation:

5. positively charged particles : D

6. dense center of the atom: C

7. negatively charged particles: A

8. articles with NO charge: B

Answer:

Newton's second law states that the acceleration of an object is directly related to the net force and inversely related to its mass.

Answer:

a = v²/r

Explanation:

The acceleration of a body moving in a circular path is known as the centripetal acceleration. This is the acceleration of a body that keeps the body within the circular path. It is written in terms of the linear velocity v and the radius of the circle of rotation as shown;

a = v²/r where

v is the linear velocity

r is the radius

a is the centripetal acceleration

Answer:

Intercellular communication refers to the communication between cells. Membrane vesicle trafficking has an important role in intercellular communications in humans and animals, e.g., in synaptic transmission, hormone secretion via vesicular exocytosis.

Elements of art are stylistic features that are included within an art piece to help the artist communicate. The seven most common elements include line, shape, texture, form, space, colour and value, with the additions of mark making, and materiality.

You would create a darker value if you shaded it, shading it makes it darker.

Answer:

✨ love ✨

words words words words words words words words words words words

Answer:

butter

Explanation:

Answer:

As the Speed increases , the potential energy decreases

Answer:

Since the object only has so much total internal energy, and the faster an object moves the larger the kinetic energy is, hence the potential energy will decrease as a result of speed increasing.

Explanation:

Hope this helped!

Answer:

Opposite

Explanation:

Waves that hit a fixed boundary return to the starting point on the opposite side that it began.

After striking, the wave gets reflected and it moves in the direction opposite to the initial direction.

Hence, it would mean that waves that hit a fixed boundary return to the starting point on the opposite direction.

Answer:

27km/h that's the answer

Answer:

if a force does a negative amount of work on an object ,so the speed of the object got  decrease  ,because the acceleration  becomes negative  so if the acceleration becomes negative so the speed is decreased.

Explanation:

Answer: a) 2,508

Explanation:

Answer:

The force P required  is 1759.22 N

Explanation:

The missing diagram is seen in the first image below.

From the second image, we can see the schematic diagram of the engine hanging over the pulley.

To start with determining the value of the angle ∝;

[tex]tan \ \alpha = \dfrac{CD}{BD}[/tex]

where;

BD = AB-AD

Then;

[tex]tan \ \alpha = \dfrac{CD}{AB-AD}[/tex]

[tex]\alpha = tan^{-1} \bigg(\dfrac{CD}{AB-AD} \bigg )[/tex]

replacing their respective values, where;

CD = 2 sin 30° m,  AB = 2m and AD = 2 cos 30° m

[tex]\alpha = tan^{-1} \bigg(\dfrac{2 \ sin \ 30^0}{2-2 \ cos \ 30^0} \bigg )[/tex]

[tex]\alpha = tan^{-1} \bigg(\dfrac{1}{2-1.732} \bigg )[/tex]

[tex]\alpha = tan^{-1} \bigg(\dfrac{1}{0.268} \bigg )[/tex]

[tex]\alpha = tan^{-1} \bigg(3.73\bigg )[/tex]

[tex]\alpha \simeq 75^0[/tex]

From the third diagram attached below:

The tension occurring in the thread BC is equal to force P

[tex]T_{BC} = P[/tex]

Using the force equilibrium expression along the horizontal direction.

[tex]\sum F_x = 0\\\\ -T_{AC} \ cos \ 30^0 + Pcos \alpha = 0[/tex]

replacing the value of [tex]\alpha \simeq 75^0[/tex]

[tex]-T_{AC} \ cos 30^0 + P cos 75^0 = 0[/tex]

[tex]P \ cos \ 75^0 = T_{AC} \ cos \ 30^0[/tex]

[tex]P =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0} \ \ \ - - - (1)[/tex]

Along the vertical direction, the force equilibrium equation can be expressed as:

[tex]\sum F_y =0[/tex]

[tex]-W + P \ sin \alpha + T_{AC} \ sin \ 30^0 = 0[/tex]

[tex]W = P \ sin \ \alpha + T_{AC} \ sin \ 30^0[/tex]

replacing [tex]\alpha \simeq 75^0[/tex] and [tex]P =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0}[/tex]

[tex]W =\dfrac{T_{AC} \ cos \ 30^0}{cos \ 75^0}\times sin \ 75^0 + T_{AC} \ sin \ 30^0[/tex]

Also, replacing W for (200 × 9.81) N

[tex]200 \times 9.81 =\dfrac{T_{AC} \ cos \ 30^0}{cos \ 75^0}\times sin \ 75^0 + T_{AC} \ sin \ 30^0[/tex]

[tex]200 \times 9.81 = T_{AC} \ cos \ 30^0 \ tan \ 75^0 + T_{AC} \ sin \ 30^0[/tex]

[tex]1962= T_{AC} \ ( cos \ 30^0 \ tan \ 75^0 + \ sin \ 30^0)[/tex]

[tex]1962= T_{AC} \ (0.8660\times 3.732 + 0.5)[/tex]

[tex]1962= T_{AC} \ (3.231912 + 0.5)[/tex]

[tex]1962= T_{AC} \ (3.731912)[/tex]

[tex]T_{AC} = \dfrac{1962}{ \ (3.731912)}[/tex]

[tex]T_{AC} = 525.736 \ N[/tex]

From [tex]P =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0}[/tex]

[tex]P =\dfrac{ 525.736 \ cos \ 30^0}{\ cos \ 75^0}[/tex]

[tex]P =\dfrac{ 525.736 \times0.866}{0.2588}[/tex]

P = 1759.22 N

Thus, the force P required  is 1759.22 N

Answer:

M₂ = M  then L₂ = L

M₂> M  then L₂ = \frac{M}{M_{2}} L

Explanation:

This is a static equilibrium exercise, to solve it we must fix a reference system at the turning point, generally in the center of the rod. By convention counterclockwise turns are considered positive

          ∑ τ = 0

The mass of the rock is M and placed at a distance, L the mass of the rod M₁, is considered to be placed in its center of mass, which by uniform e is in its geometric center (x = 0) and the triangular mass M₂, with a distance L₂

The triangular shape of the second object determines that its mass can be considered concentrated in its geometric center (median) that tapers with a vertical line if the triangle is equilateral, the most used shape in measurements.

         M L + M₁ 0 - m₂ L₂ = 0

         M L - m₂ L₂ = 0

         L₂ = [tex]\frac{M}{M_{2}}[/tex] L

From this answer we have several possibilities

* if the two masses are equal then L₂ = L

* If the masses are different, with M₂> M then L₂ = \frac{M}{M_{2}} L

The triangular object should be placed at the distance of   [tex]\dfrac{M}{M_{2}} \times L[/tex]  from the left end of the stick.

The given problem can be resolved using the concept of static equilibrium. We must consider a reference system at the turning point, generally in the center of the rod. By convention counterclockwise turns are considered positive. So, static equilibrium says,

[tex]\sum \tau = 0[/tex]

Here, [tex]\tau[/tex] is the torque.

Let the  mass of the rock is M and it is placed at a distance L. Such that the mass of the rod is [tex]M_{1}[/tex] which is considered to be placed in its center of mass.  By uniformity in its geometric center (x = 0) and the triangular mass [tex]M_{2}[/tex] lies at the distance [tex]L_{2}[/tex].

Also, triangular shape of the second object determines that its mass can be considered concentrated in its geometric center (median) that tapers with a vertical line if the triangle is equilateral, the most used shape in measurements.

Then,

[tex]ML+M_{1} \times 0 -M_{2} \times L_{2}=0\\\\ML=M_{2}L_{2}\\\\L_{2}=\dfrac{M}{M_{2}} \times L[/tex]

Thus, we can conclude that triangular object should be placed at the distance of  [tex]\dfrac{M}{M_{2}} \times L[/tex]  from the left end of the stick.

Learn more about the equilibrium of forces here:

https://brainly.com/question/24018969

Answer:

h' = 603.08 m

Explanation:

First, we will calculate the initial velocity of the pellet on the surface of Earth by using third equation of motion:

2gh = Vf² - Vi²

where,

g = acceleration due to gravity on the surface of earth = - 9.8 m/s² (negative sign due to upward motion)

h = height of pellet = 100 m

Vf = final velocity of pellet = 0 m/s (since, pellet will momentarily stop at highest point)

Vi = Initial Velocity of Pellet = ?

Therefore,

(2)(-9.8 m/s²)(100 m) = (0 m/s)² - Vi²

Vi = √(1960 m²/s²)

Vi = 44.27 m/s

Now, we use this equation at the surface of moon with same initial velocity:

2g'h' = Vf² - Vi²

where,

g' = acceleration due to gravity on the surface of moon = 1.625 m/s²

h' = maximum height gained by pellet on moon = ?

Therefore,

2(1.625 m/s²)h' = (44.27 m/s)² - (0 m/s)²

h' = (1960 m²/s²)/(3.25 m/s²)

h' = 603.08 m

Answer:

D. The momentum of Car B is three times as great in magnitude as that of car A.

Explanation:

I majored in Physics

Answer:

The water molecules start moving faster.

Explanation:

As the water is heated, the water molecules absorb heat and moves faster. The longer the water is kept on the stove, the more heat and kinetic energy the molecules gain. Soon, the hydrogen bonds break down completely because of the high kinetic energy and the water molecules escape into the air as gas.

The temperature that causes this to happen is known as the boiling point.

Answer:

When light gets in contact with water, it emerges into the air, and then it speeds up, which makes it look shallower than it actually is.

Explanation:

Answer:

Refraction

Explanation:

Light rays reflected from the fish are refracted at the surface of the water, but the eyes and brain trace the light rays back into the water as thought they had not refracted, but traveled away from the fish in a straight line. This effect creates a "virtual" image of the fish that appears at a shallower depth.

Answer:

The efficiency of the heat engine is 33.33%.

Explanation:

Given;

input energy of the heat engine, = 6000 J

energy lost by the heat engine, = 4000 J

output energy of the heat engine = 6000 J - 4000 J = 2000 J

The efficiency of the heat engine is given as;

Efficiency = output energy / input energy

Efficiency = 2000 / 6000

Efficiency = 0.3333

Efficiency (%) = 33.33%

Therefore, the efficiency of the heat engine is 33.33%.

Answer:

A: Opposite

Explanation:

From Newton's third law of motion, to every action there is an equal an opposite reaction. Thus, when the wave hits a fixed boundary, it is returned as a reflected wave to the starting point albeit on the other side.

Answer:

I would choose refrigerator.

Explanation:

In my opinion, such a battery could be invented by changing the device as a low power mode. Plus, it is a must in order to decrease the CFC gas spreading through the atmosphere. That's all, thank you.

Answer:

the kinetic energy of A will be twice that of B

Explanation:

The formula for calculating kinetic energy is expressed using the formula

KE = 1/2mv²

m is the mass of the object

v is the velocity

For object A:

KEA = 1/2mAvA²

For object B:

KEB = 1/2mBvB² ... 1

If object A is twice the mass of object B, then mA = 2mB

From 1:

KEA = 1/2mAvA²

Substitute mA = 2mB

KEA = 1/2(2mB)vA² .... 2

Divide 1 by 2

KEB/KEA = 1/2mBvB²/mBvA²

KEB/KEA = 1/2vB²/vA²

Assuming they have the same velocities then vA ,= VB

The equation becomes:

KEB/KEA = 1/2vB²/vA²

KEB/KEA = 1/2vB²/vB²

KEB/KEA = 1/2

KEA = 2EB

Hence the kinetic energy of A will be twice that of B

Answer:

a) The position vector of P is [tex]\vec P =(0, 2,4)[/tex].

b) The distance vector from P to Q is [tex]\overrightarrow{PQ} = (-3,-1,1)[/tex].

c) The distance between P and Q is [tex]\|\overrightarrow{PQ}\|=\sqrt{11}[/tex].

d) A vector parallel to PQ with magnitude of 10 is [tex]\vec v = \left(-\frac{30\sqrt{11}}{11},-\frac{10\sqrt{11}}{11}, \frac{10\sqrt{11}}{11} \right)[/tex].

Explanation:

a) The position vector of a point is the vector displacement from the origin to the location of the point. That is:

[tex]\vec P = (0,2,4)-(0,0,0)[/tex]

[tex]\vec P = (0-0, 2-0, 4-0)[/tex]

[tex]\vec P =(0, 2,4)[/tex]

The position vector of P is [tex]\vec P =(0, 2,4)[/tex].

b) First, we calculate the position vector of point Q:

[tex]\vec Q = (-3,1,5)-(0,0,0)[/tex]

[tex]\vec Q = (-3-0,1-0,5-0)[/tex]

[tex]\vec Q =(-3,1,5)[/tex]

The distance vector from P to Q is define by the following vectorial expression:

[tex]\overrightarrow{PQ} = \vec Q - \vec P[/tex] (1)

[tex]\overrightarrow{PQ} = (-3,1,5)-(0,2,4)[/tex]

[tex]\overrightarrow{PQ} =(-3-0,1-2,5-4)[/tex]

[tex]\overrightarrow{PQ} = (-3,-1,1)[/tex]

The distance vector from P to Q is [tex]\overrightarrow{PQ} = (-3,-1,1)[/tex].

c) There are two approaches to calculate the distance between P and Q:

First Method - Pythagorean Theorem:

[tex]\|\overrightarrow{PQ}\| = \sqrt{(-3)^{2}+(-1)^{2}+1^{2}}[/tex]

[tex]\|\overrightarrow{PQ}\|=\sqrt{11}[/tex]

Second Method - Dot Product:

[tex]\|\overrightarrow{PQ}\| = \sqrt{\overrightarrow{PQ}\,\bullet\,\overrightarrow{PQ}}[/tex] (2)

[tex]\|\overrightarrow{PQ}\| = \sqrt{(-3,-1,1)\,\bullet (-3,-1,1)}[/tex]

[tex]\|\overrightarrow{PQ}\|=\sqrt{11}[/tex]

The distance between P and Q is [tex]\|\overrightarrow{PQ}\|=\sqrt{11}[/tex].

d) To determine a vector parallel to PQ with a given magnitude is determined by the following expression:

[tex]\vec v = \frac{k}{\|\overrightarrow{PQ}\|} \cdot \overrightarrow{PQ}[/tex] (3)

Where [tex]k[/tex] is the scale factor.

If we know that [tex]\overrightarrow{PQ} = (-3,-1,1)[/tex], [tex]\|\overrightarrow{PQ}\|=\sqrt{11}[/tex] and [tex]k = 10[/tex], then the vector is:

[tex]\vec v = \frac{10}{\sqrt{11}}\cdot (-3,-1,1)[/tex]

[tex]\vec v = \left(-\frac{30}{\sqrt{11}},-\frac{10}{\sqrt{11}},\frac{10}{\sqrt{11}}\right)[/tex]

[tex]\vec v = \left(-\frac{30\sqrt{11}}{11},-\frac{10\sqrt{11}}{11}, \frac{10\sqrt{11}}{11} \right)[/tex]

A vector parallel to PQ with magnitude of 10 is [tex]\vec v = \left(-\frac{30\sqrt{11}}{11},-\frac{10\sqrt{11}}{11}, \frac{10\sqrt{11}}{11} \right)[/tex].

Answer:

true

Explanation: