What is the elastic potential energy of a spring that is compressed a distance of 0.35 m and has a spring constant of 71.8 N/m?

Answer: 1700

Explanation:

Linear expansivity, area expansivity and volume or cubic expansivity are

The correct option is option C.

Doubling the charge on the second particle produces the smallest increase in force.

Let the given charges be Q and q such that:

Q = 8.1C and q = 2.6C

separated by a distance of R = 0.03m

Now, we have to calculate the minimum increase in electric force.

The electrostatic force is given by:

F  = kQq/R²

Option A is wrong since reducing the charge will reduce the force.

Option B is wrong since the force is inversely proportional to the distance, so increasing the distance will decrease the force.

Option C ⇒ Double the charge on the second particle:

F = 2kQq/R²

Option D ⇒ Reduce the distance between the particles by one half:

So, the new distance between the charges is R/2, then

F = kQq/(R/2)²

F = 4kQq/R²

Clearly, the operation in option C gives the smallest increase in the force.

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Answer:

Range of projectile, R

For projection above ground surface, the range of the angle of projection with respect to horizontal direction, θ, is 0° ≤ θ ≤ 90° and the corresponding range of 2θ is 0° ≤ 2θ ≤ 180°.S

hope you're looking for this.

Answer:

50m; 0m/s.

Explanation:

Given the following data;

Initial velocity = 20m/s

Acceleration, a = - 4m/s²

Time, t = 5secs

To find the displacement, we would use the second equation of motion;

[tex] S = ut + \frac {1}{2}at^{2}[/tex]

Substituting into the equation, we have;

[tex] S =20*5 + \frac{1}{2}*(-4)*5^{2}[/tex]

[tex] S =100 + (-2)*25[/tex]

[tex] S =100 - 50[/tex]

S = 50m

Next, to find the final velocity, we would use the third equation of motion;

[tex] V^{2} = U^{2} + 2aS [/tex]

Where;

Substituting into the equation, we have;

[tex] V^{2} = 20^{2} + 2(-4)*50 [/tex]

[tex] V^{2} = 400 - 400[/tex]

[tex] V^{2} = 0[/tex]

V = 0m/s

Therefore, the displacement of the bus is 50m and its final velocity is 0m/s.

Answer:

c. Having product appeal

Explanation:

Answer:

c

Explanation:

The coefficient of kinetic friction between the box and the floor is 0.323.

By Newton's Law of Motion we understand that an object is at equilibrium if and only if it is at rest or it is moving at constant velocity. If a horizontal force is applied on the box and it is moving, then a force with equal magnitude and opposed to that force must exists, which corresponds to the kinetic friction. Let suppose that the box is moving on a horizontal ground.

The equation of equilibrium for the box is described below:

[tex]\Sigma F = P - \mu_{k}\cdot m\cdot g = 0[/tex] (1)

Where:

An expression for the kinetic coefficient of friction is derived by clearing the variable in (1):

[tex]\mu_{k} = \frac{P}{m\cdot g}[/tex] (2)

If we know that [tex]P = 38\,N[/tex], [tex]m = 12\,kg[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the kinetic coefficient of friction is:

[tex]\mu_{k} = \frac{38\,N}{(12\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]

[tex]\mu_{k} = 0.323[/tex]

The coefficient of kinetic friction between the box and the floor is 0.323.

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Answer:

A) Vo = 137.34 [m/s]

B) Y = 961.38 [m]

C) y = 367.8[m] for t = 3 [s] and y = 470.88[m] for t = 4[s]

Explanation:

To solve this problem we must use the following equation of kinematics. We must keep in mind that the negative sign of the equation means that the acceleration of gravity acts in the opposite direction to the movement of the balloon.

A)

[tex]v_{f} =v_{o} -g*t[/tex]

where:

Vf = final velocity = 0 (the maximum hang time, maximum elevation)

Vo = initial velocity [m/s]

g = gravity acceleration = 9.81 [m/s²]

t = time = 14 [s]

0 = Vo - (9,81*14)

Vo = 137.34 [m/s]

B)

Now we use the following equation.

[tex]y=y_{o}+v_{o} *t -(\frac{1}{2} )*g*t^{2}[/tex]

where:

Yo = initial position = 0

Y = final position [m]

Vo = initial velocity = 137.34 [m/s]

Now replacing

Y = (137.34*14) - (0.5*9.81*14²)

Y = 961.38 [m]

C)

With the above equation, we can calculate the distances between t = 3 and t = 4 [s]

[tex]y=y_{o} +v_{o} *t - (1/2)*g*t^{2}[/tex]

y = (137.34*3) - (0.5*9.81*3²)

y = 367.8[m] for t = 3 [s]

for t = 4[s]

y =  (137.34*4) - (0.5*9.81*4²)

y = 470.88[m]

Answer:

hope this helps

Explanation:

solid particles are arranged according to the size of the object

while liquid particles are arranged how they move freely together in the object they are in.

while gas particles are arranged scatterly like they spread.

Answer:Geography, Earth Science, Astronomy

Rotation describes the circular motion of an object around its center.

Explanation:

Answer:

it is Cyclic and predictable

Answer:

Vf₁ = 30 m/s

s₂ = 90 m

car wins the race.

Explanation:

To find the speed of car after 5 s, we use 1st equation of motion:

Vf₁ = Vi₁ + a₁t₁

where,

Vf₁ = Final Speed of Car = ?

Vi₁ = Initial Speed of Car = 0 m/s

a₁ = acceleration of car = 5 m/s²

t₁ = time = 5 s

Therefore,

Vf₁ = 0 m/s + (5 m/s²)(5 s)

Vf₁ = 25 m/s

To find the distance of motorcycle after 6 s, we use 2nd equation of motion:

s₂ = Vi₂ t₂ + (1/2)a₂t₂²

where,

s₂ = distance covered by the motorcycle = ?

Vi₁ = Initial Speed of motorcycle = 0 m/s

a₂ = acceleration of motorcycle = 8 m/s²

t₂ = time = 6 s

Therefore,

s₂ = (0 m/s)(6 s) + (1/2)(5 m/s²)(6 s)²

s₂ = 90 m

We can use 2nd equation of motion to find time taken by each car and motorcycle to reach the finish point:

For Car:

s₁ = Vi₁ t₁ + (1/2)a₁t₁²

where,

s₁ = 200 m - 50 m = 150 m (since, car starts 50 m ahead)

Therefore.

150 m = (0 m/s)(t₁) + (1/2)(5 m/s²)t₁²

t₁ = √60 s²

t₁ = 7.74 s

For Motorcycle:

s₂ = Vi₂ t₂ + (1/2)a₂t₂²

where,

s₁ = 200 m

Therefore.

200 m = (0 m/s)(t₂) + (1/2)(5 m/s²)t₂²

t₂ = √80 s²

t₂ = 8.94 s

Since, the car takes less time to reach finish line.

Therefore, car wins the race.

Answer:

a = 6 [m/s²]

Explanation:

To solve this problem we must use the following equation of kinematics

[tex]v_{f} =v_{o} +a*t[/tex]

where:

Vf = final velocity = 58 [m/s]

Vo = initial velocity = 10 [m/s]

a = acceleration [m/s²]

t = time = 8 [s]

Now replacing:

58 = 10 + (a*8)

58 - 10 = 8*a

48 = 8*a

a = 6 [m/s²]

Answer:

1) 2197.44 J

2) 0 J

3) 2197.44 J = Constant

4) 2197.44 J

5) Approximately 8.86 m/s

Explanation:

The given parameters are;

The mass of the diver, m = 56 kg

The height of the cliff, h = 4.0 m

The speed with which the diver is moving, vₓ = 8.0 m/s

The gravitational potential energy = Mass, m × Height of the cliff, h × Acceleration due to gravity, g

1) Her gravitational potential energy = 56 × 4.0 × 9.81 = 2197.44 J

2) The kinetic energy = 1/2·m·u²

Where;

u = Her initial velocity = 0 when she just leaves the cliff

Therefore;

Her kinetic energy when she just leaves the cliff = 1/2 × 56 × 0² = 0 J

3) The total mechanical energy = Kinetic energy + Potential energy

The total mechanical energy is constant

Her total mechanical energy relative to the water surface when she leaves the cliff = Her gravitational potential energy = 2197.44 J = Constant

4) Her total mechanical energy relative to the water surface just before she enters the water = 2197.44 J

5) The speed with which she enters the water, v, is given from, v² = u² + 2·g·h

Where;

u = The initial velocity at the top of the cliff before she jumps= 0 m/s

∴ v² = 0² + 2 × 9.81 × 4 = 78.48

v = √78.48 ≈ 8.86 m/s

The speed with which she enters the water, v ≈ 8.86 m/s

1) Her gravitational potential energy relative to the water surface when she leaves the cliff is; GPE(leaves cliff) = 2195.2 J

2) Her kinetic energy when she leaves the cliff is; KE = 0J

3) Her total mechanical energy relative to the water surface when she leaves the cliff is; ME_total = 2195.2 J

4) Her total mechanical energy relative to the water surface just before she enters the water is; ME_total = 2195.2 J

5) The speed at which she enters the water is; v = 8.85 m/s

We are given;

Mass of the diver; m = 56 kg

Height of the cliff; h = 4 m

Speed at which she is moving; vₓ = 8 m/s

1) Formula for gravitational potential energy is;

GPE = mgh

where;

m is mass

g is acceleration due to gravity

h is height

Thus;

GPE = 56 × 4 × 9.8

GPE(leaves cliff) = 2195.2 J

2) The formula for kinetic energy when she leaves the cliff is;

KE = ¹/₂mu²

Where;

m is mass

u = initial velocity = 0 m/s

Thus;

KE = ¹/₂ × 56 × 0²

KE(leaves cliff) = 0 J

3) The total mechanical energy relative to the water surface when she leaves the cliffis;

ME_total = GPE(leaves cliff) + KE(leaves cliff)

Thus;

ME_total = 2195.2 + 0

ME_total = 2195.2 J  

4) Her total mechanical energy relative to the water surface just before she enters the water is same as that when she leaves the cliff = 2195.2 J

5) The speed with which she enters the water, v, is gotten from newtons third equation of motion;

v² = u² + 2gh

Thus;

v² = 0² + (2 × 9.8 × 4)

v² = 78.4

v = √78.4

v = 8.85 m/s

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Answer:

2,200

Explanation:

truck=35800,car=33600

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Answer:

same i just love helping people out

god bless you have a blessed day

Explanation:

Answer:

I do too

Explanation:

I like to help others especially if it has been filled in by struggling in a certain concept of a subject.

Do u agree?

Answer:

1. Aristotle

2. Thales of Miletus

Explanation:

1. Answer: The first professor who implemented a scientific approach in order to investigate a problem was Aristotle(384-322 BC)

Aristotle was a student under Plato, he pioneered the scientific method in ancient Greece.

2. Answer: The father of scientific mythology was

Thales of Miletus  (c. 624/623-c. 548/545 BC)

Thales was a Greek mathematician, astronomer, and pre-socratic philosopher, Asia minor.

Answer:

Rt = 908.25 [ohm]

Explanation:

In order to solve this problem, we must remember that the resistors connected in series are added up arithmetically.

In this case, R2 and R3 are in series therefore.

R₂₃ = 200 + 470

R₂₃ = 670 [ohm]

Now this new resistor (R₂₃) is connected in parallel with the resistor R4. therefore we must use the following arithmetic expression, to add resistances in parallel.

[tex]\frac{1}{R_{4-23} }= \frac{1}{R_{4}}+\frac{1}{R_{23} } \\\frac{1}{R_{4-23} }=\frac{1}{1800}+\frac{1}{670} \\R_{4-23}=488.25[ohm][/tex]

In this way R₁, R₅ and R₄₋₂₃ are connected in series.

Rt = R₁ + R₅ + R₄₋₂₃

Rt = 150 + 270 + 488.25

Rt = 908.25 [ohm]

Answer:

Explanation:

Force = mass * acceleration (Newton's second law of motion)

Given

Mass = 32kg

Get the acceleration using the expression d = 1/2at^2

3.5 = 1/2 (a)2.5^2

3.5 = 3.125a

a = 3.5/3.125

a = 1.12m/s^2

Get the net force;

F =  3.2 * 1.12

F = 35.84N

Hence the net force used to accelerate is 35.84N

Answer:

I'm not sure what your asking for

reaction and law

Explanation:

the newton third law of motion

Answer:

I'm pretty sure its radiation and reflection. Some of the Sun's energy gets reflected off into space. (I could be wrong though)

Answer:

radiation and reflection

hope helps you

have a nice day

Explanation:

hope this will surely be helpful to you

Answer:

Explanation:. The vending machine holds 10 bags of pretzels and 8 packs of cheese crackers.  Both the pretzels and crackers dispenseat 2 bags per hour. How many hours will elapse when there is the same number of snacks remaining in the vending machine?  How many of each item will there be at this point?Equation1______________________________Equation2_______________________________Hours: ___________________________ Items: ____________________________What does each equation have n common? ______________________How does this affect the solution?

The target heart rate zone is where you want your exercise heart rate to be. Option B is correct.

The target heart rate zone is a specific range of heart rates that you should aim to achieve during exercise to maximize the benefits of your workout. This range is typically calculated based on a percentage of your maximum heart rate. For example, if your maximum heart rate is 180 beats per minute, your target heart rate zone might be between 120 and 150 beats per minute.

The objective pulse is 50 to 85 percent of your greatest pulse. It is the level at which your heart is pulsating with moderate to extreme focus. To decide your greatest pulse, take 220 and deduct your age. Supporting an exercise at this speed works on cardiorespiratory perseverance.

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