Describe which relationships in Ohm’s law are DIRECT and which are INVERSE. Use examples to support your answer. You can use calculations, drawings, or graphs to make your point more clear

Answer:

The volume would become 2.34 L if the temperature dropped to 11.0 °C

Explanation:

From the question, A balloon was filled to a volume of 2.50 L when the temperature was 30.0 °C.

To determine what the volume will become if the temperature dropped to 11.0 °C, we will use one of the Gas laws that relates Volume and Temperature. The Gas law that relates Volume and Temperature is the Charlse' law which states that, "the Volume of a fixed mass of gas is directly proportional to its Temperature ( in Kelvin) provided that the Pressure remains constant"

That is

V ∝ T ( at constant pressure)

Where V is the Volume

and T is the Temperature in Kelvin

Then, we can write that

V = kT

Where k is the constant of proportionality

Then,

[tex]\frac{V}{T} = k[/tex]

Hence, we can write that

[tex]\frac{V_{1} }{T_{1} } = \frac{V_{2} }{T_{2} } = \frac{V_{3} }{T_{3} } ...[/tex]

∴ [tex]\frac{V_{1} }{T_{1} } = \frac{V_{2} }{T_{2} }[/tex]

Where [tex]{V_{1} }[/tex] is the initial volume

[tex]{T_{1} }[/tex] is the initial temperature

[tex]{V_{2} }[/tex] is the final volume

and [tex]{T_{2} }[/tex] is the final temperature

From the question,

[tex]{V_{1} }[/tex] = 2.50 L

[tex]{T_{1} }[/tex] = 30.0 °C = 30.0 + 273.15 K = 303.15 K

[tex]{V_{2} }[/tex] = ??

[tex]{T_{2} }[/tex] = 11.0 °C = 11.0 + 273.15 K = 284.15 K

Putting the values into the equation, we get

[tex]\frac{2.50}{303.15} = \frac{V_{2} }{284.15}[/tex]

∴ [tex]V_{2} = \frac{2.50 \times 284.15}{303.15}[/tex]

[tex]V_{2} = 2.34 L[/tex]

Hence, the volume would become 2.34 L if the temperature dropped to 11.0 °C.

The final volume of the balloon is equal to 0.92 Liter.

Given the following data:

To determine the final volume of the balloon, we would apply Charles's law:

Mathematically, Charles law is given by the formula;

[tex]\frac{V}{T} =k\\\\\frac{V_1}{T_1} = \frac{V_2}{T_2}[/tex]

Where;

Substituting the given parameters into the formula, we have;

[tex]\frac{2.5}{30} = \frac{V_2}{11} \\\\2.5 \times 11 = 30V_2\\\\27.5 = 30V_2\\\\V_2 = \frac{27.5}{30}[/tex]

Final volume = 0.92 Liter.

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Answer:

I'm going to guess 15 N

Explanation:

Answer:

big bang theory

Explanation:

Edwin Hubble is credited for the initial development of the Big Bang theory, an idea which helps to explain the formation of the universe over 15 billion years ago.

Answer:

I believe none

Explanation:

Answer:

261.84 N.

Explanation:

From the question given above, the following data were obtained:

Energy (E) = work (W) = 741 J

Distance (d) = 2.83 m

Force (F) =?

Work is simply defined as the product of force and the distance moved in the direction of the the force. Mathematically, it is expressed as:

Work (W) = Force (F) × distance (d)

W = F × d

Thus, we obtained the weight of the object using the above formula.

Work (W) = 741 J

Distance (d) = 2.83 m

Force (F) =?

W = F × d

741 = F × 2.83

Divide both side by 2.83

F = 741 / 2.83

F = 261.84 N

Since force and weight has the same unit of measurement i.e Newton (N), the weight of the object is 261.84 N

Answer:

False

Explanation:

Answer:

False

Explanation:

Hope this helped, Have a Wonderful Day/Night!!

Answer:

Explanation:

The total work done by the wave is expressed as;

Workdone = Potential energy + Kinetic energy

Workdone = mgh + 1/2mv²

m is the mass = 77kg

g is the acceleration due to gravity = 9.8m/s²

v is the velocity = 8.2m/s

h is the height = 1.65m

Substitute into the formula;

Workdone = 77(9.8)(1.65) + 1/2(77)8.2²

Workdone = 1245.09 + 2588.74

Workdone = 3833.83Joules

Hence the amount of non conservative work done on the sofa is 3833.83Joules

Given:

We know,

→ [tex]Work \ done = Potential \ energy +Kinetic \ energy[/tex]

or,

                     [tex]= mgh +\frac{1}{2} mv^2[/tex]

By putting the values,

                     [tex]= 77\times 9.8\times 1.65+\frac{1}{2}\times 77\times (8.2)^2[/tex]

                     [tex]= 1245.09+2588.74[/tex]

                     [tex]= 3833.83 \ Joules[/tex]

Thus the above approach is right.

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Answer:

ok

Explanation:

ok ok I will thanks for free

Explanation:

1) the diagonal force times the sine of the angle it makes will give you the vertical component or y component . use this to get the vertical components of all the diagonal forces . add all the vertical components as well as the vertical forces together

2) the diagonal force times the cosine of the angle it makes gives the horizontal component or x component. do this to get the x component of all the diagonal forces.

add all the x components as well as the horizontal forces together to get the final x component

3)using the triangle of vectors, the resultant force is calculated

sum of y component = 3.2N

sum of x component= 5.7N

resultant force = 6.5N

At angular speed of 3 rev/s, the object moves a distance equal to 3 times the circumference of the circle each second, or a distance of 3 • 2π (1.20 cm) ≈ 22.6 cm.

So, with a linear speed of 22.6 cm/s = 0.226 m/s, the object has a centripetal acceleration a of

a = (0.226 m/s)² / (0.012 m) ≈ 18.8 m/s²

directed toward the center of the circle.

Answer:

[tex]\mid \vec C\mid=31.9[/tex]

Explanation:

Consider the vectors:

[tex]\vec A=9.4\mathbf{\hat{i}}-3.6\mathbf{\hat{j}}[/tex]

[tex]\vec B=-9.5\mathbf{\hat{i}}-13.4\mathbf{\hat{j}}[/tex]

Calculate the magnitude of

[tex]\vec C=-2\vec B-\vec A[/tex]

Substitute the values of the vectors:

[tex]\vec C=-2(-9.5\mathbf{\hat{i}}-13.4\mathbf{\hat{j}})-(9.4\mathbf{\hat{i}}-3.6\mathbf{\hat{j}})[/tex]

Operate and remove parentheses:

[tex]\vec C=19\mathbf{\hat{i}}+26.8\mathbf{\hat{j}}-9.4\mathbf{\hat{i}}+3.6\mathbf {\hat{j}}[/tex]

Operating both components separately:

[tex]\vec C=9.6\mathbf{\hat{i}}+30.4\mathbf{\hat{j}}[/tex]

Now find the magnitude of C:

[tex]\mid \vec C\mid=\sqrt{9.6^2+30.4^2}[/tex]

[tex]\mid \vec C\mid=\sqrt{1016.32}[/tex]

[tex]\mathbf{\mid \vec C\mid=31.9}[/tex]

Answer:

9,800 watts

Explanation:

The first step is to calculate the force

F= mg

= 500 × 9.8

= 4,900 N

The next step is to calculate the work done

= 4,900 × 100

= 490,000 joules

Therefore the power can be calculated as follows

Power= work done /time

= 490,000/50

= 9,800 watts

Answer:

[tex]\mathbf{F_{thrust} \simeq 2.3 \times 10^4 \ N}[/tex]

Explanation:

For a lunar lander acceleration in an upward direction, the net force that acts on it can be expressed as:

[tex]F_{net} = F_{thrust} - F_{g(moon)}[/tex]

The equation for the net force that acts on the lunar lander is:

[tex]F_{net} = m_La[/tex]

For the lander touch down to be zero acceleration, it implies that the acceleration of the moving rocket is zero(a free body fall)

i.e. a = 0

We can now regard the Apollo 12 lunar as a freely falling body

However; the force of gravity as a result of the moon acting on the lunar rocket is:

[tex]F_{g(moon)} = m_Lg_m[/tex]

Then; the equation for the thrust force of the lunar rocket is:

[tex]F_{net} = F_{thrust} - F_{g(moon)}[/tex]

[tex]m_La = F_{thrust}-m_Lg_m[/tex]

[tex]m_L(0)= F_{thrust}-m_Lg_m[/tex]

[tex]0= F_{thrust}-m_Lg_m[/tex]

[tex]-F_{thrust}= -m_Lg_m[/tex]

[tex]F_{thrust}= m_Lg_m[/tex]

Finally; the thrust force of the lunar rocket can be determined as:

[tex]F_{thrust}= m_Lg_m[/tex]

Acceleration due to gravity ot the surface of the moon = 1.625 m/s²

[tex]F_{thrust}=(1.4 \times 10^{4} \ kg) (1.625 \ m/s^2)[/tex]

[tex]F_{thrust}=2.275 \times 10^4 \ N[/tex]

[tex]\mathbf{F_{thrust} \simeq 2.3 \times 10^4 \ N}[/tex]

A thrust force of 22680N will be necessary to  to have the lander touch down with zero acceleration.

Given the data in the question;

Mass of Apollo 12 lunar lander; [tex]m_L = 1.4 * 10^4 kg[/tex]

Using the expression for the net-force that acts on the lander when it is accelerated in the upward direction:

[tex]F_{net} = F_{thrust} - F_{g(moon)}[/tex] ------- Let this be equation 1

Also, the net-force that acts on the lunar lander can be expressed as:

[tex]F_{net} = m_L*a[/tex] ------- Let this be equation 2

Where [tex]m_L[/tex] is the mass of the lunar lander and a is the acceleration ( 0 ),

( the acceleration of the moving rocket is zero and its freely falling )

Also, Force of gravity due to moon that acts on the lunar lander is expressed as:

[tex]F_{g(moon)} = m_L * g_m[/tex]  ------- Let this be equation 3

Where [tex]m_L[/tex] is the mass of the lunar lander and [tex]g_m[/tex] is the moon's gravity ( [tex]1.62m/s^2[/tex] )

Lets substitute equation 2 and 3 into 1

[tex]m_L * a = F_{thrust} - (m_L * g_m)\\\\m_L * 0= F_{thrust} - (m_L * g_m)\\\\0=F_{thrust} - (m_L * g_m)\\\\ F_{thrust} = m_L * g_m[/tex]

We substitute our values into the equation

[tex]F_{thrust} = (1.4*10^4kg) * 1.62m/s^2\\\\F_{thrust} = 22680 kg.m/s^2\\\\F_{thrust} = 22680N[/tex]

Therefore, a thrust force of 22680N will be necessary to  to have the lander touch down with zero acceleration.

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Answer:

d

Explanation:

Answer:

D

Explanation:

Answer:

[tex]3.24\times 10^{-7}\ \text{m}[/tex]

Explanation:

m = Mass of bullet = 4.8 g

v = Velocity of bullet = 170 m/s

B = Magnetic field of Earth = [tex]5\times 10^{-5}\ \text{T}[/tex]

q = Charge of bullet = [tex]1.06\times 10^{-8}\ \text{C}[/tex]

a = Acceleration

Time the bullet will be in the air for is [tex]t=\dfrac{1000}{170}=5.88\ \text{s}[/tex]

Force is given by

[tex]F=ma[/tex]

Magnetic force is given by

[tex]F=qvB[/tex]

So

[tex]ma=qvB\\\Rightarrow a=\dfrac{qvB}{m}\\\Rightarrow a=\dfrac{1.06\times 10^{-8}\times 170\times 5\times 10^{-5}}{4.8\times 10^{-3}}\ \text{m/s}^2[/tex]

From the linear equations of motion we have

[tex]s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0+\dfrac{1}{2}\times \dfrac{1.06\times 10^{-8}\times 170\times 5\times 10^{-5}}{4.8\times 10^{-3}}\times 5.88^2\\\Rightarrow s=3.24\times 10^{-7}\ \text{m}[/tex]

The defelection of the bullet is [tex]3.24\times 10^{-7}\ \text{m}[/tex]

Answer: 30 N (in the direction of the 45 N force)

Explanation:

You have to choose a positive and negative direction since the question says the forces are in opposite directions.

Lets assume:

- <----------> +

<------ 15 N

45 ------->

45-15=30

Answer:

d = 7.5 [cm]

Explanation:

To be able to solve this problem we must understand well the concept of a simple lever of First Class, which consists of a support point (fulcrum) and a beam where a force is applied at one end, while at the other end the force exerted is multiplied.

In the attached image we can see an image that will help us better understand the concept of this first-class lever.

The moment created by the effort and the distance is equal to:

M = 100*30 = 3000 [N*cm]

Now the balance is created because we have the same moment exerted by the effort and the one created by the load.

3000 = 400*d

d = 7.5 [cm]

Given :

A weight lifter lifts a 280-N set of weights from ground level to a position over his head, a vertical distance of 1.60 m.

To Find :

How much work does the weight lifter do, assuming he moves the weights at constant speed.

Solution :

We know, work done is given by :

[tex]W= Fdcos\ \theta[/tex]

Here, F = 280 N and d = 1.60 m

Since, weight lifer is applying and upward force and displacement is also in upward direction.

[tex]\theta = 0^{\circ}\\cos\ 0^{\circ} = 1[/tex]

Putting all these in above equation, we get :

[tex]W = 280\times 1.6\times 1\\\\W = 448\ J[/tex]

Therefore, work done by weight lifter 448 J.

Hence, this is the required solution.

We are given:

The tuning fork vibrates at 15660 oscillations per minute

Period of one back-and forth movement:

the given data can be rewritten as:

1 minute / 15660 oscillations

60 seconds / 15660 oscillations          (1 minute  = 60 seconds)

dividing the values

0.0038 seconds / Oscillation

Therefore, one back and forth vibration takes 0.0038 seconds

Answer:

J

Explanation:

The daughter moves with greater acceleration backwards because of her weight.

Both mother and daughter move backward, but the daughter moves with greater acceleration. Therefore, option (J) is correct.

This law states that the acceleration of a body depends on two factors. The first one is the mass of the body and the net force acting on the body. The acceleration is inversely proportional to the mass and is directly proportional to the net force acting on the body and

Newton’s second law can be written in an expression as follows:

F = ma

or,   a = F/m

According to Newton's third law of motion, there is an equal and opposite reaction for every action. So when the two skaters push each other. They both will move in a backward direction.

But the mass of the daughter is 50 Kg while the mass of the mother is 100 Kg. As the mass of the daughter is less. Therefore daughter moves in a backward direction with greater acceleration.

Therefore, both move backward but the daughter moves with greater acceleration in comparison to the mother.

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#SPJ2

Answer:

1 hour

Explanation:

t = 6/6

6/6 = 1

You just said that his speed enables him to cover 6 miles in each hour. From this information, I calculated that it will take him one hour to travel 6 miles.

Answer:

The minimum downward force on the box that will keep it from slipping is 63.14 N

Explanation:

Given;

mass of the object, m = 30 kg

applied force, f = 125 N

coefficient of static friction, μ = 0.35

Normal reaction (R) is acting upwards, weight of the box (mg) is acting downwards and the minimum downward force (F) on the box that will keep it from slipping is also acting downwards.

The net vertical forces on the box is given by;

R - mg - F = 0

F = R - mg

Now, determine normal reaction, R

f = μR

R = f / μ

R = 125 / 0.35

R = 357.14 N

Finally, determine the minimum downward force on the box that will keep it from slipping;

F = R - mg

F = 357.14 - (30 x 9.8)

F = 357.14 - 294

F = 63.14 N

Therefore, the minimum downward force on the box that will keep it from slipping is 63.14 N

We are given:

Mass of the rocket = 10 kg

Weight of the Rocket = 100 N

Upward thrust applied by the rocket = 400 N

Net upward force on the rocket:

We are given that gravity pulls the rocket with a force of 100 N

Also, the rocket applied a force of 400N against gravity

Net upward force = Upward thrust - Force applied by gravity

Net upward force = 400 - 100

Net upward force = 300 N

Upward Acceleration of the Rocket:

From newton's second law:

F = ma

replacing the variables

300 = 10 * a

a = 30 m/s²

Answer:

real world examples

Explanation:

concave wave rock- The wave rock is formed by weathering of the surrounding area, which helps in proving the deposition part, as the wave rock was below the ground, occurred due to deposition of rock years over years. It is made from very tough material from its surrounding. The weathering reduced the surrounding terrain, while the bedrock remained to witness for the history.  --Also, volcanic eruptions have changed earth greatly. EX- Ash and sulfur went into Earth's atmosphere because of the Tambora eruption which dimmed incoming sunlight. It lowered global temperatures by about 3°F. The Mount Pinatubo of 1991 that erupted in the Philippines cooled the planet by about 1°F.

Answer:

The ball will hit the ground 108.25 m down the field

Explanation:

To determine how far down the field the ball will hit the ground, that is the Range of the ball. From formula to calculate Range in projectile motion,

[tex]R = \frac{u^{2} sin2\theta }{g}[/tex]

Where R is the Range

u is the initial velocity

θ is the angle of projection

and g is acceleration due to gravity (Take g = 9.8 m/s²)

From the question,

Initial velocity, u = 35 m/s

Angle, θ = 60°

Putting these values into the equation, we get

[tex]R = \frac{35^{2} ( sin2(60)) }{9.8}[/tex]

[tex]R = \frac{1225 \times sin(120)}{9.8}[/tex]

[tex]R = \frac{1225 \times 0.8660}{9.8}[/tex]

[tex]R = \frac{1060.85}{9.8}[/tex]

[tex]R = 108.25 m[/tex]

Hence, the ball will hit the ground 108.25 m down the field.