Is any force exerted on the vertical sides of the loop that you used in the experiment and how does it affect the apparent mass?

a. There is a force exerted on each side of the loop that are equal in magnitude and opposite in direction.
b. No force is exerted on the sides of the loop.The lengths of the sides of the loop are much smaller than the bottom section and therefore the force on the sides are negligible.
c. The force on the sides are not in the same direction in which the mass is measured and therefore it does not affect the results.
d. The force exerted on the sides of the loop cancel each other out.

This question is incomplete, the complete question is;

A flat coil is in a uniform magnetic field. What angle between the magnetic field and the plane of the coil produces the maximum flux, and what angle produces 90% of the maximum flux

A) max: 0°

90% of max: 90°

B) max: 90°

90% of max: 45°

C) max: 90°

90% of max: 64°

D) max: 0°

90% of max: 26°

E) max: 90°

90% of max: 80°

Answer:

Option C) max: 90°

90% of max: 64°  is the correct Answer

Explanation:

from the Suppose Area is A.

then flux at angular position O is

Ф = BAsin∅

⇒ Ф = βmaxSin∅

flux will be max when sin∅ = 1

therefore sin∅ = 1

∅ = sin⁻¹ 1

∅ = 90°

Now at 90% of max flux

Ф ⇒ 0.9βmax = βmax sin∅

0.9 = sin∅

∅ = sin⁻¹ (0.9)

∅ = 64.15° ≈ 64°

Therefore Max ∅ = 90°

90% flux = 64°

Option C) max: 90°

90% of max: 64°  is the correct Answer

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

Answer: True

Explanation: When light is reflected off lets say a mirror it is bent and changes direction to bounce off of another wall or object. For example if you take a flash light and shine it into a mirror the light reflects into a different direction your welcome

When light is reflected, the incident rays return back in straight direction and never bends and make the material gleaming. Hence, the statement is false.

Reflection is the phenomenon that, when waves incident on a material it returns back in straight direction. Both sound wave and light wave can be reflected. But, lightwaves are only reflected from transparent materials.

Reflection of light ray  make the material surface gleaming like in a mirror. Similarly reflection of sound waves produces echos. The phenomenon in which the light wave bends in its direction when moving from one medium to the other is called refraction.

The measure of bending of light in a medium is called the refractive index of that medium. Hence, bending of light is not reflection and it is termed as refraction. Thus, the statement is false.

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Answer:

yes

Explanation:

the brain perform two separate tasks at the same time

Answer:

Please mark as Brainliest!!

Explanation:

Strain energy is defined as the energy stored in a body due to deformation. The strain energy per unit volume is known as strain energy density and the area under the stress-strain curve towards the point of deformation. When the applied force is released, the whole system returns to its original shape.

Answer:

Watt

Explanation:

Power is defined as the work done by an object per unit time. Its mathematical form is given by :

[tex]P=\dfrac{W}{t}[/tex]

The SI unit of work done is Joules (J) and that of time is seconds (s).

J/s is equal to watts. Watt is the metric unit of power.

Answer:

d

Explanation:

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.

Answer:

NEWTON

Explanation:

ISSAC NEWTON

HOPE IT HELP U

Answer:

hans bethe

Explanation:

Hans Bethe

In 1939, in a paper entitled "Energy Production in Stars", Hans Bethe analyzed the different possibilities for reactions by which hydrogen is fused into helium. He defined two processes that he believed to be the sources of energy in stars.

i looked it up and thats what it said

Answer:

I like to memorize excerpt from articles to solve and answer questions like these. I hope this can help, it's from study.com: "The relationship between voltage, current, and resistance is described by Ohm's law. This equation, i = v/r, tells us that the current, i, flowing through a circuit is directly proportional to the voltage, v, and inversely proportional to the resistance, r."

Answer:

Kinetic energy is converted to internal energy

Explanation:

We are required to explain what happens to this decreased kinetic energy in this question.

Here is what a diffuser does, this device increases the pressure of a fluid and in so doing it slows it down. The lost kinetic energy would then be converted to internal energy. This is the correct answer to the question.

Thank you

Answer:

can I see like the actual activity on a computer or book?

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:

Leave onions in cold water for about 15 minutes! Takes out the chemical reaction in the onion's defense system.

Explanation:

This is what people NEED to know for cooking... Lol :)

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:

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:

Equilibrium is when the rate of the forward reaction equals the rate of the reverse reaction. All reactant and product concentrations are constant at equilibrium.

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: 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:

False

Explanation:

Answer:

False

Explanation:

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

Answer:

Explanation:

Coordinate system is one that describe the location of an object in a given plane. It implies the use of axes (coordinates) and points.

Given that the man in the question walks 400 m due north of east. The cardinal points can be used in this case, with the north and east cardinals as the required axis.

scale = [tex]\frac{length on drawing}{original length}[/tex]

         = [tex]\frac{10}{400}[/tex]

         = [tex]\frac{1}{40}[/tex]

scale = 1:40

This is a reduced scale which implies that 1 cm on the drawing is equal to 40 m on the original length.

The man's direction is [tex]45^{o}[/tex] north of east.

The graphical drawing of the vector is herewith attached to this answer.

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:

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:

0.438kg/ms-¹

Explanation:

Momentum, denoted by p, can be calculated by using the formula;

p = mv

Where;

m = mass (kg)

v = velocity (m/s)

Momentum (p) of bird = 0.216 kg × 5.87 m/s = 1.268kg/ms-¹

Momentum (p) of crawling baby = 7.29 kg kg × 0.234 m/s = 1.706kg/ms-¹

Having calculated the momentum of the bird to be 1.268kg/ms-¹, and the momentum of the baby to be 1.706kg/ms-¹, the difference in momentum between the flying bird and the crawling baby is:

{1.706kg/ms-¹ - 1.268kg/ms-¹} = 0.438kg/ms-¹

Answer:

b and c

Explanation: