a stage light uses a white lamp wgat color of light will be seen when
a yellow filter is used
an orange filter is used
​

The magnetic field strength of the solenoid of radius 2 m is 2.0096 T.

Magnetic field is the region or space around which magnetic force is felt or experienced.

To calculate the magnetic field strength inside a solenoid, we use the formula below.

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

Hence, the magnetic field strength is 2.0096 T.

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Lenses, which are pieces of curved, clear glass, were employed in early telescopes to focus light.

Curved mirrors are used by the majority of telescopes nowadays to collect light from the night sky. Light is focused by a telescope's mirror or lens' shape.

Astronomers use a telescope to observe distant things. Curved mirrors are used by the majority of telescopes, including all large telescopes, to collect and concentrate light from the night sky.

The original telescopes employed lenses, which are simply curved pieces of clear glass, to focus light. The "optics" of a telescope are the mirrors or lenses. Strong telescopes may view objects that are extremely faint and incredibly far away.

Therefore, Lenses, which are pieces of curved, clear glass, were employed in early telescopes to focus light.

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I think the answer is b but im not really sure

The distance traveled by the horse pulling the wagon with a force of 1450N is 115.57 metres.

Power is defined as the amount of energy used or transferred in a certain amount of time. This means that power can be calculated by dividing the work done by the time taken.

P = W/t

When a force is applied to move an object, work is done on the object. Work done = force × distance

Power = (Force × distance)/time

According to this question, a horse pulls a wagon with a force of 1450 Newtons.

798 = 1450 × d/210

167,580 = 1450d

Distance = 167,580 ÷ 1450

Distance = 115.57 metres

Therefore, 115.57 metres is the distance traveled by the wagon.

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Answer:

b> infinity

Explanation:

i think b is the correct one

Answer:

(B) Infinity

Explanation:

Gravity can basically never become zero except hypothetically at infinity.

Answer:

Explanation:

b)  Fg = W = mg = (104 kg)(9.8 m/s²) = 1019.2 N  (weight of the box)

Ff = (coeff. friction)(N)

N = normal force = Fg = 1019.2 N

c)  coeff. friction = Ff/N = 45 N/1019.2 N = 0.044

d)  Fnet = 245 N - 45 N = 200 N

It's hard to draw a free body diagram with the available symbols, but here's a try at it.  Just add a box between the arrows!

                            W

                            ↓

     Ff  →                                 ←  Fapplied

                             ↑

                             N

The mass of ball B and the final velocity of ball A can complement the conservation of linear momentum. The answer is yes.

Momentum can simply be defined as the product of mass and velocity. It is a vector quantity.

Given that ball A with a mass of 0.280kg makes an elastic head-on collision with ball B initially at rest. After collision, ball B moves off with half the original speed of ball A.

In an elastic head-on collision, momentum is mostly always conserved. That is, the sum of the momentum before collision will be equal to the sum of the momentum after collision.

Mathematically, MaUa = MaVa + MbVb

Is the momentum conserved in the collision?

The answer is yes!

Why?

Because we need to consider the mass of the ball B and the final velocity of the ball A.

Therefore, In consideration of the mass of the ball B and the final velocity of the ball A, we can say that the momentum is conserved.

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Since the efficiency of the pulley is 75% and the single movable pulley is used to lift 4000N load to the height of 50 cm, the effort distance, effort applied, output work and mechanical advantage are as follows:

Effort Distance = 50 cm

Effort Applied = 5333.33 N

Output work = 2000 J

Mechanical Advantage = 0.75.

When a single movable pulley is used to lift a load, the effort applied is equal to the weight of the load being lifted, and the effort distance is the distance the effort is applied.

Load (output force) = 4000 N

Height lifted = 50 cm = 0.5 m

Efficiency of the pulley = 75%

Given the above information, we can calculate the effort applied, output work, and mechanical advantage as follows:

Effort applied (input force) = Load ÷ Efficiency

= 4000 N ÷ 0.75

= 5333.33 N

Output work = Load x Height lifted

= 4000 N x 0.5 m

= 2000 J

Effort distance = Height lifted

= 50 cm

Mechanical advantage = Load ÷ Effort applied

= 4000 N ÷ 5333.33 N

= 0.75

It is worth noting that pulley efficiency is the ratio of output work to the input work. And since the efficiency of the pulley is 75% so the input work is greater than the output work, which means some of the energy is wasted in overcoming friction and other losses.

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Since Taylor has set this fitness goal: I will go to the YMCA and lift weights for 30 minutes every other day for the entire month of January. The option that MOST likely TRUE about this goal is option A:It is a short-term goal.

A short-term goal is a goal that can be achieved within a relatively short period of time, usually less than a year. In this case, the goal is to go to the YMCA and lift weights for 30 minutes every other day for the entire month of January.

In this case, Taylor's goal is to go to the YMCA and lift weights for 30 minutes every other day for the entire month of January. This goal is specific, as it clearly states what Taylor wants to accomplish, and it is measurable.

This goal can be achieved within the month of January and it's a specific, measurable and time-bound goal, which are characteristics of short-term goals.

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See full question below

Taylor has set this fitness goal: I will go to the YMCA and lift weights for 30 minutes every other day for the entire month of January. What is MOST likely TRUE about this goal?

It is a short-term goal.

It is a long-term goal.

It is an unrealistic goal.

It is a mental health goal.

Answer:it’s a short-term goal

Explanation:

because it’s only for the month of January she will be doing this goal.

Answer:ssssw

Explanation:

(True) Synthetic pesticides and fertilizers were used extensively during the Green Revolution, but the negative health and environmental impacts were not well-known since the effects added up over time.

Chemicals created by humans, synthetic pesticides are meant to kill or deter pests. They are mostly utilized in agriculture, although they are also employed in other fields of endeavor and in household settings. In the 1930s, synthetic pesticides were first used in the United States of America (USA).

Chemicals known as synthetic pesticides are employed to manage and control plant pests like weeds, insects, and/or fungal infections.

Human health problems associated with exposure to synthetic pesticides include

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When two objects collide in two dimensions (for example, x and y), the momentum will be conserved separately in each direction (as long as there isn't an external impulse in that direction).

The momentum before and after a direct impact between two pieces of clay is maintained. The amount of initial kinetic energy lost during the impact as a fraction is 0.961.

The total momentum of the two objects prior to the collision equals the total momentum of the two following the collision when object 1 and object 2 collide in an isolated system. The momentum obtained by object 2 equals the momentum lost by object 1, and vice versa.

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The magnitude of the minimum force needed to move the wagons from rest is 474320 N

We have come to know that the force and coefficient of friction have a simple relationship as shown by the equation below:

Frictional force (N) = coefficient of friction (μ) × normal reaction (N)

F = μN

Applying the above formula, we can determine the force needed to move the wagons from rest. Details below:

F = μN

F = 0.220 × 2156000

F = 474320 N

Thus, the force needed is 474320 N

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A bar magnet's magnetic field lines flow from its north pole toward its south pole, just like they do in all magnets. Field lines that begin close to a pole's edges stay nearby the bar magnet longer than those that begin closer to the pole's centre.

Because the molecules in magnets are arranged so that their electrons spin in the same direction, magnets are unique. This configuration and motion produces a magnetic force that emanates from a north- and south-seeking pole, respectively.

At the poles, it is strongest.

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The two charges, each of magnitude 6.0 μC, exert a force of 0.70 N at separation  of 1.47 meters.

Electric force is the attracting or repulsive interaction between any two charged things. Similar to any force, Newton's laws of motion describe how it affects the target body and how it does so. One of the many forces that affect objects is the electric force.

We know that electric force can be defined as:

Force: F = kQq/r²

0.70 = 9.0 × 10⁹ × (6.0×10⁻⁶)²/r²

r = 1.47 meter.

Hence, the separation between them is 1.47 meters.

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The nodal voltage is V and the currents are as defined in the diagram.

At the upper node, Kirchhoff's current law equation is  I1+I5=I2+I3

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To find the normal force, we can use the equation: normal force = weight + friction force * cos(incline angle).

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Answer:

2 nitrogen and 1 oxygen,

Explanation:

N2=2 nitrogen

O= a single element

Answer:

800J

Explanation:

W = Fs, Work equals force times displacement

in this case, the force is 400N and the displacement is 2 meters.

The regular SI unit for work is joules

The evaluation of the capacitor (in series) network is as follows;

Part A

Q = 3.2 μC

Part B

Q₁ = 3.2 μC

Part C

Q₂ = 3.2 μC

Part D

U = 76.8 μJ

Part E

U₁ = 34 2/15 μJ

Part F

U₂ = 53 1/3 μJ

Part G

V₁ = 21 1/3 V

Part H

V₂ = 26 2/3 V

A capacitor consists of pairs of conductors separated by insulators. Capacitors are used to store electric charge.

The specified parameters are;

The voltage across ab = 48 V

The capacitance of the first capacitor, C₁ = 150 nF

Capacitance of the second capacitor, C₂ = 120 nF

Part A

The total charge in a capacitor network can be found as follows;

[tex]C_{eq} = \left(\dfrac{1}{150} + \dfrac{1}{120} \right)^{-1} nF = \left(\dfrac{3}{200} \right)^{-1}nF[/tex]

[tex]C_{eq} =\left(\dfrac{3}{200} \right)^{-1}nF=66\frac{2}{3} \, nF[/tex]

[tex]Q_{eq} = C_{eq}\times V_{ab}[/tex]

Therefore;

[tex]Q_{eq}[/tex] = 66 2/3 nF × 48 V = 3,200 × 10⁻⁹ C = 3.2 μC

Part B

The charge in the 150 nF capacitor is obtained from the formula for the charge in a capacitor; Q = C × V as follows;

Q = C₁V₁ = C₂V₂

The charge in the capacitors, C₁ and C₂ are the same as the total charge of 3.2 μC

The charge, Q₁ on the 150 nF capacitor, C₁ is therefore, 3.2 nC

Part C

The capacitors, C₁ and C₂ are in series, therefore, the charge in each capacitor is equivalent to the charge in the circuit, which is 3.2 μC.

Therefore, the charge, Q₂, in the 120 nF capacitor, C₂ is 3.2 μC

Q₂ = 3.2 μF

Part D

The total energy stored in the network can be obtained using the formula;

U = (1/2)·C·V²

Where;

U = The energy in the capacitor

C = The equivalent capacitance of the network = 66 2/3 nF

V = The voltage

Therefore;

[tex]U = \dfrac{1}{2} \times C_{eq}\times V^2[/tex]

[tex]U = \dfrac{1}{2} \times 66\frac{2}{3} \times 10^{-9}\times 48^2 = 76.8[/tex]

Part E

The energy stored in the 150 nF capacitor is found as follows;

[tex]Q_{eq}[/tex] = Q₁ = C₁ × V₁

V₁ = [tex]Q_{eq}[/tex] ÷ C₁

Therefore;

V₁ = 3.2 μC ÷ 150 nF = [tex]21\frac{1}{3}[/tex] V

U₁ = 0.5×C₁×V₁²

Part F

The energy stored in the 120 nF capacitor, U₂, can be found as follows;

V₂ = 3.2 μC ÷ 120 nF = [tex]26\frac{2}{3}[/tex] V

U₂ = 0.5 × 150 nF × [tex]\left(26\frac{2}{3} \, V\right)^2[/tex] = [tex]53\frac{1}{3}\, \mathrm{ \mu J}[/tex]

Part G;

The potential difference across the 150 nF, obtained in Part E, is 21 1/3 V

Part H

The potential difference across the 120 nF, obtained in part F, is 26 2/3 V

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From conservation of linear momentum, the final velocity is 15.7 m/s and the time taken is 1.6 s

Velocity can be defined as a distance travel in a specific direction per time taken. It is a vector quantity.

Given that 30kg metal ball is dropped from a height of 12.5 m.  

a.  Find the final velocity when the ball  hits the ground.

The maximum K.E of the ball at it final velocity will be equal to its maximum P.E at height 12.5 m. That is,

K.E = P.E

1/2mv² = mgh

Where

Substitute all the parameters

1/2 × 30 × v² = 30 × 9.8 × 12.5

v² = 245

v = √245

v = 15.65 m/s

b. The time it takes for the ball to hit the ground can be found through

h = ut + 1/2gt²

but u = 0

h = 1/2gt²

Substitute all the necessary parameters

12.5 = 1/2 × 9.8 × t²

12.5 = 4.9t²

t² = 12.5/4.9

t² = 2.55

t = √2.55

t = 1.6 s

Therefore, the final velocity when the ball hits the ground is 15.65 m/s and the time it takes for the ball to hit the ground is 1.6 s

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1. The rank of  the vector combinations on the basis of their magnitude

a+b = a +d > a+c > f +c = d > a + e.

2. The  rank of  the vector combinations on the basis of their magnitude

a+d  = a +b > a+c > f +c =d > a+e.

Any physical quantity having both magnitude and direction is called vector quantity.

According to figure:

a = 2 unit b = 2 unit c = 1 unit d = 2 unit e = 1 unit and f = √5 unit.

a + c = 2+1 unit = 3 unit  

a + b = 2√2 unit

a + e = 2 -1 = 1 unit

a +d = 2√2 unit

f+ c = 2 unit

a+d = 2√2 unit

f+c = 2 unit

a+b = 2√2 unit

a+e = 1 unit

a+c = 3 unit

Hence, rank of  the vector combinations on the basis of their magnitude

a+b = a +d > a+c > f +c = d > a + e.

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a) The average velocity at 4s is 12 m/s

b) The instantaneous velocity at t= 6s is 12 m/s

c) The average acceleration is; 12  - 0/ 4 - 0 n= 3 m/s^2

d) The time for the race is 10 s

We know that the velocity time graph is the kind of set up that has two axis and we can be able to use it to be read off the velocity of the acceleration  of the object. I would want us to recall  that the acceleration is the rate of change of the velocity of the object with time.

Now we have that;

The graph for the velocity of the object can be obtained by looking at the image that has been shown and then we read off the points that are on the graph. The  vertical axis shows the velocity of the graph while the horizontal axis shows the time that is covered in the graph.

The acceleration would be the ratio of the change in the velocity to the change in the time of the object.

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The initial horizontal velocity of the football if a player kicks a football from ground level with a velocity of 27/ms at an angle of 30 degrees above the horizontal is 23.662 m/s.

To find the initial horizontal velocity (Vx) of the football, you can use the horizontal component of the velocity vector. The horizontal component of velocity is found by multiplying the velocity by the cosine of the angle of launch.

[tex]V_{x}[/tex] = V × cos(angle)

where V is the initial velocity, the angle is the angle of launch (in this case, 30 degrees), and cos is the cosine function.

So, in this case, the initial horizontal velocity of the football would be:

[tex]V_{x}[/tex] = 27 m/s × cos(30)

[tex]V_{x}[/tex] = 27 m/s × 0.866

[tex]V_{x}[/tex] = 23.662 m/s

So the initial horizontal velocity of the football, which moves in the x-direction, is approximately 23.662 m/s.

This means the football will move 23.662 m/s in x direction initially and if no air resistance acts upon the football, it will move at the same velocity in x direction throughout the motion.

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The bullet stops moving on hitting on a surface. Hence, the impulse here is equal to the momentum. Therefore, the mass of the bullet is 0.1 Kg.

Impulse in physics is the change in momentum. It is the product of the force and change in time.

hence, impulse = f. dt

When the bullet is travelling with  a velocity of 500 m/s it has a momentum. When it brought to rest, momentum become zero. Thus, the momentum is equal to the impulse here.

Therefore, f.dt = m. v

f.dt = 50 N s

v = 500 m/s

m = 50 N s/500 m/s = 0.1 Kg

Therefore, the mass of the bullet is 0.1 Kg.

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To solve this we must be knowing each and every concept related to velocity. Therefore, the amplitude and maximum velocity are 0.23 m and 2.75 m/s respectively.

V is the velocity measurement of an object's rate of motion and direction of motion. As a result, in order to calculate velocity using this definition, we must be familiar with both magnitude and direction.

For example, if an item travels west with 5 meters a second (m/s), its velocity to the west will be 5 m/s. The most frequent and simplest approach to determine velocity is using the formula shown below.

v = √(k / m) ×A

v = velocity of the mass

k= spring constant

m =mass of the object

A= amplitude of the oscillation.

substituting all the given values in the above equation, we get

2.3 m/s = √(200 N/m / 3.00 kg)×A

A = 2.3 m/s / √(200 N/m / 3.00 kg)

    = 0.23 m

v =√(200 N/m / 3.00 kg) ×0.23 m

 = 2.75 m/s

Therefore, the amplitude and maximum velocity are 0.23 m and 2.75 m/s respectively.

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The points in order of increasing value of the velocity can be ranked as:

B < C < A ≡ F < E < D.

The rate at which a body's displacement changes in relation to time is known as its velocity. Velocity is a vector quantity with both magnitude and direction. SI unit of velocity is meter/second.

According to the figure:

The points in order of increasing value of the velocity can be written as:

B < C < A ≡ F < E < D.

As the displacement becomes from positive to negative near point B, it has the lowest velocity and  as the displacement becomes from negative to positive near point D, it has the highest velocity

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The potential energy of a rock of mass 500 g, held at a height of 2 m above ground is 9.8J.

The formula for the gravitational potential energy of an object is:

PE = mgh

where m is the mass of the object, g is the acceleration due to gravity (approximately 9.8 m/s^2 on the Earth's surface), and h is the height of the object above the reference point.

So, in this case, we can plug in the given values to find the potential energy of the rock:

PE = (500 g) x (9.8 m/s^2) x (2 m)

The weight of the rock is m*g = 500 g * 9.8 m/s^2 = 4.9 N

Therefore,

PE = 4.9 N x 2 m = 9.8 J

Therefore, the potential energy of a rock of mass 500 g, held at a height of 2 m above ground is 9.8J.

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The normal force on the second tier of the cake is 14N. It is the sum of the weight of the last and the second tier of the cake.

The normal force is the force which the surfaces exert to prevent solid objects from passing through each other. Normal force is a contact force. If two objects or surfaces are not in contact, they cannot exert a normal force on each other.

Fₙ = mg

Fₙ = Normal force

Fₙ on the second tier is the sum of the first and second tier of the cake is:

Fₙ = 8N + 6N

Fₙ = 14N

The normal force on the second tier is 14N.

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