please answer asap!! thank you for the help, will mark brainliest if correct!


The color of visible light depends on the wavelength of the light.

Please select the best answer from the choices provided

T
F

Answer:

Force is an external agency that changes or tends to change the state of body from rest to motion or motion to rest.

The SI unit of force is newton(N)

Answer:

8.45 m

Explanation:

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

Mass (m) = 90 Kg

Initial velocity (u) = 13 m/s

Final velocity (v) = 0 m/s

Height (h) =?

NOTE: Acceleration due to gravity (g) = 10 m/s²

The height of the hill can be obtained as follow:

v² = u² – 2gh (since the cart is going against gravity)

0² = 13² – (2 × 10 × h)

0 = 169 – 20h

Rearrange

20h = 169

Divide both side by 20

h = 169/20

h = 8.45 m

Therefore, the height of the hill is 8.45 m

Answer:

The correct answers are: A, C, D, E

Explanation:

The magnetic field is a solenoid is given by

         B = μ₀  [tex]\frac{N}{L}[/tex]   I

where N is the number of turns, I the current and L length of the solenoid.

Using this equation let's examine the different responses to permute increasing the magnetic field

A) True. a thicker wire decreases the resistance and the current can increase the system.

B) False. If there is no voltage source there is no current in the system

C) True. the field is proportional to the number of turns

D) True. the magnetic moments of the core align with the field increasing its value

E) True. When the loops are closer together, more of them can fit per unit length

F) False. If the wire is shorter the number of turns decreases.

The correct answers are: A, C, D, E

This question is incomplete, the missing image is uploaded along this answer.

Answer:

the axial component of the anchoring force required to hold the contraction in place is 365.6 lb

Explanation:

Given the data in the question and as illustrated in the image below;

first we calculate the area at section 1

A₁ = (πD²)/4

we substitute

A₁ = (π(3 in)²)/4

A₁ = 7.06858 in²

we know that; 1 ft = 12 in

A₁ = ( 7.06858 / (12²) ) ft²

A₁ = ( 7.06858 / 144 ) ft²

A₁ = 0.0491 ft²

now, we write the elation for area at section 2

A₂ = πd²/4

here, d is the diameter at section 2

next, we use the conservation of mass equation between the two section;

m" = pV₁A₁ = pV₂A₂

we calculate the mass flow rate;

m" = pV₁A₁

= (1.94[tex]\frac{slug}{ft^2}[/tex]) × 30[tex]\frac{ft}{s}[/tex] × 0.0491 ft²

=  2.8576 slug/s

Now, Apply the linear momentum along the horizontal direction for the control volume between 1 - 2

-pV₁A₁V₁ = pV₂A₂V₂ = P₁A₁ - F[tex]_A[/tex] - P₂A₂

m"( V₂ - V₁ ) = P₁A₁ - F[tex]_A[/tex] - P₂A₂

F[tex]_A[/tex] = P₁A₁ - P₂A₂ - m"( V₂ - V₁ )

we substitute

F[tex]_A[/tex] = ((80×[tex]\frac{144 in^2}{1 ft^2}[/tex])×0.0491 ft²) - (0×(πd²/4)) - 2.8576( 100 - 30 )ft/s

F[tex]_A[/tex] =  565.632 - 0 - 200.032

F[tex]_A[/tex] = 565.632 - 200.032

F[tex]_A[/tex] = 365.6 lb

Therefore, the axial component of the anchoring force required to hold the contraction in place is 365.6 lb

Answer:  I'm not sure, but I think it would be a total lunar eclipse

When the moon passes through the penumbra of Earth’s shadow it is referred to as partial lunar eclipse. The correct option is C.

A partial lunar eclipse occurs when the moon is not completely immersed in the umbra of the earth's shadow.

During a partial solar eclipse, the Moon, Sun, and Earth do not align perfectly straight, and the Moon casts only the penumbra of its shadow on Earth. From our vantage point, it appears that the Moon has eaten the Sun.

A shadow's penumbra is the lighter outer edge. Partial solar eclipses are caused by the Moon's penumbra, while penumbral lunar eclipses are caused by the Earth's penumbra. The penumbra is a type of lighter shadow.

The Penumbra is a half-shadow region that occurs when an object only partially covers a light source.

Thus, the correct option is C.

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Answer:

320 J

Explanation:

From the question,

KE = 1/2mv².................. Equation 1

Where KE = Kinetic Energy, m = mass of the rock, v = velocity of the rock

Given: m = 40 kg, v = 4m/s

Substitute these values into equation 1

KE = 1/2(40)(4²)

KE = 20×16

KE = 320 J

Hence the kinetic energy of the rock is 320 J

Answer:

this is a lateral fault

Explanation:

Answer:

the amplitude of the wave

the energy of the wave

the type of wave

the type of medium

Explanation:

रात में घूमने वाला arthaarat निशाचर

Answer:

-11.600N

Explanation:

lol

Answer:

  F = 49 N

Explanation:

For this exercise we must use Newton's second law. Let's set a reference frame with the x axis parallel to the floor.

As they indicate that the box is going at constant speed, its acceleration is zero

Y axis y

       N-W = 0

       N = mg

X axis

        F-fr = 0

        F = fr

the friction force has the formula

      fr = μ N

      fr = μ mg

we substitute

            F = μ m g

let's calculate

            F = 0.1 50 9.8

            F = 49 N

Answer:

electric force

Explanation:

its a contact and noncontact force

Answer:

[tex]3.52\times 10^{25}\ \text{J}[/tex]

Explanation:

G = Gravitational constant = [tex]6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2[/tex]

M = Mass of Sun = [tex]1.989\times 10^{30}\ \text{kg}[/tex]

m = Mass of Earth = [tex]5.972\times 10^{24}\ \text{kg}[/tex]

[tex]r_i[/tex] = Initial radius of orbit = [tex]1.5\times 10^{11}\ \text{m}[/tex]

[tex]r_f[/tex] = Final radius of orbit = [tex]((1.5\times 10^{11})+2\times 10^3)\ \text{m}[/tex]

Energy required is given by

[tex]E=\dfrac{1}{2}\Delta U\\\Rightarrow E=\dfrac{GMm}{2}(\dfrac{1}{r_i}-\dfrac{1}{r_f})\\\Rightarrow E=\dfrac{6.674\times 10^{-11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{2}(\dfrac{1}{1.5\times 10^{11}}-\dfrac{1}{(1.5\times 10^{11})+2\times 10^3})\\\Rightarrow E=3.52\times 10^{25}\ \text{J}[/tex]

The energy required would be [tex]3.52\times 10^{25}\ \text{J}[/tex].

Answer:

OMG IM ON THE SAME QUESTION

Explanation:

Answer:

If the angular acceleration is a constant A, then this constant times the time will be equal to the angular velocity.

We know that after accelerating for 10 seconds, the angular velocity is 0.4 rad/s

Then:

A*10s = 0.4 rad/s

If we solve this for A we get:

A = (0.4 rad/s)/10s = 0.04 rad/s^2

This is the angular acceleartion.

Now we assume that the angular speed remains constant at 0.4 rad/s, and we know that the radius of the carousel is 20m, we want to find the linear velocity of the children.

We know that for an angular velocity W and for a radius R, the linear velocity is:

V = R*W

Then in this case the linear velocity is:

V = 20m*(0.4 s^(-1)) = 8 m/s

Answer:

the oscillations of the electrons must be in the 10⁸ Hz = 100 MHz range

Explanation:

The speed of a wave of radio, television, light, heat, all are manifestations of electromagnetic waves that are oscillations of electric and magnetic fields that support each other, the speed of all these waves is the same and the vacuum is equal to c = 3 108 m / s

All waves have a relationship between the speed of the wave, its frequency and wavelength

          c = λ f

          f = c /λ

for this case lam = 1 m

          f = 3 10⁸/1

          f = 3 10⁸ Hz

the oscillations of the electrons must be in the MHz range

It should be clarified that the speed of light in air is a little lower

          n = c / v

          v = c / n

the refractive index of vacuum is n = 1 and the refractive index of air is n = 1.000002

Answer:

See explanations below

Explanation:

According to Newtons second law of motion

F = mass * acceleration

F = ma

If mass of an object is decreased to half, then m₂ = 1/2 m

If acting force is reduced by quarter, then F₂ = 3/4 F

F₂ = m₂a₂

3/4F = 1/2m a₂

Divide both expressions

(3/4F)/F = (1/2m)a₂/ma

3/4 = 1/2a₂/a

3/4 = a₂/2a

4a₂ = 6a

2a₂ = 3a

a₂ = 3/2 a

Hence the acceleration of its motion will be one and a half of its original acceleration.

Answer:

Is that your answer

Answer:

hydrogen, helium, lithium, beryllium, boron, carbon, nitrogen, oxygen, fluorine, neon, sodium, magnesium, aluminum, silicon, phosphorus, sulfur, chlorine, argon, potassium, calcium, scandium, titanium, vanadium, chromium, maganese, iron, cobalt, nickel, copper, zinnc, gallium, germanium, aresnic, selenium, bromine, krypton, rubidium, strongtium, yttrium, zirconium, niobium, molybdenum, technetium, ruthenium, rhodium, palladium, silver, cadmium, indium, tin, atimony, tellurium, iodine, xenon, cesium, barium, lanthanum, cerium, praseodyumium, neodymim, promethium, samarium, europium, gadolinium, terbium, dysprosium, holmium, erbium, thulium, ytterbium, luteium, hafnium, tantalum, tungsten, rhenium, osmium, irdium, platinum, gold, mercury, thallium, lead bismuth, polnium, astatine, radon, rancium, radium, actinum, thorium, protactinium

Explanation:

Answer:

answer is 3

Explanation:

by using s= [(v+ u)/2] x t

28= (v+ 0)/2 x 11

v= 5.09 ms^-1

Explanation:

In train's rest frame, the speed of photon is [tex]c[/tex] and the proper length of the train is [tex]L[/tex]. The time taken by the photon to cross the train is [tex]t=\frac{L}{c}[/tex]

In ground frame, the speed of the photon is given as follows:

[tex]v_{x}=\frac{v_{x}+v}{1+\frac{v_{x} \cdot v}{c^{2}}}[/tex]

[tex]=\frac{c+v}{1+\frac{c v}{c^{2}}} \\=c[/tex]

The speed of light or photon remains same in every frame of reference.

Now, the speed of train is very less as compared to the speed of photon so that [tex]v<c[/tex] So that, [tex]\frac{v}{c} \ll 1[/tex]

The length contraction in the ground frame is given as follows:

[tex]L^{\prime}=L \sqrt{1-\frac{v^{2}}{c^{2}}}[/tex]

[tex]=L[/tex]

Time taken by the photon to travel the length of the train in ground frame is .

Answer:

a = 72 m/s^2

Explanation:

Since the object is moving in circle, the acceleration points TOWARDS the centre of the circle (it is centripetal acceleration)

a = v^2 / r

a = 60^2 / 50

a = 72 m/s^2

Kinetic energy decreases if either the mass or the velocity of the object decreases or if both decrease. and  Kinetic energy increases if either the mass or the velocity of the object increases or if both increase.

Explanation:

These statements explain the relationship between mass, velocity, and kinetic energy .i.e KE = ½mv², (3) Kinetic energy decreases if either the mass or the velocity of the object decreases or if b decrease. (4)Kinetic energy increases if either the mass or the velocity of the object decreases or if both decrease.

When a body has a mass of m kg and travels a velocity of v m/s then the kinetic energy is given by,

KE=1/2mv²

In the above relation, it is clearly shown that,

Kinetic energy is directly proportional to the mass of the body,

i.e

KE∝m

and also Kinetic energy directly varies with the square of the velocity of the body.

I.e

KE∝v²

Therefore, From the given expression we can conclude that when mass and velocity increased either one of them or both, the kinetic energy also gets increased.

And when mass and velocity decreased either one of them or both, the kinetic energy also gets decreased. Hence options (3) and (4) are correct.

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Answer:

Regular reflection

Explanation:

Regular reflection occurs when light reflects off a very smooth surface and forms a clear image.

i hope this helps a bit.

According to the context, a sharp image is formed when light reflects from a regular reflection.

It is reflection without diffusion that obeys the laws of geometrical optics, as in mirrors.

This reflection of light happens when the angles that the two rays determine with the surface are equal.

Therefore, we can conclude that according to the context, a sharp image is formed when light reflects from a regular reflection.

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Answer:

When you add energy to a substance?

One change of state happens when you add energy to the substance. This change of state is called melting. By adding energy to the molecules in a solid the molecules begin to move quicker and can break away from the other molecules. The temperature at which a substance goes from a solid to a liquid is it melting point.

When you remove energy from a substance?

When a substance is heated, it gains thermal energy. Therefore, its particles move faster and its temperature rises. When a substance is cooled, it loses thermal energy, which causes its particles to move more slowly and its temperature to drop.

Explanation:

The addition of energy increases the kinetic energy of the particles, which reduces the intermolecular forces between the particles. Freezing occurs when a liquid becomes a solid and energy is released.

Atoms lose energy when they change from solid to liquid or gas and quid to gas.

Answer:

average speed of the fist during the hook = 15 m/s or 1500 cm/s

Explanation:

We are given;

Speed of shoulder contraction, v_s = 75 cm/s = 0.75 m/s

Distance moved through the arc length by shoulder, d_s = 5 cm = 0.05 m

Distance moved by the fist, d_f = 100 cm = 1 m

Now, we are to find the average speed of the fist during the hook; v_f

Thus can be gotten from proportion;.

d_f/d_s = v_f/ v_s

Making V_f the subject, we have;

v_f = (d_f × v_s)/d_s

Thus;

v_f = (1 × 0.75)/0.05

v_f = 0.75/0.05

v_f = 15 m/s

Answer:

J = -0.522 m/s

Explanation:

Given that,

The mass of the baseball, m = 0.145 kg

Initial velocity, u = 14.5 m/s

Final velocity, v = 10.9 m/s

aWe need to find the direction of the impulse that the glass imparts to the baseball. Impulse is equal to the change in momentum such that,

[tex]J=m(v-u)[/tex]

Substitute all the values,

[tex]J=0.145\times (10.9-14.5)\\\\=-0.522\ kg-m/s[/tex]

The direction of impulse is opposite to the direction of velocity.

Answer:

261

Explanation:

Acellus

Answer:

the magnitude of the induced current is 182.86 A.

Explanation:

Given;

number of turns, N = 240 turns

cross sectional area of the loop, A = 0.2 m²

uniform magnetic field strength, B = 1.6 T

resistance of the loop, R = 21 ohms

time, Δt = 20.0 ms

The magnitude of the induced emf is calculated as;

[tex]emf = \frac{NA B}{t} \\\\emf = \frac{240 \times 0.2 \times 1.6}{20 \times 10^{-3}} \\\\emf = 3,840\ V[/tex]

The induced current in the loop is calculated as;

[tex]I = \frac{emf}{R} \\\\I = \frac{3840}{21} \\\\I= 182.86 \ A[/tex]

Therefore, the magnitude of the induced current is 182.86 A.