Light rays travel from one medium into another and refract away from the boundary. What changes about the light to cause this refraction?

A. Its speed increases.
B. Its frequency increases.
C. Its frequency decreases.
D. Its speed decreases.​

Answer:

The warm seas create a large humid air mass. The warm air rises and forms a low pressure cell, known as a tropical depression.

Explanation:

Hurricanes arise in the tropical latitudes (between 10 degrees and 25 degrees N) in summer and autumn when sea surface temperature are 28 degrees C (82 degrees F) or higher.

Answer:

air

Explanation:

Answer:

The first choice, one block per minute.

Answer:

The height of the hill is 0.46 m.

Explanation:

Given;

mass of the child and sled, m = 50 kg

initial velocity of the sled, u = 0

final velocity of the sled, v = 3 m/s

The height of the high is calculated from the law of conservation of energy;

P.E at top = K.E at bottom

mgh = ¹/₂mv²

gh = ¹/₂v²

[tex]h = \frac{v^2}{2g} \\\\h = \frac{3^2}{2\times 9.8} \\\\h = 0.46 \ m[/tex]

Therefore, the height of the hill is 0.46 m.

Answer:

In a scientific experiment, a constant error -- also known as a systematic error -- is a source of error that causes measurements to deviate consistently from their true value.

Explanation:

Answer:

Depending, on how much it's push against together.

Explanation:

Since, the two objects are getting in contact. But, if it's a type of item/thing there's a different frictions, but I know it's normal friction when two objects comes in contact. But, its depending on how much you push it against the two items.

Answer:

Explanation:

a ) The volume of blood flowing per second throughout the vessel is constant .

a₁ v₁ = a₂ v₂

a₁ and a₂ are cross sectional area at two places of vessel and v₁ and v₂ are velocity of blood at these places .

2A x v₁ = A x .40

v₁ = .20 m /s

b )

Let normal pressure be P₁ when cross sectional area is 2A and at cross sectional area A , pressure is P₂

Applying Bernoulli's theorem

P₁ + 1/2 ρv₁² = P₂ + 1/2 ρv₂²

P₁ - P₂ = 1/2  ρ(v₂² - v₁² )

= .5 x 1060 ( .4² - .2² )

= 63.6 Pa .

Answer:

Displacement = 30 miles due east.

Explanation:

Let the distance due west be A

Let the distance due east be B

Given the following data;

A = 50 miles

B = 80 miles

To find the displacement;

Displacement can be defined as the change in the position of a body or an object. It is a vector quantity because it has both magnitude and direction.

Thus, the displacement would be calculated by subtracting distance A from distance B because the rider rode in opposite directions.

Displacement = B - A

Displacement = 80 - 50

Displacement = 30 miles due east.

The energy required to move 2 electrons through a potential difference of 1.0 x 10² volts is -3.2 x 10⁻¹⁷ joules.

The potential difference, also known as voltage, is the difference in electrical potential energy per unit of charge between two points in an electrical circuit or an electric field.

In simpler terms, the potential difference is the amount of energy required to move a unit of electric charge from one point to another in an electric field or an electrical circuit. It is measured in volts (V) and can be calculated using the equation:

V = W/Q

Where V = is the potential difference,

W =  is the work done in moving the charge Q from one point to another,

Q =is the amount of charge that is moved.

Potential difference is an essential concept in electrical engineering and physics, as it governs the flow of electric current in a circuit and determines the behavior of electrical devices such as resistors, capacitors, and batteries.

Here in this question,

The energy required to move an electron through a potential difference is given by the formula:

E = qV

Where

E = is the energy required,

q = is the charge of the electron,

V = is the potential difference.

The charge of one electron is -1.6 x 10⁻¹⁹coulombs.

Therefore, for two electrons, the total charge is:

q = 2 x (-1.6 x 10⁻¹⁹coulombs) = -3.2 x 10⁻¹⁹ coulombs

The potential difference is given as 1.0 x 10² volts.

Now, the energy required to move 2 electrons through a potential difference of 1.0 x 10² volts is:

E = (-3.2 x 10⁻¹⁹ coulombs) x (1.0 x 10² volts) = -3.2 x 10⁻¹⁷ joules.

Note: that the negative sign indicates that the electrons lose potential energy as they move through the potential difference.

Therefore, The required energy is -3.2 x 10⁻¹⁷ joules.

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Answer:

A. Coefficient of kinetic friction, μ = 0.34

B. The box moves a distance of 9.64 m before coming to a stop

C. The heavier box will stop after 2.1 seconds

Explanation:

a. The coefficient of kinetic or sliding friction is given as: μ = F/R

where applied force, F = m × (∆v)/t

∆v = v - u

∆v = 0 - 8 m/s = -8 m/s; t = 2.4 s

R = normal reaction = m×g

where g = 9.8 m/s²

Substituting in the kinetic friction formula; μ = m∆v/t ÷ 1/m×g

μ = ∆v/g×t

μ = 8 / 9.8 × 2.4

μ = 0.34

b. Using the equation v² = u² + 2as to calculate the distance travelled by the box

where v = 0 m/s; u = 8.0 m/s; a = ? s = ?

From F = ma = μR

a = μR/m = (μ × m × g)/m

a = μg

a = 0.34 × 9.8

a = 3.32 m/s²

This is negative acceleration or deceleration

Substituting in the equation of motion

8² + 2 × -3.32 × s = 0

-6.64s = -64

s = 9.64 m

Therefore, the box moves a distance of 9.64 m before coming to a stop

c. The coefficient of friction is independent of mass.

Using the formula in (a): μ = ∆v/g×t

t = ∆v/μg

t = 7/0.34 × 9.8

t = 2.10 s

Therefore, the heavier box will stop after 2.1 seconds

The  coefficient of kinetic friction of the box is 0.34.

The distance traveled by the box is 9.6 m.

The time taken for the heavier box to stop is 2.1 s.

The  coefficient of kinetic friction of the box is calculated as follows;

[tex]\mu mg = ma\\\\\mu g = a\\\\\mu g = \frac{v}{t} \\\\\mu = \frac{v}{gt} \\\\\mu = \frac{8}{9.8 \times 2.4} \\\\\mu = 0.34[/tex]

The distance traveled by the box is calculated as follows;

[tex]v^2 = u^2 + 2as\\\\2as = -u^2\\\\s = \frac{-u^2}{2a} \\\\s = \frac{-u^2}{2\mu g} \\\\s = \frac{-(8)^2}{2\times 0.34 \times 9.8} \\\\s = -9.6 \ m\\\\|s| = 9.6 \ m[/tex]

The time taken for the heavier box to stop is calculated as;

[tex]\mu = \frac{v}{gt} \\\\t = \frac{v}{\mu g} \\\\t = \frac{7}{0.34 \times 9.8}\\\\ t = 2.1 \ s[/tex]

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The impulse exerted on the beach ball is 2.75 kgm/s.

The impulse exerted on the exercise ball is - 2.75 kgm/s.

This is the force applied to an object that acts over a period of time.

The impulse exerted on the beach ball is the change in the momentum of the ball and it is calculated as follows;

J = ΔP

J = m(v - u)

J = 0.25(7 - (-4))

J = 0.25(7 + 4)

J = 2.75 kgm/s

The impulse exerted on the exercise ball is equal in magnitude but opposite in direction to the beach ball.

Thus, the impulse exerted on the exercise ball is - 2.75 kgm/s.

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Answer:

B

Explanation:

The crest of all three waves are of equal height

Answer:

D. equal to 1.2

Explanation:

on edg

The length of the power cord will be equal to 1.2 m.

The length of the power cord when shanti gets off the train is equal to 1.2 m.

The Correct answer is Option D.

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Answer:

The mass of the fish is 18.1 kg.

Explanation:

Given;

weight of the fish, F = 108 N

extension of the spring by the given weight, x = 14 cm = 0.14 m

First, determine the elastic constant of the spring by applying Hook's law;

F = kx

where;

k is the spring constant

k = F/x

k = 108 / 0.14

k = 771.43 N/m

When the spring is stretched to 23cm, the mass of the fish is calculated as follows;

[tex]F = mg = Kx\\\\m = \frac{Kx}{g} \\\\m = \frac{771.43\times 0.23}{9.8} \\\\m = 18.1 \ kg[/tex]

Therefore, the mass of the fish is 18.1 kg.

Answer:

ΔT = 17.11 °C

Explanation:

In this case, we have a ship standing on a place with a given mass and it's about to be launched to a lock containing water.

At first, before launch, the ship has a potential energy, and when the ship hits the water after being launched, this potential energy is transformed into kinetic energy.

So, let's calculate first the potential energy of the ship:

E = mgh   (1)

We have the mass, gravity and height, so we need to replace the given data here. Before we do that, let's remember to use the correct units. A ton is 1000 kg, so replacing and converting we have:

E = (55000 ton * 1000 kg/ton) * (9.8 m/s²) * 10 m

E = 5.39x10⁹ J

Now this energy will be the same when the ship hits the water, only that is kinetic energy that will result in the rise of temperature. To get this rise we use the following expression:

E = m * C * ΔT   (2)

We have the energy, the mass of water (assuming density of water as 1 kg/m³) and the specific heat, so, replacing in (2) and solving for ΔT we have:

ΔT = E / m * C    (3)

ΔT = 5.39x10⁹ / 4200 * 75000

Hope this helps

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

Answer:

a) the required angular acceleration magnitude α is  π rad/s² or 3.14 rad/s²

b) the time at which the contact occur is 8 seconds

Explanation:

Given the data in the question;

first we convert the given angular velocity to rad/s

angular velocity = 240 rpm = ((240/60) × 2π ) = 8π rad/s

so

ωA = 8π rad/s

next we determine angular acceleration at point A; so

ωA = at

8π rad/s = at -------let this be equation

thus, angular acceleration of disk A is ωA and rotates in counter clockwise direction.

Next we determine the velocity of point C;

Vc = rA × ωA

where Vc is velocity at point C, rA is radius of A ( 150/1000)m,  { from the diagram }

so we substitute

Vc = 0.15m × 8π

Vc = 1.2π m/s

for angular velocity at point B;

Vc = rB × ωB

where rB is the radius of B ( 200/1000)m

we substitute

1.2π = 0.2 × ωB

ωB = 1.2π / 0.2

ωB = 6π rad/s

Thus, the wheel B rotates with an angular velocity of 6π rad/s in clock wise direction.

Now,

a) Determine the required angular acceleration magnitude α

we find the the angular acceleration of disk B after 2 seconds, using the expression;

ωB = at

where angular acceleration is a and t is time ( t - 2)

we substitute

ωB = at

6π = a( t - 2) -------- let this be equation 2

now, lets substract equation 1 form equation 2

(6π = a( t - 2)) - (8π = at)

(6π = at - 2a) - ( 8π = at)

-2π = 0 + -2a

2π = 2a

a = 2π/2

a = π rad/s² or 3.14 rad/s²

Therefore, the required angular acceleration magnitude α is  π rad/s² or 3.14 rad/s²

b) determine the time at which the contact occurs;

from equation 1

8π = at

we substitute in the value of a

8π = π × t

t = 8π / π

t = 8 seconds

Therefore, the time at which the contact occur is 8 seconds

Answer:

10 mm

Explanation:

We'll begin by calculating the spring constant of the spring. This can be obtained as follow:

Extention (e) = 5 mm

Force (F) = 125 N

Spring constant (K) =?

F = Ke

125 = K × 5

Divide both side by 5

K = 125 / 5

K = 25 N/mm

Finally, we shall determine how much the spring will stretch when a 250 N force is applied. This can be obtained as follow:

Force (F) = 250 N

Spring constant (K) = 25 N/mm

Extention (e) =?

F = Ke

250 = 25 × e

Divide both side by 25

e = 250 / 25

e = 10 mm

Thus, the spring will stretch 10 mm when a 250 N force is applied.

Answer:

The answer is "83.1%".

Explanation:

Given:

[tex]\text{Mass of the earth}\ (M_E) = 5.97 \times 10^{24}\ \ kg\\\\\text{Radius of the earth}\ (R_E) = 6380 \ km = 6.38 \times 10^{6}\ \ m\\\\\text{acceleration of gravity}\ (g_E) = 608 \ \ km= 608,000 \ \ m \\\\G= 6.67 \times 10^{-11} \ \ \frac{N \cdot n^2}{kg^2}[/tex]

Using formula:

[tex]\to g_E = G \frac{M_E}{(R_E +h)^2}[/tex]

[tex]\to g_{608 \ km}=6.67\times10^{-11}\frac{5.97 \times 10^{24}}{(6.38 \times 10^{6}+ 608,000)^2}\\\\\to g_{608 \ km}=6.67\times10^{-11}\frac{5.97 \times 10^{24}}{(6,380,000+ 608,000)^2}\\\\\to g_{608 \ km} =6.67\times10^{-11}\frac{5.97 \times 10^{24}}{(6,380,000+ 608,000)^2}\\\\\to g_{608 \ km}=6.67\times10^{-11}\frac{5.97 \times 10^{24}}{(6,988,000)^2}\\\\\to g_{608 \ km}=6.67\times10^{-11}\frac{5.97 \times 10^{24}}{4.88 \times 10^{13}}\\\\[/tex]

[tex]\to g_{608 \ km}=6.67\times10^{-24}\frac{5.97 \times 10^{24}}{4.88}\\\\\to g_{608 \ km}=6.67\times \frac{5.97}{4.88}\\\\\to g_{608 \ km}=6.67\times 1.22336066\\\\\to g_{608 \ km}= 8.15 \ \frac{m}{s^2}[/tex]

Calculating the gravity on the Earth’s surface:

[tex]\to \frac{g_{608 \ km} }{ g_{\ earth \ surface}}[/tex] [tex]= \frac{8.15}{9.8} \times 100=83.1 \%[/tex]

Answer:

warm front .,

Explanation:

...........

Answer:

The cutoff wavelength is the minimum wavelength in which a particular fiber still acts as a single mode fiber. Above the cutoff wavelength, the fiber will only allow the LP01 mode to propagate through the fiber (fiber is a single mode fiber at this wavelength).

Explanation:

Answer:

A. shin guards is that answer

Answer:

Shin Guards Brainlyiest please

Explanation:

Answer:

the  frequency of the oscillation is 2.5 Hz.

Explanation:

Given;

time to complete the oscillation, t = 0.4 s

number of oscillations, n = 1

The frequency of the oscillation is calculated as;

[tex]F = \frac{n}{t} \\\\F = \frac{1}{0.4} \\\\F = 2.5 \ Hz[/tex]

Therefore, the  frequency of the oscillation is 2.5 Hz.

Answer:

The right solution is "0.50 m/s²". A further explanation is provided below.

Explanation:

The given values are:

Mass,

m = 50 kg

Speed,

v = 10 m/s

Rolling friction acting backward (south),

f = 10 N

Air resistance acting backward (south),

[tex]F_T[/tex] = 15 N

The total force acting will be:

⇒  [tex]F = -f-F[/tex]

On substituting the given values, we get

⇒      [tex]=-10-15[/tex]

⇒      [tex]=-25 \ N[/tex]

Now,

⇒  [tex]a = \frac{F}{m}[/tex]

On substituting the given values, we get

⇒     [tex]=\frac{-25}{50}[/tex]

⇒     [tex]=-0.50 \ m/s^2[/tex]

The horizontal acceleration will be "0.50 m/s²" because the (-)ve sign indicates it in south direction.

Answer:

797700000 J

Explanation:

From the question,

The work done by the rocket, is given as,

W = Ek+Ep............. Equation 1

Where Ek and Ep are the potential and the kinetic energy of the rocket respectively.

Ep = mgh............ Equation 2

Ek = 1/2mv²............. equation 3

Substitute equation 2 and equation 3 into equation 1

W = mgh+1/2mv².............. Equation 4

Where m = mass of the rocket, h = height, v = velocity of the rocket, g = acceleration due to gravity.

Given: m = 2000 kg, h = 12 km = 12000 m, v = 750 m/s, g = 9.8 m/s²

Substitute into equation 4

W = 2000(12000)(9.8)+1/2(2000)(750²)

W = 235200000+562500000

W = 797700000 J

Answer:

I. 100 Bg

II. 50 Bg

III. 25 Bg

Explanation:

I. Determination of the activity after 26 hours.

We'll begin by calculating the number of half-lives that has elapse. This can be obtained as follow:

Half-life (t½) = 13 hours

Time (t) = 26 hours

Number of half-lives (n) =?

n = t / t½

n = 26 / 13

n = 2

Finally, we shall determine remaining activity

Original activity (N₀) = 400 Bg

Number of half-lives (n) = 2

Activity remaining (N) =?

N = 1/2ⁿ × N₀

N = 1/2² × 400

N = 1/4 × 400

N = 100 Bg

II. Determination of the activity after 39 hours.

We'll begin by calculating the number of half-lives that has elapse. This can be obtained as follow:

Half-life (t½) = 13 hours

Time (t) = 39 hours

Number of half-lives (n) =?

n = t / t½

n = 39 / 13

n = 3

Finally, we shall determine remaining activity.

Original activity (N₀) = 400 Bg

Number of half-lives (n) = 3

Activity remaining (N) =?

N = 1/2ⁿ × N₀

N = 1/2³ × 400

N = 1/8 × 400

N = 50 Bg

III. Determination of the activity after 52 hours.

We'll begin by calculating the number of half-lives that has elapse. This can be obtained as follow:

Half-life (t½) = 13 hours

Time (t) = 52 hours

Number of half-lives (n) =?

n = t / t½

n = 52 / 13

n = 4

Finally, we shall determine remaining activity

Original activity (N₀) = 400 Bg

Number of half-lives (n) = 4

Activity remaining (N) =?

N = 1/2ⁿ × N₀

N = 1/2⁴ × 400

N = 1/16 × 400

N = 25 Bg

Answer:

A plant

Explanation:

because animals don't have cell walls, and fungus and bacteria dont have chloroplasts

Answer:b

Explanation:

Guess