Someone please help!!

Answer:

8.333C 15F

Explanation:

15R in C - 0R in C = 8.3333C

15R in F - 0R in F = 15F

Answer:

294.68

Explanation:

I hope you right answer

The velocity of light travelling in that medium = v= 107.9 m/s

It is a ratio of the velocity of light in vacuum to velocity of light in a specific medium

given

refractive index of material( u) : 2.78 * 10 ^6

c = 3* 10 ^8 m/s

velocity of light travelling in that medium(v) = ?

u = speed of light / v

v = speed of light / u

v= 3* 10 ^8 / 2.78 * 10 ^6

v= 107.9 m/s

velocity of light travelling in that medium = v= 107.9 m/s

learn more about refractive index :

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Answer:

(a) The list of angles based on amount of time in the air; 15° < 25° < 45° < 60°, < 80°

(b) The list of angles based on distance = 80° < 15° < 25° < 60° < 45°

Explanation:

(a) The given parameters are;

The angles in which the ball is thrown, θ = 45°, 15°, 60°, 25°, and 80°

The velocity with which the ball is thrown each time = The same velocity

The amount of time the projectile is in the air given as follows;

[tex]2 \cdot t = \dfrac{2\cdot u \cdot sin (\theta) }{g}[/tex]

Where;

t = Half the amount of time the projectile is in the air

θ = The angle of flight of the projectile

u = The initial velocity of the projectile = Constant for all angles in which the ball is thrown

g = The acceleration due to gravity = 9.8 m/s² = Constant

Therefore, we have;

When θ = 45°;

[tex]2 \cdot t = \dfrac{2\cdot u \cdot sin (45^ {\circ}) }{g} = \dfrac{2\cdot u \cdot \dfrac{\sqrt{2} }{2} }{g} = \dfrac{\sqrt{2} \cdot u }{g}[/tex]

When θ = 15°;

[tex]2 \cdot t = \dfrac{2\cdot u \cdot sin (15^ {\circ}) }{g} \approx \dfrac{0.517638\cdot u }{g}[/tex]

When θ = 60°;

[tex]2 \cdot t = \dfrac{2\cdot u \cdot sin (60^ {\circ}) }{g} = \dfrac{2\cdot u \cdot \dfrac{\sqrt{3} }{2} }{g} = \dfrac{\sqrt{3} \cdot u }{g}[/tex]

When θ = 25°;

[tex]2 \cdot t = \dfrac{2\cdot u \cdot sin (25^ {\circ}) }{g} = \dfrac{0.8452365\cdot u }{g}[/tex]

When θ = 80°;

[tex]2 \cdot t = \dfrac{2\cdot u \cdot sin (80^ {\circ}) }{g} = \dfrac{1.9696155\cdot u }{g}[/tex]

Therefore, the list of the angles based on the amount of time in the air from the shortest time to the longest time is given as follows;

List of angles based on amount of time in the air; 15° < 25° < 45° < 60°, < 80°

(b) The horizontal range, R is given as follows;

[tex]R = \dfrac{u^2 \cdot sin(2 \cdot \theta) }{g}[/tex]

When θ = 45°

[tex]R = \dfrac{u^2 \cdot sin(2 \times 45 ^{\circ}) }{g} = \dfrac{u^2 \cdot sin(90 ^{\circ}) }{g} = \dfrac{u^2 }{g}[/tex]

When θ = 15°

[tex]R = \dfrac{u^2 \cdot sin(2 \times 15 ^{\circ}) }{g} = \dfrac{u^2 \cdot sin(30 ^{\circ}) }{g} = \dfrac{1}{2} \cdot \dfrac{u^2 }{g} = 0.5 \cdot \dfrac{u^2 }{g}[/tex]

When θ = 60°;

[tex]R = \dfrac{u^2 \cdot sin(2 \times 60 ^{\circ}) }{g} = \dfrac{u^2 \cdot sin(120 ^{\circ}) }{g} = \dfrac{u^2 }{g} \cdot \dfrac{\sqrt{3} }{2} =0.866025 \cdot \dfrac{u^2 }{g}[/tex]

When θ = 25°;

[tex]R = \dfrac{u^2 \cdot sin(2 \times 25 ^{\circ}) }{g} = \dfrac{u^2 \cdot sin(50 ^{\circ}) }{g} = \dfrac{0.7660444 \cdot u^2 }{g}[/tex]

When θ = 80°

[tex]R = \dfrac{u^2 \cdot sin(2 \times 80 ^{\circ}) }{g} = \dfrac{u^2 \cdot sin(160 ^{\circ}) }{g} = \dfrac{0.34202 \cdot u^2 }{g}[/tex]

Therefore the list of angles based on the horizontal range the baseball travels from the shortest distance to the longest distance is given as follows;

List of angles based on distance = 80° < 15° < 25° < 60° < 45°.

Answer:

Explanation:

Mass = 50kg +7kg +4kg =61 kg

Velocity = 3m/s

Momentum = mass × velocity

[tex]p =61\times 3\\\\p =183MLT^-^1[/tex]

Answer:

Mass= 61 kg. Success to The homework.

Answer:

Potential energy is the stored energy of position possessed by an object. In the above equation, m represents the mass of the object, h represents the height of the object and g represents the gravitational field strength (9.8 N/kg on Earth) - sometimes referred to as the acceleration of gravity.

Explanation:

Answer:

9194.4278 seconds

Explanation:

Orbital period is given by :

T² = (4π²r³) /GM

G = Gravitational constant = 6.67 * 10^-11

M = 5.9801 * 10^24 kg

r = (3.110 * 10^6 m + 6378000 m) =3110000 + 6378000 = 9488000 m

T² = (4π² * 9488000^3) / (6.67* 10^-11 * 5.9801*10^24)

T² = 3.37197 * 10^22 / 398872670000000

T² = 84537504.161415

T = sqrt(84537504.161415)

T = 9194.4278 seconds

Answer:

i cant see it just put the question

Explanation:

Answer:

Force = 2.4  10^(-17) N

Explanation:

Recall that if you have the field, the force is easily obtained by multiplying E times the charge:

F = E * q = 150 * 1.6  10^(-19)  N = 240  * 10^(-19) N = 2.4  10^(-17) N

which agrees with answer C in your list of possible answers.

Answer:

Explanation:

10

Answer:

3,200,000,000 J

Explanation:

Work is defined as the amount of energy transferred as an object is moved a certain distance with a certain force. Mathematically, we express this with the equation

[tex]W=Fs[/tex]

where W is work (measured in joules), F is the force applied (in Newtons), and s is the distance, also called the displacement (in meters).

Here, we have F = 1,600,000 N and s = 2000 m, so our work will be

[tex]W=1.600.000(2.000)=3.200.000.000[/tex] J

Answer:

v = 5.4 m/s

Explanation:

We use conservation of mechanical energy, that is all potential energy in the initial state, and all kinetic energy in the final state:

[tex]m*g*h=\frac{1}{2} m*v^2\\g*1.5=\frac{1}{2} v^2\\v^2=3*g=29.4\\v=\sqrt{29.4} \\v=5.42\,m/s[/tex]

which rounded to one decimal is

v = 5.4 m/s

(in agreement with the third answer on the list)

Answer:

D) 720 kmph

Explanation:

First, let's find what distance this aeroplane covered. Distance (d) is the product of speed and time - here, we have a speed of 240 kmph and a time of 5 hours, gives us

[tex]d=240\cdot5=1200[/tex] km

Using that same fact, we can set up a new equation to solve for speed (s) when we have a distance of 1200 km and a time of 1 2/3 hours. For the sake of cleanliness, I'm gonna rewrite 1 2/3 as the improper fraction 5/3:

[tex]1200=(5/3)s[/tex]

Multiplying both sides of the equation by 3/5:

[tex]s=(3/5)(1200)=720[/tex] kmph

So our answer is D.

Answer: 720km/h

Explanation:

Answer:

a. Please find the attached velocity time graph of the car's motion created with Microsoft Excel

b. The total distance traveled by the car is 8,725 meters

c. The average speed of the car is 22.9605263 m/s

Explanation:

The given parameters of the motion are;

The initial speed of the car, v₁ = 15 m/s

The time during which the car runs at the initial speed, t₁ = 30 seconds

The new speed the car then accelerates at 'a₁' to, v₂ = 25 m/s

The duration it takes for the car to accelerate to the new speed = 20 seconds

The time during which the car runs at the initial speed = 300 seconds

The time it takes the car to be brought to rest with a deceleration, 'a₂' from the new speed (20 m/s) = 30 seconds

The final speed of the car at rest, v₃ = 0 m/s

The acceleration, a₁ = (v₂ - v₁)/t₁ = (25 - 15)/20 = 1/2 m/s²

The deceleration , a₂ = (v₃ - v₂)/t₁ = (0 - 25)/30 = -5/6 m/s²

a. Please find attached the drawing of the velocity time graph of the motion created with Microsoft Excel

b. The total distance traveled by the car, 'Δx', is given b the area under the velocity time graph as follows;

Area of trapezoid, A₁ = (320 + 300)/2 × 10 = 3,100

Area of rectangle, A₂ = 15 × 350 = 5,250

Area of triangle, A₃ = 1/2×30×25 = 375

The total area under the velocity time graph = A₁ + A₂ + A₃ = 3,100 + 5,250 + 375 = 8,725

The total area under the velocity time graph = The total distance traveled by the car, Δx = 8,725 meters

c. The average speed of the car is given as follows;

[tex]The \ average \ speed \ of \ the \ car, \overline v =\dfrac{\Delta x}{\Delta t} = \dfrac{The \ total \ distance \ traveled by \ the \ car}{The \ total \ time \ the \ car \ travels}[/tex]

Where;

Δt = The total time during which the car travels

∴ The average speed of the car = 8,725 m/(380 s) = 22.9605263 m/s

Answer:

The Rutherford Model shows an atom with electrons orbiting a fixed, positively charged nucleus in set, predictable paths. The Bohr model shows electrons travel in defined circular orbits around the nucleus.

Explanation:

Answer:

it take 15 seconds

Explanation:

it take 15 seconds to increase it time

Answer:

Explanation:

In addition to the 'daughter' products, two or three neutrons also explode out of the fission reaction and these can then collide with other uranium nuclei which are unstable so these cause further fission reactions. This is known as a chain reaction .

Hope this helps

Answer:

B

Explanation:

Answer:

the DTS triangle

Explanation:

DTS stands for distance, time, and speed, which you can use to calculate any of the three if you have the other two.

Answer:

C) plate movement at fault lines.

Explanation:

Earthquakes are vibrations that develop when there is a build-up of stress in the Earth's crust resulting in plate movement at fault lines. Most often, earthquakes occur at subduction zones.

Answer:

hi have a nice day

Explanation:

i think forms of space

Answer:

Transverse waves

Explanation:

EM waves are made up of photons, which are massless particles. These waves propogate (travel) through space at the speed of light as transverse waves, waves that move back and forth perpendicular to the direction they're travelling in, as opposed to longitudinal waves, which move back and forth along the path of travel.

Correct question:

You have a transformer with 50 turns in the primary coil if you want to increase the voltage from 5V to 6V how many turns should be in the secondary coil ?

Answer:

the number of turns in the secondary coil is 60 turns.

Explanation:

Given;

number of turns in the primary coil, P = 50 turns

primary voltage, Ep = 5V

secondary voltage, Es = 6V

The number of turns in the secondary coil is calculated as;

[tex]\frac{N_s}{N_p} = \frac{E_s}{E_p} \\\\\frac{N_s}{50} = \frac{6}{5} \\\\N_s = \frac{50\times 6}{5} \\\\N_s = 60 \ turns[/tex]

Therefore, the number of turns in the secondary coil is 60 turns.

The man's resultant velocity is

8.5 m/s up .

Answer:

1.

Astronomers define star brightness in terms of apparent magnitude — how bright the star appears from Earth — and absolute magnitude — how bright the star appears at a standard distance of 32.6 light-years.

2.

Sirius, also known as the Dog Star or Sirius A, is the brightest star in Earth's night sky. The name means "glowing" in Greek — a fitting description, as only a few planets, the full moon and the International Space Station outshine this star.

Answer:

Sirius is the brightest star in the night sky

Explanation:

The name means 'glowing' in greek

Answer:

C.

Explanation:

Answer:

1.4

Explanation: