Ultraviolet light from a distant star is traveling at 3.0 × 108 m/s. how long will it take for the light to reach earth if it must travel 4.0 × 1013 km? 2.7 × 10–2 hours 2.2 × 103 hours 3.7 × 104 hours 1.3 × 105 hours

Answer:

The correct answer is the Convex lens. An image is formed when a ray of light coming from a point intersects at another point. The image is formed by the real intersection of light. The image is formed by the virtual intersection of Light.

here is the site : textbook.com

Answer:

1/3 kg

Explanation:

Concepts

The formula p = mv tells us that momentum is mass times velocity. We can easily rearrange this equation to solve for m by dividing both sides by v:

Application

We are asked to find the mass of a beach ball, given the velocity as 9 m/s and the momentum as 3 kg · m/s. We can use the formula m = v/p to find the mass of the ball in kilograms.

Solution

Answer:

Explanation:

It is given that five cells of 2V are connected in series, so total voltage of the battery:

[tex]\dashrightarrow \: \: \sf V = 2 \times 5 = 10 V [/tex]

Three resistor of 5[tex]\Omega[/tex], 10[tex]\Omega[/tex], 15[tex]\Omega[/tex] are connected in Series, so the net resistance:

[tex]\dashrightarrow \: \: \sf R_{n} = R_{1} + R_{2} + R_{3}[/tex]

[tex]\dashrightarrow \: \: \sf R = 5 + 10 + 15 [/tex]

[tex]{ \pink{\dashrightarrow \sf \: \: { \underbrace{R = 30 \: \Omega}}}}[/tex]

According to ohm's law:

[tex]\dashrightarrow \sf\: \: V = IR [/tex]

[tex]\dashrightarrow \sf \: \: I = \dfrac{V}{R}[/tex]

On substituting resultant voltage (V) as 10 V and resultant resistant, as 30 [tex]{\pmb{\sf{\Omega}}}[/tex] we get:

[tex]\dashrightarrow \sf \: \: I = \dfrac{10V}{30\Omega}[/tex]

[tex]{ \pink{\dashrightarrow \sf \: \: { \underbrace{I = 0.33 A}}}}[/tex]

[tex]\therefore[/tex]The electric current passing through the above circuit when the key is closed will be 0.33 A

Answer:

Resistance = 5 + 10 + 15 = 30

Explanation:

Current= 0.33 Amperes

Answer:

I think the object with 500 N of force applied.

[tex]\qquad\qquad\huge\underline{{\sf Answer}}[/tex]

As we know, period (t)of a wave and it's frequency (f) are Reciprocal of each other, so if one is know then we can just reverse the order to get the other one.

that is :

[tex]\qquad \sf  \dashrightarrow \: frequency = \dfrac{1}{period} [/tex]

Now, let's move on to problems ~

# 11. f = 15 hertz, t = ?

[tex]\qquad \sf  \dashrightarrow \: t = \dfrac{1}{15} \: \: sec[/tex]

[tex]\qquad \sf  \dashrightarrow \: t \approx0.067 \: \: sec[/tex]

#12. f = 292 hertz, t = ?

[tex]\qquad \sf  \dashrightarrow \: t = \dfrac{1}{292} \: \: sec[/tex]

[tex]\qquad \sf  \dashrightarrow \: t \approx0.0034 \: \: sec[/tex]

#13. t = 12 seconds, f = ?

[tex]\qquad \sf  \dashrightarrow \: f= \dfrac{1}{12} \: \: hertz[/tex]

[tex]\qquad \sf  \dashrightarrow \: f \approx0.083 \: \: hertz[/tex]

We know that, frequency = reciprocal of the period of time an object swings or vibrates at. Therefore,

[tex]\tt\:FORMULA \downarrow\\\\\star\boxed{\mathfrak{Frequency = \frac{1}{period \: of \: time}}}[/tex]

Now, let's solve the questions by using this formula.

[tex]\rule{150pt}{2pt}[/tex]

11. Frequency = 15 Hz

Period = ?

[tex]\sf\:Frequency = \frac{1}{period}\\\sf\:15 = \frac{1}{period}\\\boxed{\bf\:period = \frac{1}{15} = 0.67 \: s}[/tex]

[tex]\rule{150pt}{2pt}[/tex]

12. Frequency = 292 Hz

Period = ?

[tex]\sf\:Frequency = \frac{1}{period}\\\sf\:292 = \frac{1}{period}\\\boxed{\bf\:period = \frac{1}{292} = 0.0034 \: s}[/tex]

[tex]\rule{150pt}{2pt}[/tex]

13. Period = 12 s

Frequency = ?

[tex]\sf\:Frequency = \frac{1}{period}\\\sf\:Frequency = \frac{1}{12}\\\boxed{\bf\:Frequency = 0.083 \: Hz}[/tex]

[tex]\rule{150pt}{2pt}[/tex]

The SI units of,

[tex]\rule{150pt}{2pt}[/tex]

The available energy ea when the proton and the antiproton have equal energies of 450gev is 29 GeV.

The available energy can be found out with the following formula.

[tex]E_a=\sqrt{2E_tE_k++(m_1c^2)^2+(m_2c^2)^2}[/tex]

Here C is the speed of light (3×10⁸ m/s), Et is the energy of proton, Ek is the rest energy.

A proton and an antiproton have equal energies of 450gev. The particles collide head-on. The rest energy of the proton is 938mev

Thus, put the given values in the above formula,

[tex]E_a=\sqrt{2(450\times\dfrac{10^9eV}{1MeV})(938\times\dfrac{10^6eV}{1MeV})+(938\times\dfrac{10^6eV}{1MeV})^2+(938\times\dfrac{10^6eV}{1MeV})}\\E_a=29\rm\; Gev[/tex]

Thus, the available energy ea when the proton and the antiproton have equal energies of 450gev is 29 GeV.

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The downward force on the 2. 67-cm piston that is required to lift a mass of 2000 kg supported by the 20-cm piston is 350 N.

Pressure applied to a closed system of fluid will be transfer at each point of the fluid and the boundaries of the system.

Let suppose at the two point of such system the input force apllied is F₁ and output force we get is F₂. The area of this points is A₁ and A₂ respectively. As the pressure at both point is same. Then by the pascal law,

[tex]\dfrac{F_1}{A_1}=\dfrac{F_2}{A_2}\\[/tex]

In case of radius,

[tex]\dfrac{F_1}{\pi r_1^2}=\dfrac{F_2}{\pi r_2^2}\\\\\dfrac{F_1}{r_1^2}=\dfrac{F_2}{ r_2^2}[/tex]

Two pistons of a hydraulic lift have radii of 2. 67 cm and 20. 0 cm, respectively. Thus,

[tex]r_1=2.67\rm\; cm[/tex]

[tex]r_2=20.0\rm\; cm[/tex]

The downward force on the 2. 67-cm piston that is required to lift a mass of 2000 kg supported by the 20-cm piston is The force applied on second piston is,

[tex]F_2=mg\\F_2=2000\times9.81\\F_2=19620 \rm\; N[/tex]

Put the values in the above formula,

[tex]\dfrac{F_1}{2.67^2}=\dfrac{19620}{20^2}\\F_1=\dfrac{19620\times2.67^2}{20^2}\\F_1\approx 350\rm\; N[/tex]

Thus, the downward force on the 2. 67-cm piston that is required to lift a mass of 2000 kg supported by the 20-cm piston is 350 N.

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The value of distance covered by the particle during the first 3 second which moves along a line so that its acceleration is 12t, is 54 ft.

Acceleration of an object is the rate of change of velocity of the object per unit time.

A particle moves along a line so that at time t0 its acceleration is given by [tex]a(t) = 12t\rm\; ft/sec^2[/tex]

The value of distance covered by the particle during the first 3 second has to be obtained.

Acceleration is the change in velocity (v) per unit time. Thus, the above equation can be written as,

[tex]a(t) = 12t\rm\; ft/sec^2\\\dfrac{dv}{dt}= 12t\rm\; ft/sec^2[/tex]

The equation can be solved as further.

[tex]\dfrac{dv}{dt}= 12t\\v= 6t^2+v_0\rm[/tex]

Here, the vo is the initial velocity which is equal to zero.

[tex]v=6t^2[/tex]

Velocity is the change in distance(s) per unit time. Thus, the above equation can be written as,

[tex]\dfrac{ds}{dt}=6t^2\\\dfrac{ds}{dt}=2t^3+s_0[/tex]

Here, the so is the initial position which is equal to zero. Thus,

[tex]s=2t^3[/tex]

Put the value of time t=3.

[tex]s=2(3)^3\\s=54\rm\; ft[/tex]

Thus, the value of distance covered by the particle during the first 3 second which moves along a line so that  its acceleration is given by a(t) = 12t is 54 ft.

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Answer:

it would have to be 1 or maybe 2

The formulas used to analyze the horizontal and vertical motion of projectiles launched at an angle involve the use of tangent, cosine and sine.

The vertical motion of a projectile is affected by gravity and the velocity of vertical motion given by the following formula;

Vy = Vsinθ

The horizontal velocity of a projectile is given by the following formula;

Vx = Vcosθ

The direction of the motion is calculated as follows;

tanθ  = Vy/Vx

Thus, the formulas used to analyze the horizontal and vertical motion of projectiles launched at an angle involve the use of tangent, cosine and sine.

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Answer:

d

Explanation:

edg 2022

The acceleration of the center of mass of the blocks-string-pulley system has magnitude equal to the ratio of difference to the sum of masses of block multiplied with gravitational acceleration.

The center of mass of a object or for a system of object is the point, where the center of distribution of mass in space. The center of mass for the two bodies can be given as,

It can be given as,

[tex]x_{cm}=\dfrac{m_1x_1+m_2x_2}{m_1+m_2}[/tex]

Here, (m) represents the mass and (x) represents the distance.

The acceleration of the center of mass of the blocks-string-pulley system has magnitude of a_cm has to be find out.

The image of the blocks-string-pulley system is attached below. The acceleration of this system is,

[tex]\vec a_{cm}=\left(\dfrac{m_1-m_2}{m_1+m_2}\right)g[/tex]

Thus, the acceleration of the center of mass of the blocks-string-pulley system has magnitude equal to the ratio of difference to the sum of masses of block multiplied with gravitational acceleration.

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Resistance for 165 metres is 3 ohms.

Final speed of the tennis ball, moving with a speed of 5. 82 m/s , has a head-on collision with a 0. 090-kg ball is 2.964 m/s.

Momentum of an object is the force of speed of it in motion. Momentum of a moving body is the product of mass times velocity. By the law of conservation of momentum,

[tex]m_1u_1 + m_2u_2 = (m_1+m_2)v[/tex]

Here, (m) is the mass, (u) is initial velocity before collision, v is final velocity after collision and (subscript 1, and 2) are used for body 1 and 2 respectively. Rewrite the formula for final velocity as,

[tex]v=\dfrac{m_1u_1 + m_2u_2}{(m_1+m_2)}[/tex]

A 0. 060-kg tennis ball, moving with a speed of 5. 82 m/s, has a head-on collision with a 0. 090-kg ball, initially moving in the same direction at a speed of 3.44 m/s. Thus, the initial velocity of the second ball is,

[tex]v_{2f}=5.82+3.44+v_{1f}\\v_{2f}=2.38+v_{1f}[/tex]

Let v1f is the final velocity of first ball. Thus, the initial velocity of the first ball is,

[tex]v_{1f}=\dfrac{(0.060)(5.82) + (0.090)(3.44-2.38)}{(0.060)+(0.090)}\\v_{1f}=2.964\rm\; m/s[/tex]

Thus, final speed of the tennis ball, moving with a speed of 5. 82 m/s , has a head-on collision with a 0. 090-kg ball is 2.964 m/s.

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The minimum energy of the two oppositely directed gamma rays in a PET procedure is : 0.51 MeV

In a PET procedure when an electron and positron collide the energy equivalent of their individual masses are converted into gamma rays which will move in directions opposite to each other.

Given that an electron and a positron posses equal masses the minimum energy of the gamma ray will be 0.51MeV

Hence we can conclude that the minimum energy of the two oppositely directed gamma rays in a PET procedure is 0.51MeV.

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Answer:

4 m/s

Explanation:

you moved 140 meters in 35 seconds

v = 140 / 35 = 4 m/s

Answer:

b 1000

if you multiply 2m/s by 500 you get 1000 it should look like this.

500 x 2m/s

Answer:

More Than 1.3 million.

Hope it helps :))

When the narrow beam of ultrasound waves reflects off a liver tumor, the depth of the tumor is found to be 6.30 cm.

Snell's law states that the ratio of the sin of the angle of incidence to the angle of reflection is equal to the ratio of the speed of the wave in the first medium to the speed of the wave in the second medium.

[tex]\dfrac{sin\:i}{sin\:r} =\dfrac{v_{1} }{v_{2} }[/tex]

Where, i represents the angle of incidence and r is the angle of reflection.

Given the speed of sound in the liver is 10% less than in the medium, so

[tex]\dfrac{v_{liver} }{v_{medium} }=0.900[/tex]

According to the figure, the angle of incidence is 50°, then from the Snell's law, we have

[tex]sin\:r=(\dfrac{v_{liver} }{v_{medium} } )\times sin\:i\\\\r=sin^{-1} [(0.900)sin 50][/tex]

The angle of reflection comes out to be

r = 43.6°

From the law of reflection,

[tex]d =\dfrac{12\:\rm cm}{2} =6\:\rm cm[/tex]

The depth of the tumor is

[tex]h=\dfrac{d}{tan\:r} =\dfrac{6}{tan\:43.6} \\\\h=6.30\:\rm cm[/tex]

The depth of the tumor in the liver is 6.30 cm.

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Motion is a change in position measured by distance and time.

Answer:

Cat with 6 kg sitting on a 30 meter hill has greater potential energy.

Explanation:

gravitational potential energy (J) = mass (kg) × gravitational field × height (m)

gravitational field in Earth is 10 N/kg = fixed unit

1st Cat: 4 kg

2nd Cat: 6 kg

Both of their height is 30 meters.

1st cat = mgh = 4 × 10 × 30 = 1200 Joules

2nd cat = mgh = 6 × 10 × 30 = 1800 Joules

Answer:

If the length of the pendulum string increases, the period of time increases

Explanation:

The period of a simple pendulum is directly proportional to the square root of length of the pendulum

Answer:

reflection

Explanation:

Answer: Reflection

Explanation: I have my ways :>

Answer:

.60 A

Explanation:

The RMS (Root Mean Square) voltage (the effective voltage) of the  PEAK voltage  is  .7 * peak

.7 * 425 = 297.5 v

Now,   V = IR        re-arranged   to   V/R = I

                                                         297.5 v  /  500 ohm = .595 A = ~ .6 A

Due to concentration difference, electric charge flows from a region of high electric potential to a region of low electric potential.

Electric charge flows from a region of high electric potential to a region of low electric potential because of concentration difference. This movement is same as the diffusion in which molecules move from high concentration to lower concentration.

So we can conclude that due to concentration difference, electric charge flows from a region of high electric potential to a region of low electric potential.

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When white light shines on a thin film of soap, light is reflected from back surface of the film only.

Doppler Shift is the change in frequency of sound based on the observer’s point of view.

When white light falls on a thin film of soap, light gets reflected and refracted inside the soap. This results in refraction of the reflected light rays.

When white light shines on a thin film of soap, light is reflected from the back surface of the film only.

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The universe is the collection of galaxies, The Andromeda Galaxy has spectral lines with blue shift. The conclusion is that the Universe has stopped expanding.

The Andromeda Galaxy is the nearest spiral neighbor that has spectral lines showing a blue shift.

Therefore, this concludes that, the Universe has stopped expanding. This galaxy is slowly shifting towards us.

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The period of the pendulum of length 3. 171 m when acceleration of gravity is 9. 832 m/s, is 3.57 seconds.

Pendulum is the body which is pivoted to a point and perform back and forth motion around that point by swinging due to the influence of gravity.

The time period of a pendulum is the time taken by it to complete one cycle of swing left to right and right to left.

It can be given as,

[tex]T=2\pi \sqrt{\dfrac{L}{g}}[/tex]

Here, (g) is the gravitational force of Earth and (L) is the length of the pendulum.

A pendulum of length 3.171 m. The acceleration of gravity is 9.832 m/s2. The period at the north pole is,

[tex]T=2\pi \sqrt{\dfrac{3.171}{9.832}}\\T=3.57\rm\; s[/tex]

Thus, the period of the pendulum of length 3. 171 m when acceleration of gravity is 9. 832 m/s, is 3.57 seconds.

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Answer:

frost wedging

Explanation:

When glaciers melt and the rock beneath is weathered and broken down, this is called frost wedging.

Answer:

mb

Explanation:

m b