If 2000 kg cannon fires 2 kg projectile having muzzle velocity 200 m/s then the KINETIC ENERGY of the projectile will be ("E4" means "*10^4") *
1 point
4E4 J
1E4 J
1E3 J
None of the above

The speed at aphelion is mathematically given as

v2 = 21.43 km/s

Question Parameter(s):

At perihelion a planet in another solar system is 175 x 106 km from its sun at traveling at 40 km/s.

At aphelion it is 250 x 106 km distant.

Generally, the equation for the angular momentum conservation is mathematically given as

I1*w1 = I2*w2

Therefore

(0.5*m*R1^2)*v1/R1 = (0.5*m*R2^2)*v2/R2

Where

v1*R1 = v2*R2

v2 = v1*(R1/R2) = (30*10^3)*(2.50*10^11)/(3.50*10^11)

v2 = 2.1428*10^4 m/s

v2 = 21.43 km/s

In conclusion, speed is

v2 = 21.43 km/s

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Answer:

Below

Explanation:

0-1 sec descends at constant rate from 10 to 6 m

1-2 sec stops at 6m

2-3 sec descends at constant rate to 2 m

3-4 sec stops at 2 m

4-5 sec descends at another constant rate to 0 m

To calculate the resultant force on the aircraft as it accelerates from the catapult, we can use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy.

Since all of the kinetic energy at take-off is from the work done on the aircraft by the catapult, we can equate the work done by the catapult to the kinetic energy of the aircraft at take-off, that is:

Work done by catapult = kinetic energy of aircraft at take-off

The work done by the catapult is given by the force provided by the catapult multiplied by the distance over which it acts, that is:

Work done by catapult = force × distance

The distance over which the force provided by the catapult acts is given as 150 m, so we have:

Work done by catapult = force × 150

The kinetic energy of the aircraft at take-off is given by:

(1/2) × mass × velocity^2

Since the aircraft is initially at rest, its initial velocity is zero, so we have:

kinetic energy of aircraft at take-off = 0.5 × mass × (final velocity)^2

Using the work-energy principle, we can equate the two expressions for work done and kinetic energy, that is:

force × 150 = 0.5 × mass × (final velocity)^2

Solving for force, we get:

force = 0.5 × mass × (final velocity)^2 / 150

Therefore, the resultant force on the aircraft as it accelerates is given by:

force = 0.5 × mass × (final velocity)^2 / 150

Note that we need to know the mass and final velocity of the aircraft in order to calculate the resultant force.

this is the correct answer

Answer:

To find the drift velocity, we can use the equation:I = nAvqWhere I is the current, n is the number of charge carriers per unit volume, A is the cross-sectional area of the wire, v is the drift velocity of the charge carriers, and q is the charge of an electron.We are given the electric field E as 4.5×10^−4 V/m. The current I is 2.8×10^17 electrons/s. The diameter of the wire is 1.9 mm, so the radius r is 0.95 mm or 0.00095 m. The cross-sectional area A is then πr^2 = 2.83×10^-6 m^2.The charge of an electron is q = 1.6×10^-19 C.We can rearrange the equation to solve for v:v = I/(nAq)To find n, we can use the density of aluminum and its atomic weight to calculate the number of atoms per unit volume, and then multiply by the number of electrons per atom:n = (density of aluminum * Avogadro's number) / (atomic weight of aluminum * volume per atom) * number of electrons per atomPlugging in the values, we get:n = (2.7 g/cm^3 * 6.02×10^23) / (26.98 g/mol * 4.05×10^-23 m^3/mol) * 3 = 1.85×10^29 electrons/m^3Now we can calculate the drift velocity:v = I/(nAq) = (2.8×10^17 electrons/s) / (1.85×10^29 electrons/m^3 * 2.83×10^-6 m^2 * 1.6×10^-19 C/electron) = 0.0025 m/sSo the drift velocity is 0.0025 m/s.To find the mean time between collisions, we can use the equation:τ = mv/eEwhere τ is the mean time between collisions, m is the mass of an electron, v is the drift velocity, e is the charge of an electron, and E is the electric field.The mass of an electron is 9.11×10^-31 kg. Plugging in the values, we get:τ = (9.11×10^-31 kg) * (0.0025 m/s) / (1.6×10^-19 C) / (4.5×10^-4 V/m) = 3.54×10^-14 sSo the mean time between collisions is 3.54×10^-14 s.

The ball will strike the ground after 8.04 seconds.

Using the kinematic equation for displacement:

y = y0 + v0t - 1/2g*t^2

where:

We want to findt, let's substitute the values:

0 = 58.52 + 19.50t - 1/2(9.81)*t^2

4.905t^2 - 19.50t - 58.52 = 0

Using the quadratic formula:

t = (-b ± sqrt(b^2 - 4ac)) / 2a

where:

t = (-(-19.50) ± sqrt((-19.50)^2 - 4(4.905)(-58.52))) / 2(4.905)

t = (19.50 ± 31.37) / 9.81

The two possible solutions are:

t1 = 5.61 s (ball on the way up)

t2 = 8.04 s (ball on the way down)

Since the question is asking for when the ball strikes the ground, we take the larger solution, which is:

t = 8.04 s

In other words, the ball will strike the ground after 8.04 seconds.

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The object's distance from the lens is 11.4 cm.

The thin lens equation can be used to determine how a converging lens's object distance (p), image distance (q), and focal length (f) relate to one another:

1/p + 1/q = 1/f

where p denotes the distance to the object, q is the distance to the picture, and f is the focal length.

Let's solve for the object distance using the thin lens equation:

1/p + 1/59.6 = 1/9.56

Combining both sides with p59.69.56 results in:

59.69.56 + p9.56 = p*59.6

Adding and subtracting:

570.176 + 9.56p = 59.6p

50.04p = 570.176

p = 11.4 cm

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A solid, uniform sphere of mass M and radius R revolving about a diameter has a moment of inertia of (2/5) * M * R2.

I=21MR2 is the ring's moment of inertia with respect to the Z-axis. Ix=Iy because the disc's uniformity makes all of its distances equal, and Iz=Ix+Iy because of the perpendicular axes theorem, which leads to Ix=2I=4MR.

In linear motion, the moment of inertia plays the same role as mass does. The resistance of a body to a change in its rotaional motiton is measured in this way. For a particular rigid frame and rotational axis, it is fixed.

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Answer:

Explanation:

Force = mass*acceleration

15=mass*5

mass=3kg

Answer:

The starting velocity.

Explanation:

We must understand that this equation comes from the following equation of kinematics.

\(v_{f}=v_{o}+a*t\)

where:

Vf = final velocity = 33 [m/s]

Vo = starting velocity [m/s]

a = acceleration = 3 [m/s²]

t = time = 30 [s]

So, these values can be assembly in the following way:

\(v_{f}=v_{o}+a*t\\a*t=v_{f}-v_{o}\\3=\frac{33-v_{o}}{30}\)

Answer:

53.8Joule

Explanation:

hope it is helpful

Answer:

approximate 153.1J

Explanation:

W= 1/2k(x^2) = 1/2x725x(0.650)^2 = 153.15625 (J)

a) The total surface area of tin G in terms of π is 170π cm².

b) The mass of tin H is 336 g.

To solve the given problem, we need to determine the total surface area of tin G in terms of π and the mass of tin H. Since the mass of an empty cylindrical tin is proportional to its surface area, we can use the given information to find the solutions.

a) Total surface area of tin G in terms of π:

The surface area of a cylinder consists of two circular bases and the lateral surface area. The formula for the lateral surface area of a cylinder is given by:

Lateral surface area = 2πrh

where r is the radius of the base and h is the height of the cylinder.

In the case of tin G, the given dimensions are a radius of 5 cm and a height of 12 cm. Substituting these values into the formula, we can calculate the lateral surface area:

Lateral surface area = 2π(5 cm)(12 cm)

Lateral surface area = 120π cm²

Since the total surface area of the cylinder includes the two circular bases as well, we need to add their areas. The area of a circle is given by:

Area of a circle = πr²

The radius of the circular base of tin G is 5 cm, so the area of each circular base is:

Area of each circular base = π(5 cm)²

Area of each circular base = 25π cm²

To find the total surface area of tin G, we sum the lateral surface area and the areas of the two circular bases:

Total surface area of tin G = Lateral surface area + 2 × Area of each circular base

Total surface area of tin G = 120π cm² + 2 × 25π cm²

Total surface area of tin G = 120π cm² + 50π cm²

Total surface area of tin G = 170π cm²

Therefore, the total surface area of tin G in terms of π is 170π cm².

b) Mass of tin H:

We are given that the surface area of tin H is 792π cm². We can assume that the same proportionality factor applies as in tin G, so we can set up the following proportion:

(surface area of tin G) / (mass of tin G) = (surface area of tin H) / (mass of tin H)

Using the given values, we have:

(170π cm²) / (72 g) = (792π cm²) / (mass of tin H)

Cross-multiplying and solving for the mass of tin H, we get:

(170π cm²) × (mass of tin H) = (72 g) × (792π cm²)

mass of tin H = (72 g) × (792π cm²) / (170π cm²)

mass of tin H = 336 g

Therefore, the mass of tin H is 336 g.

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Answer:

all of the above I think??

The absolute pressure at an ocean depth of 1.0 x 10^3 m is 1.002 x 10^8 Pa.

Hydrostatic pressure is the pressure that a fluid exerts on a surface due to the weight of the fluid above it. It is the result of the force of gravity acting on a column of fluid, and it is directly proportional to the height of the column of fluid and the density of the fluid.

The absolute pressure at an ocean depth of 1.0 x 10^3 m can be calculated using the hydrostatic pressure equation:

P = ρgh + Po

where:

P is the absolute pressure at the given depth

ρ is the density of the water

g is the acceleration due to gravity (assumed to be 9.81 m/s²)

h is the depth of the ocean

Po is the atmospheric pressure at the surface (assumed to be 1.01 x 10^5 Pa)

Substituting the given values, we get:

P = (1.025 x 10^3 kg/m³) x (9.81 m/s²) x (1.0 x 10^3 m) + 1.01 x 10^5 Pa

P = 1.025 x 9.81 x 10^6 Pa + 1.01 x 10^5 Pa

P = 1.002 x 10^8 Pa.

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If a type ii cepheid has an apparent magnitude of 12 and a pulsation period of 3 days. the  approximate distance to this cepheid variable star is 0.14 parsecs.

To determine the distance to a Cepheid variable star, you can use the period-luminosity relationship, which states that the intrinsic brightness (luminosity) of a Cepheid is directly proportional to the length of its pulsation period. In other words, the longer the period, the brighter the Cepheid.

To use this relationship to determine the distance to a Cepheid, you need to know its apparent magnitude (how bright it appears from Earth) and its pulsation period. You also need a reference value for the luminosity of a Cepheid with a known period, which can be obtained from observations of nearby Cepheid variables.

Using these values, you can calculate the distance to the Cepheid using the following formula:

distance = 10 ^ ((reference luminosity - apparent magnitude + 5) / 5)

Plugging in the values you provided, we get:

Distance = 10 ^ ((-2.5 - 12 + 5) / 5)

= 10 ^ (-9.5 / 5)

= 10 ^ (-1.9)

= 0.14 parsecs

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Sound waves are a type of mechanical wave that propagate through a medium, typically air but also other materials such as water or solids.

Frequency: the frequency of a sound wave refers to the number of cycles or vibrations it completes per second and is measured in Hertz (Hz).

Amplitude: the amplitude of a sound wave refers to the maximum displacement or intensity of the wave from its equilibrium position. It represents the loudness or volume of the sound, with larger amplitudes corresponding to louder sounds and smaller amplitudes corresponding to softer sounds.

Wavelength: the wavelength of a sound wave is the distance between two consecutive points in the wave that are in phase, such as from one peak to the next or one trough to the next. It is inversely related to the frequency of the wave.

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Answer:

0.002 m/s

Explanation:

27.13(60) = 1,627.8 seconds

3.12/1,627.8 = 0.00191 ≈ 0.002 = s

Are you sure you're not looking for cm/s?

Answer:

The mass will stay the same if the bag stays sealed.

Explanation:

The  speed of the car after 20 seconds are  20 m/s.

The speed of an object is a measure of how quickly the object is moving in a particular direction. It is typically measured in units such as meters per second (m/s). Speed is a scalar quantity, which means it is only concerned with the magnitude of the velocity (or speed) and not the direction

The graph shows that the speed of the car was 20 m/s after 20 seconds.

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Here is your answer mate,

Question,

\(Determine\: the\: amount\\ \: of\: power\:used\: in\\\: holding\: a\: 25\: kg\: box\:\\ , \: 1.5\: meters \: above\: the\: floor\\\: for\: 60\: seconds\)

Answer,

\(\)

Solution,

\(\)

Given,

\(MASS \: IS\: 25\: KG\: \\ and \: HEIGHTIS\: 1.5m\: \)

\(\)

WORK DONE (done against gravity) =

mass×acceleration due to gravity ×height

WORK = 25× 10× 1.5

\(\: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \)= 375 Nm

\(\)

Now

POWER =

\( \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\frac{work}{time} \)

POWER

\(\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:= \frac{375}{60} Watt \)

\(\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: =6.25\)

\(Therfore\: your \: answer\: is\: 6.25\)

\(\)

Check this,

\(Acceleration\: due\: to \: gravity\\\: can\: be\: 9.8\: m/s²\: \\As\: nothing\: mentioned\\\: in\: question\: \\I \: took \: it \: as \: 10\)

\(\)

Have a good day

(a) To find the horizontal distance d that the ball travels before landing, we can use the equation d = (Vx)t, where Vx is the horizontal velocity of the ball and t is the time it takes for the ball to fall from the ramp to the floor.

First, we need to find the horizontal velocity of the ball when it leaves the ramp. Since the ball is rolling without slipping, we can use the equation V = ωR, where V is the linear velocity, ω is the angular velocity, and R is the radius of the ball. The radius of the ball is 0.20 m / 2 = 0.10 m.

Next, we need to find the angular velocity of the ball. We can use the equation ω = √(2gh/R), where g is the acceleration due to gravity (9.8 m/s^2), h is the vertical height that the ball drops (0.49 m), and R is the radius of the ball (0.10 m). Plugging in the values gives us:

ω = √(2(9.8)(0.49)/0.10) = 31.3 rad/s

Now we can find the horizontal velocity of the ball:

V = ωR = (31.3)(0.10) = 3.13 m/s

Finally, we can find the time it takes for the ball to fall from the ramp to the floor using the equation h = 0.5gt^2, where h is the vertical height (1.31 m) and g is the acceleration due to gravity (9.8 m/s^2). Rearranging the equation and plugging in the values gives us:

t = √(2h/g) = √(2(1.31)/9.8) = 0.52 s

Now we can find the horizontal distance d that the ball travels before landing:

d = (Vx)t = (3.13)(0.52) = 1.63 m

Answer: The ball travels 1.63 m before landing.

(b) To find the number of revolutions that the ball makes during its fall, we can use the equation θ = ωt, where θ is the angular displacement, ω is the angular velocity, and t is the time. The angular displacement is equal to the number of revolutions times 2π, so we can rearrange the equation to solve for the number of revolutions:

θ / 2π = ωt / 2π = (31.3)(0.52) / 2π = 2.59 rev

Answer: The ball makes 2.59 revolutions during its fall.

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Answer:

Carbon dioxide and water vapour

Explanation:

So the products of a combustion reaction are primarily:

carbon dioxide + water vapour, however other gases such as nitrogen, methane, and sulphur dioxide are also produced in smaller concentrations.

Carbon dioxide and water vapour are the main byproducts

Answer:

Carbon dioxide

Explanation:

The major byproduct of fossil fuel combustion is carbon dioxide. When fossil fuels such as coal, oil, and natural gas are burned, they release carbon dioxide into the atmosphere. This is because fossil fuels are made up of hydrocarbons, which are compounds made up of carbon and hydrogen. When these compounds are burned, they react with oxygen in the air to produce carbon dioxide \(\rm (CO_2)\) and water vapor \(\rm (H_2O)\).

Methane is also produced during fossil fuel combustion, but in smaller amounts compared to carbon dioxide. Sulfuric acid is not a byproduct of fossil fuel combustion, but rather a product of the reaction between sulfur dioxide \(\rm (SO_2)\) and water vapor in the atmosphere. While water vapor is also produced during fossil fuel combustion, it is not considered a major byproduct, as it is a natural component of the air and atmosphere.

The skater's final angular velocity is approximately 9.86 rad/s.

The skater's final angular velocity can be calculated using the principle of conservation of angular momentum. The equation for angular momentum is given by:

L = Iω

where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

Initially, the skater has an angular momentum of:

L_initial = I_initial * ω_initial

Substituting the given values:

L_initial = 2.12 kg m² * 3.25 rad/s

The skater's final angular momentum remains the same, as angular momentum is conserved:

L_final = L_initial

The final moment of inertia is given as 0.699 kg m². Therefore, the final angular velocity can be calculated as:

L_final = I_final * ω_final

0.699 kg m² * ω_final = 2.12 kg m² * 3.25 rad/s

Solving for ω_final:

ω_final = (2.12 kg m² * 3.25 rad/s) / 0.699 kg m²

Hence, the skater's final angular velocity is approximately 9.86 rad/s.

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Answer:

The ancient Greeks counted the Earth's Moon and Sun as planets along with Mercury, Venus, Mars, Jupiter, and Saturn. Earth was not considered a planet, but rather was thought to be the central object around which all the other celestial objects orbited.   ......     Jupiter is composed of gases — hydrogen and helium, mostly — all ... the largest planet in the solar system, the gas giant just doesn't have ... But unlike the sun, it lacks the necessary amount to begin fusion, the process that fuels a star. ... to 80 times more massive than it is at present to be considered a star.

Answer:

Explanation:

The empirical equation y = 9.8x represents the relationship between the displacement y of an object and the time x it has been falling under the influence of gravity.

On the other hand, the equation V = gT + V0 represents the relationship between the velocity V of an object, the time T, the initial velocity V0, and the acceleration due to gravity g.

To compare the two equations, we can equate the displacement y in the first equation with the expression for displacement in terms of velocity and time, which is y = (1/2)gt^2 + V0t, where t is the time.

Substituting this into the empirical equation, we get:

9.8x = (1/2)gt^2 + V0t

We can see that this equation has three variables: g, V0, and t. We can't determine all three variables from this equation alone.

However, if we know the time it takes for an object to fall a certain distance, we can use this equation to solve for g and V0. For example, if we know that an object falls 1 meter in 0.45 seconds, we can substitute x=1 and t=0.45 into the equation:

9.8(1) = (1/2)g(0.45)^2 + V0(0.45)

Simplifying this equation, we get:

g = 19.62 m/s^2

V0 = 0.45(9.8) = 4.41 m/s

So the acceleration due to gravity is 19.62 m/s^2 and the initial velocity is 4.41 m/s. Note that these values may not be exactly equal to the true values, as the empirical equation y=9.8x is only an approximation and there may be other factors affecting the motion of the object.

Answer:

what is heat and transfer