An airplane was 300 km [N] of Toronto airport, 3 hours later the airplane 600 km [S]
of Toronto airport, calculate.
a. The airplane’ total displacement relative the Toronto airport.
b. The airplane average velocity.

Answer:

1 L = 1000 cm^3

1 m^3 = 1000 L = 10^6 cm^3

1 m^3 = 10^6 g = 10^3 kg  density of water

M = Density * Volume

V = 2 * 4 * 5 = 40 m^3

M = 10^3 kg / m^3 * 40 m^3 = 4.0 * 10^4 kg

Answer:

The moment of inertia of the motor is 0.0823 Newton-meter-square seconds.

Explanation:

From Newton's Laws of Motion and Principle of Motion of D'Alembert, the net torque of a system (\(\tau\)), measured in Newton-meters, is:

\(\tau = I\cdot \alpha\) (1)

Where:

\(I\) - Moment of inertia, measured in Newton-meter-square seconds.

\(\alpha\) - Angular acceleration, measured in radians per square second.

If motor have an uniform acceleration, then we can calculate acceleration by this formula:

\(\alpha = \frac{\omega - \omega_{o}}{t}\) (2)

Where:

\(\omega_{o}\) - Initial angular speed, measured in radians per second.

\(\omega\) - Final angular speed, measured in radians per second.

\(t\) - Time, measured in seconds.

If we know that \(\tau = 3\,N\cdot m\), \(\omega_{o} = 0\,\frac{rad}{s }\), \(\omega = 145.875\,\frac{rad}{s}\) and \(t = 4\,s\), then the moment of inertia of the motor is:

\(\alpha = \frac{145.875\,\frac{rad}{s}-0\,\frac{rad}{s}}{4\,s}\)

\(\alpha = 36.469\,\frac{rad}{s^{2}}\)

\(I = \frac{\tau}{\alpha}\)

\(I = \frac{3\,N\cdot m}{36.469\,\frac{rad}{s^{2}} }\)

\(I = 0.0823\,N\cdot m\cdot s^{2}\)

The moment of inertia of the motor is 0.0823 Newton-meter-square seconds.

Answer:

gravitational potenetial energy =ℎ

Explanation:

none

Answer:

0.00129rad/s

Explanation:

The angular velocity is expressed as;

v = wr

w is the angular velocity

r is the radius

Given

v =  20,000 mph

r = 4300mi

Get w;

w = v/r

w = 20000* 0.44704/4300*1609.34

w = 8940.8/6,920,162

w = 0.00129rad/s

Hence the angular velocity generated is 0.00129rad/s

a) The flux is, Φ = 16.4 N·m²/C

b) If n is perpendicular to E then, Φ = 0

c) If parallel to E then, Φ = 62.8 N·m²/C

The flux through a surface is given by the equation:

Φ = E * A * cos(θ)

where E is the electric field strength, A is the area of the surface, and θ is the angle between the electric field and the surface normal.

a) If the normal vector to the surface is at 30 degrees to E, then the flux would be:

Φ = (2.0 x \(10^3\) N/C) * π*(0.10 m)² * cos(30°) = 16.4 N·m²/C

b) If n is perpendicular to E, then the angle between them is 90 degrees and cos(90°) = 0. Therefore, the flux would be zero.

Φ = (2.0 x \(10^3\) N/C) * π*(0.10 m)² * cos(90°) = 0

c) If n is parallel to E, then the angle between them is 0 degrees and cos(0°) = 1. Therefore, the flux would be:

Φ = (2.0 x \(10^3\) N/C) * π*(0.10 m)² * cos(0°) = 62.8 N·m²/C

Note that in this case, the flux is maximum because the electric field is perpendicular to the surface.

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Explanation:

the answer is 1835N that how best I can help

Answer:

am pretty sure u have to measure like my name is Justin am cracked at fortnite my Guy

Answer:

umm of which subject ?? science ?? physics?? chemistry?? biology??ooook

Therefore, the volume of the ore is 0.0069 cubic meters.

Volume is a measure of the amount of space that an object or substance occupies. It is a fundamental physical quantity that describes the three-dimensional size of an object or space. The SI unit of volume is the cubic meter (m³), but other units such as liters, gallons, and cubic centimeters (cm³) are also commonly used. Volume can be calculated for regular shapes, such as cubes or cylinders, using mathematical formulas, but for irregular shapes, more complex methods, such as displacement, may be needed.

Here,

To calculate the volume of the ore, we need to convert all the measurements to the same unit. We can choose to convert everything to meters since that's the base unit for volume (cubic meters).

8.8 dm = 0.88 m (1 dm = 0.1 m)

92.4 mm = 0.0924 m (1 mm = 0.001 m)

0.007 dam = 0.07 m (1 dam = 10 m)

Now we can use the formula for volume of a rectangular block:

Volume = length x width x height

Volume = 0.88 m x 0.0924 m x 0.07 m

Volume = 0.0069 cubic meters

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The resultant of the displacement is 336.5m

The process of splitting a vector into its components is called resolution of the vector. The vectors are splitted into vertical and horizontal component.

For the first displacement;

The vertical component = - 150 sin45 = -106.1 m

The horizontal component = - 150 cos 45° = -106.1 m

For the second displacement;

The vertical displacement = - 85sin90 = -85

The horizontal component = 0

For the third displacement;

The vertical displacement = 550 sin55 = 450.5

The horizontal displacement = 550 cos 55 = 315.5

Sum of vertical component = 450.5-85-106.1 = 263.4

sum of horizontal component = 315.5 -106.1 = 209.4

Using Pythagorean theorem

R = √ 263.4² + 209.4²

R = √113227.92

R = 336.5m

The resultant angle = tan^-1( 263.4/209.4)

= tan^-1(1.26)

= 51.56°

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The type of power plant burns material to make electricity is Fossil fuel. The correct option is C.

Energy is the ability to do work.

There are different kind of energies like kinetic energy, potential energy, gravitational energy and  heat energy.

The energy obtained from burning different type of materials like Coal, gas and oil and produces heat.

Fossil fuel power plants burn coal or oil to create heat and generate steam to run turbines which generate electricity.

Thus, the correct option is C.

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Using the conversion factor that we have seen in the solution, the value obtained is 40.25 kilometers per hours

To convert miles per hour (mph) to kilometers per hour (kph), you need to multiply the speed value in mph by a conversion factor of 1.61, which is the number of kilometers in one mile.

Given that;

1 miles/hour = 1.61 kilometer per hour

25 miles per hour = 25 * 1.61/1

= 40.25 kilometers per hours

Hence the value that we get is 40.25 kilometers per hours

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The convert unit of  25 miles per hour is equivalent to 40.2335 kilometers per hour

Unit conversion is the process of converting a value expressed in one unit of measurement to another unit of measurement that is equivalent in terms of its value or quantity. This is done by using a conversion factor, which is a numerical factor that relates the two units of measurement.

To convert miles per hour (mph) to kilometers per hour (km/h), we need to multiply the speed in mph by 1.60934, which is the conversion factor from miles to kilometers.

So, to convert 25 mph to km/h:

25 mph × 1.60934 = 40.2335 km/h

Therefore, 25 miles per hour is equivalent to 40.2335 kilometers per hour.

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Answer:

(1.8 × 10^9) meters in one minute

Explanation:

To determine how far light can travel in one minute, we need to multiply its speed by the number of seconds in a minute.

The speed of light is 3 × 10^8 meters per second.

There are 60 seconds in a minute.

Therefore, the distance light can travel in one minute is:

Distance = Speed × Time

Distance = (3 × 10^8 meters per second) × (60 seconds)

Calculating this, we get:

Distance = 3 × 10^8 meters/second × 60 seconds

Distance = 18 × 10^8 meters

Distance = 1.8 × 10^9 meters

So, light can travel approximately 1.8 billion (1.8 × 10^9) meters in one minute.

Answer:

~4%

Explanation:

% = |(7.6 - 7.9)|/7.9

= 0.3/7.9 ≈ 0.04 = 4%

Answer:

Refer to the step-by-step Explanation.

Step-by-step Explanation:

Simplify the equation with given substitutions,

Given Equation:

\(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2\)

Given Substitutions:

\(\omega=v/R\\\\ \omega_{_{0}}=v_{_{0}}/R\\\\\ I=(2/5)mR^2\)\(\hrulefill\)

Start by substituting in the appropriate values: \(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2 \\\\\\\\\Longrightarrow mgh+(1/2)mv^2+(1/2)\bold{[(2/5)mR^2]} \bold{[v/R]}^2=(1/2)mv_{_{0}}^2+(1/2)\bold{[(2/5)mR^2]}\bold{[v_{_{0}}/R]}^2\)

Adjusting the equation so it easier to work with.\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the left-hand side of the equation:

\(mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

Simplifying the third term.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2}\cdot \dfrac{2}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

\(\\ \boxed{\left\begin{array}{ccc}\text{\Underline{Power of a Fraction Rule:}}\\\\\Big(\dfrac{a}{b}\Big)^2=\dfrac{a^2}{b^2} \end{array}\right }\)

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2 \cdot\dfrac{v^2}{R^2} \Big]\)

"R²'s" cancel, we are left with:

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5}mv^2\)

We have like terms, combine them.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{7}{10} mv^2\)

Each term has an "m" in common, factor it out.

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)\)

Now we have the following equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the right-hand side of the equation:

\(\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac12\cdot\dfrac25\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\cdot\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15mv_{_{0}}^2\Big\\\\\\\\\)

\(\Longrightarrow \dfrac{7}{10}mv_{_{0}}^2\)

Now we have the equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

\(\hrulefill\)

Now solving the equation for the variable "v":

\(m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

Dividing each side by "m," this will cancel the "m" variable on each side.

\(\Longrightarrow gh+\dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2\)

Subtract the term "gh" from either side of the equation.

\(\Longrightarrow \dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2-gh\)

Multiply each side of the equation by "10/7."

\(\Longrightarrow v^2=\dfrac{10}{7}\cdot\dfrac{7}{10}v_{_{0}}^2-\dfrac{10}{7}gh\\\\\\\\\Longrightarrow v^2=v_{_{0}}^2-\dfrac{10}{7}gh\)

Now squaring both sides.

\(\Longrightarrow \boxed{\boxed{v=\sqrt{v_{_{0}}^2-\dfrac{10}{7}gh}}}\)

Thus, the simplified equation above matches the simplified equation that was given.  

The energy lost to friction and air resistance is 186 J.

option B.

The energy lost to friction and air resistance is calculated from the change in the mechanical energy of the pendulum.

The initial potential energy of the pendulum at the initial position is calculated as;

PEi = mghi

where;

P.Ei = 12 kg x 9.8 m/s² x 20 m

P.Ei = 2,352 J

The final kinetic energy of the pendulum is calculated as follows;

K.Ef = 0.5 x 12 kg x (19 m/s)²

K.Ef = 2,166 J

ΔE = 2,166 J - 2,352 J

ΔE = -186 J

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Answer:

B, which is why ice skaters often keep their arms close to their body when doing spins and jumps to minimize resistance.

\(\text{Given that,}\\\\\text{Mass, m = 1 kg and height, h = 0.90 m}\\\\\text{Acoording to the condition,}\\\\mgh = \dfrac 12 mv^2\\\\\implies g h = \dfrac{v^2}2\\\\\implies v^2 =2gh\\\\\implies v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 0.90} = \sqrt{17.658} = 4.203 ~ \text{ms}^{-1}\)

Can you please fill in whatever goes in the blanks ?

Without them, the question makes no sense and has no answer.

Two forces (___) and (___) are applied to an object whose mass is 11.8 kg. The larger force is (___) . When both forces point due east, the object's acceleration has a magnitude of 0.408 m/s2. However, when (___) points due east and (___) points due west, the acceleration is 0.227 m/s2, due east. Find (a) the magnitude of (___) and (b) the magnitude of (___) .

The resistance of the wire is is 36.55 ohms.

The induced emf in the loop is calculated by Faraday's law as follows;

emf = A(B₂- B₁)/t

emf = 0.606 (7.97 - 2.91)/0.856

emf = 3.58 V

The resistance of the wire is calculated by using Ohms law as follows;

emf = IR

R = emf/I

R = 3.58/0.098

R = 36.55 ohms

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In nuclear fission, the resulting daughter nuclei are called fission fragments.

Nuclear fission is a process whereby heavy nuclei elements split into smaller, less heavy nuclei elements known as daughter nuclei. When this happens, energy is usually released in the form of heat.

The daughter nuclei are referred to as fission fragments. these fragments can themselves be either stable or unstable.

Nuclear fission differs from nuclear fusion. The latter is a process whereby two light nuclei combine fuses to become one heavy nucleus.

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Answer:

Obviously the answer is Sun...

Answer:

\(r=200m\)

Explanation:

From the question we are told that:

Charges:

\(Q_1=8.0mC\)

\(Q_2=2.0mC\)

\(Q_3=8.mC\)

Distance \(d=300m\)

Generally the equation for Force is mathematically given by

\(F=\frac{kq_1q_2}{r^2}\)

Therefore

\(F_{32}=F_{31}\)

\(\frac{q_2}{(300-r)^2}=\frac{q_1}{r^2}\)

\(\frac{2*10^{-3}}{(300-r)^2}=\frac{8*10^{-3}}{r^2}\)

\(r=2(300-r)\)

\(r=200m\)

To catch a 14.715N ball traveling at a velocity of 37.5m/s, required mechanical work is 1056.10 joule.

Physics' definition of work makes clear how it is related to energy: anytime work is performed, energy is transferred.

In a scientific sense, a work requires the application of a force and a displacement in the force's direction. Given this, we can state that

Work is the product of the component of the force acting in the displacement's direction and its magnitude.

Weight of the ball = 14.715 N.

Mass of the ball = 14.715 N ÷ (9.8 m/s²) = 1.502 kg.

Velocity of the ball = 37.5 m/s

Kinetic energy of the ball = 1/2 × 1.502 × 37.5² Joule = 1056.10 Joule.

Hence, to catch a 14.715N ball traveling at a velocity of 37.5m/s, required work is 1056.10 joule.

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The motorcycle had covered a distance of 16 meters in half the total time.

a) To calculate the acceleration, we can use the formula:

a = (v - u) / t

where a is the acceleration, v is the final velocity, u is the initial velocity (which is 0 since the motorcycle starts from rest), and t is the time.

Given:

u = 0 m/s (initial velocity)

v = ? (final velocity)

t = 4 s (time)

s = 64 m (distance)

Using the equation of motion:

s = ut + 1/2at^2

We can rearrange the equation to solve for acceleration:

a = 2s / t^2

a = 2(64) / (4)^2

a = 128 / 16

a = 8 m/s^2

Therefore, the acceleration of the motorcycle is 8 m/s^2.

b) To find the final velocity, we can use the formula:

v = u + at

v = 0 + (8)(4)

v = 32 m/s

Therefore, the final velocity of the motorcycle is 32 m/s.

c) To determine the time at which the motorcycle had covered half the total distance, we divide the total distance by 2 and use the formula:

s = ut + 1/2at^2

32 = 0 + 1/2(8)t^2

16 = 4t^2

t^2 = 4

t = 2 s

Therefore, the motorcycle had covered half the total distance at 2 seconds.

d) To calculate the distance covered in half the total time, we use the formula:

s = ut + 1/2at^2

s = 0 + 1/2(8)(2)^2

s = 0 + 1/2(8)(4)

s = 0 + 16

s = 16 m

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