An aircraft flying at a steady velocity of 70 m/s eastwards at a height of 800 m drops a package of supplies.
a) Express the initial velocity of the package as a vector. What
observer on the ground.
b) How long will it take for the package to reach the ground?
c) How fast will it be going as it lands? Express your answer as
a vector.
d) Describe the path of the package as seen by a stationary
assumptions have you made about the frame of reference?
e) Describe the path of the package as seen by someone in the
aeroplane.
​

Both masses will have the same acceleration. The cart accelerates to the right with a magnitude of 4.9 m/[tex]s^{2}[/tex]. The correct answer is 4.90 m/[tex]s^{2}[/tex]

Given that a massless string connects a 1.00 kg mass to a 3.00 kg cart which is resting on a frictionless horizontal surface.

Let M = 1kg and m = 3 kg

Since the horizontal surface is frictionless, the tension in the string will be the same. when the mass is hanged over a frictionless pulley, the tension will also be the same.

When the mass is released, the cart accelerates to the right can be calculated  from Newton' second law of motion. That is,

M( g + a) = m(g - a)

1(9.8 + a) = 3( 9.8 - a)

9.8 +a = 29.4 - 3a

collect the like terms

4a = 19.6

a = 19.6/4

a = 4.9 m/[tex]s^{2}[/tex]

Therefore, the cart accelerates to the right with a magnitude of 4.9 m/[tex]s^{2}[/tex]. The correct answer is 4.90 m/[tex]s^{2}[/tex]

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[tex] = \frac{1}{2} \times 0.008 \times 420J \\ = 1.68J[/tex]

[tex]1.68J = \frac{1}{2} \times 75 \: kg \times v2 \\ = > v2 = \frac{1.68 \times 2}{75} m {s}^{ - 1} \\ = 0.0448 \: m \: {s}^{ - 1} [/tex]

Answer:

0.0448 m/s

Hope you could get an idea from here.

Doubt clarification - use comment section.

Answer:

1176.798 J

Explanation:

[] Use the following equation:

U = mgh

m - mass

g - gravitational field

h - height

Have a nice day!

     I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)

- Heather

Answer:

500 J.

Explanation:

Work done = force * distance

= 25 * 20

= 500 Joules.

The final temperature of water is equal to 50.9999°C

Given the following data:

To determine the final temperature of water:

Mathematically, quantity of heat is given by the formula;

[tex]Q=mc\theta[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]5000=2.38892 \times 10^{-2}\times 4186 \times \theta\\\\5000=100.0001912 \theta\\\\ \theta=\frac{5000}{100.0001912} \\\\ \theta=49.9999^{\circ}C[/tex]

For the final temperature:

[tex]\theta = T_2 - T_1\\\\T_2 = \theta+T_1\\\\T_2 = 49.9999 + 50[/tex]

Final temperature = 50.9999°C

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Answer:

here's your answer below

Explanation:

sorry something went wrong

.

Ek = 1/2 mv^2

9 × 10^4 = 1/2 × 800 × v^2

9 × 10^4/400 = 400 v^2 / 400

9 × 10^4/400 = v^2

√225 = v

15 ms⁻¹ = v

That's the only way I know how to work it out

I think in this case velocity and speed would be considered the same because me

s = d/t and v=d/t

one is distance travelled and the other is displacement of a body

Answer:

try 2.4136 as an answer/ good luck

The relationship of the photoelectric effect and the de Broglie expression allows us to find the result for the wavelength of the ejected electrons is:

The photoelectric effect was explained by Einstein assuming that the light rays behave like particles called photons, therefore the

                 [tex]E_{photon} = K + \Phi[/tex]  

where [tex]E_{photon}[/tex] is the energy of the photon given by the Planck relation, K is the kinetic energy of the ejected electrons and Ф the work function of the material.

The Planck relationship states that the energy of the photons is proportional to the frequency.

            [tex]E_{photon} = h f[/tex]

Where h is Planck's constant and f is the frequency of the photons.

They indicate The work function is Ф= 2.25 10⁻¹⁹ J, the energy of the photon [tex]E_{photon}[/tex] = 4.52 10⁻¹⁹ J, let's find the kinetic energy of the ejected electrons.

            [tex]K = E_{photon} - \Phi[/tex]  

Let's calculate.

           K = (4.52-2.25) 10⁻¹⁹

           K = 2.27 10⁻¹⁹ J

Kinetic energy is the energy of motion and is given by the relationship.

          [tex]K = \frac{p^2}{2m}[/tex]  

The wave-particle duality was established by de Broglie with the relation.

         [tex]p = \frac{h}{\lambda }[/tex]  

Let's replace.

         [tex]K = \frac{1}{2m} (\frac{h}{\lambda} )^2 \\\lambda^2 = \frac{h^2}{2m K}[/tex]

let's calculate.

        [tex]\lambda^2 = \frac{(6.63 \ 10^{-34})^2 }{2 \ 9.1 \ 10^{-31 } \ 2.27 \ 10^{-19}}[/tex]

        [tex]\lambda = \sqrt{1.06397 \ 10^{-18}}[/tex]

        λ  = 1.03 10⁻⁹m

In conclusion with the relationship of the photoelectric effect and the de Broglie expression we can find the result for the wavelength of the ejected electrons is:

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The output work is always less than the input work because some of the input work is used to overcome friction.

Answer:

Formula = D = m/v

Given value of D = 19.3 and m = 13g

19.3 = 13/v

V × 19.3 = 13

v = 13/19.3

v = 0.67

Answer:

Muscle memory is found in many everyday activities that become automatic and improve with practice, such as riding bicycles, driving motor vehicles, playing ball sports, typing on keyboards, entering PINs, playing musical instruments, poker, martial arts, and dancing.

Explanation:

Answer:

Explanation:

The kinetic energy of an object can be found by using the formula

[tex]k = \frac{1}{2} m {v}^{2} \\ [/tex]

m is the mass in kg

v is the velocity in m/s

From the question

m = 1500 kg

v = 35 m/s

We have

[tex]k = \frac{1}{2} \times 1500 \times {35}^{2} \\ = 750 \times 1225 \\ = 918750[/tex]

We have the final answer as

Hope this helps you

Answer:

a 5 kilogram mass, at a height of 3 meters, while acted on by Earth's gravity would have 147.15 Joules of potential energy, PE = 3kg * 9.81 m/s2 * 5m = 147.15 J.

Explanation:

the independent variable and the dependent variable.

Answer:

it is B

Explanation:

Answer:

CO and H20

Explanation:

Applying Charles law

[tex]\\ \sf\Rrightarrow \dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}[/tex]

[tex]\\ \sf\Rrightarrow \dfrac{100}{313}=\dfrac{V_2}{263}[/tex]

[tex]\\ \sf\Rrightarrow V_2=\dfrac{26300}{313}[/tex]

[tex]\\ \sf\Rrightarrow V_2=84.02ml[/tex]

Start from the top and go 7 down then go to the right 9 times and there you should fine the letter "one"

Answer:

Distance is 800 m and Displacement is 0 m

Explanation:

Total Distance

= 400(2)

= 800 m

Total Displacement

= 0 m since she returns to the same spot

Answer:

1. 60 kg m/s

2. 2.4 kg

3. none they both have same momentum

The amount of work done by the car engine if it is 100% efficient is 366.67 Joules.

Given the following data:

To determine the amount of work done by the car engine if it is 100% efficient:

Mathematically, the work done by an object with respect to power and time is given by the formula:

[tex]Work\;done = \frac{Power}{time}[/tex]

Substituting the given parameters into the formula, we have;

[tex]Work\;done = \frac{22000}{60}[/tex]

Work done = 366.67 Joules.

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The final velocity of an object that starts from rest and travels for 5 seconds at an acceleration of 4.3 m/s² is 21.5m/s.

EQUATION OF MOTION:

The final velocity of a moving object can be calculated by using one of the equations of motion as follows:

V = u + at

Where;

According to this question,

v = 0 + 4.3(5)

v = 21.5m/s.

Therefore, the final velocity of an object that starts from rest and travels for 5 seconds at an acceleration of 4.3 m/s² is 21.5m/s.

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Answer:

Density of the mixture = 855.56kgm-3

Explanation:

Density = Mass / Volume

Volume of Liquid X = 0.5m³

Density of Liquid X = 900kgm-3

Mass of Liquid X = Density × Volume

= 900kgm-3 × 0.5m³ = 450kg

Volume of Liquid Y = 0.4m³

Density of Liquid Y = 800kgm-3

Mass of Liquid Y = Density × Volume

= 800kgm-3 × 0.4m³= 320kg

As X and Y are mixed, we add their masses and volumes together:

Mass = 770kg

Volume = 0.9m³

Now we can find the density of the mixture:

Density = 770kg / 0.9m³ = 855.56kgm-3 (rounded to the 2nd decimal)

The momentum of the water balloon at the given speed is 7.2 kgm/s.

The given parameters:

The momentum of the water balloon is calculated as follows;

P = mv

Substitute the given values of mass and velocity as follows;

P = 0.4 x 18

P = 7.2 kg m/s.

Thus, the momentum of the water balloon at the given speed is 7.2 kgm/s.

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The type of bond found in pure gold is metallic bonding, indicated by metallic bonds. Therefore option A is correct.

Metallic bonding occurs between metal atoms, such as gold, and is characterized by the sharing of electrons among a sea of delocalized electrons.

In metallic bonds, the valence electrons of metal atoms are not strongly bound to any particular atom but are free to move throughout the metal lattice.

This sharing of electrons gives rise to properties such as high electrical and thermal conductivity, malleability, and ductility, which are characteristic of metals like gold.

Unlike ionic or covalent bonds, metallic bonds do not involve the transfer or sharing of electrons between different elements.

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Assuming both billiard balls have the same mass, conservation of momentum says

[tex]m\vec v_1 + m\vec v_2 = m{\vec v_1}\,' + m{\vec v_2}\,'[/tex]

where m = mass of both billiard balls, and v₁ and v₂ = their initial velocities, and v₁' and v₂' = their final velocities. The masses are the same so the exact value of m is irrelevant. The first ball has initial speed 5 m/s and the second is at rest, so

[tex]\left(5 \dfrac{\rm m}{\rm s}\right) \, \vec\imath = {\vec v_1}\,' + {\vec v_2}\,'[/tex]

After the collision, the first ball has speed 4.35 m/s and is moving at angle of 30° below the original path, so

[tex]{\vec v_1}\,' = \left(4.35\dfrac{\rm m}{\rm s}\right)\left(\cos(30^\circ) \, \vec\imath + \sin(30^\circ) \, \vec\jmath\right) \approx \left(3.77 \dfrac{\rm m}{\rm s}\right) \vec\imath + \left(-2.18 \dfrac{\rm m}{\rm s}\right) \vec\jmath[/tex]

Then the second ball has final velocity vector

[tex]{\vec v_2}\,' \approx \left(1.23 \dfrac{\rm m}{\rm s}\right) \vec\imath + \left(2.18 \dfrac{\rm m}{\rm s}\right) \vec\jmath[/tex]

so it moves with speed

[tex]\left\|{\vec v_2}\,'\right\| \approx \sqrt{\left(1.23\dfrac{\rm m}{\rm s}\right)^2 + \left(2.18\dfrac{\rm m}{\rm s}\right)^2} \approx \boxed{2.50 \dfrac{\rm m}{\rm s}}[/tex]

at an angle of

[tex]\theta \approx \tan^{-1}\left(\dfrac{2.18}{1.23}\right) \approx \boxed{60.5^\circ}[/tex]

or about 60.5° above the original line of motion.

Answer:

When an object is balanced, about a pivot, the total clockwise moment must be equal to the total anticlockwise moment about that pivot.

Hope that helps.

When an object is balanced about a pivot, the total clockwise moment must be equal to the total anticlockwise moment.

Clockwise involves turning to the right while going in the direction of a clock's hands. The moment is referred to as a clockwise moment if the force acting on a body rotates the body clockwise with regard to the axis of rotation.

On the other hand, the moment is referred to as an anti-clockwise moment if the force rotates the body in the opposite direction. The clock face rule can be used to determine this magnet's polarity.

The face of the loop will display the North Pole if the current is flowing counterclockwise. On the other hand, if the current is moving counterclockwise, the South Pole is visible on the loop's face.

Therefore, the total clockwise and total counterclockwise moments must match when an object is balanced about a pivot.

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Answer: It eventually loses its stamina, and would come to a stop.

Explanation: