The motion of the package can be described as the motion of a projectile,
given that it has an horizontal velocity and it is acted on by gravity.
a) [tex]\vec{v}[/tex] = 70·ib) The package will reach ground in approximately 12.77 seconds.c) The speed of the package as it lands is approximately 145.51 m/s.d) The path of the package based on a stationary frame of reference is parabolice) The path of the package as seen from the plane is directly vertical downwards
Reasons:
Velocity of the aircraft = 70 m/s
Direction of flight of the aircraft = Eastward
Height from which the aircraft drops the package, h = 800 m
a) The initial velocity of the package, [tex]\vec{v}[/tex] = 70·i
b) The time it will take the package to reach the ground, t, is given by the formula;
[tex]\displaystyle h = \mathbf{\frac{1}{2} \cdot g \cdot t^2}[/tex]
Where;
g = The acceleration due to gravity ≈ 9.81 m/s²
Therefore;
[tex]\displaystyle t = \mathbf{\sqrt{ \frac{2 \cdot h}{g} }}[/tex]
Which gives;
[tex]\displaystyle t = \sqrt{ \frac{2 \times 800}{9.81} } \approx \mathbf{12.77}[/tex]
The time it will take the package to reach the ground, t ≈ 12.77 seconds
c) The vertical velocity just before the package reaches the ground, [tex]v_y[/tex], is given as follows;
[tex]v_y^2[/tex] = 2·g·h
Therefore;
[tex]v_y[/tex] = √(2·g·h)
Which gives;
[tex]v_y[/tex] = √(2 × 9.81 × 800) ≈ 125.28
[tex]v_y[/tex] ≈ 125.28 m/s
Which gives; [tex]\vec{v}[/tex] = 70·i - 125.28·j
Therefore, |v| = √(70² + (-125.28)²) ≈ 143.51
The speed of the package as it lands, |v| ≈ 143.51 m/s
d) The motion of the package that includes both horizontal and vertical motion is a projectile motion.
Therefore;
The path of the package is the path of a projectile, which is a parabolic shape.
e) As seen by someone on the aeroplane, the horizontal velocity will be
zero, therefore, the package will appear as accelerating directly vertical
downwards.
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Resolve the weight of the box to find the component of the weight acting parallel to the slope.
W = 50N
30
Answer:
here's your answer below
Explanation:
sorry something went wrong
.
When an object is balanced about a pivot, the total clockwise moment must be equal to the total __________ __________. What two words complete this sentence?
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.
What is balancing?
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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what are two variables that are always in every experiment?
the independent variable and the dependent variable.
Can someone please give me the (Answers) to this? ... please ...
Answer:
1. 60 kg m/s
2. 2.4 kg
3. none they both have same momentum
plssssss look at the picture i need it right hurry pls
Answer:
it is B
Explanation:
Answer:
CO and H20
Explanation:
how to tell a girl you like her?
What is the speed of an 800 kg automobile if it has a kinetic energy of 9.00 x 10^J?
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
How fast would a(n) 75 kg man need to run in order to have the same kinetic energy as an 8.0 g bullet fired at 420 m/s ?
[tex] = \frac{1}{2} \times 0.008 \times 420J \\ = 1.68J[/tex]
So, the kinetic energy of the bullet is 1.68 J. It is said that the man's kinetic energy should be same as that of the bullet.Mass of the man (m2) = 75 KgLet the velocity of the man be v2.Therefore,[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.
What is the energy of a 5 kg object that is held at a height of 3 m above the ground?
I really need the Formula, substitute, answer
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 density of gold is 19.3 g/cm3. What is the volume of a 13 g gold nugget? (Density: D = m/v)
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
Stacey runs around a 400 meter track 2 times in 5 minutes. How does her displacement differ from her distance? *
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
what happens to the state of motion of the object, when you don't exert a force on it???
Answer: It eventually loses its stamina, and would come to a stop.
Explanation:
2. Think about an activity you may have learned that involves muscle memory. Consider when you first learned the activity, how easy or difficult it was the first time, and if you can do it now without thinking. What happened in your brain during practices that resulted in muscle memory for you?
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:
What is the final velocity of an object that starts from rest and travels for 5 seconds at an acceleration of 4.3 m/s2? ____________________
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;
V = final velocity (m/s)u = initial velocity (m/s)a = acceleration (m/s²)t = time (s)According to this question,
t = 5sa = 4.3m/s²u = 0 m/s (at rest)v = ?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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Liquid X of volume 0.5m3 and density 900kgm-3 was mixed with liquid Y of volume 0.4m3 and density 800kgm-3. What was the density of the mixture?
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)
What is the change in velocity of a 1068 kg truck that experiences an impulse of 440 N ⋅ s? Include 2 decimal places in your answer.
Answer:
try 2.4136 as an answer/ good luck
a water balloon is thrown a target at 18 m/s. if the water balloon has a mass of 0.4 kg, what is the momentum?
The momentum of the water balloon at the given speed is 7.2 kgm/s.
The given parameters:
Speed of the water balloon, v = 18 m/sMass of the balloon, m = 0.4 kgThe 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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Determine the kinetic energy of a 1500 kg roller coaster that is moving with a
speed of 35 m/s.
A) 200,000 J
B) 25,000
C) O 918,750
D) 920,000J
Answer:
918,750 JExplanation:
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
918,750 JHope this helps you
why efficiency of energy converters is less than 100%
The output work is always less than the input work because some of the input work is used to overcome friction.
What type of bond is found in pure gold?
A) Metallic
B) Ionic
C)Covalent
D)Diatomic
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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A 25 N force acts on a box moving along a plane at a distance of 20 meters. What is the work it does with this force?
Help!!
Answer:
500 J.
Explanation:
Work done = force * distance
= 25 * 20
= 500 Joules.
A concrete block (B-36 x10 °C-') of volume 100 mat 40°C is cooled to
-10°C. What is the change in volume? *
A. It will increase by 0.18 m
B. It will decrease by 0.18 m'
C.It will increase by 0.05 m
D. It will decrease by 0.05 mº
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]
How much work can a 22 kW (22000W) car engine do in 60 s if it is 100% efficient?
The amount of work done by the car engine if it is 100% efficient is 366.67 Joules.
Given the following data:
Time = 60 secondsPower = 22,000 WattTo 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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What is the final temperature if it requires 5000 J of heat to warm 2.38892 x10-2 kg of water that starts at 5oC? Remember Cp for water is 4186 J/kgC
The final temperature of water is equal to 50.9999°C
Given the following data:
Mass = [tex]2.38892 \times 10^{-2}\;kg[/tex]Quantity of heat = 5000 J Specific heat capacity of water = 4186 J/kg°CTo determine the final temperature of water:
Mathematically, quantity of heat is given by the formula;
[tex]Q=mc\theta[/tex]
Where:
Q represents the quantity of heat.m represents the mass of an object.c represents the specific heat capacity.∅ represents the change in temperature.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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A roller coaster goes from 2.00 m/s [forward] to 10.0 m/s [forward) in 4.50 s. What is its acceleration?
A massless string connects a 1.00 kg mass to a 3.00 kg cart which is resting on a frictionless horizontal surface. The mass hangs over a
frictionless pulley. When the mass is released, the cartaccelerates to the right
2.45 m/s2
4.90 m/s
9.80 m/s?
19.6 m/s
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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A billiard ball with a speed of 5 m/s strikes another stationary billiard ball in a perfectly elastic collision.
After the collision, the first ball has a speed of 4.35 m/s and is traveling at an angle of 30° below its original line of motion.
Find the speed and angle of the second ball, with respect to the initial line of motion, after the collision.
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.
Halpinium metal (Jh(s)) can lose electrons via the photoelectric effect. The binding energy for Jh is 2.25x10-19 J. If a photon having an energy of 4.52x10-19 J strikes the surface of halpinium, calculate the de Broglie wavelength (in nm) of the electron that is ejected.
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:
Wavelength of de Broglie is λ= 1.03 10⁻⁹m
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:
Wavelength of de broglie is: λ = 1.03 10⁻⁹m
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A cat that has a mass of 6 kg climbs to the top of a tree that is 20 m high. How much potential energy does the cat have at the top of the tree?
Have a great day!
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
I couldn't find one .
Start from the top and go 7 down then go to the right 9 times and there you should fine the letter "one"