Answer:
3.176hours
Explanation:
270/85=3.126 hours DISTANCE / SPEED = TIME
It is the final seconds of an ice hockey game between the Flyers and the Bruins. The Bruins are down by 1 point. With 20 s left in the game, the Bruins pull the goalie and have him play as a forward in an attempt to tie the game.
The Flyers successfully defend their goal for 9 s. With only 1.25 s remaining on the game clock, a Flyer shoots the puck on the ice past the skates and sticks of the other players and toward the Bruins' goal. The puck is 37 m from the goal when it leaves the stick with an initial horizontal velocity of 30 m/s. The shot is perfectly directed toward the empty goal, but the ice slows the puck down at a constant rate of 0.50 m/s2 as it slides toward the goal. None of the Bruins can stop the puck before it reaches the goal.
a. Where is the puck when the game clock reaches zero and the horn sounds to end the game?
b. Do the Flyers win the game by 1 or 2 points?
Answer:
a
[tex]P =0.1094 \ m[/tex] past the goal post
Yes the Flyers won the match
Explanation:
Generally the distance covered by the pluck during the last 1.25 seconds is mathematically represented as
[tex]D = ut + \frac{1}{2} * a * t^2[/tex]
=> [tex]D =30 * 1.25 + \frac{1}{2} * (-0.5) * (1.25)^2[/tex]
=> [tex]D =37.1094 \ m [/tex]
Generally the position of the puck when the game clock reaches zero is mathematically represented as
[tex]P = 37.1094 -37[/tex]
[tex]P =0.1094 \ m[/tex] past the goal post
Given that the Bruins where one point down and Flyers scored another goal it means that the flyer are now two point up hence they won the match
A gymnast on the uneven parallel bars is at rest, tipped at a 45∘ angle from the vertical. The distance from her hands to her feet is 1.8 m. If we model her body as having a uniform cross section and assume that her center of gravity is midway between her hands and her feet, what is her initial angular acceleration?
The initial angular acceleration is 5.78 rad/s^2.
Calculation of the angular acceleration;Since
A gymnast on the uneven parallel bars is at rest, tipped at a 45∘ angle from the vertical. The distance from her hands to her feet is 1.8 m.
So,
[tex]Torque \tau = r \times F\\\\ \tau = l\alpha\\\\so, r \times F = l\alpha\\\\mg\ cos\ 45 \times r = (1/3)mR^2 \times \alpha\\\\9.8\times cos\ 45 \times 0.9 = (1/3)\times (1.8)^2 \times \alpha\\\\\alpha = 5.78\ rad/s^2[/tex]
hence, The initial angular acceleration is 5.78 rad/s^2.
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A 10.0-cm-diameter and a 20.0-cm-diameter charged ring are arranged concentrically (so they share the same axis). Assume both are centered at the origin. The 10 cm ring is charged to 30.0nC, while the 20 cm ring is charged to 60.0nC. Using the library electric field formula for Ering, compute the total electric field strength numerically at:
a, The origin
b. 40.0 cm along the axis of the 10 cm ring.
Answer:
a) Eₓ = 0 , b) c) Ex_total = 3.96 10⁻⁶ N
Explanation:
For this problem we will look for the expression for the electric field of a ring of charge, at a point on its central axis.
From the symmetry of the ring, the resulting field is in the direction of the axis, we suppose that it corresponds to the x and y axis and the perpendicular direction but since the two sides of the annulus are offset.
dE = k dq / r²
The component of this field in the direction to x is
d Eₓ = dE cos θ
the distance of ring to the point
r² = x² + a²
where a is the radius of each ring
the cosine is the adjacent leg between the hypotenuse
Cos θ = x /r
we substitute
d Ex = k / (x² + a²) dq x /√ (x² + a²)
we integrate
Ex = ∫ k / (x² + a²)^{3/2} x dq
Ex = k x/ (x² + a²)^{1.5} Q
the total field is the sum of the fields created by each ring, since they are on the same line (x-axis), you can perform an algebraic sum
Eₓ_total = Eₓ₁ + Eₓ₂
a) at the origin x = 0
in field is zero, since the numerator e makes zero
Eₓ = 0
b) for point x = 40.0cm = 0.400 m
we substitute the values in our equation
Ex_total = k x [Q (x² + 0.1²)^{1.5} + 2Q / (x² + 0.2²2) ^3/2]
Ex_total = k Q x [1 / (x² + 0.01)^1.5 + 2 / (x² + 0.04)^1.52]
let's calculate
Ex_total = 9 10⁹ 30 10⁻⁹ 0.40 [1 / (0.4 2 + 0.01) 3/2 + 2 / (0.4 2 + 0.04) 3/2] 10⁻⁹
Ex_total = 108 [14.2668 + 22.36] 10⁻⁹
Ex_total = 3.96 10⁻⁶ N
a motorbike and its rider have a mass of 274kg. if they experience an acceleration of 4.50 m/s^2 what force was exerted on them
Answer:
1233 N
Explanation:
force=mass×accerelation
3. What would happen if Earth did not rotate?
Answer:
If Earth were to stop spinning on its axis, gradually the oceans would migrate towards the poles from the equator. The atmosphere would still be in motion with the Earth's original 1100 mile per hour rotation speed at the equator. And this means, rocks, topsoil, trees, buildings, and so on, would be swept away into the atmosphere.
Hope that helped <3
Q1: An object with a charge of 1.2 C is located 4.5 m away from a second object that has a charge of 0.36 C. Find the electrical force that the objects exert on each other. Use 8.988100 N·m2/C2 for the value of Coulomb's constant. Give your answer to two decimal places.
Answer:
a) F= 0,19 [N] according to problem statement
b) F = 0,19*10⁹ [N] using the right value of K
Explanation:
The force between two electric charges is according to Coulomb´s law is:
F = K * q₁*q₂ / d² where q₁ and q₂ are the charges on body one and body 2 respectively, d is the distance between the two bodies and K is a constant K = 8,988100*10⁹ N.m²/C². The problem establishes to use K = 8,988100 N.m²/C².
NOTE: To value of is : K = 8,988100*10⁹ N.m²/C². I am going to solve the problem using K = 8,988100 N.m²/C² if that information was an error, all we need to get the right answer is multiply the result by 10⁹
Then:
F = 8,988100 * 1,2* 0,36 / (4,5)² [ N*m²/C² ] * [ C*C*/m²]
F = 3,882859/ 20,25 [N]
F= 0,19 [N]
The force is of repulsion since the two charges are positive and in the direction of the straight line which passes through the centers of the bodies
Use Percentages Given that the molecular mass of magnesium
hydroxide (MOH.) is 58.32 amu and the atomic mass of an atom of
oxygen is 15.999 amu, what percentage of this compound is oxygen?
Answer:
27%
Explanation:
15.999 divided by 58.32 = .27433128
Move the decimal place over 2 places.
27%
Make use of percentages. Given that the molecular mass of magnesium hydroxide (MOH.) is 58.32 amu and the atomic mass of an oxygen atom is 15.999 amu, oxygen accounts for 27% of this compound.
Molecular versions of many substancesChemical substances known as molecular compounds assume the shape of distinct molecules. Examples include common compounds like carbon dioxide (CO2) and water (H2O) (Figure 3.1.1 ).
These substances differ significantly from ionic substances like sodium chloride (NaCl). When metal atoms give up one or more of their electrons to non-metal atoms, ionic compounds are created. The cations and anions that result
Therefore,
58.32 / 15.999 (27.3%) =.27433128
Add two places to the decimal place.
27%
The molar mass, which is stated in g/mol, is defined as the mass of a specific material divided by the amount of a substance. In light of this, the molar mass is average. Amu, short for atomic mass units, is the unit used to quantify molecular mass.
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A cannonball has a speed of 320 m/s at an altitude of 122 m above the ground. What is the total mechanical energy of the cannonball assuming that the potential energy at ground level is zero
Answer:
Each kilogram of the cannonball has a total energy of 52396.454 joules.
Explanation:
From Principle of Energy Conservation we understand that energy cannot be destroyed nor created, but transformed. In this case non-conservative forces can be neglected, so that total energy of the cannonball ([tex]E[/tex]) is the sum of gravitational potential ([tex]U_{g}[/tex]) and translational kinetic energies ([tex]K[/tex]), all measured in joules. That is:
[tex]E = U_{g}+K[/tex] (Eq. 1)
By applying definitions of gravitational potential and translational kinetic energies, we proceed to expand the expression:
[tex]E = m\cdot g \cdot y + \frac{1}{2} \cdot m \cdot v^{2}[/tex] (Eq. 2)
Where:
[tex]m[/tex] - Mass of the cannonball, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]y[/tex] - Height of the cannonball above ground level, measured in meters.
[tex]v[/tex] - Speed of the cannonball, measured in meters per second.
As we do not know the mass of the cannonball, we must calculated the unit total energy ([tex]e[/tex]), measured in joules per kilogram, whose formula is found by dividing (Eq. 1) by the mass of the cannonball. Then:
[tex]e = g\cdot y + \frac{1}{2}\cdot v^{2}[/tex]
If we know that [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y =122\,m[/tex] and [tex]v = 320\,\frac{m}{s}[/tex], the unit total energy of the cannonball is:
[tex]e = \left(9.807\,\frac{m}{s^{2}}\right)\cdot (122\,m)+\frac{1}{2}\cdot \left(320\,\frac{m}{s} \right)^{2}[/tex]
[tex]e = 52396.454\,\frac{J}{kg}[/tex]
Each kilogram of the cannonball has a total energy of 52396.454 joules.
What is a lever and mention its types
Answer:
Three different types of levers exist, depending on where the input force, fulcrum, and load are. A class 1 lever has the fulcrum between the input force and load. A class 2 lever has the load between the fulcrum and input force. A class 3 lever is a lever that has the input force in between the fulcrum and the load.
Explanation:
A 0.10 g honeybee acquires a charge of +23 pC while flying, due to friction of the wings against the air. We will try and understand why this is advantageous. A. The electric field near the surface of the earth is typically 100 N/C, directed downward. What is the ratio of the magnitude of the electric force on the bee to the magnitude of the bee's weight?
Answer:
The ratio of the magnitude of the electric force on the bee to the magnitude of the bee's weight is 2.35 x 10⁻⁶
Explanation:
Given;
mass of the honeybee, m = 0.1 g
charge acquired by the honeybee, Q = 23pC = 23 x 10⁻¹² C
the electric field near the earth's surface, E = 100 N/C
The magnitude of the electric force on the bee is given by;
F = QE
F = (23 x 10⁻¹²)(100) = 23 x 10⁻¹⁰ N
The weight of the bee is given by;
W = mg
W = 0.1 x 10⁻³ x 9.8
W = 9.8 x 10⁻⁴ N
The the ratio of the magnitude of the electric force on the bee to the magnitude of the bee's weight is given by;
[tex]\frac{F}{W} = \frac{23*10^{-10}}{9.8*10^{-4}} = 2.35 *10^{-6}[/tex]
Therefore, the ratio of the magnitude of the electric force on the bee to the magnitude of the bee's weight is 2.35 x 10⁻⁶
What is the answwr I need help
Read each scenario and then answer the questions.
Scenario A: A shopping cart is pushed a distance of 3 m by a force of 15 N.
Scenario B: A shopping cart is pushed a distance of 4 m by a force of 12 N.
Scenario C: A shopping cart is pushed a distance of 6 m by a force of 10 N.
Which scenario requires the most work?
Which scenario requires the least work?
Answer:
C requires the most work and A requires the least work
Answer:
Which scenario requires the most work?
✔ C
Which scenario requires the least work?
✔ A
Each shot of the laser gun most favored by Rosa the Closer, the intrepid vigilante of the lawless 22nd century, is powered by the discharge of a 1.33 F capacitor charged to 77.9 kV. Rosa rightly reckons that she can enhance the effect of each laser pulse by increasing the electric potential energy of the charged capacitor. She could do this by replacing the capacitor's filling, whose dielectric constant is 427, with one possessing a dielectric constant of 983.
A. Find the electric potential energy of the original capacitor when it is charged.
B. Calculate the electric potential energy of the upgraded capacitor when it is charged.
Answer:
A) 4.035 × 10^(9) J
B) 9.29 × 10^(9) J
Explanation:
We are given;
Capacitance of the original capacitor; C = 1.33 F
Potential difference given to the original capacitor; V = 77.9 kV = 77.9 × 10³ V
A) The formula for Potential energy (U) for the original capacitor is given as:
U = ½CV²
Plugging in the relevant values, we have;
U = ½ × 1.33 × (77.9 × 10³)²
U = 4.035 × 10^(9) J
B) We are told that the capacitor with dielectric constant of 427, was replaced with one possessing a dielectric constant of 983.
Thus;
U = ½ × 1.33 × (983/427) × (77.9 × 10³)²
U = 9.29 × 10^(9) J
a car with a mass of 2000kg is traveling at a velocity of 50 m/s. the engine supplies a force of 1000N forward.the force of friction is 1000 N backwards. What is the velocity after 5 seconds
Answer:
no
Explanation:
ok
Pain during sports is completely normal and should not be a concern.
Please select the best answer from the choices provided.
OT
OF
Answer:
F
Explanation: Its common sence
Answer:
It’s false
Explanation:
2020 (WORST YEAR EVER) on Edge ;-)
A man throws a football straight into the air Which type of energy is the
football gaining as it rises
Answer:
It is gaining potential energy
Answer: it could also be gravitational potential energy
This is because as it rises in the air it will gain more energy to fall back down
Which of the following is NOT a unit of measurement for speed?
I don't see any units of measurement or speed I don't see anything
A coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged conductor (with charge −8.5 µC and radius 9.249 mm).
Required:
What is the magnitude of the electric field halfway between the two cylindrical conductors? The Coulomb constant is 8.98755 × 10^9 N.m^2 . Assume the region between the conductors is air, and neglect end effects. Answer in units of V/m.
Complete question:
A 50 m length of coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged conductor (with charge −8.5 µC and radius 9.249 mm).
Required:
What is the magnitude of the electric field halfway between the two cylindrical conductors? The Coulomb constant is 8.98755 × 10^9 N.m^2 . Assume the region between the conductors is air, and neglect end effects. Answer in units of V/m.
Answer:
The magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m
Explanation:
Given;
charge of the coaxial capable, Q = 8.5 µC = 8.5 x 10⁻⁶ C
length of the conductor, L = 50 m
inner radius, r₁ = 1.304 mm
outer radius, r₂ = 9.249 mm
The magnitude of the electric field halfway between the two cylindrical conductors is given by;
[tex]E = \frac{\lambda}{2\pi \epsilon_o r} = \frac{Q}{2\pi \epsilon_o r L}[/tex]
Where;
λ is linear charge density or charge per unit length
r is the distance halfway between the two cylindrical conductors
[tex]r = r_1 + \frac{1}{2}(r_2-r_1) \\\\r = 1.304 \ mm \ + \ \frac{1}{2}(9.249 \ mm-1.304 \ mm)\\\\r = 1.304 \ mm \ + \ 3.9725 \ mm\\\\r = 5.2765 \ mm[/tex]
The magnitude of the electric field is now given as;
[tex]E = \frac{8.5*10^{-6}}{2\pi(8.85*10^{-12})(5.2765*10^{-3})(50)} \\\\E = 5.793*10^5 \ V/m[/tex]
Therefore, the magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m
If the point B was initially at the same level as the center of the pulley C, what is the direction of the velocity of point B after the pulley at A has undergone 2 revolutions? Give this in degrees from the horizontal (positive in the counterclockwise direction, negative in the clockwise direction)
Answer:
θ = 720º
Explanation:
Let's use the rotational kinematics relations for this exercise, remember that all angles must be given in radians.
Let's reduce the magnitudes to the SI system
θ = 2 rev (2π rad / 1 rev) = 4π rad
we assume that the pulley has a radius r
linear and angular coordinates are related
y = θ r
let's calculate
y = 4π r
to convert the distance in angles to radians
θ = y / r
θ = 4π
let's change this angle to degrees
π rad = 180º
θ = 4π rad (180º /π rad)
θ = 720º
How is Earth's surface most likely to change in a place that experiences extensive animal activity?
Sand dunes will be formed due to deposition of sediments.
Soil will loosen due to vast network of underground tunnels.
Cracks will appear in rocks due to freezing and thawing of water.
Rocks will break due to repeated abrasion and mechanical stress.
Answer:
B
Explanation:
Answer:
Soil will loosen due to vast networks of underground tunnels.
Inkjet printers can be described as either continuous or drop-on-demand. In a continuous inkjet printer, letters are built up by squirting drops of ink at the paper from a rapidly moving nozzle. You are part of an engineering group working on the design of such a printer. Each ink drop will have a mass of 1.5×10−8g. The drops will leave the nozzle and travel toward the paper at 50 m/s in a horizontal direction, passing through a charging unit that gives each drop a positive charge q by removing some electrons from it. The drops will then pass between parallel deflecting plates, 2.0 cm long, where there is a uniform vertical electric field with magnitude 8.0×104N/C. Your team is working on the design of the charging unit that places the charge on the drops.
(a) If a drop is to be deflected 0.30 mm by the time it reaches the end of the deflection plates, what magnitude of charge must be given to the drop? How many electrons must be removed from the drop to give it this charge?
(b) If the unit that produces the stream of drops is redesigned so that it produces drops with a speed of 25 m/s, what q value is needed to achieve the same 0.30-mm deflection?
Answer:
A) q = 7.03 × 10^(-13) C
n ≈ 4388265 electrons
B) q = 1757.8 × 10^(-16) C
n ≈ 1097253 electrons
Explanation:
We are given;
Mass; m = 1.5 × 10^(−8)g = 1.5 × 10^(-11) kg
Speed; v = 50 m/s
Electric Field; E = 8 × 10⁴ N/C
Distance between plates; s = 2 cm = 0.02 m
A) Now, speed = distance/time
So, time(t) = distance/speed = 0.02/50 = 0.0004 s
From Newton's first law of motion, we know that;
d = ut + ½at²
But u is initial velocity and in this case it's zero.
But we are told that a drop is to be deflected by 0.30 mm. So, d = 0.3 × 10^(-3) m
Thus;
0.3 × 10^(-3) = 0 + ½a(0.0004)²
a = 3750 m/s²
Now, we know that force in motion normally can be expressed as;
F = ma
But in electric field, it's;
F = qE
Thus;
qE = ma
So, charge is; q = ma/E
Plugging in the relevant values;
q = (1.5 × 10^(−11) × 3750)/(8 × 10⁴)
q = 7.03 × 10^(-13) C
Now, number of electrons is given by the formula;
n = q/e
Where e is charge on electron with a value of 1.602 × 10^(-19) C/electron
So; n = (7.03 × 10^(-13))/(1.602 × 10^(-19))
n ≈ 4388265 electrons
B) We are told speed is now 25 m/s.
Thus;
time(t) = distance/speed = 0.02/25 = 0.0008 s
From Newton's first law of motion, we know that;
d = ut + ½at²
But u is initial velocity and in this case it's zero.
d remains 0.3 × 10^(-3) m
Thus;
0.3 × 10^(-3) = 0 + ½a(0.0008)²
a = 937.5 m/s²
Now, we know that force in motion normally can be expressed as;
F = ma
But in electric field, it's;
F = qE
Thus;
qE = ma
So, charge is; q = ma/E
Plugging in the relevant values;
q = (1.5 × 10^(−11) × 937.5)/(8 × 10⁴)
q = 1757.8 × 10^(-16) C
Now, number of electrons is given by the formula;
n = q/e
Where e is charge on electron with a value of 1.602 × 10^(-19) C/electron
So; n = (1757.8 × 10^(-16) )/(1.602 × 10^(-19))
n ≈ 1097253 electrons
The lower the angle of the slope, ________ the acceleration along the ramp, therefore, the speed at the bottom of a slope will be_______, and, consequently, the control will be better.
The acceleration along the ramp, therefore, the speed at the bottom of a slope will be:__________
a. lower
b. gcostheta
c. higher
d. gsintheta
Answer:
Lower
Lower
gsintheta (gsinθ)
Explanation:
The sum of forces resolved parallel to the inclined plane is given by;
F - mgsinθ = 0
ma - mgsinθ = 0
ma = mgsinθ
a = gsinθ
Acceleration is proportional to angle of inclination, thus the lower the angle of the slope, lower the acceleration along the ramp.
therefore, the speed at the bottom of a slope will be lower, (velocity is directly proportional to acceleration) and, consequently, the control will be better.
The acceleration along the ramp, is gsintheta (gsinθ)
What is the mass of a 2 kg object on the Earth and on the moon?
Answer:
Same
Explanation:
Mass is the quantity of matter in a certain object.
WHEREVER you take a 2kg object, the mass will remain 2kg. All that changes is the Weight ..Weight the force which the centre of a Planet uses to pull everything towards itself.
On earth, it is 9.81 whereas on the Moon it is 1.6
Here, we are required to determine the mass of a 2kg object on the Earth and on the moon.
The mass of the 2kg object is the same on earth and on the moon.The mass of an object is simply defined as the quantity of matter present in the object and this quantity remains constant.
The mass of an object is simply defined as the quantity of matter present in the object and this quantity remains constant.In essence, if the mass of an object is 2kg, this mass remains constant on planet earth and the moon.
PS: The weight of an object which is a measure of the acceleration due to gravity which differs with location and position is the constant which may be different about the object on the earth and the moon respectively.
The acceleration due to gravity on earth and the moon is 9.8m/s² and 1.6m/s² respectively.
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The rectangular boat shown below has base dimensions 10.0 cm × 8.0 cm. Each cube has a mass of 40 g, and the liquid in the tank has a density of 1.0 g/mL. How far has the boat sunk into the water?
A. 3.2 cm
B. 4.0 cm
C. 4.8 cm
D. 8.0 cm
Answer:
A that the answer I think I'm not tryna do u bad thi
Explanation:
but try it
A bungee cord is 30.0 mm long and, when stretched a distance xx, it exerts a restoring force of magnitude kxkx. Your father-in-law (mass 97.0 kgkg) stands on a platform 45.0 mm above the ground, and one end of the cord is tied securely to his ankle and the other end to the platform. You have promised him that when he steps off the platform he will fall a maximum distance of only 41.0 mm before the cord stops him. You had several bungee cords to select from, and you tested them by stretching them out, tying one end to a tree, and pulling on the other end with a force of 420 NN.
Part A When you do this, what distance will the bungee cord that you should select have stretched?
Answer:
0.65 m
Explanation:
See attachment for calculation
PLS HELP
Given the following blackbody curve graph, what color will Star A be?
Blackbody radiation graph for Star A, which has a surface temperature of 5,700 degrees Celsius. At the point where the curve peaks, the wavelength is approximately 550 nanometers, in the middle of the visible light spectrum.
Blue
Red
White
Yellow
Answer:
Yellow
Explanation:
It is for sure not red
Blackbody radiation graph for Star A, which has a surface temperature of 5,700 degrees Celsius. At the point where the curve peaks, the wavelength is approximately 550 nanometers, in the middle of the visible light spectrum. The color will Star A be Yellow.
What is Wavelength?A waveform signal that is carried in space or down a wire has a wavelength, which is the separation between two identical places (adjacent crests) in the consecutive cycles. This length is typically defined in wireless systems in meters (m), centimeters (cm), or millimeters (mm).
A blackbody's spectrum is continuous (it emits some light at all wavelengths), and it peaks at a particular wavelength. For hotter objects, the peak of the blackbody curve shifts to shorter wavelengths in a spectrum. If you conceive of a blackbody in terms of visible light, the hotter it is, the bluer its peak emission wavelength will be. For instance, the temperature of the sun is about 5800 Kelvin. With this temperature, a blackbody's peak is roughly 500 nanometers away, or the wavelength of yellow.
The color will Star A be Yellow.
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Which of the following subatomic particles determine the identify of an atom?
A. Protons
B. Neutrons
C. Electrons
D. Nucleus
LOL SOMEONE PLS HELP THIS IS PAST DUE SINCE A WEEK AGO IM STRESSING SO BAD LAMISJSKAKAO
How much force is needed to accelerate an object with a mass of 90 kg at a rate of 1.2 ms
Chief Boolie,
the jungle dweller, is out hunting for dinner when a coconut falls from a
tree and lands on his toe. If the nut fell for 1.4s, how fast was it traveling
when it hit Chief Boolie's toe? *
Answer:
13.72 m/s
Explanation:
Here, we are to determine the final velocity of the coconut. Applying first equation of free fall;
v = u + gt
where: v is the final velocity of the object, u is the initial velocity, g is the force of gravity and t is the time.
Since the coconut falls in the direction of gravity, then g = 9.8 m/[tex]s^{2}[/tex].
t = 1.4 s, u = 0 m/s
So that;
v = 0 + 9.8(1.4)
= 13.72
The coconut hits Chief Boolie's toe with a velocity of 13.72 m/s
Please help me! I will give you a thanks!
Answer:
The answer is D
Explanation:
Because it is big and can fit the truck.