Answer:
See explanation
Explanation:
The electric force exerted on the electron is given by Coulomb's law;
F= KqQ/r^2
F= force on the electron
K= constant of Coulomb's law
q= charge on the electron
Q=charge on the spherical object
r= distance between the charges
F= 9×10^9 × -1.602 × 10^-19 × (-7.2)/(0.03)^2
F= 103.8 × 10^-10/9×10^-4
F= 11.53 ×10^-6
F= 1.153 ×10^-5 N
The force in this case is repulsive
E= F/q
E= 1.153 ×10^-5 N/ 1.602 * 10-19 C
E= 0.719 × 10^14
E= 7.19 × 10^13 NC-1
The electro field around the electron is very strong.
c)
V= Kq/r
V= 9×10^9 × 1.602 × 10^-19/0.001
V= 14418 × 10^-10
V=1.4418 ×10^-6 V
W= qV
W= 1.602×10^-19 C × 1.4418 ×10^-6 V
W= 2.31 ×10^-25 J
Replacing the electron with a positron
F= 9×10^9 × 1.602 × 10^-19 × (-7.2)/(0.03)^2
F= 103.8 × 10^-10/9×10^-4
F= 11.53 ×10^-6
F= -1.153 ×10^-5 N
The force in this case is attractive
The maximum energy a bone can absorb without breaking is surprisingly small. Experimental data show that both leg bones, together, of a healthy human adult can absorb about 200 J before breaking. From what maximum height could a 75 kg person jump and land rigidly upright on both feet without breaking their legs
Answer:
0.21m
Explanation:
Note that the 200 J before breaking legs is Kinetic energy
So if potential energy= kinetic energy
Then 200J = mgh
h= 200/75x 9.8
= 0.21m so the maximum height will be 0.21m
Match the story events on the left to the correct element of plot structure on the right
Victor pretends he can speak French
climax
Victor gets his school schedule.
resolution
Victor tries to get Teresa's attention
after homeroom and at lunch.
exposition
Teresa asks Victor if he will help her in
French
rising action
Victor checks out books to learn French
and help Teresa
falling action
Explanation:
- Victor pretends he can speak French > Rising action.
- Victor gets his school schedule > Exposition.
- Victor tries to get Teresa's attention after homeroom and at lunch > Rising action.
- Teresa asks Victor if he will help her in French > Falling action.
- Victor checks out books to learn French and help Teresa > Climax
Answer:
In "Seventh Grade" by Gary Soto, the story reaches its climax when Mr. Bueller stays quiet about Victor not knowing French. When Mr. Bueller asks if anyone in the class knows French and then Victor raises his hand, although he doesn't speak the language, Mr. Bueller decides not to make fun of it, and instead, he continues with the class normally. This action had a positive effect on Victor, who considers Mr. Bueller to be a good person and motivates him to do well in French, despite of his previous attempt to impress Teresa. Regarding the other options, although they occur at the beginning (Teresa sees Victor in the lunch area and smiles at him and Victor raises his hand in French to impress Teresa) and at the end (Victor assures Teresa that helping her will not be a bother), they aren't considered to be the highest point of the conflict in the storyThe homogeneous beam of mass 5 kg indicated in the figure is in equilibrium and supported at points A and B. Calculate the reactions at the supports.
Explanation:
Sum of moments about point A:
∑τ = Iα
-mg (L/2) + Rb x = 0
-(5 kg) (10 m/s²) (0.75 m) + Rb (0.70 m) = 0
Rb = 53.6 N
Sum of forces in the y direction:
∑F = ma
Ra + Rb − mg = 0
Ra = mg − Rb
Ra = (5 kg) (10 m/s²) − 53.6 N
Ra = -3.6 N
Having established that a sound wave corresponds to pressure fluctuations in the medium, what can you conclude about the direction in which such pressure fluctuations travel
Answer:
the pressure fluctuation is LONGITUDINAL
Explanation:
Sound waves are an oscillating movement of air particles, this can be analyzed in two different, but equivalent ways, as an air oscillation and with a pressure wave due to these oscillations.
The expression for the wave is
ΔP = Δo sin (kx - wt)
Therefore, the pressure variation is in the same direction as the displacement variation, consequently the pressure fluctuation is LONGITUDINAL
When the k. E of
the object
object is increases
by 100% the momentin
the body is
increased by
how to solve plz
[tex]\sqrt{2}[/tex]Answer:
KE2 = 2 KE1
1/2 M V2^2 = 2 * (1/2 M V1^2)
V2^2 = 2 V1^2
V2 = [tex]\sqrt{2}[/tex] V1
Since momentum = M V the momentum increases by [tex]\sqrt{2}[/tex]
In this picture the bike rider starts at point A rides to his friend's house at point B(4 miles away),rides to point C (3 miles away)
and then returns to point A (5 miles away).
Explain-- What is his displacement and why? What is his total distance and how did you calculate it? In general, what is the
difference between distance and displacement?
BONUS QUESTION-Why can displacement never be greater than distance?
Answer:
bonus questions:becz displacement work only with the help of any force or object and distance is the totl length of two points
Gravel is __ than clay.
Answer:
more permeable
Explanation:
no idea, i just remember learning this in school lol.
... noisier when it's in a paper bag ...
A boy who exerts a 300-N force on the ice of a skating rink is pulled by his friend with a force of 75 N, causing the boy to accelerate across the ice. If drag and the friction from the ice apply a force of 5 N on the boy, what is the magnitude of the net force acting on him? (a) 70 N (b) 370 N (c) 80 N (d) 380 N
Answer:
Choice a. [tex]70\; \rm N[/tex], assuming that the skating rink is level.
Explanation:
Net force in the horizontal directionThere are two horizontal forces acting on the boy:
The pull of his friend, andFrictions.The boy should be moving in the direction of the pull of his friend. The frictions on this boy should oppose that motion. Therefore, the frictions on the boy would be in the opposite direction of the pull of his friend.
The net force in the horizontal direction should then be the difference between the pull of the friend, and the friction on this boy.
[tex]\text{Net force, horizontal} = 75\; \rm N - 5\; \rm N = 70\; \rm N[/tex].
Net force in the vertical directionThe net force on this boy should be zero in the vertical direction. Consider Newton's Second Law of motion. The net force on an object is proportional to its acceleration. In this question, the net force on this boy in the vertical direction should be proportional to the vertical acceleration of this boy.
However, because (by assumption) the ice rink is level, the boy has no motion in the vertical direction. His vertical acceleration will be zero. As a result, the net force on him should also be zero in the vertical direction.
Net forceTherefore, the (combined) net force on this boy would be:
[tex]\sqrt{(70\; \rm N)^2 + (0\; \rm N)^2} = 70\; \rm N[/tex].
a 16.0 kg cart is being pulled by a 95.4 N force to the right and a 36.0 N force to the left. What is the acceleration of the cart?
Answer:
The cart's acceleration is [tex]\approx 3.71\,\,\frac{m}{s^2}[/tex]
Explanation:
Let's start by finding the net force acting on the cart, and then find its acceleration using Newton's 2nd Law.
Net force = 95.4 N -36.0 N = 59.4 N
Now, since we know the cart's mass, we can use Newton's 2nd Law to find the cart's acceleration:
[tex]F=m\,*\,a\\a=\frac{F}{m} \\a=\frac{59.4}{16} \,\,\frac{m}{s^2} \\a\approx 3.71\,\,\frac{m}{s^2}[/tex]
Answer:
3.71 m/s²
Explanation:
As a professional teacher who has been assigned to teach science in an elementary school class design activities to teach source of energy to your learners
Answer:
For solar energy, I would show them how a magnifying glass works when exposed to the sun.
For wind energy, I would teach them how to make a paper windmill and explain how it works.
For the hydroelectric energy, I would have them make a plastic turbine and explain to them how to use it in rivers or streams.
For electromagnetic energy, I would tell them to rub a balloon until their hairs stand on end.
And for electricity, I would teach them how the other energy sources create electricity and what electricity works for in these times.
Explanation:
To explain something so complicated to a child is not as easy as it would be with a teenager or an adult.
To make the children learn about the forms of energy, I would use the nemotechnique rule, using short and easy-to-remember sentences and explaining with many examples about how to get each type of energy and its use, in addition to adding didactic, visual and auditory content, which are the most common types of learning in children.
A discus thrower achieves a high throw of 100m with the discus released at an angle of 30° calculate the initial speed of the discuss.
Answer:
u = 88.54 m/s
Explanation:
Given that,
A discus thrower achieves a high throw of 100 m.
Angle of projection is 30°
We need to find the initial speed of the discuss. It is a cse of projectile motion. The maximum height reached by the discus is given by :
[tex]H=\dfrac{u^2\sin^2\theta}{2g}[/tex]
u is the initial speed of the discus
So,
[tex]u^2=\dfrac{2\times 9.8\times 100}{\sin^2(30)}\\\\u=\sqrt{7840}\\\\u=88.54\ m/s[/tex]
So, the initial speed of the discus is 88.54 m/s.
Type your answer in the box.
An organ is a group of two or more
function.
that work together to perform a common function
Base of wall of water dam is made wider.Give reason.
IN A FORCE COMPRESSION GRAPH, WHAT IS THE STORED POTENTIAL ENERGY OF THE SPRING WHEN IT'S COMPRESS 0.60M ?
Answer:
La energía potencial elástica es la energía asociada con los materiales elásticos. Por ejemplo, un resorte al ser comprimido o elongado almacena energía potencial elástica y, al ser soltado, puede realizar trabajo sobre un objeto.
Para mantener el resorte comprimido o alargado una cierta longitud x, a partir de su largo natural, es necesario que, en este caso, la mano aplique una fuerza F_{M} sobre el resorte; esta fuerza es directamente proporcional a x.
Explanation:
ón conocida como ley de Hooke.
Para encontrar una expresión que describa la energía potencial asociada con la fuerza del resorte, se determina el trabajo que se requiere para comprimir el resorte desde su posición de equilibrio hasta cierta posición final arbitraria x. Debido a que la fuerza varía desde O hasta kx, se utiliza la fuerza promedio \frac{(F_{0}+F_{X})}{2}.
\[ \bar{F}=\frac{0+K X}{2}=\frac{1}{2}kx \]
fuerza-sobre-un-resorte
Fuerza sobre un resorte. La fuerza para estirar un resorte aumenta linealmente con su elongación .
El trabajo realizado por la fuerza aplicada será: W=\bar{Fx}=\frac{1}{2}kx^{2}
El trabajo realizado se almacena en el resorte comprimido en forma de energía potencial elástica como:
\[ \boxed{ Ep_{elas}=\frac{1}{2}kx^{2}} \]
Una vez que se ha comprimido o estirado el resorte respecto a su posición de equilibrio, la energía potencial elástica se puede considerar como la energía almacenada en el resorte deformado. Esta energía siempre es positiva en un objeto deformado al depender de x^{2}.
Por ejemplo, en la figura se observa que un resorte realiza trabajo sobre un bloque. El resorte que se encuentra sin deformar (a) cuando es empujado por un bloque de masa m, se comprime una distancia x (b). Cuando el bloque se suelta (c), partiendo del reposo, la energía potencial plástica almacenada en el sistema se transforma en energía cinética del bloque.
energia-potencial
A man weighing 700 NN and a woman weighing 440 NN have the same momentum. What is the ratio of the man's kinetic energy KmKmK_m to that of the woman K
Answer:
The ratio of the man's kinetic energy to that of the woman's kinetic energy is 0.629.
Explanation:
Given;
weight of the man, W = 700 N
Weight of the woman, W = 440 N
momentum is given by;
[tex]P = mv\\\\v = \frac{P}{m}[/tex]
Kinetic energy of the man;
[tex]K_m = \frac{1}{2}m_m(\frac{P_m}{m_m})^2 \\\\K_m = \frac{P_m^2}{2m_m}[/tex]
Momentum of the man is calculated as;
[tex]P_m^2 = 2m_mK_m[/tex]
The kinetic energy of the woman is given by;
[tex]K_w = \frac{P_w^2}{2m_w}[/tex]
The momentum of the woman is given;
[tex]P_w^2 = 2m_wK_w[/tex]
Since, momentum of the man = momentum of the woman
[tex]P_m^2 = P_w^2[/tex]
[tex]2m_mK_m = 2m_wK_w\\\\\frac{K_m}{K_w} = \frac{2m_w}{2m_m}\\\\\frac{K_m}{K_w} = \frac{m_w}{m_m}[/tex]
mass of the mas = 700 / 9.8 = 71.429
mass of the woman is = 440 / 9.8 = 44.898
[tex]\frac{K_m}{K_w} = \frac{44.898}{71.429}\\\\\frac{K_m}{K_w} =0.629[/tex]
Therefore, the ratio of the man's kinetic energy to that of the woman's kinetic energy is 0.629.
A helicopter goes straight up 500m from a landing pad. It then goes north 20m. Then it goes down 452m. a) What is the displacement of the helicopter?
Express as components of a vector.
x-component_____________________
y-component_____________________
b) What is the displacement of the helicopter? Express as a vector (magnitude and direction).
Answer_____________________
Answer:
a
x-component [tex]20 \ m[/tex]
y-component [tex]500 - 452 = 48 \ m[/tex]
b
Magnitude [tex]d = 52 \ m[/tex]
direction is [tex]\theta = 67.4^o[/tex]
Explanation:
From the question we are told that
The first vertical distance is [tex]y_1 = 500 \ m[/tex]
The first horizontal distance is [tex]x = 20 \ m[/tex]
The second vertical distance is [tex]y_2 = 452 \ m[/tex]
Generally the displacement is
x-component [tex]20 \ m[/tex]
y-component [tex]500 - 452 = 48 \ m[/tex]
Generally the helicopters displacement is mathematically evaluated as
[tex]d = \sqrt{ x- component ^2 + y- component ^2 }[/tex]
[tex]d = \sqrt{ 20t ^2 + 48 ^2 }[/tex]
[tex]d = 52 \ m[/tex]
The direction is the angle the displacement of the helicopter makes with the horizontal which is mathematically evaluated as
[tex]\theta = tan ^{-1}[ \frac{48}{20}][/tex]
=> [tex]\theta = tan ^{-1}[ 2.4 ][/tex]
=> [tex]\theta = 67.4^o[/tex]
An electron moves along the z-axis with vz=4.5×10^7m/s. As it passes the origin, what are the strength and direction of the magnetic field at the following (x, y, z) positions?
a. (1 cm , 0 cm, 0 cm)
b. (0 cm, 0 cm, 1 cm )
c. (0 cm, 2 cm , 1 cm )
A 2.0 moles of a monatomic ideal gas expands isothermally from state a to state b, Pa = 600 Pa, Va = 3.0 m3, and Vb = 9.0 m3.
a. Determine the pressure Pb.
b. Determine the work done on the gas during the process ab.
c. Determine the change in internal energy of the gas during the process ab.
d. Determine the heat transferred to the gas during the process ab.
Answer:
a) Pb= 200 PA
b).work done= -3600 joules
c).3600joules
D).the system works under isothermal condition so no heat was transferred
Explanation:
2.0 moles of a monatomic ideal gas expands isothermally from state a to state b, Pa = 600 Pa, Va = 3.0 m3, and Vb = 9.0 m3.
a). PbVb= PaVa
Pb= (PaVa)/VB
Pb= (600*3)/9
Pb= 1800/9
Pb= 200 PA
b). work done= n(Pb-Pa)(Vb-Va)
Work done= 2*(200-600)(9-3)
Work done= -600(6)
Work done=- 3600 Pam³
work done= -3600 joules
C). Change in internal energy I the work done on the system
= 3600joules
D).the system works under isothermal condition so no heat was transferred
An electron from a Ti ^ + 2 hydrogen ion leaps from one orbit with radius 13.25 angstrom to another orbit with radius 2.12 angstrom. determine the energy (Joule) e produced in said transition and the wavelength (in cm)
Answer:
ΔE = 59.75 A,
Explanation:
Titanium has 3 electrons in its last shell, as it is doubly ionized, it is left with a single electron in this shell, which is why it behaves like a hydrogen-type atom, consequently we can use Bohr's atomic theory
rₙ = a₀ /Z n²
Eₙ = 1k e² / 2a₀ (Z² / n²)
Where a₀ is Bohrd's atomic radius so = 0.529 núm
Let's find out what quantum number n has each orbit
rn = 13.25 A = 1.325 nm
for Titanium with atomic number 22
n² = Z rₙ / a₀
n = √ (22 (1.325 / 0.529))
n = 7.4
since N is an entry we take
n = 7
rn = 2.12 A = 0.212 nm
n = √ (22 / 0.529) 0.212
n = 3
With these values we can calculate the energy of the transition from level ne = 7 to level no = 3
ΔE = ka e2 Z2 / 2ao (1n02 - 1 / nf2)
ΔE = 9 10⁹ 1.6 10⁻¹⁹ 22² (2 0.529 10⁻⁹) (1/3² - 1/7²)
ΔED = 6.5875 10² (0.111 - 0.0204)
ΔE = 59.75 A
let us be the Planck relation between energy and frequency
E = h f
the frequency is related to the speed of light
c = λ f
f = c / λ
we substitute
E = h c /y
E = ΔE
h c /λ = E
λ = 6.63 10-34 3 108 / 59.75
λ= 3.01939 10⁻²⁴ m
λ = 3.01939 10⁻²² cm
(I) A car slows down from 28 m????s to rest in a distance of 88 m. What was its acceleration, assumed constant?
Answer:
The value is [tex]a = - 4.45 m/s^2[/tex]
Explanation:
From the question we are told that
The initial speed is [tex]u = 28 \ m/s[/tex] at a distance of [tex]s_1 = 0 \ m[/tex]
The final speed is [tex]v = 0 \ m/s[/tex] at a distance of [tex]s_2 = 88 \ m[/tex]
Generally from the kinematic equation we have that
[tex]v^2 = u^2 +2as[/tex]
=> [tex]a = \frac{v^2 - u^2 }{ 2(s_2 - s_1 )}[/tex]
=> [tex]a = \frac{0 - 28^2 }{ 2(88 - 0 )}[/tex]
=> [tex]a = - 4.45 m/s^2[/tex]
The negative sign shows that it is decelerating
A student submits the following work on reference frames and centripetal force, but she has made a few mistakes. Select all sentences that contain mistakes. All non-inertial reference frames exhibit “fictitious forces.” One of these fictitious forces is the centripetal force. For example, consider a car moving in a straight line. When the car turns to the right, the passengers experience a “force” to the right. However, there is no actual force applied. The passenger is merely continuing in a straight direction. When the car is turning, the reference frame of the car is an inertial reference frame. Hence, the passenger experiences this fictitious force, even though there is no actual force there.
Answer:
"However, there is no actual force applied."
"The passenger is merely continuing in a straight direction."
Explanation:
Am AP Phys student
A 2.0 kg block on a horizontal frictionless surface is attached to a spring whose force constant is 590 N/m. The block is pulled from its equilibrium position at x = 0 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the x-axis (horizontal). The velocity of the block at time t = 0.10 s is closest to:________.
Answer:
The value is [tex]v = -0.04 \ m/s[/tex]
Explanation:
From the question we are told that
The mass of the block is [tex]m = 2.0 \ kg[/tex]
The force constant of the spring is [tex]k = 590 \ N/m[/tex]
The amplitude is [tex]A = + 0.080[/tex]
The time consider is [tex]t = 0.10 \ s[/tex]
Generally the angular velocity of this block is mathematically represented as
[tex]w = \sqrt{\frac{k}{m} }[/tex]
=> [tex]w = \sqrt{\frac{590}{2} }[/tex]
=> [tex]w = 17.18 \ rad/s[/tex]
Given that the block undergoes simple harmonic motion the velocity is mathematically represented as
[tex]v = -A w sin (w* t )[/tex]
=> [tex]v = -0.080 * 17.18 sin (17.18* 0.10 )[/tex]
=> [tex]v = -0.04 \ m/s[/tex]
The velocity of the block at the given time is -0.04 m/s.
Angular speed of the blockThe angular speed of the block is determined by using the following wave equation;
[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega = \sqrt{\frac{590}{2} } \\\\\omega = 17.176 \ rad/s[/tex]
Velocity of the blockThe velocity of the block at the given time is calculated as follows;'
[tex]v = - A sin(\omega t)\\\\v =- 0.08 \times sin(17.176 \times 0.1)\\\\v = -0.04 \ m/s[/tex]
Learn more about velocity of simple harmonic motion here: https://brainly.com/question/17315536
Which statement is true regarding the waves shown?
A)
Doubling the frequency of the bottom waves by will cause it to match the
top waves.
B)
Cutting the frequency of the bottom waves in half will cause it to match
the top waves
C)
There is no way to match the bottom and top waves to the same frequency.
D)
Decreasing the frequency of the top waves by half will cause it to match
the bottom waves.
Answer:
The correct answer is B)
Cutting the frequency of the bottom waves in half will cause it to match top waves.
Explanation:
USATESTPREP
Cutting the frequency of the bottom waves will cause it to match the top waves. So the correct option is B.
What are waves?A propagation of disturbance, from one point to another point is called a wave. Waves are either mechanical or non-mechanical. Electromagnetic waves are non-mechanical waves.
The mechanical waves cannot travel without a medium e.g sound waves. Non-mechanical waves do not require a medium to travel, this means that they can even travel through a vacuum.
Waves are of two types. Transverse waves and longitudinal waves.
If the direction of propagation of wave is perpendicular to the direction of movement of particles of the medium, it results in a transverse wave.
If the direction of propagation of wave is parallel to the direction of movement of particles in a medium, it results in a longitudinal wave.
Five properties of waves are amplitude, frequency, wavelength, time period and speed.
Maximum displacement from the mean position is the amplitude. The number of vibrations in a fixed point in unit time is the frequency. The distance between two identical points is the wavelength. Time taken by a wave to pass through a point is the time period and the distance travelled between particular points in a unit of time is the speed.
Therefore the correct option is B.
Read more about waves, here
https://brainly.com/question/3639648
#SPJ5
Which of the following is NOT an observation? a.The apple tastes sour b.The apple weighs about 38 g c.The apple is light green in color d.Apples are the best fruit
Answer:
d apple's are the best fruit
How fast must a meter stick be moving if its length is observed to shrink to 0.57 m?
Answer:
0.8216c
Explanation:
Using the relationship
L' = L√(1 - v²/c²)
where
L = original length,
L' = observed length,
v = velocity,
c =speed.
L'/L = 0.57
Then
0.57 = √(1 - v²/c²)
1 - v²/c² = 0.57² = 0.3249
v²/c² = 1 - 0.3249 = 0.6751
v² = 0.6751c²
v = c√0.6751 = 0.8216c
Explanation:
A box is sitting on a board. The coefficient of static friction between the box and the board is 0.830216. The coefficient of kinetic friction between the box and the board is 0.326245. One side of the board is raised until the box starts sliding. Give a variable legend for this problem.
a) What is the angle at which the box starts sliding? The model for this problem:
θ=__________________________________ Answer________________________________
b) What is the magnitude of its acceleration after it starts sliding? The model for this problem:
a=__________________________________ Answer________________________________
Answer:
Explanation:
Coefficient of static friction μs = .830216
Coefficient of kinetic friction μk = .326245
a ) The angle at which the box starts sliding depends upon coefficient of static friction . If θ be the required angle
tanθ = μs
tanθ = .830216
θ = 39.7°
b )
When the box starts sliding , kinetic friction will be acting on it .
frictional force on the box = μk mg cos 39.7
net force on the box
= mg sin39.7 - μk mg cos 39.7
Applying Newton's law of motion
mg sin39.7 - μk mg cos 39.7 = m a
a = g sin39.7 - μk g cos 39.7
= 9.8 x sin 39.7 - .326245 x 9.8 x cos 39.7
= 6.26 - 2.46
= 3.8 m /s² .
A plate having an area of 0.6 m2 is sliding down the inclined plane at 300 to the horizontal with a velocity of 0.36 m/s. There is a cushion of fluid 1.8 mm thick between the plane and the plate. Calculate the viscosity of the fluid if the weight of the plate is 280 N.
Answer:
The viscosity of the fluid is 1.16 N-s/m²
Explanation:
Given that,
Area = 0.6 m²
Angle = 30°
Velocity = 0.36 m/s
Thickness = 1.8 mm
Weight = 280 N
We need to calculate the viscosity of the fluid
Using balance equation
[tex]w\sin\theta=\dfrac{\mu\times v}{t}\times A[/tex]
Put the value in the equation
[tex]280\sin30=\dfrac{\mu\times0.36}{1.8\times10^{-3}}\times(0.6)[/tex]
[tex]140=\mu\times120[/tex]
[tex]\mu=1.16\ N-s/m^2[/tex]
Hence. The viscosity of the fluid is 1.16 N-s/m².
A cannonball is fired on flat ground at 420 m/s at a 53.0° angle. how far away does it land?
Answer:
17,300 m
Explanation:
Using kinematic equations, first find the time it takes to land.
Δy = v₀ t + ½ at²
0 m = (420 sin 53.0° m/s) t + ½ (-9.8 m/s²) t²
t = 0 s or 68.5 s
The horizontal distance it moves in that time is:
Δx = v₀ t + ½ at²
Δx = (420 cos 53.0° m/s) (68.5 s) + ½ (0 m/s²) (68.5 s)²
Δx = 17,300 m
Alternatively, you can use the range equation:
R = v₀² sin(2θ) / g
R = (420 m/s)² sin(2 × 53.0°) / (9.8 m/s²)
R = 17,300 m
The distance a cannonball will land if it is fired on flat ground at 420 m/s at a 53.0° angle is 17,300 meters.
What is the distance?The complete movement of an object, regardless of direction, is referred to as distance. The amount of ground a thing travels from its starting point to its destination is also referred to as distance.
Given:
A cannonball is fired on flat ground at 420 m/s at a 53.0° angle,
Calculate the time to land on the ground as shown below,
[tex]\Delta y = v_o t +1/2 at^2[/tex]
0 m = (420 sin 53.0° m/s) t + ½ (-9.8 m/s²) t²
t = 0 s or 68.5 s
Calculate the distance as shown below,
[tex]\Delta x = v_o t +1/2 at^2[/tex]
Δ x = (420 cos 53.0° m/s) (68.5 s) + ½ (0 m/s²) (68.5 s)²
Δ x = 17,300 m
Thus, the total distance covered by the cannonball fired with a speed of 420 m/s is 17300 meters.
To know more about distance:
https://brainly.com/question/15256256
#SPJ2
A man walks 7 km, east in 2 hours and 2 km in 1 hour in the same direction. a) what is
the man's average speed for the whole journey? b) what is the man's average velocity
for the whole journey?
Explanation:
Average speed = distance / time
|v| = (7 km + 2 km) / (2 hr + 1 hr)
|v| = 3 km/hr
Average velocity = displacement / time
v = (7 km east + 2 km east) / (2 hr + 1 hr)
v = 3 km/hr east
he cans have essentially the same size, shape, and mass. Which can has more energy at the bottom of the ramp
Answer:
c. both have same energy
Explanation:
The complete question is
suppose you have two cans, one with milk, and the other with refried beans. The cans have essentially the same size, shape, and mass. If you release both cans at the same time, on a downhill ramp, which can has more energy at the bottom of the ramp? ignore friction and air resistance..
a. can with beans
b. can with milk
c. both have same energy
please explain your answer
Since both cans have the same size, shape, and mass, and they are released at the same height above the ramp, they'll possess the same amount of mechanical energy. This is because their mechanical energy, which is the combination of their potential and kinetic energy are both dependent on their mass. Also, having the same physical quantities like their size and shape means that they will experience the same environmental or physical factors, which will be balanced for both.