Answer:
One can write F = K d^-2 where K = G M m
So dF/dd = -2 K d^-3 = -2 K / d^3 (As d increases F decreases - it is opposite to the direction of F)
A battery charges a parallel-plate capacitor fully and then is removed. The plates are then slowly pulled apart. What happens to the potential difference between the plates as they are being separated?
A) It increases.
B) It decreases.
C) It remains constant.
D) It cannot be determined from the information given.
two electrons are separated by 1.10m, What is the magnitude of the electric force each electron exerts on the other?
Answer:
4.56×10¯⁷¹ N
Explanation:
From the question given above, the following data were obtained:
Distance apart (r) = 1.10 m
Force (F) =?
NOTE:
Gravitational constant (G) = 6.67×10¯¹¹ Nm² /Kg²
Mass of electron = 9.1×10¯³¹ Kg
Mass of the two elections = M₁ = M₂ = 9.1×10¯³¹ Kg
Thus, we can obtain the force of attraction between the two elections as illustrated below:
F = GM₁M₂ / r²
F = 6.67×10¯¹¹ × (9.1×10¯³¹)² / (1.1)²
F = 4.56×10¯⁷¹ N
Thus, the force of attraction between the two elections is 4.56×10¯⁷¹ N
An airplane which intends to fly due south at 250 km/hr experiences a wind blowing westward at 40 km/hr. What is the actual speed of the airplane relative to the ground?
Answer:
simple is rumple a daily ok I'll be
g A small object of mass 2.5 g and charge 18 uC is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground. What is the magnitude and direction of the electric field
Answer: [tex]1361.11\ N/C,\text{upward}[/tex]
Explanation:
Given
Mass of particle is [tex]m=2.5\ gm[/tex]
Charge of particle is [tex]q=18\ \mu C[/tex]
Electrostatic force must balance the weight of the particle
[tex]\lim_{n \to \infty} a_n \Rightarrow mg=qE\\\\\Rightarrow E=\dfrac{2.5\times 9.8\times 10^{-3}}{18\times 10^{-6}}\\\\\Rightarrow E=1361.1\ N/C[/tex]
Direction of the electric field is in upward direction such that it opposes the gravity force.
In December of 2011 they announced that a planet has been discovered in a habitable zone around a
star! It has clouds! It has twice the radius of the earth, but with the same density as earth, about 5.515 × 10^3kg/m3
. Find the new acceleration of gravity on the surface of this planet.
Explanation:
The density of earth [tex]\rho_E[/tex] is given by
[tex]\rho_E = \dfrac{M_E}{\left(\frac{4\pi}{3}R_E^3\right)}[/tex]
and in terms of this density, we can write the acceleration due to gravity on earth as
[tex]g_E =G\dfrac{M_E}{R_E^2} = \dfrac{4\pi G}{3}\rho_ER_E[/tex]
Similarly, the acceleration due to gravity [tex]g_P[/tex] on this new planet is given by
[tex]g_P = G\dfrac{M_P}{R_P^2} = G\dfrac{\frac{4\pi}{3}R_p^3\rho_P}{R_P^2}[/tex]
[tex]\:\:\:\:\:= \dfrac{4\pi G}{3}\rho_PR_P[/tex]
We know that this planet has the same density as earth and has a radius 2 times as large. We can then rewrite [tex]g_P[/tex] as
[tex]g_P = \dfrac{4\pi G}{3}\rho_E(2R_E)[/tex]
[tex]\:\:\:\:\:= 2\left(\dfrac{4\pi G}{3}\rho_ER_E\right) = 2g_E[/tex]
[tex]\:\:\:\:\:= 2(9.8\:\text{m/s}^2) = 19.6\:\text{m/s}^2[/tex]
The Michelson-Morley experiment was designed to measure Group of answer choices the velocity of the Earth relative to the ether. the relativistic momentum of the electron. the relativistic mass of the electron. the acceleration of gravity on the Earth's surface. the relativistic energy of the electron.
Answer:
The Michelson-Morley was designed to detect the motion of the earth through the ether.
No such relation was found and the speed of light is assumed to be the same in all reference frames.
The Michelson-Morley experiment was designed to measure: A. the velocity of the Earth relative to the ether.
Michelson-Morley experiment is an experiment which was first performed in Germany by the American physicist named, Albert Abraham Michelson between 1880 to 1881.
However, the experiment was later modified and refined by Michelson and Edward W. in 1887.
The main purpose of the Michelson-Morley experiment was to measure the velocity of planet Earth relative to the luminiferous ether, which is a medium in space that is hypothetically said to carry light waves.
In conclusion, the Michelson-Morley experiment was designed to measure the velocity of the Earth relative to the hypothetical luminiferous ether.
Read more: https://brainly.com/question/13187705
1. A body moving with uniform acceleration of 10 m/s2 covers a distance of 320 m. if its initial velocity was 60 m/s. Calculate its final velocity.
Answer:
v² = u² + 2as
v² = 3600 + 6400
v² = 10000
v = 100
Explanation:
final velocity is 100 m/s
According to third equation of kinematics
[tex]\boxed{\sf v^2-u^2=2as}[/tex]
[tex]\\ \sf\longmapsto v^2=u^2+2as[/tex]
[tex]\\ \sf\longmapsto v^2=(60)^2+2(10)(320)[/tex]
[tex]\\ \sf\longmapsto v^2=3600+3400[/tex]
[tex]\\ \sf\longmapsto v^2=10000[/tex]
[tex]\\ \sf\longmapsto v=\sqrt{10000}[/tex]
[tex]\\ \sf\longmapsto v=100m/s[/tex]
With respect to a right handed Cartesian coordinate system and given that . A = 4i + k and B = 2i + j _ 3k find A cross B
Using the left-hand rule,
[tex](4\,\vec\imath+\vec k)\times(2\,\vec\imath+\vec\jmath-3\,\vec k) = \begin{vmatrix}\vec\imath&\vec\jmath&\vec k\\4&0&1\\2&1&-3\end{vmatrix} = -\vec\imath+14\,\vec\jmath+4\,\vec k[/tex]
Then in the right-handed rectangular coordinates, the cross product is the negative of this,
[tex]\boxed{\vec\imath-14\,\vec\jmath-4\,\vec k}[/tex]
Which one of the following physical quantities has its S.I. unit m/s?
(i) Acceleration
(ii) Velocity
(iii) Force
(iv) Density
Answer:
velocity is the answer of this question.
Answer:
Velocity is the right answer ok
You are to connect resistors R1 andR2, with R1 >R2, to a battery, first individually, then inseries, and then in parallel. Rank those arrangements according tothe amount of current through the battery, greatest first. (Useonly the symbols > or =, for exampleseries>R1=R2>parallel.)
Answer:
The current is more in the parallel combination than in the series combination.
Explanation:
two resistances, R1 and R2 are connected to a battery of voltage V.
When they are in series,
R = R1 + R2
In series combination, the current is same in both the resistors, and it is given by Ohm's law.
V = I (R1 + R2)
[tex]I = \frac{V}{R_1 + R_2}[/tex]..... (1)
When they are connected in parallel.
the voltage is same in each resistor.
The effective resistance is R.
[tex]R = \frac{R_1R_2}{R_1 + R_2}[/tex]
So, the current is
[tex]I = \frac{V(R_1+R_2)}{R_1 R_2}[/tex]..... (2)
So, the current is more is the parallel combination.
What is the energy of a photon with a frequency of 3.6 × 1015 Hz? Planck’s constant is 6.63 × 10–34 J•s.
1.8 × 10–49 J
2.4 × 10–19 J
1.8 × 10–18 J
2.4 × 10–18 J
We know
[tex]\boxed{\sf E=hv}[/tex]
[tex]\\ \sf\longmapsto E=6.63\times 10^{-34}J\times 3.6\times 10^{15}s^{-1}[/tex]
[tex]\\ \sf\longmapsto E=23.86\times 10^{-19}J[/tex]
[tex]\\ \sf\longmapsto E=2.38\times 10^{-18}J[/tex]
[tex]\\ \sf\longmapsto E=2.4\times 10^{-18}J[/tex]
Answer:
D!!!!!
Explanation:
From the given picture What's the force? And where did it happen? (at least 2 forces)
Answer:
the force happens on the wall and couch
Explanation:
she is using her arm strength to lift and hold
While standing at the edge of the roof of a building, a man throws a stone upward with an initial speed of 6.63 m/s. The stone subsequently falls to the ground, which is 14.5 m below the point where the stone leaves his hand.
At what speed does the stone impact the ground? Ignore air resistance and use =9.81 m/s2 for the acceleration due to gravity.
How much time is the stone in the air?
elapsed time:
Answer:
Speed=28.1m/s(to 3s.f.) , Time=2.19s(to 3s.f.)
Explanation:
Time=Distance/Speed
=14.5/6.63
=2.19s(to 3s.f.)
Acceleration=Final Velocity(v)-Initial Velocity(u)/Time
9.81=v-6.63/2.19
v-6.63=21.5
v=28.1m/s
What is the incorrect statement regarding the isotopes of the same element?
1) Electronic configuration is equal
2) Mass number is equal
3) Number of protons are equal
4) Number of electrons are equal
Answer:
1231
Explanation:
Cho dòng điện xoay chiều trong sản xuất và sinh hoạt ở nước ta có tần số f = 50Hz. Tính chu kỳ T và tần số góc ω?
Answer:
T = 1/f = 1/50(s)
ω = 2πf = 100π (rad/s)
(vote 5 sao nhó :3 )
1 Poin Question 4 A 85-kg man stands in an elevator that has a downward acceleration of 2 m/s2. The force exerted by him on the floor is about: (Assume g = 9.8 m/s2) А ON B 663 N C) 833 N D) 1003 N
Answer:
D) 1003 N
Explanation:
Given the following data;
Mass of man = 85 kg
Acceleration of elevator = 2 m/s²
Acceleration due to gravity, g = 9.8 m/s²
To find the force exerted by the man on the floor;
Force = mg + ma
Describe how the words Science and optics would appear when viewed in a plane mirror?
Answer:
Lateral inversion will occur in a plane mirror.
Explanation:
When words are displayed in a plane or flat mirror, the result is that if the words are displayed left, they change to right and if they were normally displayed right, they change to left. This phenomenon is known as lateral inversion. So, this will apply to the words, Science and optics. Only the sides will be interchanged.
A plane mirror reflects light, therefore, the image that is produced by it remains the same size. The image produced will not appear upside down. Only the sides will be interchanged.
Difference between uniform motion and non uniform motion
Answer:
When an object covers equal distance in an equal interval of time, it is uniform motion but when an object covers unequal distance in an equal interval of time, it is called non uniform motion.
The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way out of the capacitor. Find the new energy of the capacitor, U3. Express your answer numerically in joules.
The question is incomplete. The complete question is :
A dielectric-filled parallel-plate capacitor has plate area A = 10.0 cm2 , plate separation d = 10.0 mm and dielectric constant k = 3.00. The capacitor is connected to a battery that creates a constant voltage V = 15.0 V . Throughout the problem, use ϵ0 = 8.85×10−12 C2/N⋅m2 .
Find the energy U1 of the dielectric-filled capacitor. I got U1=2.99*10^-10 J which I know is correct. Now I need these:
1. The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy U2 of the capacitor at the moment when the capacitor is half-filled with the dielectric.
2. The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way out of the capacitor. Find the new energy of the capacitor, U3.
Solution :
Given :
[tex]A = 10 \ cm^2[/tex]
[tex]$=0.0010 \ m^2$[/tex]
d = 10 mm
= 0.010 m
Then, Capacitance,
[tex]$C=\frac{k \epsilon_0 A}{d}$[/tex]
[tex]$C=\frac{8.85 \times 10^{12} \times 3 \times 0.0010}{0.010}$[/tex]
[tex]$C=2.655 \times 10^{12} \ F$[/tex]
[tex]$U_1 = \frac{1}{2}CV^2$[/tex]
[tex]$U_1 = \frac{1}{2} \times 2.655 \times 10^{-12} \times (15V)^2$[/tex]
[tex]$U_1=2.987 \times 10^{-10}\ J$[/tex]
Now,
[tex]$C_k=\frac{1}{2} \frac{k \epsilon_0}{d} \times \frac{A}{2}$[/tex]
And
[tex]$C_{air}=\frac{1}{2} \frac{\epsilon_0}{d} \times \frac{A}{2}$[/tex]
In parallel combination,
[tex]$C_{eq}= C_k + C_{air}$[/tex]
[tex]$C_{eq} = \frac{1}{2} \frac{\epsilon_0 A}{d}(1+k)$[/tex]
[tex]$C_{eq} = \frac{1}{2} \times \frac{8.85 \times 10^{-12} \times 0.0010}{0.01} \times (1+3)$[/tex]
[tex]$C_{eq} = 1.77 \times 10^{-12}\ F$[/tex]
Then energy,
[tex]$U_2 =\frac{1}{2} C_{eq} V^2$[/tex]
[tex]$U_2=\frac{1}{2} \times 1.77 \times 10^{-12} \times (15V)^2$[/tex]
[tex]$U_2=1.99 \times 10^{-10} \ J$[/tex]
b). Now the charge on the [tex]\text{capacitor}[/tex] is :
[tex]$Q=C_{eq} V$[/tex]
[tex]$Q = 1.77 \times 10^{-12} \times 15 V$[/tex]
[tex]$Q = 26.55 \times 10^{-12} \ C$[/tex]
Now when the capacitor gets disconnected from battery and the [tex]\text{dielectric}[/tex] is slowly [tex]\text{removed the rest}[/tex] of the way out of the [tex]\text{capacitor}[/tex] is :
[tex]$C_3=\frac{A \epsilon_0}{d}$[/tex]
[tex]$C_3 = \frac{0.0010 \times 8.85 \times 10^{-12}}{0.01}$[/tex]
[tex]$C_3=0.885 \times 10^{-12} \ F$[/tex]
[tex]$C_3 = 0.885 \times 10^{-12} \ F$[/tex]
Without the dielectric,
[tex]$U_3=\frac{1}{2} \frac{Q^2}{C}$[/tex]
[tex]$U_3=\frac{1}{2} \times \frac{(25.55 \times 10^{-12})^2}{0.885 \times 10^{-12}}$[/tex]
[tex]$U_3=3.98 \times 10^{-10} \ J$[/tex]
3) A lead bullet initially at 30 C just melts upon striking a target. Assuming that all of the initial kinetic energy of the bullet goes into the internal energy of the bullet to raise its temperature and melt it, calculate the speed of the bullet upon impact. (Specific heat of lead is 0.128 kJ/kg K and lead latent heat of fusion is 24.7 kJ/Kg and melting point of lead is 600 K).
Answer:
The speed of bullet is 354.2 m/s
Explanation:
initial temperature, T = 30 degree C
specific heat, c = 128 J/kg K
Latent heat, L = 24.7 x 1000 J/kg
melting point = 600 K = 327 degree C
Let the mass is m and the speed is v.
Kinetic energy = heat used to increase the temperature + Heat used to melt
[tex]\frac{1}{2} mv^2 = m c (T' - T) + m L\\\\0.5 v^2 = 128 \times (327 - 30) + 24.7\times 1000\\\\0.5 v^2 = 38016 + 24700 \\\\0.5 v^2 = 62716\\\\v = 354.2 m/s[/tex]
The lines in the emission spectrum of hydrogen result from __________.
a. energy given off in the form of visible light when an electron moves from a higher energy state to a lower energy state
b. protons given off when hydrogen burns
c. electrons given off by hydrogen as it cools
d. electrons given off by hydrogen when it burns
e. decomposing hydrogen atoms.
Answer:
Option (a) is correct.
Explanation:
The lines in the emission spectrum of hydrogen is due to the transfer of electrons form higher energy levels to the lower energy levels.
When the electrons transfer from one level of energy that is higher level of energy to the other means to the lower level of energy then they emit some photons which having the frequency or the wavelength in the visible region.
Electromagnetic radiation with a wavelength of 525 nm appears as green light to the human eye. Calculate the frequency of this light. Be sure to include units in your answer.
Answer:
5.71×10¹⁴ Hz
Explanation:
Applying,
v = λf................. Equation 1
Where v = speed of the electromagnetic radiation, λ = wavelength of the electromagnetic radiation, f = frequency
make f the subject of the equation
f = v/λ............. Equation 2
From the question,
Given: λ = 525 nm = 5.25×10⁻⁷ m,
Constant: Speed of electromagnetic wave (v) = 3.0×10⁸ m/s
Substitute these values into equation 2
f = (3.0×10⁸)/(5.25×10⁻⁷)
f = 5.71×10¹⁴ Hz
Hence the frequency of light is 5.71×10¹⁴ Hz
A rock is suspended by a light string. When the rock is in air, the tension in the string is 37.8 N. When the rock is totally immersed in water, the tension is 32.0 N. When the rock is totally immersed in an unknown liquid, the tension is 20.2 N. What is the Density of the unknown liquid?
When the rock is suspended in the air, the net force on it is
∑ F₁ = T₁ - m₁g = 0
where T₁ is the magnitude of tension in the string and m₁g is the rock's weight. So
T₁ = m₁g = 37.8 N
When immersed in water, the tension reduces to T₂ = 32.0 N. The net force on the rock is then
∑ F₂ = T₂ + B₂ - m₁g = 0
where B₂ is the magnitude of the buoyant force. Then
B₂ = m₁g - T₂ = 37.8 N - 32.0 N = 5.8 N
B₂ is also the weight of the water that was displaced by submerging the rock. Let m₂ be the mass of the displaced water; then
5.8 N = m₂g ==> m₂ ≈ 0.592 kg
If one takes the density of water to be 1.00 g/cm³ = 1.00 × 10³ kg/m³, then the volume of water V that was displaced was
1.00 × 10³ kg/m³ = m₂/V ==> V ≈ 0.000592 m³ = 592 cm³
and this is also the volume of the rock.
When immersed in the unknown liquid, the tension reduces further to T₃ = 20.2 N, and so the net force on the rock is
∑ F₃ = T₃ + B₃ - m₁g = 0
which means the buoyant force is
B₃ = m₁g - T₃ = 37.8 N - 20.2 N = 17.6 N
The mass m₃ of the liquid displaced is then
17.6 N = m₃g ==> m₃ ≈ 1.80 kg
Then the density ρ of the unknown liquid is
ρ = m₃/V ≈ (1.80 kg)/(0.000592 m³) ≈ 3040 kg/m³ = 3.04 g/cm³
A physics instructor wants to project a spectrum of visible-light colors from 400 nm to 700 nm as part of a classroom demonstration. She shines a beam of white light through a diffraction grating that has 600 lines per mm, projecting a pattern on a screen 2.9 m behind the grating.
Required:
How wide is the spectrum corresponding to m=1?
Answer:
Dr = 263 10⁻⁶ m
Explanation:
The diffraction pattern for constructive interference is described by
a sin θ = m λ
in this it indicates that the order of diffraction is m = 1
Let's use a direct proportion rule to find the separation of two slits. If there are 600 lines in 1 me, what is the distance between 2 slits
a = 2 lines 1/600
a = 2/600
a = 3.33 10⁻³ mm = 3.33 10⁻⁴ cm
let's use trigonometry
tan θ = y / L
as the measured angles are small
tan θ = sin θ / cos θ sin θ
sin θ = y / L
we substitute
a y/L = λ
y = λ L / a
for λ = 400 10-9 m
I = 400 10⁻⁹ 2.9 / 3.33 10⁻³
i = 346.89 10⁻⁶ m
f
or λ = 700 nm
y_f = 700 10⁻⁻⁹ 2.9 / 3.33 10⁻³
y_f = 609.609 10⁻⁶ m
the separation of this spectrum
Δr = v_f - i
Dr = (609.609 - 346) 10 ⁻⁶
Dr = 263 10⁻⁶ m
A car is traveling at 118 km/h when the driver sees an accident 85 m ahead and slams on the brakes. What minimum constant deceleration is required to stop the car in time to avoid a pileup
Answer:
The constant minimum deceleration required to stop the car in time to avoid pileup is 6.32 m/s²
Explanation:
From the question, the car is traveling at 118 km/h, that is the initial velocity, u = 118km/h
The distance between the car and the accident at the moment when the driver sees the accident is 85 m, that is s = 85 ,
Since the driver slams on the brakes and the car will come to a stop, then the final velocity, v = 0 km/h = 0 m/s
First, convert 118 km/h to m/s
118 km/h = (118 × 1000) /3600 = 32.7778 m/s
∴ u = 32.7778 m/s
Now, to determine the deceleration, a, required to stop,
From one of the equations of motion for linear motion,
v² = u² + 2as
Then
0² = (32.7778)² + 2×a×85
0 = 1074.3841 + 170a
∴ 170a = - 1074.3841
a = - 1074.3841 / 170
a = - 6.3199
a ≅ - 6.32 m/s²
Hence, the constant minimum deceleration required to stop the car in time to avoid pileup is 6.32 m/s²
Develop a hypothesis regarding one factor you think might affect the period of a pendulum or an oscillating mass on a spring. Potential factors include the mass, the spring constant, and the length of the pendulum's string. Write down your hypothesis. 2. Design a controlled experiment to test your hypothesis. Take extreme care to keep all factors constant except the variable you are testing.
Answer:
A hypothesis for the period of a pendulum is:
"The period of the pendulum varies with its length"
Explanation:
A hypothesis for the period of a pendulum is:
"The period of the pendulum varies with its length"
To test this hypothesis we can carry out a measurement of a simple pendulum keeping the angle fixed, in general the angle used is about 5º since when placing this value in radiand and the sine of this angle they differ little <5%. therefore measured the time of some oscillations, for example about 10 oscillations, changing the length of the pendulum to test the hypothesis.
If the hypothesis and the model used is correct, the relationship to be tested is
T² =(4π² /g) L
by making a graph of the period squared against the length if obtaining, os a line, the hypothesis is tested.
In the lab room, you are sitting in an office chair with wheels while holding onto a force sensor, and the chair is at rest. One end of a lightweight string is attached to the force sensor, and your lab partner is holding the other end of the string. Your lab partner then moves away from you, pulling on the string. Describe how your lab partner must move for the force sensor to read a constant force. Explain
Answer:
a circular motion a constant force can be measured
Explanation:
The force is expressed by the relation
F = m a
The bold are vectros.
Therefore, when your partner moves away, he has a reading of a force, so that this force remains constant there must be an acceleration at all times, one way to achieve these is with a circular motion with constant speed, in this case the module of the velocity is constant, but the direction changes at each point and there is an acceleration at each point.
Consequently with a circular motion a constant force can be measured
The force depends on the acceleration, hence the force will be constant during the circular motion for constant acceleration.
What is Force?A force can be defined as an influence that can change the motion of an object. The force is expressed by the relation
[tex]F = m a[/tex]
The force is dependent on the mass and acceleration of the object. The acceleration is a vector quantity, so the force will be a vector quantity.
Given that, in a lab room, you are sitting on a wheelchair at one end and at the other end, lab partner then moves away from you, pulling on the string that is attached to the force sensor.
When the lab partner moves away and pulled the string, there must be an acceleration during the motion. If the lab partner moves in a circular motion, then the velocity is constant but the direction changes at each point. There is an acceleration at each point that will be constant.
As the force depends on the acceleration, hence the force will be constant during the circular motion for constant acceleration.
To know more about the force, follow the link given below.
https://brainly.com/question/26115859.
Two charged objects attract each other with a force 1.0 N. What happens to the force between them if one charge is increased by a factor of 2, the other charge is increased by a factor of 4, and the separation distance between their centers is reduced to 1/4 its original value
Answer:
F' = 128 N
Explanation:
The electrostatic force of attraction between two charges is given by Colomb's Law, as follows:
[tex]F = \frac{kq_1q_2}{r^2}\\\\[/tex]
where,
F = Force of attraction = 1 N
G = universal gravitational constant
q₁ = magnitude of the first charge
q₂ = magnitude of the second charge
r = distance between charges
Therefore,
[tex]1\ N = \frac{kq_1q_2}{r^2}[/tex] --------------------- eq(1)
Now, we apply the changes given in the question:
[tex]F' = \frac{k(2q_1)(4q_2)}{(\frac{1}{4}r)^2}\\\\F' = 128(\frac{kq_1q_2}{r^2})[/tex]
using eq (1):
F' = 128(1 N)
F' = 128 N
b. A bird in air looks a fish vertically below it inside the water from a distance 5m from surface of water and fish lies at depth 4m from the surface of water. IF Mw= 4/3, what is the distance of fish as observed by bird?
Answer:
the distance of the fish (as seen by the bird) is greater than the actual distance.
Reason-
it is due to the apparent depth and differences between the refractive indices.
Have a nice day!
A skateboarder travels on a horizontal surface with an initial velocity of 3.6 m/s toward the south and a constant acceleration of 1.8 m/s^2 toward the east. Let the x direction be eastward and the y direction be northward, and let the skateboarder be at the origin at t=0.
a. What is her x position at t=0.60s?
b. What is her y position at t=0.60s?
c. What is her x velocity component at t=0.60s?
d. What is her y velocity component at t=0.60s?
Answer:
a) The x-position of the skateboarder is 0.324 meters.
b) The y-position of the skateboarder is -2.16 meters.
c) The x-velocity of the skateboard is 1.08 meters per second.
d) The y-velocity of the skateboard is -3.6 meters per second.
Explanation:
a) The x-position of the skateboarder is determined by the following expression:
[tex]x(t) = x_{o} + v_{o,x}\cdot t + \frac{1}{2}\cdot a_{x} \cdot t^{2}[/tex] (1)
Where:
[tex]x_{o}[/tex] - Initial x-position, in meters.
[tex]v_{o,x}[/tex] - Initial x-velocity, in meters per second.
[tex]t[/tex] - Time, in seconds.
[tex]a_{x}[/tex] - x-acceleration, in meters per second.
If we know that [tex]x_{o} = 0\,m[/tex], [tex]v_{o,x} = 0\,\frac{m}{s}[/tex], [tex]t = 0.60\,s[/tex] and [tex]a_{x} = 1.8\,\frac{m}{s^{2}}[/tex], then the x-position of the skateboarder is:
[tex]x(t) = 0\,m + \left(0\,\frac{m}{s} \right)\cdot (0.60\,s) + \frac{1}{2}\cdot \left(1.8\,\frac{m}{s^{2}} \right) \cdot (0.60\,s)^{2}[/tex]
[tex]x(t) = 0.324\,m[/tex]
The x-position of the skateboarder is 0.324 meters.
b) The y-position of the skateboarder is determined by the following expression:
[tex]y(t) = y_{o} + v_{o,y}\cdot t + \frac{1}{2}\cdot a_{y} \cdot t^{2}[/tex] (2)
Where:
[tex]y_{o}[/tex] - Initial y-position, in meters.
[tex]v_{o,y}[/tex] - Initial y-velocity, in meters per second.
[tex]t[/tex] - Time, in seconds.
[tex]a_{y}[/tex] - y-acceleration, in meters per second.
If we know that [tex]y_{o} = 0\,m[/tex], [tex]v_{o,y} = -3.6\,\frac{m}{s}[/tex], [tex]t = 0.60\,s[/tex] and [tex]a_{y} = 0\,\frac{m}{s^{2}}[/tex], then the x-position of the skateboarder is:
[tex]y(t) = 0\,m + \left(-3.6\,\frac{m}{s} \right)\cdot (0.60\,s) + \frac{1}{2}\cdot \left(0\,\frac{m}{s^{2}}\right)\cdot (0.60\,s)^{2}[/tex]
[tex]y(t) = -2.16\,m[/tex]
The y-position of the skateboarder is -2.16 meters.
c) The x-velocity of the skateboarder ([tex]v_{x}[/tex]), in meters per second, is calculated by this kinematic formula:
[tex]v_{x}(t) = v_{o,x} + a_{x}\cdot t[/tex] (3)
If we know that [tex]v_{o,x} = 0\,\frac{m}{s}[/tex], [tex]t = 0.60\,s[/tex] and [tex]a_{x} = 1.8\,\frac{m}{s^{2}}[/tex], then the x-velocity of the skateboarder is:
[tex]v_{x}(t) = \left(0\,\frac{m}{s} \right) + \left(1.8\,\frac{m}{s} \right)\cdot (0.60\,s)[/tex]
[tex]v_{x}(t) = 1.08\,\frac{m}{s}[/tex]
The x-velocity of the skateboard is 1.08 meters per second.
d) As the skateboarder has a constant y-velocity, then we have the following answer:
[tex]v_{y} = -3.6\,\frac{m}{s}[/tex]
The y-velocity of the skateboard is -3.6 meters per second.
