Answer:
Explanation:
Formula of lateral displacement
[tex]S_{lateral}=\frac{t}{cosr} \times sin(i-r)[/tex]
t is thickness of slab , i and r are angle of incidence and refraction respectively .
Given t = .45 m
sin i / sin r = 2.2
sin 46 / sin r = 2.2
sin r = .719 / 2.2 = .327
r = 19°
[tex]S_{lateral}=\frac{t}{cosr} \times sin(i-r)[/tex]
[tex]S_{lateral}=\frac{.45}{cos19} \times sin(46-19)[/tex]
= .45 x .454 / .9455
= .216 m
= 21.6 cm .
The displacement, D, of the beam when it exits the slab is; 21.65 cm.
We are given;
Refractive index of slab material; nm = 2.2
Thickness of slab; t = 0.45 m
Refractive index of air; na = 1
Angle of incidence; θa = 46°
From snell's law, we can calculate the angle of refraction from;
na × sin θa = nm × sin θm
Thus;
1 × sin 46 = 2.2 × sin θm
0.7193 = 2.2 × sin θm
sin θm = 0.7193/2.2
θm = sin^(-1) 0.32695
θm = 19.08°
Formula for the displacement of the beam is;
D = (t/cos θm) × sin (θa - θm)
Plugging in the relevant values gives;
D = (0.45/cos 19.08) × sin (46 - 19.08)
D = 0.4783 × 0.4527
D = 0.2165m = 21.65 cm
Read more at; https://brainly.com/question/24875145
Four equal masses m are located at the corners of a square of side L, connected by essentially massless rods. Find the rotational inertia of this system about an axis (a) that coincides with one side and (b) that bisects two opposite sides.
Answer:
Explanation:
a )
Moment of inertial of four masses about axis that coincides with one side :
Out of four masses . location of two masses will lie on the axis so their moment of inertia will be zero .
Moment of inertia of the two remaining masses
= m L² + m L²
= 2 mL²
b )
Axis that bisects two opposite sides
Each of the four masses will lie at a distance of L / 2 from this axis so moment of inertia of the four masses
= 4 x m x ( L/2 )²
= 4 x mL² / 4
= m L² .
Balls A and B attract each other gravitationally with a force of magnitude F at distance R. If we triple the mass of ball B and triple the separation of the balls to 3R, what is the magnitude of their attractive force now
Answer:
F₂ = 1/3 FExplanation:
Using the law of gravitation of force to solve this question. The law states that the Force of attraction between two masses is directly proportional to the product of their masses and inversely proportional to the square of the distances between them.
Mathematically, F = GMaMb/R² ... 1
G is the gravitational constant
Ma and Mb are the masses of the balls
R is the distance between the balls
If the mass of ball B is tripled and the magnitude of the separation of the balls is increased to 3R, the force between them will be;
F₂ = GMa(3Mb)/(3R)²
F₂ = 3GMaMb/9R² ... 2
Dividing equation 1 by 2 we will have;
F₂/F = (3GMaMb/9R²)/GMaMb/R²
F₂/F = 3GMaMb/9R² * GMaMb/R²
F₂/F = 3/9
F₂/F = 1/3
F₂ = 1/3 F
This shows that the magnitude of the new attractive force is one-third that of the initial attractive force
Consider the following three objects, each of the same mass and radius:
(1) a solid sphere
(2) a solid disk
(3) a hoop
All three are released from rest at the top of an inclined plane. The three objects proceed down the incline undergoing rolling motion without slipping. Use work-kinetic energy theorem to determine which object will reach the bottom of the incline first.
a) 1, 2, 3
b) 2, 3, 1
c) 3, 1, 2
d) 3, 2, 1
e) All three reach the bottom at the same time.
Answer:
Explanation:a 1
Rays that pass through a lens very close to the principle axis are more sharply focused than those that are very far from the axis. This spherical aberration helps us understand why:_______
Answer: it is easier to read in bright light than dim light.
Explanation:
The ray of light is the direction that is used by light in travelling through a medium. Rays that pass through a lens very close to the principle axis are more sharply focused than those that are very far from the axis.
Because of the fact that the rays are close to the principle axis, the spherical aberration helps us to understand the reason why it is easier for people to read in bright light than readin iin dim light.
An electric field can be created by a single charge or a distribution of charges. The electric field a distance from a point charge has magnitude E = k|q'|/r^2.
The electric field points away from positive charges and toward negative charges. A distribution of charges creates an electric field that can be found by taking the vector sum of the fields created by individual point harges. Note that if a charge is placed in an electric field created by q', q will not significantly affect the electric field if it is small compared to q'. Imagine an isolated positive point charge with a charge Q (many times larger than the charge on a single electron).
1. There is a single electron at a distance from the point charge. On which of the following quantities does the force on the electron depend?
a. the distance between the positive charge and the electron
b. the charge on the electron
c. the mass of the electron
d. the charge of the positive charge
e. the mass of the positive charge
f. the radius of the positive charge
g. the radius of the electron
2. For the same situation as in Part A, on which of the following quantities does the electric field at the electron's position depend?
a. the distance between the positive charge and the electron
b. the charge on the electron
c. the mass of the electron
d. the charge of the positive charge
e. the mass of the positive charge
f. the radius of the positive charge
g. the radius of the electron
Answer:
a) true.
b) True
c) False. In the equation above the mass does not appear
d) True
e) False. Mass does not appear in the equation
f) False. The load even when distributed in the space can be considered concentrated in the center
Explanation:
1. The electric force is given by the relation
F = k Q e / r2
where k is the Coulomb constant, Q the charge used, e the charge of the electron and r the distance between the two.
The strength depends on:
a) true.
b) True
c) False. In the equation above the mass does not appear
d) True
e) False. Mass does not appear in the equation
f) False. The load even when distributed in the space can be considered concentrated in the center
two.
a) True
b) Treu
c) Fail
f) false
For a single electron located at a distance from a positive charge, we have:
1. The force on the electron depends on the distance between it and the positive charge (option a) and the charge of both particles (option b and d).
2. The electric field at the electron's position depends on the distance between the positive charge and it (option a) and the charge of the positive particle (option d).
Part 1
The force on a single electron at a distance from the point charge is given by Coulomb's law:
[tex] F = \frac{Kq_{1}q_{2}}{r^{2}} [/tex] (1)
Where:
K: is the Coulomb's constant q₁: is the charge of the positive chargeq₂: is the charge of the electrond: is the distance between the positive charge and the electronAs we can see in equation (1), the force on the electron by the positive charge depends on both charges q₁ and q₂, and the distance, so the correct options are:
a. The distance between the positive charge and the electron
b. The charge on the electron
d. The charge of the positive charge
The other options (c, e, f, and g) are incorrect because the electric force does not depend on the particles' masses or their radii.
Part 2The electric field (E) at a distance "r" from a point charge is given by:
[tex] E = \frac{Kq_{1}}{r^{2}} [/tex] (2)
From equation (2), we can see that the electric field is directly proportional to the charge and inversely proportional to the distance of interest (r).
The electric field at the electron's position is given by the one produced by the positive charge, so the correct options are:
a. The distance between the positive charge and the electron
d. The charge of the positive charge
The other options (b, c, e, f, and g) are incorrect because the electric field is independent of the mass of the charges involved and their radii.
Therefore, the correct options for part 1 are a, b, and d and for part 2 are a and d.
Learn more about the electric field here:
brainly.com/question/13308086
I hope it helps you!
On a certain planet a body is thrown vertically upwards with an initial speed of 40 m / s. If the maximum height was 100 m, the acceleration due to gravity is
a) 15 m / s 2
b) 12.5 m / s 2
c) 8 m / s 2
d) 10 m / s 2
Answer:
C) 8 m/s²
Explanation:
Given:
v₀ = 40 m/s
v = 0 m/s
Δy = 100 m
Find: a
v² = v₀² + 2aΔy
(0 m/s)² = (40 m/s)² + 2a (100 m)
a = -8 m/s²
A 0.2 kg rubber ball is dropped from the window of a building. It strikes the sidewalk below at 30 m/s and rebounds at 20 m/s. The magnitude of the change in momentum of the ball as a result of the collision with the sidewalk is _______.
Answer:
10 kgm/s
Explanation:
Change in momentum: This can be defined as the product of mass and change in velocity. The S.I unit of change in momentum is Kgm/s.
From the question,
ΔM = m(v-u)...................... Equation 1
Where ΔM = change in momentum, u = initial velocity, v = final velocity.
Note: Let upward direction be negative, and downward direction be positive.
Given: m = 0.2 kg, v = -20 m/s, u = 30 m/s
Substitute into equation 1
ΔM = 0.2(-20-30)
ΔM = 0.2(-50)
ΔM = -10 kgm/s.
The negative sign shows that the change in momentum is Upward
The magnitude of the change in momentum of the ball as a result of the collision with the sidewalk is -10 kg-m/s.
Given data:
The mass of rubber ball is, m = 0.2 kg.
The initial speed of ball is, u = 30 m/s.
The final rebounding speed of ball is, v = - 20 m/s ( Negative sign shows that during the rebounding, the ball changes its direction)
The momentum of any object is defined as the product of mass and change in velocity. The S.I unit of momentum is Kg-m/s. And the expression for the change in momentum is given as,
[tex]p= m ( v-u)[/tex]
Solving as,
[tex]p= 0.2 \times ( -20-30)\\\\p=-10 \;\rm kg.m/s[/tex]
Thus, we can conclude that the magnitude of the change in momentum of the ball as a result of the collision with the sidewalk is -10 kg-m/s.
Learn more about the change in momentum here:
https://brainly.com/question/904448
A scientist is testing the seismometer in his lab and has created an apparatus that mimics the motion of the earthquake felt in part (a) by attaching the test mass to a spring. If the test mass weighs 13 N, what should be the spring constant of the spring the scientist use to simulate the relative motion of the test mass and the ground from part (a)?
Complete Question
The complete question is shown on the first uploaded image
Answer:
a
[tex]a_{max} = 0.00246 \ m/s^2[/tex]
b
[tex]k =722.2 \ N/m[/tex]
Explanation:
From the question we are told that
The amplitude is [tex]A = 1.8 \ cm = 0.018 \ m[/tex]
The period is [tex]T = 17 \ s[/tex]
The test weight is [tex]W = 13 \ N[/tex]
Generally the radial acceleration is mathematically represented as
[tex]a = w^2 r[/tex]
at maximum angular acceleration
[tex]r = A[/tex]
So
[tex]a_{max} = w^2 A[/tex]
Now [tex]w[/tex] is the angular velocity which is mathematically represented as
[tex]w = \frac{2 * \pi }{T}[/tex]
Therefore
[tex]a_{max} = [\frac{2 * \pi}{T} ]^2 * A[/tex]
substituting values
[tex]a_{max} = [\frac{2 * 3.142}{17} ]^2 * 0.018[/tex]
[tex]a_{max} = 0.00246 \ m/s^2[/tex]
Generally this test weight is mathematically represented as
[tex]W = k * A[/tex]
Where k is the spring constant
Therefore
[tex]k = \frac{W}{A}[/tex]
substituting values
[tex]k = \frac{13}{0.018}[/tex]
[tex]k =722.2 \ N/m[/tex]
An air-filled capacitor is formed from two long conducting cylindrical shells that are coaxial and have radii of 42 mm and 74 mm. The electric potential of the inner conductor with respect to the outer conductor is -308 V ( = 1/4πε0 = 8.99 × 10^9 N · m^2/C^2).
The maximum energy density of the capacitor is closest to:_______
Correct answer is 2.7 x 10^-3 J/m3
I hope that helps ! <33
The maximum energy density of the capacitor is closest to: 2.7 x 10^-3 J/m3.
What is meant by the energy density of a capacitor?Energy density is defined as the total energy per unit volume of the capacitor. Since, Now, for a parallel plate capacitor, A × d = Volume of space between plates to which electric field E = V / d is confined. Therefore, Energy is stored per unit volume.
How do you calculate energy density?All Answers (14) Energy density is equal to 1/2*C*V2/weight, where C is the capacitance you computed and V should be your nominal voltage (i.e 2.7 V). Power Density is V2/4/ESR/weight, where ESR is the equivalent series resistance.
Learn more about energy density at
https://brainly.com/question/13035557
#SPJ2
Kevin is a black high school senior. While walking home from a sporting event at school, he sees a police car and decides to take another street to avoid it. He worries that the police will stop and question him even though he has not done anything wrong. Which theory explains this thought process? Dramaturgy Social construction of reality Social exchange theory Ethnomethodology
Answer:
Ethnomethodology theory
Explanation:
Take note of the fact that we are told Kevin worries that the police will stop and question him even though he has not done anything wrong.
This statement shows us that Kevin already understood his society from past experiences, and thus he tries to avoid social interactions with particular member of his society (the police) who may be show discrimination towards him.
The intensity of sunlight at the Earth's distance from the Sun is 1370 W/m2. (a) Assume the Earth absorbs all the sunlight incident upon it. Find the total force the Sun exerts on the Earth due to radiation pressure. N (b) Explain how this force compares with the Sun's gravitational attraction.
Answer:
F= 3.56e22N
Explanation:
Using the force of radiation acting on the earth which is
force = radiation pressure x area = (intensity/c)xpi R^2
force = 1370W/m^2 x pi x( 6.37x10^6m)^2/3x10^8m/s
force = 5.82x10^8 N
But the sun's gravitational attraction means the magnitude of the solar gravitational force on earth: If that's the case, the answer is approx 10^22 N:
F=GMm/r^2
G=6.67x10^(-11)=6.67e-11
M=mass sun = 2x10^30kg=2e30
m=mass earth = 6x10^24kg
r=earth sun distance = 1.5x10^11m
F=(6.6e-11)(2e30)(6e24)/(1.5e11)^2 =
F= 3.56e22N
Asteroid A has 3.5 times the mass and 2.0 times the velocity of Asteroid B. If
Asteroid B has a kinetic energy of 2,300,000 J then what is the kinetic energy of
Asteroid A?
Answer:
K_A = 32.2 10⁶ J
Explanation:
In this exercise we must relate the quantities given to find the kinetic energy
Asteroid A data
m_A = 3.5 m_B
v_A = 2.0 v
they also give the value of the kinetic energy of asteroid A
K_B = 2.3 10⁶ J
the expression for scientific energy is
K = ½ m v²
let's replace
K_A = ½ m_a V_a2
K_A = ½ 3.5 m_B (2.0 v_B)^2
K_A = 3.5 2² (½ m_B v_B²)
K_A = 14 K_B
K_A = 32.2 10⁶ J
Si se deja caer una piedra desde un helicóptero en reposo, entonces al cabo de 20 s cual será la rapidez y la distancia recorrida por la piedra
Answer:
La piedra alcanza una rapidez de 196.14 metros por segundo y una distancia recorrida de 1961.4 metros en 20 segundos.
Explanation:
Si se excluye los efectos del arrastre por la viscosidad del aire, la piedra experimenta un movimiento de caída libre, es decir, que la piedra es acelerada por la gravedad terrestre. La distancia recorrida y la rapidez final de la piedra pueden obtenerse con la ayuda de las siguientes ecuaciones cinemáticas:
[tex]v = v_{o} + g\cdot t[/tex]
[tex]y - y_{o} = v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2}[/tex]
Donde:
[tex]v[/tex], [tex]v_{o}[/tex] - Rapideces final e inicial de la piedra, medidas en metros por segundo.
[tex]t[/tex] - Tiempo, medido en segundos.
[tex]g[/tex] - Aceleración gravitacional, medida en metros por segundo al cuadrado.
[tex]y[/tex]. [tex]y_{o}[/tex] - Posiciones final e inicial de la piedra, medidos en metros.
Si [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{o} = 0\,m[/tex], entonces:
[tex]v = 0\,\frac{m}{s} +\left(-9.807\,\frac{m}{s^{2}} \right) \cdot (20\,s)[/tex]
[tex]v = -196.14\,\frac{m}{s}[/tex]
[tex]y-y_{o} = \left(0\,\frac{m}{s} \right)\cdot (20\,s) + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (20\,s)^{2}[/tex]
[tex]y-y_{o} = -1961.4\,m[/tex]
La piedra alcanza una rapidez de 196.14 metros por segundo y una distancia recorrida de 1961.4 metros en 20 segundos.
In your own words, discuss how energy conservation applies to a pendulum. Where is the potential energy the most? Where is the potential energy the least? Where is kinetic energy the most? Where is kinetic energy the least?
Answer:
Explanation:
Energy conservation applies to the swinging of pendulum . When the bob is at one extreme , it is at some height from its lowest point . So it has some gravitational potential energy . At that time since it remains at rest its kinetic energy is zero or the least . As it goes down while swinging , its potential energy decreases and kinetic energy increases following conservation of mechanical energy . At the At the lowest point , its potential energy is least and kinetic energy is maximum .
In this way , there is conservation of mechanical energy .
As more energy from fossil fuels and other fuels is released on Earth, the overall temperature of Earth tends to rise. Discuss how temperature equilibrium explains why Earth’s temperature cannot rise indefinitely.
Answer:
processes are competitive and reach a thermal equilibrium where the absorbed energy is equal to the energy emitted, this is the equilibrium temperature of the planet.
Explanation:
The temperature of planet Earth is due to two main types of process, internal and external.
Internal processes are all chemical processes that occur that release heat into the environment or due to gases that trap heat on the planet, greenhouse effect
External processes is heating due to energy coming from the Sun. This includes direct heating of the surface by the absorption of energy and reflects of energy in different atmospheric layers.
These are the two terms that heat the Earth
In addition there are several processes so the planet loses energy,
* energy radiation to outer space that is a few degrees kelvin, for which there is a permanent emission
* endothermic processes that need to absorb heat to perform, this lowers the temperature of the system
* liquid (water) system that absorbs large amounts of heat to change state and temperature.
These processes are competitive and reach a thermal equilibrium where the absorbed energy is equal to the energy emitted, this is the equilibrium temperature of the planet.
Therefore it is impossible for the temperature to increase indefinitely since the emission would increase by decreasing the value
Each of the boxes starts at rest and is then pulled for 2.0 m across a level, frictionless floor by a rope with the noted force. Which box has the highest final speed
Answer:
Explanation:
d
A 269-turn solenoid is 102 cm long and has a radius of 2.3 cm. It carries a current of 3.9 A. What is the magnetic field inside the solenoid near its center?
Answer:
Magnitude of the magnetic field inside the solenoid near its centre is 1.293 x 10⁻³ T
Explanation:
Given;
number of turns of solenoid, N = 269 turn
length of the solenoid, L = 102 cm = 1.02 m
radius of the solenoid, r = 2.3 cm = 0.023 m
current in the solenoid, I = 3.9 A
Magnitude of the magnetic field inside the solenoid near its centre is calculated as;
[tex]B = \frac{\mu_o NI}{l} \\\\[/tex]
Where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
[tex]B = \frac{4\pi*10^{-7} *269*3.9}{1.02} \\\\B = 1.293 *10^{-3} \ T[/tex]
Therefore, magnitude of the magnetic field inside the solenoid near its centre is 1.293 x 10⁻³ T
Two unknown resistors are connected together. When they are connected in series their equivalent resistance is 15 Ω. When they are connected in parallel, their equivalent resistance is 3.3 Ω. What are the resistances of these resistors?
Explanation:
Let x and y are two unknown resistors. When they are connected in series their equivalent resistance is 15 Ω. When they are connected in parallel, their equivalent resistance is 3.3 Ω.
For series combination,
[tex]x+y=15[/tex] ......(1)
For parallel combination,
[tex]\dfrac{1}{x}+\dfrac{1}{y}=3.3[/tex] ....(2)
We need to find the resistances of these resistors. Solving equation (1) and (2) we get :
x = 0.29 and y = 14.7
Hence, the resistances of these resistors are 0.29 ohms and 14.7 ohms.
A generator rotates at 95 Hz in a magnetic field of 0.025 T. It has 550 turns and produces an rms voltage of 170 V and an rms current of 60.0 A.
Required:
a. What is the peak current produced?
b. What is the area of each turn of the coil?
Answer:
Peak current= 84.86 A
Area of each turn = 0.029 m^2
Explanation:
The peak value of current can be obtained from Irms= 0.707Io. Where Io is the peak current.
Hence;
Irms= 60.0A
Io= Irms/0.707
Io = 60.0/0.707
Io= 84.86 A
Vrms= 0.707Vo
Vo= Vrms/0.707= 170/0.707 = 240.45 V
From;
V0 = NABω
Where;
Vo= peak voltage
N= number of turns
B= magnetic field
A= area of each coil
ω= angular velocity
But ω= 2πf = 2×π×95= 596.9 rads-1
Substituting values;
A= Vo/NBω
A= 240.45/550×0.025×596.9
A= 0.029 m^2
g A tube open at both ends, resonated at it's fundamental frequency, to a sound wave traveling at 330m/s. If the length of the tube is 4cm, find the frequency of the sound wave.
Answer:
frequency =4125Hz
Explanation:
L = 4cm = 0.04m
f =v/2L
f = 330/2 x 0.04
f = 4125Hz
An automotive air conditioner produces a 1-kW cooling effect while consuming 0.75 kW of power. What is the rate at which heat is rejected from this air conditioner
Answer:
The rejected by the air conditioning system is 1.75 kilowatts.
Explanation:
A air conditioning system is a refrigeration cycle, whose receives heat from cold reservoir with the help of power input before releasing it to hot reservoir. The First Law of Thermodynamics describes the model:
[tex]\dot Q_{L} + \dot W - \dot Q_{H} = 0[/tex]
Where:
[tex]\dot Q_{L}[/tex] - Heat rate from cold reservoir, measured in kilowatts.
[tex]\dot Q_{H}[/tex] - Heat rate liberated to the hot reservoir, measured in kilowatts.
[tex]\dot W[/tex] - Power input, measured in kilowatts.
The heat rejected is now cleared:
[tex]\dot Q_{H} = \dot Q_{L} + \dot W[/tex]
If [tex]\dot Q_{L} = 1\,kW[/tex] and [tex]\dot W = 0.75\,kW[/tex], then:
[tex]\dot Q_{H} = 1\,kW + 0.75\,kW[/tex]
[tex]\dot Q_{H} = 1.75\,kW[/tex]
The rejected by the air conditioning system is 1.75 kilowatts.
A positively charged particle has a velocity in the negative z direction at a certain point P. The magnetic force on the particle at this point is in the negative y direction. Which one of the following statements about the magnetic field at point P can be determined from this data?
a. Bx is positive
b. Bz is positive
c. By is negative
d. By is positive
e. Bx is negative
Answer:
a. Bx is positive
Explanation:
See attached file
Consider a block of mass equal to 10kg sliding on an inclined plane of 30°, as shown in the figure below. The coefficient of kinetic friction between the block and the plane surface is c = 0.4 (a) Determine the value of the horizontal and vertical acceleration of the block. (b) If the block starts from rest in t=0s and when it is in the X=0 and Y=5m position, calculate what its horizontal and vertical position will be at the instant t=1s. (C) How long does the LM block take to reach the base of the tilted plane?
Answer:
(a) aₓ = 1.33 m/s² and aᵧ = -0.770 m/s²
(b) x = 0.665 m and y = 4.62 m
(c) 3.61 s
Explanation:
(a) There are two ways we can solve this. The first way is to sum the forces in the x and y direction, then use the relation tan 30° = -aᵧ/aₓ, where aᵧ is the acceleration in the +y direction (up) and aₓ is the acceleration in the +x direction (right).
The second way is to sum the forces in the parallel and perpendicular directions to find the acceleration parallel to the incline, a. Then, use the relations aᵧ = -a sin 30° and aₓ = a cos 30°.
Let's try the first method. Sum of forces in the +y direction:
∑F = ma
N cos 30° + Nμ sin 30° − mg = maᵧ
N cos 30° + Nμ sin 30° − mg = -maₓ tan 30°
Sum of forces in the +x direction:
∑F = ma
N sin 30° − Nμ cos 30° = maₓ
Substituting:
N cos 30° + Nμ sin 30° − mg = -(N sin 30° − Nμ cos 30°) tan 30°
N cos 30° + Nμ sin 30° − mg = -N sin 30° tan 30° + Nμ sin 30°
N cos 30° − mg = -N sin 30° tan 30°
N (cos 30° + sin 30° tan 30°) = mg
N = mg / (cos 30° + sin 30° tan 30°)
N = (10 kg) (10 m/s²) / (cos 30° + sin 30° tan 30°)
N = 86.6 N
Now, solving for the accelerations:
N sin 30° − Nμ cos 30° = maₓ
aₓ = N (sin 30° − μ cos 30°) / m
aₓ = (86.6 N) (sin 30° − 0.4 cos 30°) / 10 kg
aₓ = 1.33 m/s²
N cos 30° + Nμ sin 30° − mg = maᵧ
aᵧ = N (cos 30° + μ sin 30°) / m − g
aᵧ = (86.6 N) (cos 30° + 0.4 sin 30°) / 10 kg − 10 m/s²
aᵧ = -0.770 m/s²
Now let's try the second method.
Sum of forces in the perpendicular direction:
∑F = ma
N − mg cos 30° = 0
N = mg cos 30°
Sum of forces in the parallel direction:
∑F = ma
mg sin 30° − Nμ = ma
mg sin 30° − mgμ cos 30° = ma
a = g (sin 30° − μ cos 30°)
a = (10 m/s²) (sin 30° − 0.4 cos 30°)
a = 1.536 m/s²
Solving for the accelerations:
aₓ = a cos 30°
aₓ = 1.33 m/s²
aᵧ = -a sin 30°
aᵧ = -0.770 m/s²
As you can see, the second method is faster and easier, but both methods will give you the same answer.
(b) In the x direction:
Given:
x₀ = 0 m
v₀ = 0 m/s
aₓ = 1.33 m/s²
t = 1 s
Find: x
x = x₀ + v₀ t + ½ at²
x = 0 m + (0 m/s) (1 s) + ½ (1.33 m/s²) (1 s)²
x = 0.665 m
In the y direction:
Given:
y₀ = 5 m
v₀ = 0 m/s
aᵧ = -0.770 m/s²
t = 1 s
Find: y
y = y₀ + v₀ t + ½ at²
y = 5 m + (0 m/s) (1 s) + ½ (-0.770 m/s²) (1 s)²
y = 4.62 m
(c) In the y direction:
Given:
y₀ = 5 m
y = 0 m
v₀ = 0 m/s
aᵧ = -0.770 m/s²
Find: t
y = y₀ + v₀ t + ½ at²
0 m = 5 m + (0 m/s) t + ½ (-0.770 m/s²) t²
t = 3.61 s
Suppose you are looking into the end of a long cylindrical tube in which there is a uniform magnetic field pointing away from you. If the magnitude of the field is decreasing with time the direction of the induced magnetic field is
Answer:
If the magnitude of the field is decreasing with time the direction of the induced magnetic field is CLOCKWISE
Explanation
This is because If the magnetic field decreases with time, the electric field will be produced in order to oppose the change in line with lenz law. Thus The right hand rule can be applied to find that the direction of electric field is in the clockwise direction.
A car starts from rest and accelerates at a constant rate after the car has gone 50 m it has a speed of 21 m/s what is the acceleration of the car
Answer:
4.41 m/s^2
Explanation:
(v_f)^2 - (v_i)^2 = 2a * change in distance
(21)^2 - (0)^2 = 2a * 50
a = (21^2)/(2*50)
a = 4.41 m/s^2
An inductor is connected to the terminals of a battery that has an emf of 12.0 V and negligible internal resistance. The current is 4.86 mA at 0.700 ms after the connection is completed. After a long time the current is 6.80 mA.
What are
(a) the resistance R of the inductor and
(b) the inductance L of the inductor?
Answer:
a) 1764.71 ohms
b) 1.73 H
Explanation:
From the question, we can identify the following parameters;
Vo =12 V , i = 4.86 mA, t =0.700 ms, io =6.80 mA
(a) Indcued emf V = L di/dt =0
From ohms law Vo = ioR
R = 12/6.80*0.001
R=1764.71 ohms
(b) For LR circuit
i =io (1-e^-t/T)
Time constant T = L/R
4.86 = 6.80 (1-e^-0.7*10^-3/T)
divide both side by 6.8
0.715 = 0.0007/T
L/R = 0.0007/0.715
L/R = 0.000979020979
Substitute R from above
L = 0.000979020979 * 1764.71
L =1.73 H
which of the following is a physical change?
A. a newspaper burns when placed in a fire.
B.an iron chair rusts when left outside
C.a sample of water boils and releases gas.
D.a plant changes carbon dioxide and water into sugar
A 7.0-kg shell at rest explodes into two fragments, one with a mass of 2.0 kg and the other with a mass of 5.0 kg. If the heavier fragment gains 100 J of kinetic energy from the explosion, how much kinetic energy does the lighter one gain?
Answer:
39.94m/s.Explanation:
Kinetic energy is expressed as KE = 1/2 mv² where;
m is the mass of the body
v is the velocity of the body.
For the heavier shell;
m = 5kg
KE gained = 100J
Substituting this values into the formula above to get the velocity v;
100 = 1/2 * 5 * v²
5v² = 200
v² = 200/5
v² = 40
v = √40
v = 6.32 m/s
Note that after the explosion, both body fragments will possess the same velocity.
For the lighter shell;
mass = 2.0kg and v = 6.32m/s
KE of the lighter shell = 1/2 * 2 * 6.32²
KE of the lighter shell = 6.32²
KE of the lighter shell= 39.94m/s
Hence, the lighter one gains a kinetic energy of 39.94m/s.
The gain in the kinetic energy of the smaller fragment is 249.64 J.
The given parameters;
Mass of the shell, m = 7.0 kgMass of one fragment, m₁ = 2.0 kgMass of the second fragment, m₂ = 5.0 kgKinetic energy of heavier fragment, K.E₁ = 100 JThe velocity of the heavier fragment is calculated as follows;
[tex]K.E = \frac{1}{2} mv^2\\\\mv^2 = 2K.E\\\\v^2 = \frac{2K.E}{m} \\\\v= \sqrt{\frac{2K.E}{m} } \\\\v = \sqrt{\frac{2 \times 100}{5} }\\\\v = 6.32 \ m/s[/tex]
Apply the principle of conservation of linear momentum to determine the velocity of the smaller fragment as;
[tex]m_1 u_1 + m_2 u_2 = v(m_1 + m_2)\\\\-6.32(5) \ + 2u_2 = 0(7)\\\\-31.6 + 2u_2 = 0\\\\2u_2 = 31.6\\\\u_2 = \frac{31.6}{2} \\\\u_2 = 15.8 \ m/s[/tex]
The gain in the kinetic energy of the smaller fragment is calculated as follows;
[tex]K.E_2 = \frac{1}{2} mu_2^2\\\\K.E_2 = \frac{1}{2} \times 2 \times (15.8)^2\\\\K.E_2 = 249.64 \ J[/tex]
Thus, the gain in the kinetic energy of the smaller fragment is 249.64 J.
Learn more about conservation of linear momentum here: https://brainly.com/question/7538238
A person can survive a feet-first impact at a speed of about 12 m/s (27 mi/h) on concrete, 15 m/s (34 mi/h) on soil, and 34 m/s (76 mi/h) on water. What is the reason for the different values for different surfaces
Answer:
Different surfaces have different impact force during collision which depends on the time it takes the person to come to rest after collision.
Explanation:
Given;
speed on concrete = 12 m/s (27 mi/h)
speed on soil = 15 m/s (34 mi/h)
speed on water = 34 m/s (76 mi/h)
The impact force on this person during collision is rate of change of momentum;
[tex]F = \frac{\delta P}{\delta t}[/tex]
During collision, the force exerted on this person depends on how long the collision lasts; that is, how long it takes for this person to come to rest after collision with each of the surfaces.
The longer the time of collision, the smaller the force exerted by each.
It takes shorter time for the person to come to rest on concrete surface than on soil surface, also it takes shorter time for the person to come to rest on soil surface than on water surface.
As a result of the reason above, the force exerted on the person during collision by the concrete surface is greater than that of soil surface which is greater than that of water surface.
In general, how do highland climates compare with nearby areas at lower elevations? They are cooler and drier. They are cooler and wetter. They are warmer and drier. They are warmer and wetter.
Answer:
They are cooler and wetter
Explanation:
Highland areas have lower temperatures compared to low lying areas. The climate is more wetter because of more rainfalls compared to low lying areas and the wind carries moist air over the highlands.