Answer:
The current in the tube is 0.601 A
Explanation:
Given;
diameter of the fluorescent, d = 3 cm
negative charge flowing in the fluorescent tube, -e = 3 x 10¹⁸ electrons/second
positive charge flowing in the fluorescent tube, +e = 0.75 x 10¹⁸ electrons/ second
The current in the fluorescent tube is due to presence of positive and negative charges to create neutrality in the conductor (fluorescent tube).
Q = It
I = Q/t
where;
I is current in Ampere (A)
Q is charge in Coulombs (C)
t is time is seconds (s)
1 e = 1.602 x 10⁻¹⁹ C
3 x 10¹⁸ e/ s = ?
= (3 x 10¹⁸ e/s x 1.602 x 10⁻¹⁹ C) / 1e
= 0.4806 C/s
negative charge per second (Q/t) = 0.4806 C/s
positive charge per second (Q/t) = (0.75 x 10¹⁸ e/s x 1.602 x 10⁻¹⁹ C) / 1e
positive charge per second (Q/t) = 0.12015 C/s
Total charge per second in the tube, Q / t = (0.4806 C/s + 0.12015 C/s)
I = 0.601 A
Therefore, the current in the tube is 0.601 A
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.
If you have completely polarized light of intensity 125 W/m2, what will its intensity be after passing through a polarizing filter with its axis at an 89.5° angle to the light's polarization direction?
Answer:
When we have completely polarized light with intensity I0, and it passes through a polarizing filter with its axis at an angle θ with respect to the light's polarization direction, the new intensity of the light will be:
I = I0*cos^2(θ)
This is called the "Malus' law".
in this case, we have:
I0 = 125 W/m^2
θ = 89.5°
then:
I = (125 W/m^2)*cos^2(89.9°) = 0.00038 W/m^2
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 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.
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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
An object on the end of a spring is set into oscillation by giving it an initial velocity while it is at its equilibrium position. In the first trial, the initial velocity is v0 and in the second it is 4v0. In the second trial, A : the amplitude is twice as great and the maximum acceleration is half as great. B : both the amplitude and the maximum acceleration are four times as great. C : the amplitude is half as great and the maximum acceleration is twice as great. D : both the amplitude and the maximum acceleration are twice as great. E : the amplitude is four times as great and the maximum acceleration is twice as great.
Explanation:
It is given that, in the first trial, the initial velocity is [tex]v_o[/tex] and in the second it is [tex]4v_o[/tex].
The total energy of the system remains constant. So,
[tex]\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2=\text{constant}[/tex] ....(1)
x is amplitude
It means that the amplitude is directly proportional to velocity. If velcoity increases to four times, then the amplitude also becomes 4 times.
Differentiating equation (1) we get :
[tex]mv\dfrac{dv}{dt}+kx\dfrac{dx}{dt}=0[/tex]
Since,
[tex]\dfrac{dv}{dt}=a,\ \text{acceleration}[/tex] and [tex]\dfrac{dx}{dt}=v,\ \text{velocity}[/tex]
So,
[tex]mva+kxv=0[/tex]
It means that the acceleration is also proportional to the amplitude. So, acceleration also becomes 4 times.
Hence, the correct option is (B) "both the amplitude and the maximum acceleration are four times as great"
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.
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Rank the electromagnetic radiation from lowest to highest in the simulation in terms of energy, wavelength, and frequency.
a. Energy
b. Wavelength
c. Frequency
Answer:
A.ENERGY: Radio<microwaves<infrared<visible light<ultraviolet<xrays<gammarays
B. WAVELENGTH: Radio>microwaves> infrared>visible light>ultraviolet>xray> gammarays
C. FREQUENCY: Radio<microwaves<infrared<visible light<ultraviolet<xray< gammarays
Explanation:
THIS IS BECAUSE OF THE FOLLOWING EQUATIONS
1.ENERGY (E)= hX freqency
So as energy of radiation increases frequency also increases
2. Velocity (v) = wavelength x frequency
So as wavelength increases frequency decreases and vice versa
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 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.
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The principles of magnetism apply everywhere on earth. What does this tell us about God and His character?
Answer:
God is omnipresent.
Explanation:
This means God is everywhere and He works where ever we are in the world
A 12 kg box is pulled across the floor with a 48 N horizontal force. If the force of friction is 12 N, what is the acceleration of the box?
Answer:
The acceleration of the box is 3 m/s²
Explanation:
Given;
mass of the box, m = 12 kg
horizontal force pulling the box forward, Fx = 48 N
frictional force acting against the box in opposite direction, Fk = 12 N
The net horizontal force on the box, F = 48 N - 12 N
The net horizontal force on the box, F = 36 N
Apply Newton's second law of motion to determine the acceleration of the box;
F = ma
where;
F is the net horizontal force on the box
a is the acceleration of the box
a = F / m
a = 36 / 12
a = 3 m/s²
Therefore, the acceleration of the box is 3 m/s²
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
You would like to store 8.1 J of energy in the magnetic field of a solenoid. The solenoid has 620 circular turns of diameter 6.6 cm distributed uniformly along its 33 cm length.
A. How much current is needed?
_____________ A
B. What is the magnitude of the magnetic field inside the solenoid?
________________T
C. What is the energy density (energy/volume) inside the solenoid?
________________ kJ/m^3
Answer:
(a) The current needed is 56.92 A
(b) The magnitude of the magnetic field inside the solenoid is 0.134 T
(c) The energy density inside the solenoid is 7.144 kJ/m³
Explanation:
Given;
energy stored in the magnetic field of solenoid, E = 8.1 J
number of turns of the solenoid, N = 620 turns
diameter of the solenoid, D = 6.6 cm = 0.066 m
radius of the solenoid, r = D/2 = 0.033 m
length of the solenoid, L = 33 cm = 0.33 m
Inductance of the solenoid is given as;
[tex]L= \frac{\mu_o N^2 A}{l}[/tex]
where;
A is the area of the solenoid = πr² = π (0.033)² = 0.00342 m²
μ₀ is permeability of free space = 4π x 10⁻⁷ H/m
[tex]L= \frac{4\pi*10^{-7} *620^2 *0.00342}{0.33} \\\\L = 0.005 \ H[/tex]
(A). How much current needed
Energy stored in magnetic field of solenoid is given as;
[tex]E = \frac{1}{2} LI^2\\\\[/tex]
Where;
I is the current in the solenoid
[tex]E = \frac{1}{2} LI^2\\\\I^2 = \frac{2E}{L}\\\\I = \sqrt{\frac{2*8.1}{0.005}}\\\\ I = 56.92 \ A[/tex]
(B) The magnitude of the magnetic field inside the solenoid
B = μ₀nI
where;
n is number of turns per unit length
B = μ₀(N/L)I
B = (4π x 10⁻⁷)(620/0.33)(56.92)
B = 0.134 T
(C) The energy density (energy/volume) inside the solenoid
[tex]U_B = \frac{B^2}{2\mu_0} \\\\U_B = \frac{(0.134)^2}{2*4\pi*10^{-7}} \\\\U_B = 7143.54 \ J/m^3\\\\U_B = 7.144 \ kJ/m^3[/tex]
What is the relationship between the surface area of a parachute and the amount of air resistance it builds up when it is deployed by a sky diver?
Answer:
An open parachute increases the cross-sectional area of the falling skydiver and thus increases the amount of air resistance which he encounters. Once the parachute is opened, the air resistance overwhelms the downward force of gravity.
Explanation:
The larger a parachute, the greater the force.
Hope it helps you in a little way.
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 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.
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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
Question 8 of 10
On which parts of the heating curve for water does adding thermal energy
mainly cause the particles to move faster?
200
150 -
B
To
100
Temperature ('C)
A
50
С
0
-50
10
40
50
60
70
Time (min)
O A. C and D
B. A and B
O O O O
O C. Band C
OD. B and D
Answer:
The correct answer is A
Explanation:
In this exercise we are given a graph of temperature versus time.
In calorimeter processes there are two types
* one that when giving thermal energy to the system its temperature increases, this fundamentally due to the greater kinetic energy of the molecular ones, this process observes in the graphs as a straight line of constant slope
* A process donates all the thermal energy that is introduced is cracked in breaking the molecular bonds, taking matter from one thermodynamic state to another, for example: liquid to gas.
This process in curves as a horizontal line, that is, there is no temperature change,
When analyzing the graph shown, parts C and D are the one that show a change in temperature with thermal energy. The correct answer is A
Answer:
C and D
Explanation:
Just took the quiz
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
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
A 22g bullet traveling 210 m/s penetrates a 2.0kg block of wood and emerges going 150m/s. If the block were stationary on a frictionless plane before the collision, what is the velocity of the block after the bullet passes through
Answer:
The final velocity of the block after the bullet passes through is 0.66 meters per second.
Explanation:
The interaction between the bullet and the block of woods is a clear example of a perfectly inelastic collision, which can be modelled after the Principle of Momentum Conservation. There are no external forces exerted on the bullet-block system. The equation describing the collision is described below:
[tex]m_{B}\cdot v_{B,o} + m_{W}\cdot v_{W,o} = m_{B}\cdot v_{B,f} + m_{W}\cdot v_{W,f}[/tex]
Where:
[tex]m_{B}[/tex], [tex]m_{W}[/tex]- Masses of the bullet and the block of wood, measured in kilograms.
[tex]v_{B,o}[/tex], [tex]v_{W,o}[/tex] - Initial speeds of the bullet and the block of wood, measured in meters per second.
[tex]v_{B,f}[/tex], [tex]v_{W,f}[/tex]- Final speeds of the bullet and the block of wood, measured in meters per second.
The final speed of the block is cleared:
[tex]v_{W,f} = \frac{m_{B}\cdot (v_{B,o}-v_{B,f})+m_{W}\cdot v_{W,o}}{m_{W}}[/tex]
[tex]v_{W,f} = v_{W,o} + \frac{m_{B}}{m_{W}} \cdot (v_{B,o}-v_{B,f})[/tex]
If [tex]v_{W,o} = 0\,\frac{m}{s}[/tex], [tex]m_{B} = 0.022\,kg[/tex], [tex]m_{W} = 2\,kg[/tex], [tex]v_{B,o} = 210\,\frac{m}{s}[/tex] and [tex]v_{B,f} = 150\,\frac{m}{s}[/tex], then the final velocity of the block after the bullet passes through is:
[tex]v_{W,f} = 0\,\frac{m}{s}+\left(\frac{0.022\,kg}{2\,kg}\right)\cdot \left(210\,\frac{m}{s}-150\,\frac{m}{s} \right)[/tex]
[tex]v_{W,f} = 0.66\,\frac{m}{s}[/tex]
The final velocity of the block after the bullet passes through is 0.66 meters per second.
The Atwood machine consists of two masses hanging from the ends of a rope that passes over a pulley. The pulley can be approximated by a uniform disk with mass p=7.95 kg and radius p=0.89 m. The hanging masses are L=32.0 kg and R=17.8 kg. Calculate the magnitude of the masses' acceleration and the tension in the left and right ends of the rope, L and R , respectively.
Answer:
Acceleration(a) = 2.588 m/s²
TL = 230.784 N
TR = 220.5 N
Explanation:
Given:
M = 7.95 kg
mL = 32 kg
mR = 17.8 kg
g = 9.8 m/s²
Find:
Acceleration(a)
TL
TR
Computation:
Acceleration(a) = [(mL - mR)g] / [mL + mR + M/2]
Acceleration(a) = [(32 - 17.8)9.8] / [32 + 17.8 + 7.95/2]
Acceleration(a) = [139.16] / [53.775]
Acceleration(a) = 2.588 m/s²
TL = mL(g-a)
TL = 32(9.8-2.588)
TL = 230.784 N
TR = mR(g+a)
TR = 17.8(9.8+2.588)
TR = 220.5 N
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
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 .
Chameleons catch insects with their tongues, which they can rapidly extend to great lengths. In a typical strike, the chameleon's tongue accelerates at a remarkable 210 m/s^2 for 20 ms, then travels at constant speed for another 30 ms.
During this total time of 50 ms, 1/20 of a second, how far does the tongue reach?
Express your answer to two significant figures and include the appropriate units.
Answer:
x_total = 0.17m
Explanation:
We can treat this exercise with the kinematics equations, where in the first part it is accelerated and in the second it is a uniform movement.
Let's analyze accelerated motion
The time that lasts is t = 20 10⁻³ s, the initial speed is zero (v₀ = 0), let's find the length that advances
x₁ = v₀ t + ½ a t²
x₁ = ½ a t²
x₁ = ½ 210 (20 10⁻³)²
x₁ = 4.2 10⁻² m
let's find the speed for the end of this movement
v = v₀ + a t
v = 0 + 210 20 10⁻³
v = 4.2 m / s
with this speed we can find the distance that the uniform movement
x₂ = v t2
x₂ = 4.2 30 10⁻³
x₂ = 1.26 10⁻¹ m
x₂ = 0.126m
the total distance traveled is
x_total = x₁ + x₂
x_total = 0.0420 +0.126
x_total = 0.168m
Let's reduce the significant figures to two
x_total = 0.17m
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
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.
Diamagnetic materialsA) have small negative values of magnetic susceptibility.B) are those in which the magnetic moments of all electrons in each atom cancel.C) experience a small induced magnetic moment when placed in an external magnetic field.D) exhibit the property of diamagnetism independently of temperature.E)are described by all
Answer:
C) experience a small induced magnetic moment when placed in an external magnetic field.
Explanation:
Diamagnetics materials are those that experience a small induced magnetic moment when placed in an external magnetic field. These materials, such as bismuth, copper, silver and lead, have elementary magnets in their compositions. When they are exposed to an external magnetic cap, these elemental magnets tend to follow an orientation contrary to the external magnetic field. As a result, a magnetic field is created in the opposite direction to the external magnetic field.