Answer:
(a) [tex]\vec D = 2\,i - 2\,j[/tex], (b) [tex]\vec E = -6\,i + 12\,j[/tex]
Explanation:
Let be [tex]\vec A = 3\,i - 3\,j\,[m][/tex], [tex]\vec B = i - 4\,j\,[m][/tex] and [tex]\vec C = -2\,i + 5\,j \,[m][/tex], each resultant is found by using the component method:
(a) [tex]\vec D = \vec A + \vec B + \vec C[/tex]
[tex]\vec D = (3\,i - 3\,j) + (i-4\,j) + (-2\,i+5\,j)\,[m][/tex]
[tex]\vec D = (3\,i + i -2\,i)+(-3\,j-4\,j+5\,j)\,[m][/tex]
[tex]\vec D = (3 + 1 -2)\,i + (-3-4+5)\,j\,[m][/tex]
[tex]\vec D = 2\,i - 2\,j[/tex]
(b) [tex]\vec E = -\vec A - \vec B + \vec C[/tex]
[tex]\vec E = -(3\,i-3\,j)-(i - 4\,j)+(-2\,i+5\,j)[/tex]
[tex]\vec E = (-3\,i + 3\,j) +(-i+4\,j) + (-2\,i + 5\,j)[/tex]
[tex]\vec E = (-3\,i-i-2\,i) + (3\,j+4\,j+5\,j)[/tex]
[tex]\vec E = (-3-1-2)\,i + (3+4+5)\,j[/tex]
[tex]\vec E = -6\,i + 12\,j[/tex]
Which type of reaction is shown in this energy diagram?
Answer:
Option C
Explanation:
The graph shows endothermic reaction because the reactants are lower in energy and the products are higher is energy. Endothermic reactions absorb energy having products with higher energy.
Answer:
C
Explanation:
In an endothermic reaction, the energy-time graph shows reactants are at a lower energy level than the products.
What will the surface charge density be if the radius of the disk is doubled but its total charge remains the same
Answer:
the new surface charge density = Q/4πr²( initial surface charge density divided by 4)
Explanation:
charge density(surface) = Q/A = charge/area
let r be the initial radius of the disk
therefore, area A = πr²
charge density = Q/πr²
Now that the radius is doubled, let it be represented as R
∴ R = 2r
Recall, charge density = Q/A
A = πR = π(2r)² = 4πr²
the new surface charge density = Q/4πr²
the initial surface charge density divided by 4
Suppose there is a uniform electric field pointing in the positive x-direction with a magnitude of 5.0 V/m. The electric potential is measured to be 50 V at the position x = 10 m. What is the electric potential at other positions?
Position [m] = (−20)--- (0.00) ---(10)--- (11)--- (99)
Electric Potential [V]=
Answer:
Electric potential at position, x = -20 m, = -100 V
Electric potential at position, x = 0 m, = 0
Electric potential at position, x = 10 m, = 50 V
Electric potential at position, x = 11 m, = 55 V
Electric potential at position, x = 99 m, 495 V
Explanation:
Given;
magnitude of electric field, E = 5.0 V/m
at position x = 10 m, electric potential = 50 V
Electric potential at position, x = -20 m
V = Ex
V = 5 (-20)
V = -100 V
Electric potential at position, x = 0 m
V = Ex
V = 5(0)
V = 0
Electric potential at position, x = 10 m
V = Ex
V = 5(10)
V = 50 V
Electric potential at position, x = 11 m
V = Ex
V = 5(11)
V = 55 V
Electric potential at position, x = 99 m
V = Ex
V = 5(99)
V = 495 V
Following a collision between a large spacecraft and an asteroid, a copper disk of radius 28.0 m and thickness 1.20 m, at a temperature of 850°C, is floating in space, rotating about its axis with an angular speed of 20.0 rad/s. As the disk radiates infrared light, its temperature falls to 20.0°C. No external torque acts on the disk.
A) Find the change in kinetic energy of the disk.
B) Find the change in internal energy of the disk.
C) Find the amount of energy it radiates.
Answer:
A. 9.31 x10^10J
B. -8.47x10 ^ 12J
C. 8.38x 10^12J
Explanation:
See attached file pls
A container is filled with fluid 1, and the apparent depth of the fluid is 9.00 cm. The container is next filled with fluid 2, and the apparent depth of this fluid is 6.86 cm. If the index of refraction of the first fluid is 1.37, what is the index of refraction of the second fluid
Answer:
The refractive index of fluid 2 is 1.78
Explanation:
Refractive index , n = real depth/apparent depth
For the first fluid, n = 1.37 and apparent depth = 9.00 cm.
The real depth of the container is thus
real depth = n × apparent depth = 1.37 × 9.00 cm = 12.33 cm
To find the refractive index of fluid index of fluid 2, we use the relation
Refractive index , n = real depth/apparent depth.
Now,the real depth = 12.33 cm and the apparent depth = 6.86 cm.
So, n = 12.33 cm/6.86 cm = 1.78
So the refractive index of fluid 2 is 1.78
Since the same container is used, real depth of fluid 1 is equal to the real depth of fluid 2. The index of refraction of the second fluid is 1.8
Given that a container is filled with fluid 1, and the apparent depth of the fluid is 9.00 cm. The container is next filled with fluid 2, and the apparent depth of this fluid is 6.86 cm. If the index of refraction of the first fluid is 1.37,
Then,
Index of refraction = [tex]\frac{Real depth}{Apparent depth}[/tex]
Real depth = Index of refraction x apparent depth
Since the same container is used, we can make an assumption that;
real depth of fluid 1 = real depth of fluid 2
That is,
1.37 x 9 = n x 6.86
Where n = Index of refraction for the second fluid.
make n the subject of formula
n = 12.33 / 6.86
n = 1.79
Therefore, the index of refraction of the second fluid is 1.8 approximately.
Learn more about refraction here: https://brainly.com/question/10729741
A valuable statuette from a Greek shipwreck lies at the bottom of the Mediterranean Sea. The statuette has a mass of 10,566 g and a volume of 4,064 cm3. The density of seawater is 1.03 g/mL.
a. What is the weight of the statuette?
b. What is the mass of displaced water?
c. What is the weight of displaced water?
d. What is the buoyant force on the statuette?
e. What is the net force on the statuette?
f. How much force would be required to lift the statuette?
Answer:
A) W = 103.55 N
B) mass of displaced water = 4186 g
C) W_displaced water = 41.06 N
D) Buoyant force = 41.06 N.
E) ZERO
F) 62.54 N
Explanation:
We are given;
mass of statuette;m = 10,566 g = 10.566 kg
volume = 4,064 cm³
Density of seawater;ρ = 1.03 g/mL = 1.03 g/cm³
A) The dry weight of the statuette can be calculated as;
W = mg
So;
W = 10.556 × 9.81
W = 103.55 N
B) Mass of displaced water is calculated from;
Density = mass/volume
So, mass = Density × Volume
m = 1.03 × 4,064 = 4186 g
C) Weight of displaced water is given by;
W_displaced water = (m_displaced water) × g
W_displaced water = 4.186 kg × 9.81 m/s^2 = 41.06 N
D) The buoyant force is the same as the weight of the displaced water.
Thus, Buoyant force = 41.06 N.
E) The apparent weight of the statuette is calculated from;
Apparent weight = Dry weight - Weight of displaced water
Apparent weight = 103.6 N - 41.06 N = 62.54 N. It is sitting on the bottom of the sea, so the sea floor is providing an opposite force that is equal but opposite the weight so that the net force on the statuette is zero. Since It has zero acceleration, in any direction, hence the net force on it is zero.
F. From E above, The Force required to lift the statuette = 62.54 N
a 5 charge is locataed 1.25 m to the left of a -3 charge. What is the magnitude and direction of the electrostatic force on the postive charge
Answer:
The force is 86.5×10^9 N towards the negative charge (to the right)
Explanation:
The electrostatic force on the charges is given by Coulomb's law;
F= Kq1q2/r^2
This an inverse square law.
F= electrostatic force on the charges
K= constant of Coulomb's law
q1 and q2= magnitude of the charges
Since K= 9.0×10^9Nm^2C^2
F= 9.0×10^9 × 5 × 3/(1.25)^2 = 135×10^9/1.56
F= 86.5×10^9 N
The force is 86.5×10^9 N towards the negative charge.
A solenoid of length 2.40 m and radius 1.70 cm carries a current of 0.190 A. Determine the magnitude of the magnetic field inside if the solenoid cons
Complete question:
A solenoid of length 2.40 m and radius 1.70 cm carries a current of 0.190 A. Determine the magnitude of the magnetic field inside if the solenoid consists of 2100 turns of wire.
Answer:
The magnitude of the magnetic field inside the solenoid is 2.089 x 10⁻⁴ T.
Explanation:
Given;
length of solenoid, L = 2.4 m
radius of solenoid, R = 1.7 cm = 0.017 m
current in the solenoid, I = 0.19 A
number of turns of the solenoid, N = 2100 turns
The magnitude of the magnetic field inside the solenoid is given by;
B = μnI
Where;
μ is permeability of free space = 4π x 10⁻⁷ m/A
n is number of turns per length = N/L
I is current in the solenoid
B = μnI = μ(N/L)I
B = 4π x 10⁻⁷(2100 / 2.4)0.19
B = 4π x 10⁻⁷ (875) 0.19
B = 2.089 x 10⁻⁴ T
Therefore, the magnitude of the magnetic field inside the solenoid is 2.089 x 10⁻⁴ T.
Two long parallel wires separated by 4.0 mm each carry a current of 24 A. These two currents are in the same direction. What is the magnitude of the magnetic field at a point that is between the two wires and 1.0 mm from one of the two wires
Answer:
Explanation:
Magnetic field at a a point R distance away
B = μ₀ / 4π X 2I / R where I is current
Magnetic field due to one current
= 10⁻⁷ x 2 x 24 / 1 x 10⁻³
48 x 10⁻⁴ T
Magnetic field due to other current
= 10⁻⁷ x 2 x 24 / 3x 10⁻³
16 x 10⁻⁴ T
Total magnetic field , as they act in opposite direction, is
= (48 - 16 ) x 10⁻⁴
32 x 10⁻⁴ T .
Each side of a metal plate is illuminated by light of different wavelengths. The left side is illuminated by light with λ0 = 500 nm and the right side by light of unknown λ. Two electrodes A and B provide the stopping potential for the ejected electrons. If the voltage across AB is VAB=1.2775 V, what is the unknown λ?
Answer:
The wavelength is [tex]\lambda = 1029 nm[/tex]
Explanation:
From the question we are told that
The wavelength of the left light is [tex]\lambda_o = 500 nm = 500 *10^{-9} \ m[/tex]
The voltage across A and B is [tex]V_{AB } = 1.2775 \ V[/tex]
Let the stopping potential at A be [tex]V_A[/tex] and the electric potential at B be [tex]V_B[/tex]
The voltage across A and B is mathematically represented as
[tex]V_{AB} = V_A - V_B[/tex]
Now According to Einstein's photoelectric equation the stopping potential at A for the ejected electron from the left side in terms of electron volt is mathematically represented as
[tex]eV_A = \frac{h * c}{\lambda_o } - W[/tex]
Where W is the work function of the metal
h is the Planck constant with values [tex]h = 6.626 *10^{-34} \ J \cdot s[/tex]
c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
And the stopping potential at B for the ejected electron from the right side in terms of electron volt is mathematically represented as
[tex]eV_B = \frac{h * c}{\lambda } - W[/tex]
So
[tex]eV_{AB} = eV_A - eV_B[/tex]
=> [tex]eV_{AB} = \frac{h * c}{\lambda_o } - W - [\frac{h * c}{\lambda } - W][/tex]
=> [tex]eV_{AB} = \frac{h * c}{\lambda_o } - \frac{h * c}{\lambda }[/tex]
=> [tex]\frac{h * c}{\lambda } = \frac{h * c}{\lambda_o } -eV_{AB}[/tex]
=> [tex]\frac{1}{\lambda } =\frac{1}{\lambda_o } - \frac{ eV_{AB}}{hc}[/tex]
Where e is the charge on an electron with the value [tex]e = 1.60 *10^{-19} \ C[/tex]
=> [tex]\frac{1}{\lambda } = \frac{1}{500 *10^{-9} } - \frac{1.60 *10^{-19} * 1.2775}{6.626 *10^{-34} * 3.0 *10^{8}}[/tex]
=> [tex]\frac{1}{\lambda } = 9.717*10^{5} m^{-1}[/tex]
=> [tex]\lambda = 1.029 *10^{-6} \ m[/tex]
=> [tex]\lambda = 1029 nm[/tex]
A circular loop of wire has radius of 9.50 cm. A sinusoidal electromagnetic plane wave traveling in air passes through the loop, with the direction of the magnetic field of the wave perpendicular to the plane of the loop. The intensity of the wave at the location of the loop is 0.0295 W/m^2, and the wavelength of the wave is 6.40 m.
Required:
What is the maximum emf induced in the loop?
Answer:
The maximum emf induced in the loop is 0.132 Volts
Explanation:
Given;
radius of the circular loop, r = 9.5 cm
intensity of the wave, I = 0.0295 W/m²
wavelength, λ = 6.40 m
The intensity of the wave is given as;
[tex]I = \frac{B_o^2*c }{2\mu_o}[/tex]
where;
B₀ is the amplitude of the field
c is the speed of light = 3 x 10⁸ m/s
μ₀ is permeability of free space = 4π x 10⁻⁵ m/A
[tex]I = \frac{B_o^2*c }{2\mu_o}\\\\B_o^2 = \frac{I*2\mu_o}{c} \\\\B_o^2 = \frac{0.0295*2*4\pi*10^{-7}}{3*10^8} \\\\B_o^2 = 2.472 *10^{-16}\\\\B_o = \sqrt{2.472 *10^{-16}}\\\\ B_o = 1.572*10^{-8} \ T[/tex]
Area of the circular loop;
A = πr²
A = π(0.095)²
A = 0.0284 m²
Frequency of the wave;
f = c / λ
f = (3 x 10⁸) / (6.4)
f = 46875000 Hz
Angular velocity of the wave;
ω = 2πf
ω = 2π(46875000)
ω = 294562500 rad/s
The maximum induced emf is calculated as;
emf = B₀Aω
= (1.572 x 10⁻⁸)(0.0284)(294562500)
= 0.132 Volts
Therefore, the maximum emf induced in the loop is 0.132 Volts
A resistor and a capacitor are connected in series across an ideal battery having a constant voltage across its terminals. (a) At the moment contact is made with the battery the voltage across the capacitor is
Answer:
(a) D. Zero.
(b) C. Equal to the battery's terminal voltage.
Explanation:
The question is incomplete, see the complete question for your reference and information.
A resistor and a capacitor are connected in series across an ideal battery having a constant voltage
across its terminals. At the moment contact is made with the battery
(a) the voltage across the capacitor is
A) equal to the battery's terminal voltage.
B) less than the battery's terminal voltage, but greater than zero.
C) equal to the battery's terminal voltage.
D) zero.
(b) the voltage across the resistor is
A) equal to the battery's terminal voltage.
B) less than the battery's terminal voltage, but greater than zero.
C) equal to the battery's terminal voltage.
D) zero
A RC circuit is a circuit that is composed of both resistors and capacitors connect to a source of current or voltage.
basically when a voltage source is applied to an RC circuit, the capacitor, C charges up through the resistance, R
You set two parallel slits 0.1 mm apart at a distance of 2 m from a screen and illuminate them with light of wavelength 450 nm. The distance between a bright spot in the interference pattern and the dark spot adjacent to it is Group of answer choices
Answer:
Δx = 9 x 10⁻³ m = 9 mm
Explanation:
The formula for fringe spacing in Young's Double Slit Experiment is given as follows:
Δx = λL/d
where,
Δx = fringe spacing = ?
λ = wavelength of light = 450 nm = 450 x 10⁻⁹ m
L = Distance between slits and screen = 2 m
d = distance between slits = 0.1 mm = 0.1 x 10⁻³ m
Therefore,
Δx = (450 x 10⁻⁹ m)(2 m)/(0.1 x 10⁻³ m)
Δx = 9 x 10⁻³ m = 9 mm
Consider the following spectrum where two colorful lines (A and B) are positioned on a dark background. The violet end of the spectrum is on the left and the red end of the spectrum is on the right. A B 5. (1 point) What is the name for this type of spectrum? 6. (1 point) Transition A is associated with an electron moving between the n= 1 and n= 3 levels. Transition B is associated with an electron moving between the n= 2 and n= 5 levels. Which transition is associated with a photon of longer wavelength?
Answer:
Explanation:
a )
This type of spectrum is called line emission spectrum . Because it consists of lines . It is emission spectrum because it is due to emission of radiation from a source .
b ) The wavelength of a photon is inversely proportional to its energy . Photon due to transition between n = 1 and n = 3 will have higher energy than
that due to transition between n = 2 and n = 5 . So the later photon ( B) will have greater wavelength or photon due to transition between n = 2 and n = 5 will have greater wavelength .
We repeat the experiment from the video, but this time we connect the wires in parallel rather than in series. Which wire will now dissipate the most heat?
Both wires will dissipate the same amount of heat.
A. The Nichrome wire (resistance 2.7)
B. The copper wire (resistance 0.1)
Answer: B. The copper wire (resistance 0.1)
Explanation: When resistance is in parallel, voltage (V) is the same but current is different for every resistance. Current (i) is related to voltage and resistance (R) by Ohm's Law
i = [tex]\frac{V}{R}[/tex]
So, since both wires are in parallel, they have the same voltage but because the copper wire resistance is smaller than Nichrome wire, the first's current will be bigger.
Every resistor in a circuit dissipates electrical power (P) that is converted into heat energy. The dissipation can be found by:
P = [tex]i^{2}*R[/tex]
As current for copper wire is bigger than nichrome, power will be bigger and it will dissipate more heat.
In conclusion, the copper wire will dissipate more heat when connected in parallel.
A box of mass 0.8 kg is placed on an inclined surface that makes an angle 30 above
the horizontal, Figure 1. A constant force of 18 N is applied on the box in a direction 10°
with the horizontal causing the box to accelerate up the incline.
The coefficient of
kinetic friction between the block and the plane is 0.25.
Show the free body diagrams
(a) Calculate the block's
acceleration as it moves up the incline. (6 marks)
(b) If the block slides down at a constant speed, find the value of force applied.
(4 marks)
Answer:
a) a = 17.1 m / s², b) F = 3.04 N
Explanation:
This is an exercise of Newton's second law, in this case the selection of the reference system is very important, we have two possibilities
* a reference system with the horizontal x axis, for this selection the normal and the friction force have x and y components
* a reference system with the x axis parallel to the plane, in this case the weight and the applied force have x and y components
We are going to select the second possibility, since it is the most used in inclined plane problems, let's analyze the angle of the applied force (F) it has an angle 10º with respect to the x axis, if we rotate this axis 30º the new angle is
θ = 10 -30 = -20º
The negative sign indicates that it is in the fourth quadrant. Let's use trigonometry to find the components of the applied force
sin (-20) = F_{y} / F
cos (-20) = Fₓ / F
F_{y} = F sin (-20)
Fₓ = F cos (-20)
F_y = 18 sin (-20) = -6.16 N
Fₓ = 18 cos (-20) = 16.9 N
The decomposition of the weight is the customary
sin 30 = Wₓ / W
cos 30 = W_y / W
Wₓ = W sin 30 = mg sin 30
W_y = W cos 30 = m g cos 30
Wₓ = 0.8 9.8 sin 30 = 3.92 N
W_y = 0.8 9.8 cos 30 = 6.79 N
Notice that in the case the angle is measured with respect to the axis y perpendicular to the plane
Now we can write Newton's second law for each axis
X axis
Fₓ - fr = m a
Y Axis
N - [tex]F_{y}[/tex] - Wy = 0
N =F_{y} + Wy
N = 6.16 + 6.79
They both go to the negative side of the axis and
N = 12.95 N
The friction force has the formula
fr = μ N
we substitute
Fₓ - μ N = m a
a = (Fₓ - μ N) / m
we calculate
a = (16.9 - 0.25 12.95) / 0.8
a = 17.1 m / s²
b) now the block slides down with constant speed, therefore the acceleration is zero
ask for the value of the applied force, we will suppose that with the same angle, that is, only its modulus was reduced
Newton's law for the x axis
Fₓ -fr = 0
Fₓ = fr
F cos 20 = μ N
F = μ N / cos 20
we calculate
F = 0.25 12.95 / cos 20
F = 3.04 N
this is the force applied at an angle of 10º to the horizontal
You shine unpolarized light with intensity 54.0 W/m^2 on an ideal polarizer, and then the light that emerges from this polarizer falls on a second ideal polarizer. The light that emerges from the second polarizer has intensity 19.0 W/m^2. Find the angle between the polarizing axes of the two polarizers.°
Answer:
The angle between the polarizing axes of the two polarizers is 54°
Explanation:
Given;
intensity of unpolarized light, I₀ = 54.0 W/m²
intensity of light that emerges from second ideal polarizer, I₁ = 19.0 W/m²
The angle between the polarizing axes of the two polarizers is dtermined by applying Malus' law for intensity of a linearly polarized light passing through a polarizer.
I₁ = I₀Cos²θ
Cos²θ = I₁ / I₀
Cos²θ = 19 / 54
Cos²θ =0.3519
Cos θ = √0.3519
Cos θ = 0.5932
θ = Cos⁻¹(0.5932)
θ = 53.6°
θ = 54°
Therefore, the angle between the polarizing axes of the two polarizers is 54°
A proton moving at 4.80 106 m/s through a magnetic field of magnitude 1.74 T experiences a magnetic force of magnitude 7.00 10-13 N. What is the angle between the proton's velocity and the field? (Enter both possible answers from smallest to largest. Enter only positive values between 0 and 360.)
Answer:
31.55° and 148.45°
Explanation:
Formula for calculating the force experiences by the proton placed in a magnetic field is as expressed below;
F = qvBsinθ where;
F is the magnetic force experienced by the proton
q is the charge on the proton
v is the velocity of the proton
B is the magnetic field
θ is the angle between the proton's velocity and the field (Required)
Given parameters
F = 7.00 * 10⁻¹³N
q = 1.602*10⁻¹⁹C
v = 4.80 * 10⁶ m/s
B = 1.74 T
θ =?
From the formula F = qvBsinθ;
sinθ = F/qvB
sinθ = 7.00 * 10⁻¹³/1.602*10⁻¹⁹* 4.80 * 10⁶*1.74
sinθ = 7.00 * 10⁻¹³/13.38*10⁻¹³
sinθ = 0.5231689 * 10⁰
sinθ = 0.5231689
θ = sin⁻¹0.5231689
θ = 31.55°
The following are the positive values of the angle between 0° and 360°
Sin is positive in the first and second quadrant. In the second quadrant the angle is equal to 180°-31.55° = 148.45°.
Hence the possible values of the angle from smallest to largest are 31.55° and 148.45°
The _____________ is the thermonuclear fusion of hydrogen to form helium operating in the cores of massive stars on the main sequence
An object on a rope is lowered steadily decreasing speed. Which is true?
A) The tope tensions is greater than the objects weight
B) the rope tension equals the objects weight
C)the rope tension is less than the objects weight
D) the rope tension can’t be compared to the objects weight
Answer:
C) the rope tension is less than the objects weight
Explanation:
According to Newton's Second Law, when an unbalanced or net force is applied to a body, it produces an acceleration in the body in the direction of the net force itself.
In this scenario, we have two forces acting on the object. First is the weight of object acting downward. Second is the tension in the rope acting upwards.
Since, the object is being lowered in the direction of weight. Therefore, weight of the object must be greater than the tension in the rope. So, the net force has the downward direction and the object is lowered. Hence, the correct option is:
C) the rope tension is less than the objects weight
A small meteorite with mass of 1 g strikes the outer wall of a communication satellite with a speed of 2Okm/s (relative to the satellite). The mass of the satellite is 200 kg.
About how much energy (in kJ) was converted to heat?
Answer:
The energy coverted to heat is 200 kilojoules.
Explanation:
GIven the absence of external forces exerted both on the small meteorite and on the communication satellite, the Principle of Linear Momentum is considered and let suppose that collision is completely inelastic and that satellite is initially at rest. Hence, the expression for the satellite-meteorite system:
[tex]m_{M}\cdot v_{M} + m_{S}\cdot v_{S} = (m_{M}+m_{S})\cdot v[/tex]
Where:
[tex]m_{M}[/tex], [tex]m_{S}[/tex] - Masses of the small meteorite and the communication satellite, measured in kilograms.
[tex]v_{M}[/tex], [tex]v_{S}[/tex] - Speeds of the small meteorite and the communication satellite, measured in meters per second.
[tex]v[/tex] - Final speed of the satellite-meteorite system, measured in meters per second.
The final speed of the satellite-meteorite system is cleared:
[tex]v = \frac{m_{M}\cdot v_{M}+m_{S}\cdot v_{S}}{m_{M}+m_{S}}[/tex]
If [tex]m_{M} = 1\times 10^{-3}\,kg[/tex], [tex]m_{S} = 200\,kg[/tex], [tex]v_{M} = 20000\,\frac{m}{s}[/tex] and [tex]v_{S} = 0\,\frac{m}{s}[/tex], the final speed is now calculated:
[tex]v = \frac{(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)}{1\times 10^{-3}\,kg+200\,kg}[/tex]
[tex]v = 0.1\,\frac{m}{s}[/tex]
Which means that the new system remains stationary and all mechanical energy from meteorite is dissipated in the form of heat. According to the Principle of Energy Conservation and the Work-Energy Theorem, the change in the kinetic energy is equal to the dissipated energy in the form of heat:
[tex]K_{S} + K_{M} - K - Q_{disp} = 0[/tex]
[tex]Q_{disp} = K_{S}+K_{M}-K[/tex]
Where:
[tex]K_{S}[/tex], [tex]K_{M}[/tex] - Initial translational kinetic energies of the communication satellite and small meteorite, measured in joules.
[tex]K[/tex] - Kinetic energy of the satellite-meteorite system, measured in joules.
[tex]Q_{disp}[/tex] - Dissipated heat, measured in joules.
The previous expression is expanded by using the definition for the translational kinetic energy:
[tex]Q_{disp} = \frac{1}{2}\cdot [m_{M}\cdot v_{M}^{2}+m_{S}\cdot v_{S}^{2}-(m_{M}+m_{S})\cdot v^{2}][/tex]
Given that [tex]m_{M} = 1\times 10^{-3}\,kg[/tex], [tex]m_{S} = 200\,kg[/tex], [tex]v_{M} = 20000\,\frac{m}{s}[/tex], [tex]v_{S} = 0\,\frac{m}{s}[/tex] and [tex]v = 0.1\,\frac{m}{s}[/tex], the dissipated heat is:
[tex]Q_{disp} = \frac{1}{2}\cdot \left[(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)^{2}+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)^{2}-(200.001\,kg)\cdot \left(0.001\,\frac{m}{s} \right)^{2}\right][/tex][tex]Q_{disp} = 200000\,J[/tex]
[tex]Q_{disp} = 200\,kJ[/tex]
The energy coverted to heat is 200 kilojoules.
A stunt driver rounds a banked, circular curve. The driver rounds the curve at a high, constant speed, such that the car is just on the verge of skidding to the outside of the curve. A front view of a car driving on a banked curve. The cross section of the banked road is constructed like a ramp. The car drives transversely to the slope of the ramp, so that the wheels of one side of the car are lower than the wheels on the other side of the car. Which forces are directly responsible for producing the car’s centripetal acceleration? Coriolis force centripetal force frictional force normal force gravitational force
Answer: C
Frictional force
Explanation:
The description of the question above is an example of a circular motion.
For a car travelling in a curved path, the frictional force between the tyres and the road surface will provide the centripetal force.
Since the road is banked, and the cross section of the banked road is constructed like a ramp. The car drives transversely to the slope of the ramp, so that the wheels of one side of the car are lower than the wheels on the other side of the car, for cornering the banked road, the car will not rely only on the frictional force.
Therefore, the correct answer is option C - the frictional force.
If a 950 kg merry-go-round platform of radius 4.5 meters is driven by a mechanism located 2.0 meters from its center of rotation, how much force must the mechanism provide to get the platform moving at 5.5 revolutions per minute after 12 seconds if it were initially at rest
Answer:
F = 213.75 N
Explanation:
First we need to calculate the angular acceleration of merry-go-round. For that purpose we use 1st equation of motion in angular form.
ωf = ωi + αt
where,
ωf=final angular velocity=(5.5 rev/min)(2π rad/1 rev)(1 min/60 s)=0.58 rad/s
ωi =initial angular velocity = 0 rad/s
t = time = 12 s
α = angular acceleration = ?
Therefore,
0.58 rad/s = 0 rad/s + α(12 s)
α = (0.58 rad/s)/(12 s)
α = 0.05 rad/s²
Now, we shall find the linear acceleration of the merry-go-round:
a = rα
where,
a = linear acceleration = ?
r = radius = 4.5 m
Therefore,
a = (4.5 m)(0.05 rad/s²)
a = 0.225 m/s²
Now, the force is given by Newton;s 2nd Law:
F = ma
where,
F = Force = ?
m = mass pf merry-go-round = 950 kg
Therefore,
F = (950 kg)(0.225 m/s²)
F = 213.75 N
A magnetic field is entering into a coil of wire with radius of 2(mm) and 200 turns. The direction of magnetic field makes an angle 25° in respect to normal to surface of coil. The magnetic field entering coil varies 0.02 (T) in every 2 seconds. The two ends of coil are connected to a resistor of 20 (Ω).
A) Calculate Emf induced in coil
B) Calculate the current in resistor
C) Calculate the power delivered to resistor by Emf
Answer:
a) 2.278 x 10^-5 volts
b) 1.139 x 10^-6 Ampere
c) 2.59 x 10^-11 W
Explanation:
The radius of the wire r = 2 mm = 0.002 m
the number of turns N = 200 turns
direction of the magnetic field ∅ = 25°
magnetic field strength B = 0.02 T
varying time = 2 sec
The cross sectional area of the wire = [tex]\pi r^{2}[/tex]
==> A = 3.142 x [tex]0.002^{2}[/tex] = 1.257 x 10^-5 m^2
Field flux Φ = BA cos ∅ = 0.02 x 1.257 x 10^-5 x cos 25°
==> Φ = 2.278 x 10^-7 Wb
The induced EMF is given as
E = NdΦ/dt
where dΦ/dt = (2.278 x 10^-7)/2 = 1.139 x 10^-7
E = 200 x 1.139 x 10^-7 = 2.278 x 10^-5 volts
b) If the two ends are connected to a resistor of 20 Ω, the current through the resistor is given as
[tex]I[/tex] = E/R
where R is the resistor
[tex]I[/tex] = (2.278 x 10^-5)/20 = 1.139 x 10^-6 Ampere
c) power delivered to the resistor is given as
P = [tex]I[/tex]E
P = (1.139 x 10^-6) x (2.278 x 10^-5) = 2.59 x 10^-11 W
If an astronomer wants to find and identify as many stars as possible in a star cluster that has recently formed near the surface of a giant molecular cloud (such as the Trapezium cluster in the Orion Nebula), what instrument would be best for her to use
Answer:
Infrared telescope and camera
Explanation:
An infrared telescope uses infrared light to detect celestial bodies. The infrared radiation is one of the known forms of electromagnetic radiation. Infrared radiation is given off by a body possessing some form of heat. All bodies above the absolute zero temperature in the universe radiates some form of heat, which can then be detected by an infrared telescope, and infrared radiation can be used to study or look into a system that is void of detectable visible light.
Stars are celestial bodies that are constantly radiating heat. In order to see a clearer picture of the these bodies, Infrared images is better used, since they are able to penetrate the surrounding clouds of dust, and have located many more stellar components than any other types of telescope, especially in dusty regions of star clusters like the Trapezium cluster.
A solid disk of radius 1.4 cm and mass 430 g is attached by a wire to one of its circular faces. It is twisted through an angle of 10 o and released. If the spring has a torsion constant of 370 N-m/rad, what is the frequency of the motion
Answer:
f= 4,186 10² Hz
Explanation:
El sistema descrito es un pendulo de torsión que oscila con con velocidad angular, que esta dada por
w = √ k/I
donde ka es constante de torsion de hilo e I es el momento de inercia del disco
El momento de inercia de indican que giran un eje que pasa por enronqueces
I= ½ M R2
reduzcamos las cantidades al sistema SI
R= 1,4 cm = 0,014 m
M= 430 g = 0,430 kg
substituimos
w= √ (2 k/M R2)
calculemos
w = RA ( 2 370 / (0,430 0,014 2)
w = 2,963 103 rad/s
la velocidad angular esta relacionada con la frecuencia por
w =2pi f
f= w/2π
f= 2,963 10³/ (2π)
f= 4,186 10² Hz
greater than: The electric potential energy of a proton at point A is _____ the electric potential energy of an proton at point B.
Answer:
[similar to]
Explanation:
it is the missing word
Jane is collecting data for a ball rolling down a hill. she measure out a set of different distances and then proceeds to use a stopwatch to find the time it takes the ball to roll each distance
Answer:
The Independent variable in this experiment is the time taken by the ball to roll down each distance.
The dependent variable is the distance through which the ball rolls
The control variables are: slope of hill, weight, of the ball, size of ball, wind speed, surface characteristics of the ball.
Explanation:
The complete question is
Jane is collecting data for a ball rolling down a hill. She measures out a set of different distances and then proceeds to use a stop watch to find the time it takes the ball to roll. What are the independent, dependent, and control variables in this experiment?
Independent variable have their values not dependent on any other variable in the scope of the experiment. The time for the ball to roll down the hill is not dependent on any other variable in the experiment. Naturally, some common independent variables are time, space, density, mass, fluid flow rate.
A dependent variable has its value dependent on the independent variable in the experiment. The value of the distance the ball rolls depends on the time it takes to roll down the hill.
The relationship between the dependent and independent variables in an experiment is given as
y = f(x)
where y is the output or the dependent variable,
and x is the independent variable.
Control variables are those variable that if not held constant could greatly affect the results of an experiment. For an experiment to be more accurate, control variables should be confined to a given set of value throughout the experiment.
An electronic stove is rated 1000w, 240v. explain this statement. please I'm in need
Answer and Explanation:
When an electronic appliance such as electric heater, electric stove e.t.c is rated, the rating actually specifies the ideal working properties of the appliance. For example if it is rated 200W, 220V, it shows that the power the appliance will consume at a voltage of 220V is 200W.
Therefore, for the electronic stove mentioned with a rating of 1000W, 240V, the stove will consume or draw a power of 1000 watts at a voltage of 240volts.
Ratings can also help determine some other properties of the appliance such as current consumption and resistance in the appliance. For the given electronic stove, the current consumed can be found by using the following relation:
P = IV -------------(i)
Where;
P = Power rating = 1000W
I = Current used
V = Voltage rating = 240V
Substituting these values into equation (i) gives;
1000 = I x 240
I = [tex]\frac{1000}{240}[/tex] = 4.17A
Therefore, the current used by the stove is 4.17A
To get the resistance R of the stove, we can use the relation;
P = [tex]\frac{V^2}{R}[/tex]
R = [tex]\frac{V^2}{P}[/tex]
R = [tex]\frac{240^2}{1000}[/tex]
R = 57.6Ω
Therefore, the resistance of the stove is 57.6Ω
Two factors that regulate (control) glandular secretion.
Answer:
The factors include age and puberty
Explanation:
Glandular secretion release chemicals such as hormones in response to the body’s metabolic needs.
As an individual ages , the metabolic rate of the body also reduces . This is due to the stress and ageing of the cells of the body. This explains why glandular secretion is optimal with young people and Lower in older people. It also explains why the immune system of a young person is mostly stronger than older people.
Puberty is another factor which affects glandular secretion as during puberty there is usually a high amount of hormonal changes due to high levels of secretions of some hormones. These hormones could however inhibit the other glandular secretions.