Answer:
As a body moving upward
T=real weight + apparent weight
T=550+490
T=1040
hope u will get the answer:)
Explanation:
A model airplane has momentum given by p=[(-0.75kg.m/s3)t2 + (3.0kg.m/s)] i + (0.25kg.m/s2)t j. Find the components Fx, Fy, and Fz of the net force on the airplane.
Answer:
[tex]F_x[/tex] = -1.5t
[tex]F_y[/tex] = 0.25
[tex]F_{z}[/tex] = 0
Explanation:
Given equation;
p = [(-0.75 kgm/s³)t² + (3.0 kgm/s)] i + (0.25 kgm/s²)t j.
From Newton's law, the rate of change of momentum of a body is the net force acting on that body. i.e
∑F = [tex]\frac{dp}{dt}[/tex] -----------(i)
Substitute the equation of p into equation (i) and differentiate with respect to t as follows;
∑F = [tex]\frac{dp}{dt}[/tex] = [tex]\frac{d| [(-0.75)t^{2} + (3.0)] i + (0.25)t j|}{dt}[/tex]
∑F = [tex]\frac{dp}{dt}[/tex] = [tex][-1.5t + 0]i + 0.25j[/tex]
∑F = [tex][-1.5t + 0]i + 0.25j[/tex]
But
∑F = [tex]F_xi + F_yj + F_zk[/tex]
Where;
[tex]F_x, F_y, F_z[/tex] are the components of the net force in the x, y and z direction respectively.
=> [tex]F_xi + F_yj + F_zk[/tex] = [tex][-1.5t + 0]i + 0.25j[/tex] = [tex]-1.5ti + 0.25j[/tex]
=> [tex]F_x[/tex] = -1.5t
=> [tex]F_y[/tex] = 0.25
=> [tex]F_{z}[/tex] = 0
The unit of current, the ampere, is defined in terms of the force between currents. Two 1.0-meter-long sections of very long wires a distance 1.5 m apart each carry a current of 1.0 A.
Required:
What is the force between them
Answer:
[tex]1.33\times 10^{-8} N[/tex]
Explanation:
According to the given scenario, the computation of force between them is shown below:-
As we know that
[tex]\mu = 4\pi\times 10^{-7}[/tex]
The force between two current carrying wires will be
[tex]F = \frac{\mu_oI_1I_2L}{2\pi r}[/tex]
[tex]= \frac{4\pi\times 10^{-7} (1 A) (1 A) (1.0 m)}{2\pi (1.5m)}[/tex]
[tex]= 1.33\times 10^{-7} N[/tex]
[tex]= 1.33\times 10^{-8} N[/tex]
Therefore for computing the force between two wires we simply applied the above formula.
So, the force between two wires carrying 1 A current [tex]= 1.33\times 10^{-8} N[/tex].
Find the net work W done on the particle by the external forces during the motion of the particle in terms of the initial and final kinetic energies. Express your answer in terms of Ki and Kf.
The net work done (W) on a particle by the external forces during the motion of the particle in terms of the initial and final kinetic energies is equal to [tex]W = K_f - K_i[/tex]
The net work done (W) can be defined as the work done in moving an object by a net force, which is the vector sum of all the forces acting on the object.
According to Newton's Second Law of Motion, the net work done (W) on an object or physical body is equal to the change in the kinetic energy possessed by the object or physical body.
Mathematically, the net work done (W) on an object or physical body is given by the formula:
[tex]W =\Delta K_E\\\\W = K_f - K_i[/tex]
Where:
W is the net work done.[tex]K_f[/tex] is the initial kinetic energy.[tex]K_i[/tex] is the final kinetic energy.Read more: https://brainly.com/question/22599382
A raspberry has a red color because it _____ red light. A. emits B. reflects C. absorbs D. transmits
Answer:
B. reflects
Explanation:
Red objects appear red because they reflect red light.
Answer:
B
Explanation:
just did the quiz
g A particle (charge = +40 mC) is located on the x axis at the point x = -20 cm, and a second particle (charge = -50 mC) is placed on the x axis at x = +30 cm. What is the magnitude of the total electrostatic force on a third particle (charge = -4.0 mC) placed at the origin (x = 0)? Group of answer choices
Answer:
Explanation:
We shall find electric field at origin due to two given charges sitting on the either side of origin .
Total field will add up due to their same direction .
Field due to a charge Q
= 9 x 10⁹ x Q / R² ; R is distance of point , Q is charge
Field due to first charge
= 9 x 10⁹ x 40 x 10⁻³ / 2² x 10⁻⁴
= 90 x 10¹⁰ N/C
Field due to second charge
= 9 x 10⁹ x 50 x 10⁻³ / 2² x 10⁻⁴
= 112.5 x 10¹⁰ N/C
Total field
= 202.5 x 10¹⁰ N/C
Force on given charge at origin
= charge x field
= 4 x 10⁻³ x 202.5 x 10¹⁰
= 810 x 10⁷ N .
g When a thin-filament light bulb is connected to two 1.6 V batteries in series, the current is 0.075 A. What is the resistance of the glowing thin-filament bulb
Answer:
R = 42.67 ohms
Explanation:
It is given that, a thin-filament light bulb is connected to two 1.6 V batteries in series, the current is 0.075 A.
It means that when two batteries are connected in series, then the total voltage is 3.2 volts
Let R is the resistance of the glowing thin-filament bulb. So, using Ohm's law we get :
[tex]V=IR\\\\R=\dfrac{V}{I}[/tex]
So,
[tex]R=\dfrac{3.2}{0.075}\\\\R=42.67\ \Omega[/tex]
So, the resistance of the bulb is 42.67 ohms.
After a long walk in the 128C outdoors, a person wearing glasses enters a room at 258C and 55 percent relative humidity. Determine whether the glasses will become fogged.
Answer:
Yes the glasses will be fogged
Explanation:
See attacher file
The number of neutrons in the nucleus of zinc 65 Zn 30 is:
35
Need more data to answer
65
30
Explanation:
proton number + neutron number = atomic mass
30 + 35 = 65
A positively charged particle has a velocity in the negative z direction at 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:
When reviewing the correct answer is A
Explanation:
The magnetic force is given by the expression
F = qv xB
where the bold letters indicate vectors, from this expression the module can be calculated
F = = q v b sin θ
the direction of the force is given by the rule of the right hand, for a positive charge the speed held by the thumb, the extended fingers point in the direction of the magnetic field and the palm points the direction of the force
in this case
the speed is in the negative part of the z axis
the force is in the negative direction of the axis and
consequently the magnetizing field is in the positive direction of the x axis
When reviewing the correct answer is A
Uses of pressure and the uses of density
Answer:
Pressure is a scalar quantity defined as per unit area.
Density is the objects ,times its the acceleration due to gravity.
Explanation:
Pressure is the alternative object increases the area of contact decrease .
Pressure is the force component to the surface used to calculate pressure.
pressure is that collisions of the gas to container as the per unit time .
pressure is an physical important quantity to play the solid and fluid .
Pressure is the expressed in a number of units depend the context use, pressure exerted by the liquid alone.
Density is the objects, times, volume of the object that times acceleration objects.
Density is the used to the system complex objects and materials.
Density force is the weight of a region or objects static fluid.
A solenoid used to produce magnetic fields for research purposes is 2.2 mm long, with an inner radius of 30 cmcm and 1200 turns of wire. When running, the solenoid produced a field of 1.4 TT in the center. Given this, how large a current does it carry?
Answer:
The current is [tex]I = 2042\ A[/tex]
Explanation:
From the question we are told that
The length of the solenoid is [tex]l = 2.2 \ m[/tex]
The radius is [tex]r_i = 30 \ cm = 0.30 \ m[/tex]
The number of turn is [tex]N = 1200 \ turns[/tex]
The magnetic field is [tex]B = 1.4 \ T[/tex]
The magnetic field produced is mathematically represented as
[tex]B = \frac{\mu_o * N * I }{l }[/tex]
making [tex]I[/tex] the subject
[tex]I = \frac{B * l}{\mu_o * N }[/tex]
Where [tex]\mu_o[/tex] is the permeability of free space with values [tex]\mu_o = 4\pi *10^{-7} N/A^2[/tex]
substituting values
[tex]I = \frac{1.4 * 2.2 }{4\pi *10^{-7} * 1200 }[/tex]
[tex]I = 2042\ A[/tex]
Write a numerical expression for the emissive intensity (in W/m^2.sr) coming out of a tiny hole in an enclosure of surface temperature 1000K and emissivity 0.6:
Answer:
6.0 × [tex]10^{11}[/tex] W/[tex]m^{2}[/tex]
Explanation:
From Wien's displacement formula;
Q = e A[tex]T^{4}[/tex]
Where: Q is the quantity of heat transferred, e is the emissivity of the surface, A is the area, and T is the temperature.
The emissive intensity = [tex]\frac{Q}{A}[/tex] = e[tex]T^{4}[/tex]
Given from the question that: e = 0.6 and T = 1000K, thus;
emissive intensity = 0.6 × [tex](1000)^{4}[/tex]
= 0.6 × 1.0 × [tex]10^{12}[/tex]
= 6.0 × [tex]10^{11}[/tex] [tex]\frac{W}{m^{2} }[/tex]
Therefore, the emissive intensity coming out of the surface is 6.0 × [tex]10^{11}[/tex] W/[tex]m^{2}[/tex].
Use Stefan's law to find the intensity of the cosmic background radiation emitted by the fireball of the Big Bang at a temperature of 2.81 K.
Complete Question
Use Stefan's law to find the intensity of the cosmic background radiation emitted by the fireball of the Big Bang at a temperature of 2.81 K. Remember that Stefan's Law gives the Power (Watts) and Intensity is Power per unit Area (W/m2).
Answer:
The intensity is [tex]I = 3.535 *10^{-6} \ W/m^2[/tex]
Explanation:
From the question we are told that
The temperature is [tex]T = 2.81 \ K[/tex]
Now According to Stefan's law
[tex]Power(P) = \sigma * A * T^4[/tex]
Where [tex]\sigma[/tex] is the Stefan Boltzmann constant with value [tex]\sigma = 5.67*10^{-8} m^2 \cdot kg \cdot s^{-2} K^{-1}[/tex]
Now the intensity of the cosmic background radiation emitted according to the unit from the question is mathematically evaluated as
[tex]I = \frac{P}{A}[/tex]
=> [tex]I = \frac{\sigma * A * T^4}{A}[/tex]
=> [tex]I = \sigma * T^4[/tex]
substituting values
[tex]I = 5.67 *10^{-8} * (2.81)^4[/tex]
[tex]I = 3.535 *10^{-6} \ W/m^2[/tex]
n oscillator is driven by a sinusoidal force. The frequency of the applied force A : must be less than the natural frequency of the oscillator. B : is independent of the natural frequency of the oscillator. C : becomes the natural frequency of the oscillator. D : must be equal to the natural frequency of the oscillator. E : must be greater than the natural frequency of the oscillator
Answer:
B : is independent of the natural frequency of the oscillator
Explanation:
You can apply any force you like to a natural oscillator. It is independent of the natural frequency of the oscillator.
The result you get will depend on how the frequency of the applied force and the natural frequency relate to each other. It will also depend on the robustness of the oscillator with respect to the applied force.
Clearly, if the force is small enough, it will have no effect on the oscillator. If it is large enough, it will overpower any motion the oscillator may attempt. For forces in the intermediate range, there will be some mix of natural oscillation and forced behavior. One may modulate the other, for example.
You need to design a clock that will oscillate at 10 MHz and will spend 75% of each cycle in the high state. You will be using a 500 pF capacitor. What values do you need to specify for R1 and R2
Answer:
Hello your question has some missing parts and the required diagram attached below is the missing part and the diagram
Digital circuits require actions to take place at precise times, so they are controlled by a clock that generates a steady sequence of rectangular voltage pulses. One of the most widely
used integrated circuits for creating clock pulses is called a 555 timer. shows how the timer’s output pulses, oscillating between 0 V and 5 V, are controlled with two resistors and a capacitor. The circuit manufacturer tells users that TH, the time the clock output spends in the high (5V) state, is TH =(R1 + R2)*C*ln(2). Similarly, the time spent in the low (0 V) state is TL = R2*C*ln(2). Design a clock that will oscillate at 10 MHz and will spend 75% of each cycle in the high state. You will be using a 500 pF capacitor. What values do you need to specify for R1 and R2?
ANSWER : R1 = 144.3Ω, R2 = 72.2 Ω
Explanation:
Frequency = 10 MHz
Time period = 1 / F = 0.1 u s
Duty cycle = 75% = 0.75
Duty cycle can be represented as : Ton / T
Also: Ton = Th = 0.75 * 0.1 u s = 75 n s
TL = T - Th = 100 ns - 75 n s = 25 n s
To find the value of R2 we use the equation for time spent in the low (0 V) state
TL = R2*C*ln(2)
hence R2 = TL / ( C * In 2 )
c = 500 pF
Hence R2 = 25 / ( 500 pF * 0.693 ) = 72.2 Ω
To find the value of R1 we use the equation for the time the clock output spends in the high (5V) state,
Th = (R1 + R2)*C*ln(2)
from the equation make R1 the subject of the formula
R1 = (Th - ( R2 * C * In2 )) / (C * In 2)
R1 = ( 75 ns - ( 72.2 * 500 pF * 0.693)) / ( 500 pF * 0.693 )
R1 = ( 75 ns - ( 25 ns ) / 500 pf * 0.693
= 144.3Ω
Often in science it is helpful to talk about a representative example of the objects or phenomena being studied. However, you must always keep in mind that the average case is not always representative. For example, our Sun is often described as an "average" star in the Milky Way. In what sense is this statement true? In what sense is this statement seriously misleading? Do you think it is useful to characterize the stars in the Milky Way by simply citing our "average" Sun?
Explanation:
The statement "our sun is an 'average' sun" is true when it is used to describe or characterize some unique physical properties of stars generally in the universe. 'Average' in this sense is used to define a typical sun such as, "stars should glow like our sun an average star."
The statement is used wrongly when used to in quantifying other stars in the universe, based on calculated values from our sun. In this case, we cannot truly say if our sun is a true representative average of other stars in the universe.
Yes! it is useful to characterize the milky way by simply citing the average sun. Properties like their ability to glow and radiate heat can be defined by citing an average star like our sun, so long as we don't translate it into citing quantitative properties of the sun as an average of our Milky Way Galaxy like the mass, temperature, etc.
Suppose a small metal object, initially at a temperature of 40 degrees, is immersed in a room which is held at the constant temperature of 70 degrees. It takes 2 minutes for the temperature of the object to reach 55 degrees.
Required:
a. Calculate the heat transfer coefficient r , i.e. the constant of proportionality in the differential equation that describes Newton's law of cooling.
r =_________
b. Suppose now that the room temperature begins to vary; i.e. Troom(t) = 90 + sin(0.1 t) . Use Newton's law of cooling and the heat transfer coefficient you calculated in the previous step to compute the temperature of the object as a function of time. Suppose that T(0) = 70 . T(t)
a)The heat transfer coefficient will be 0.231 .The concept of the Newtons law of cooling is used in the given problem.
What is newtons law of colling?The pace at which an item cools is proportional to the temperature differential between the object and its surroundings,
According to Newton's law of cooling. Simply explained, in a cold environment, a glass of hot water will cool down faster than in a hot room.
The given data in the problem is;
T₀ is the initial temperature= 40°
[tex]\rm T_S[/tex] is the constant temperature = 70°
T is the final temperature = 55°
From the given equation the Newtons law of cooling;
[tex]\rm T=T_S+(T_0-T_S)e^{-kt} \\\\ 55=70+(40-70)e^{-3k} \\\\-30 e^{-3k}=-15\\\\ e^{-3k}= -\frac{1}{2} \\\\ ln e^[3k}=ln 2 \\\\ K=\frac{1}{3} ln2 \\\\ K=0.231[/tex]
Hence the heat transfer coefficient will be 0.231 .
To learn more about the Newtons law of cooling refer to the link;
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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
A motor is designed to operate on 117 V and draws a current of 12.3 A when it first starts up. At its normal operating speed, the motor draws a current of 3.38 A. Obtain (a) the resistance of the armature coil, (b) the back emf developed at normal speed, and (c) the current drawn by the motor at one-third normal speed
Answer:
a) using
R=V/I =117/12.3
R=9.5 ohms
b)
E=V-I*R =117-3.38*9.5
E=84.8Volts
c)
at (1/3)rd of normal speed ,back emf is (1/3) of its maximum
value
E=(1/3)*84.8=28.3Volts
Current drawn
I=V-Eback/R =117-28.3/9.5
I=9.33A
Explanation:
The resistance is = 9.5 ohms
The back emf developed at normal speed is = 84.90 volts
The current drawn at one-third normal speed =9.33 A.
To calculate the resistance of the armature coil this formula is used;
V = IR
make R the subject of formula,
R = V/I
where R = resistance
V = voltage
I = Current
R = 117/12.3
R = 9.5 ohms
To calculate the back emf developed at normal speed, this formula is used;
E = V + Ir ( for normal emf)
But for back emf which is the difference between the supplied voltage and the loss from the current through the resistance, this formula is used;
E = V - Ir
where V = 117v
I = 3.38
r = 9.5
E = 117 - ( 3.38 × 9.5)
= 117 - 32.11
= 84.90 volts
To calculate the current drawn at one-third normal speed;
1/3 of Emf = 1/3 × 84.90
= 28.3volts
Therefore current (I) = V - E/ R
= 117- 28.3/9.5
= 88.7/9.5
= 9.33 A
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A scooter is traveling at a constant speed v when it encounters a circular hill of radius r = 480 m. The driver and scooter together have mass m = 159 kg.
(a) What speed in m/s does the scooter have if the driver feels weightlessness (i.e., has an apparent weight of zero) at the top of the hill?
(b) If the driver is traveling at the speed above and encounters a hill with a radius 2r,
Answer:
68.585m/sec , 779.1 N
Explanation:
To feel weightless, centripetal acceleration must equal g (9.8m/sec^2). The accelerations then cancel.
From centripetal motion.
F =( mv^2)/2
But since we are dealing with weightlessness
r = 480m
g = 9.8m/s^2
M also cancels, so forget M.
V^2 = Fr
V = √ Fr
V =√ (9.8 x 480) = 4704
= 68.585m/sec.
b) Centripetal acceleration = (v^2/2r) = (68.585^2/960) = 4704/960
= 4.9m/sec^2.
Weight (force) = (mass x acceleration) = 159kg x (g - 4.9)
159kg × ( 9.8-4.9)
159kg × 4.9
= 779.1N
A) The speed of the scooter at which the driver will feel weightlessness is;
v = 68.586 m/s
B) The apparent weight of both the driver and the scooter at the top of the hill is;
F_net = 779.1 N
We are given;
Mass; m = 159 kg
Radius; r = 480 m
A) Since it's motion about a circular hill, it means we are dealing with centripetal force.
Formula for centripetal force is given as;
F = mv²/r
Now, we want to find the speed of the scooter if the driver feels weightlessness.
This means that the centripetal force would be equal to the gravitational force.
Thus;
mg = mv²/r
m will cancel out to give;
v²/r = g
v² = gr
v = √(gr)
v = √(9.8 × 480)
v = √4704
v = 68.586 m/s
B) Now, he is travelling with speed of;
v = 68.586 m/s
And the radius is 2r
Let's first find the centripetal acceleration from the formula; α = v²/r
Thus; α = 4704/(2 × 480)
α = 4.9 m/s²
Now, since he has encountered a hill with a radius of 2r up the slope, it means that the apparent weight will now be;
F_app = m(g - α)
F_net = 159(9.8 - 4.9)
F_net = 779.1 N
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In a polar coordinate system, the velocity vector can be written as . The term theta with dot on top is called _______________________ angular velocity transverse velocity radial velocity angular acceleration
Answer:
I believe it's called rapid growth
Explanation:
that is my answer no matter what
Charge of uniform density (40 pC/m2) is distributed on a spherical surface (radius = 1.0 cm), and a second concentric spherical surface (radius = 3.0 cm) carries a uniform charge density of 60 pC/m2. What is the magnitude of the electric field at a point 4.0 cm from the center of the two surfaces?
Answer:
4.1 N/C
Explanation:
First of all, we know from maths that the surface area of a sphere = 4πr²
Charge on inner sphere ..
Q(i) = 40.0*10^-12C/m² x 4π(0.01m)²
Q(i) = 5.03*10^-14 C
Charge on outer sphere
Q(o) = 60*10^-12 x 4π(0.03m)²
Q(o) = 6.79*10^-13 C
Inner sphere has a - 5.03*10^-14C charge (-Qi) on inside of the outer shell. As a result, there is a net zero charge within the outer shell.
For the outer shell to show a NET charge +6.79*10^-13C, it's must have a +ve charge
= +6.79*10^-13C + (+5.03*10^-14C)
= +7.29*10^-13 C
Now again, we have
E = kQ /r²
E = (9.0*10^9)(+7.29*10^-13 C) / (0.04)²
E = 6.561*10^-3 / 1.6*10^-3
E = 4.10 N/C
Thus, the magnitude of the electric field is 4.1 N/C
Two long, parallel, current-carrying wires lie in an xy-plane. The first wire lies on the line y = 0.300 m and carries a current of 26.0 A in the +x direction. The second wire lies along the x-axis. The wires exert attractive forces on each other, and the force per unit length on each wire is 295 µN/m. What is the y-value (in m) of the line in the xy-plane where the total magnetic field is zero?
Answer:
The y-value is z = 0.759 m
Explanation:
From the question we are told that
The position of the first y-axis is [tex]y_1 = 0.300 \ m[/tex]
The current on the first wire is [tex]I_ 1 = 26.0 \ A[/tex]
The force per unit length on each wire is [tex]\frac{F}{l} = 295 \mu N/m = 295 * 10^{-6} \ N/m[/tex]
Generally the force per unit length on first wire is mathematically represented as
[tex]\frac{F}{l} = \frac{\mu_o * I_1 * I_2 }{2*\pi* y_1}[/tex]
Where [tex]\mu _o[/tex] is the permeability of free space with value [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]
substituting values
[tex]295 *10^{-6} = \frac{ 4\pi * 10^{-7} * 26.0 * I_2 }{2 *3.142* 0.300}[/tex]
[tex]I_2 = \frac{295 *10^{-6 } * 0.300 * 2* 3.142 }{ 4\pi * 10^{-7} * 26 }[/tex]
[tex]I_2 = 17.0 \ A[/tex]
Now the at the point where the magnetic field is zero the magnetic field of each wire are equal , let that point by z meters from the second wire on the y-axis so
[tex]\frac{\mu_o I_2}{2 * \pi * y_1} = \frac{\mu_o I_1}{2 * \pi * (y_1-z)}[/tex]
[tex]I_2 (y_1 - z) = I_1 * y_1[/tex]
substituting values
[tex]17.0 ( 0.300 - z) = 26 * 0.300[/tex]
z = 0.759 m
The refractive index n of transparent acrylic plastic (full name Poly(methyl methacrylate)) depends on the color (wavelength) of the light passing through it. At wavelength 486.1 nm (blue, designated with letter F) -> nF=1.497, and at wavelength 656.3 nm (red, designated with letter C) -> nC=1.488. Two beams (one of each wavelength) are prepared to coincide, and enter the flat polished surface of an acrylic block at angle of 45 arc degree measured from the normal to the surface. What is the angle between the blue beam and the red beam in the acrylic block?
Answer:
The angle between the blue beam and the red beam in the acrylic block is
[tex]\theta _d =0.19 ^o[/tex]
Explanation:
From the question we are told that
The refractive index of the transparent acrylic plastic for blue light is [tex]n_F = 1.497[/tex]
The wavelength of the blue light is [tex]F = 486.1 nm = 486.1 *10^{-9} \ m[/tex]
The refractive index of the transparent acrylic plastic for red light is [tex]n_C = 1.488[/tex]
The wavelength of the red light is [tex]C = 656.3 nm = 656.3 *10^{-9} \ m[/tex]
The incidence angle is [tex]i = 45^o[/tex]
Generally from Snell's law the angle of refraction of the blue light in the acrylic block is mathematically represented as
[tex]r_F = sin ^{-1}[\frac{sin(i) * n_a }{n_F} ][/tex]
Where [tex]n_a[/tex] is the refractive index of air which have a value of[tex]n_a = 1[/tex]
So
[tex]r_F = sin ^{-1}[\frac{sin(45) * 1 }{ 1.497} ][/tex]
[tex]r_F = 28.18^o[/tex]
Generally from Snell's law the angle of refraction of the red light in the acrylic block is mathematically represented as
[tex]r_C = sin ^{-1}[\frac{sin(i) * n_a }{n_C} ][/tex]
Where [tex]n_a[/tex] is the refractive index of air which have a value of[tex]n_a = 1[/tex]
So
[tex]r_C = sin ^{-1}[\frac{sin(45) * 1 }{ 1.488} ][/tex]
[tex]r_F = 28.37^o[/tex]
The angle between the blue beam and the red beam in the acrylic block
[tex]\theta _d = r_C - r_F[/tex]
substituting values
[tex]\theta _d = 28.37 - 28.18[/tex]
[tex]\theta _d =0.19 ^o[/tex]
38.A student pushes a 0.15 kg box down against a spring doing 25 J of work on the spring. The student releases the box which launches the box into the air. What is the maximum height reached by the box assuming negligible frictional forces
Answer:
Explanation:
Potential energy stored in the spring = 25 J
This energy is converted into gravitational potential energy . If h be the height attained
gravitational potential energy = mgh
mgh = 25
.15 x 9.8 x h = 25
h = 17 m
Which statement correctly describes how a bar magnet should be placed on a globe to correctly align with Earth's magnetic field?
Answer:
The answer is B. When the magnet is placed on a globe to correctly align with Earth’s magnetic field, it is considered to be suspended freely. The Earth has geographical poles as well with North and South poles. Since unlike poles attract, the South Pole of the magnet will be attracted to the geographical North.
Explanation:
B)Place the magnet vertically on the equator, with the south end facing the North pole.
What is a bar magnet?A bar magnet is a square piece of an item, made from iron, metal, or every other ferromagnetic substance or ferromagnetic composite, that indicates everlasting magnetic homes. It has two poles, a north and a south pole such that when suspended freely, the magnet aligns itself so that the northern pole factors towards the magnetic north pole of the earth.
What are the uses of a bar magnet?Bar magnets are used as stirrers in laboratories for magnetic experiments.They also find applications in medical procedures.Electronic devices such as telephones, radios, and television sets use magnets.Many industries use bar magnets for the collection of loose metals and also for retaining the magnetism of other magnets.Learn more about bar magnet here: https://brainly.com/question/18742643
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A 800 kg safe is 2.1 m above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compresses it 46 cm. What is the spring constant of the spring?
Answer:
k = 17043.5 N/m = 17.04 KN/m
Explanation:
First we need to find the force applied by safe pn the spring:
F = Weight of Safe
F = mg
where,
F = Force Applied by the safe on the spring = ?
m = mass of safe = 800 kg
g = 9.8 m/s²
Therefore,
F = (800 kg)(9.8 m/s²)
F = 7840 N
Now, using Hooke's Law:
F = kΔx
where,
K = Spring Constant = ?
Δx = compression = 46 cm = 0.46 m
Therefore,
7840 N = k (0.46 m)
k = 7840 N/0.46 m
k = 17043.5 N/m = 17.04 KN/m
In a shot-put competition, a shot moving at 15m/s has 450J of mechanical kinetic energy. What is the mass of the shot? Please help, and include the formula for the answer and a step by step explanation
Answer:
Mass of shot (m) = 4 kg
Explanation:
Given:
Velocity (v) = 15 m/s
Mechanical kinetic energy (K.E) = 450 J
Find:
Mass of shot (m) = ?
Computation:
Mechanical kinetic energy (K.E) = 1/2mv²
Mechanical kinetic energy (K.E) = [1/2](m)(15)²
450 = [1/2](m)(15)²
900 = 225 m
Mass of shot (m) = 4 kg
In a high school swim competition, a student takes 1.6 s to complete 2.0 somersaults. Determine the average angular speed of the diver, in rad/s, during this time interval.
Answer:
9.82 rads/sec
Explanation:
We are given;
Time taken; t = 1.6 secs
Number of somersaults = 2
Now, we know that,
1 revolution = 2π radians
And number of somersaults is the same thing as number of revolutions
So,
Total radians = 2π × 2 = 5π
Angular velocity = total number of revolutions/time period = 5π/1.6 = 9.82 rads/sec
Estimate the distance (in cm) between the central bright region and the third dark fringe on a screen 6.3 m from two double slits 0.49 mm apart illuminated by 739-nm light. (give answer in millimeters)
Answer:
Explanation:
distance of third dark fringe
= 2.5 x λ D / d
where λ is wavelength of light , D is screen distance and d is slit separation
putting the given values
required distance = 2.5 x 739 x 10⁻⁹ x 6.3 / .49 x 10⁻³
= 23753.57 x 10⁻⁶
= 23.754 x 10⁻³ m
= 23.754 mm .