Answer:
Part A:
[tex]E=127500N/C\\E=1.275*10^{5} N/C[/tex]
Part B:
Option B (Towards the South)
Explanation:
Part A:
Magnitude if electric field E:
E=Force/charge
Force=2.04×10−14 N
Charge=1.6×10−19 C
[tex]E=\frac{2.04*10^{-14}}{1.6*10^{-19}} \\E=127500N/C\\E=1.275*10^{5} N/C[/tex]
Part B:
Option B (Towards the South)
As electron is experiencing the force towards south,it means the direction of the electric field is towards the south because direction of field lines is from positive to negative, so proton is moving towards south it means negative charge is in south to which proton is attracted. So electric field is towards South.
Final answer:
The correct option is (b) toward the south. The magnitude of the electric field is 1.275 N/C, and the direction of the electric field is toward the south, as it attracts the positively charged proton in that direction.
Explanation:
To determine the magnitude of the electric field experienced by a proton, we use the formula for electric force: F = Eq, where F is the electric force, E is the electric field strength, and q is the charge of the particle.
Given that the force experienced by the proton is 2.04×10⁻¹⁴ N and the charge of a proton (q) is approximately 1.6×10⁻¹⁹ C, we can rearrange the formula to solve for E: E = F/q. Substituting the values, we obtain E = (2.04×10⁻¹⁴ N) / (1.6×10⁻¹⁹ C) = 1.275 N/C. Therefore, the magnitude of the electric field is 1.275 N/C.
Regarding the direction of the electric field, it is crucial to understand that electric fields exert forces on positive charges away from the source of the field and toward negative charges or areas of lower potential.
Since the proton is positively charged and experiences a force toward the south, the electric field's direction must be toward the south to attract the proton in that direction. So, the correct choice for the direction of the electric field is (b) toward the south.
A sprinter in a 100-m race accelerates uniformly for the first 71 m and then runs with constant velocity. The sprinter’s time for the first 71 m is 9.9 s. Determine his acceleration.
Answer:
The acceleration of the sprinter is 1.4 m/s²
Explanation:
Hi there!
The equation of position of the sprinter is the following:
x = x0 + v0 · t + 1/2 · a · t²
Where:
x = position of the sprinter at a time t.
x0 = initial position.
v0 = initial velocity.
t = time.
a = acceleration.
Since the origin of the frame of reference is located at the starting point and the sprinter starts from rest, then, x0 and v0 are equal to zero:
x = 1/2 · a · t²
At t = 9.9 s, x = 71 m
71 m = 1/2 · a · (9.9 s)²
2 · 71 m / (9.9 s)² = a
a = 1.4 m/s²
The acceleration of the sprinter is 1.4 m/s²
The sprinter's acceleration can calculated by solving two equations from physics: the first formula of motion (final velocity equals initial velocity plus acceleration times time) and the formula for final velocity (distance divided by time). The result is an approximate acceleration of 0.72 m/s^2.
Explanation:The question requires the use of kinematics, a topic in physics. The formula to calculate uniform acceleration is given by a = 2*(final velocity - initial velocity) / time. Here, final velocity is the velocity achieved after 71m, initial velocity is 0 (standing start), and time is 9.9s. However, we are not given final velocity. We will have to use the first formula of motion, v = u + at, in which v is the final velocity, u is the initial velocity, a is acceleration and t is time. Simultaneously, we have v = d/t where d is distance and t is time. Solving these two equations will give the acceleration (a).
From the first formula of motion and the equation v = d/t, it's clear the sprinter's final velocity (when he/she stops accelerating) is 71m/9.9s which equals approximately 7.17 m/s. Substituting into the first formula of motion gives 7.17 m/s = 0 + a*9.9s which simplifies to a = 7.17 m/s / 9.9s. This gives us an acceleration of approximately 0.72 m/s2.
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A beam of monochromatic light is diffracted by a slit of width 0.580 mm. The diffraction pattern forms on a wall 1.56 m beyond the slit. The width of the central maximum is 2.05 mm.
Calculate the wavelength of the light.
Answer:
λ = 381 nm
Explanation:
given,
width of slit,a = 0.580 x 10⁻³ m
distance of the screen from the slit, D = 1.56 m
width of central maximum,y = 2.05 x 10⁻³ m
Distance of the edge of central maximum from center of the central maximum y ‘ = y/ 2
Y ‘ = 1.025 x 10⁻³ m
for first maximum
d sin θ = λ
sin θ = tan θ = y'/D
[tex]\lambda = \dfrac{y'a}{D}[/tex]
[tex]\lambda = \dfrac{1.025\times 10^{-3}\times 0.580\times 10^{-3}}{1.56}[/tex]
λ = 381 x 10⁻⁹ m
λ = 381 nm
wavelength of light is equal to 381 nm.
A bicyclist in the Tour de France has a speed of 27.0 miles per hour (mi/h) on a flat section of the road. What is this speed in (a) kilometers per hour (km/h), and (b) meters per second (m/s)?
Answer:
(a) [tex]43.45\frac{km}{h}[/tex]
(b) [tex]12.08\frac{m}{s}[/tex]
Explanation:
(a) 1 mile is equal to 1.609344 kilometers. So, we have:
[tex]27\frac{mi}{h}*\frac{1.609344 km}{1mi}=43.45\frac{km}{h}[/tex]
(b) 1 mile is equal to 1609.34 meters. 1 hour is equal to 3600 seconds. So, we have:
[tex]27\frac{mi}{h}*\frac{1609.34m}{1mi}=4.35*10^{4}\frac{m}{h}\\4.35*10^{4}\frac{m}{h}*\frac{1h}{3600s}=12.08\frac{m}{s}[/tex]
A bicyclist's speed of 27.0 mi/h is equivalent to 43.45 km/h and 12.07 m/s.
Explanation:
(a) To convert miles per hour (mi/h) to kilometers per hour (km/h), we can use the conversion factor of 1.60934 km/h = 1 mi/h. So, multiplying the given speed of 27.0 mi/h by the conversion factor, we get:
27.0 mi/h * 1.60934 km/h = 43.45 km/h
(b) To convert kilometers per hour (km/h) to meters per second (m/s), we can use the conversion factor of 1 km/h = 0.277778 m/s. So, multiplying the given speed of 43.45 km/h by the conversion factor, we get:
43.45 km/h * 0.277778 m/s = 12.07 m/s
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One formula for a wave with a y displacement (e.g., of astring) traveling in the x direction is
y(x,t) = A\sin(kx-\omega t).
All the questions in this problem refer to this formula and tothe wave it describes.
a) Which are independent variables?
b) Which are parameters that determine the characteristics ofthe wave?
c) What is the phase phi(x,t) of the wave?
d) What is the wavelength lambda of the wave?
e) What is the period Tof this wave?
f) What is the speed of propagation v of this wave?
a) The independent variables are x and t
b) The parameters are [tex]k[/tex] (wave number), [tex]A[/tex] (the amplitude), [tex]\omega[/tex] (angular frequency)
c) The phase of this wave is zero
d) The wavelength of the wave is [tex]\lambda=\frac{2\pi}{k}[/tex]
e) The period of the wave is [tex]T=\frac{2\pi}{\omega}[/tex]
f) The speed of propagation of the wave is [tex]v=\frac{\omega}{k}[/tex]
Explanation:
a)
In physics and mathematics:
The independent variable is the variable whose value change (controlled or not controlled) and that does not depend on the other variableThe dependent variable is the variable whose value depends on the independent variableFor the wave in this problem, therefore, we have:
The time (t) and the x-position of the wave (x) are the two independent variablesThe displacement along the y-direction (y) is the dependent variable, since its value depends on the value of x and tb)
In physics and mathematics, the parameters of a function are the quantities whose value is constant (so, they do not change), and the value of the dependent variable also depends on the values of these parameters.
Therefore in this problem, for the function that represents the y-displacement of the wave, the parameters are all the constant factors in the formula that are not variables. Therefore, they are:
k, called the wave numberA, the amplitude of the wave[tex]\omega[/tex], the angular frequency of the wavec)
The phase of this wave is zero.
In fact, a general equation for a wave is in the form
[tex]y(x,t)=Asin(kx-\omega t+\phi)[/tex]
where [tex]\phi[/tex] is the phase of the wave, and it represents the initial angular displacement of the wave when x = 0 and t = 0.
However, the equation of the wave in this problem is
[tex]y(x,t)=Asin(kx-\omega t)[/tex]
Therefore, we see that its phase is zero:
[tex]\phi=0[/tex]
d)
The wavelength of a wave is related to the wave number by the following equation
[tex]k=\frac{2\pi}{\lambda}[/tex]
where
k is the wave number
[tex]\lambda[/tex] is the wavelength
For the wave in this problem, we know its wave number, [tex]k[/tex], therefore we can find its wavelength by re-arranging the equation above:
[tex]\lambda=\frac{2\pi}{k}[/tex]
e)
The period of a wave is related to its angular frequency by the following equation
[tex]\omega=\frac{2\pi}{T}[/tex]
where
[tex]\omega[/tex] is the angular frequency
T is the period of the wave
For the wave in this problem, we know its angular frequency [tex]\omega[/tex], therefore we can find its period by re-arranging the equation above:
[tex]T=\frac{2\pi}{\omega}[/tex]
f)
The speed of propagation of a wave is given by the so-called wave equation:
[tex]v=f\lambda[/tex]
where
v is the speed of propagation of the wave
f is the frequency
[tex]\lambda[/tex] is the wavelength
The frequency is related to the period of the wave by
[tex]f=\frac{1}{T}[/tex]
So, we can rewrite the wave equation as
[tex]v=\frac{\lambda}{T}[/tex]
From part d) and e), we found an expression for both the wavelength and the period:
[tex]\lambda=\frac{2\pi}{k}\\T=\frac{2\pi}{\omega}[/tex]
Therefore, we can rewrite the speed of the wave as:
[tex]v=\frac{2\pi/k}{2\pi/\omega}=\frac{\omega}{k}[/tex]
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Is it possible for a car to be accelerating to the west while it is driving to the east?
Answer:
Yes
Explanation:
If the acceleration has an opposite direction to the velocity of the car, this means that it is opposed to is motion. Therefore, it is called deceleration, since the car's velocity will decrease until it stops and then will start it moving towards the west.
A small object is attached to a horizontal spring and set in simple harmonic motion with amplitude A and period T. How long does it take for the object to travel a total distance of 6A? T/2
Answer:
1.5T
Explanation:
Since, the amplitude of SHM is A. So, in one time period T the object will travel travel a distance of 4A.
6A-4A= 2A.
Now, this 2A distance must be traveled in T/2 time period.
So, the total time taken to travel a distance of 6A is T+T/2 = 3T/2 = 1.5T
In simple harmonic motion, an object oscillates between a maximum displacement (A) and its equilibrium point, completing a full cycle within a period (T). To move a total distance of 6A, it will need 1.5 cycles, thus taking 1.5T time.
Explanation:The question you asked is about a simple harmonic oscillator, in this case, an object attached to a spring sliding on a frictionless surface. The motion of this object can be classified as simple harmonic motion (SHM).
In simple harmonic motion, the object oscillates between a maximum displacement or amplitude (A) and its equilibrium point. The time taken for the object to complete one full cycle, i.e., return to its starting point, is called the period (T).
To answer your question, when the object completes one full cycle of movement, it travels a distance of 4A (A to -A and -A to A). Therefore, if the object needs to travel a total distance of 6A, it will need 1.5 cycles. As the period, T, is the time taken for one cycle, the total time taken to travel 6A is 1.5T.
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The production of heat by metabolic processes takes place throughout the volume of an animal, but loss of heat takes place only at the surface (i.e. the skin). Since heat loss must be balanced by heat production if an animal is to maintain a constant internal temperature, the relationship between surface area and volume is relevant for physiology.
If the surface area of a cube is increased by a factor of 2, by what factor does the volume of the cube change? Give your answer to two significant figures. 1.59
To solve this problem we will apply the concepts related to the change in length in proportion to the area and volume. We will define the states of the lengths in their final and initial state and later with the given relationship, we will extrapolate these measures to the area and volume
The initial measures,
[tex]\text{Initial Length} = L[/tex]
[tex]\text{Initial surface Area} = 6L^2[/tex] (Surface of a Cube)
[tex]\text{Initial Volume} = L^3[/tex]
The final measures
[tex]\text{Final Length} = L_f[/tex]
[tex]\text{Final surface area} = 6L_f^2[/tex]
[tex]\text{Final Volume} = L_f^3[/tex]
Given,
[tex]\frac{(SA)_f}{(SA)_i} = 2[/tex]
Now applying the same relation we have that
[tex](\frac{L_f}{L_i})^2 = 2[/tex]
[tex]\frac{L_f}{L_i} = \sqrt{2}[/tex]
The relation with volume would be
[tex]\frac{(Volume)_f}{(Volume)_i} = (\frac{L_f}{L_i})^3[/tex]
[tex]\frac{(Volume)_f}{(Volume)_i} = (\sqrt{2})^3[/tex]
[tex]\frac{(Volume)_f}{(Volume)_i} = (2\sqrt{2})[/tex]
[tex]\frac{(Volume)_f}{(Volume)_i} = 2.83[/tex]
Volume of the cube change by a factor of 2.83
Two forces F1 and F2 act on an object at a point P in the indicated directions. The magnitude of F1 is 15 lb and the magnitude of F2 is 8 lb. If θ1=45â and θ2=25â, find the resultant acting force on the object as well as it's magnitude and its direction.
Answer
given,
F₁ = 15 lb
F₂ = 8 lb
θ₁ = 45°
θ₂ = 25°
Assuming the question's diagram is attached below.
now,
computing the horizontal component of the forces.
F_h = F₁ cos θ₁ - F₂ cos θ₂
F_h = 15 cos 45° - 8 cos 25°
F_h = 3.36 lb
now, vertical component of the forces
F_v = F₁ sin θ₁ + F₂ sin θ₂
F_v = 15 sin 45° + 8 sin 25°
F_v = 13.98 lb
resultant force would be equal to
[tex]F = \sqrt{F_h^2+F_v^2}[/tex]
[tex]F = \sqrt{3.36^2+13.98^2}[/tex]
F = 14.38 lb
the magnitude of resultant force is equal to 14.38 lb
direction of forces
[tex]\theta =tan^{-1}(\dfrac{F_v}{F_h})[/tex]
[tex]\theta =tan^{-1}(\dfrac{13.98}{3.36})[/tex]
θ = 76.48°
The resultant force will be 14.3844 N making an angle of 76.5° from the horizontal line.
Given to us
F1 = 15 lbF2 = 8 lb[tex]\theta_1 = 45^o[/tex][tex]\theta_2 = 25^o[/tex]We know that a force is a vector quantity and can be divided into two component a vertical and a horizontal component. As shown below in the image.
The horizontal component is the cosine component while the vertical component is the sine component.
Resultant of the horizontal components of the forces,As the forces are pointing in different directions, therefore, the force on the left will be taken as positive while the force on the right is taken as negative.
[tex]F_{H} = F_1 Cos \theta_1 + F_2 Cos \theta_2\\F_{H} = [15\times Cos(45^o)]+[-8\times Cos(25^o)]\\F_H = 3.3561\ N[/tex]
Resultant of the Vertical components of the forces,As the forces are pointing in the same direction, therefore, the net force will be in the same direction,
[tex]F_{V} = F_1 Cos \theta_1 + F_2 Cos \theta_2\\F_{V} = [15\times Sin(45^o)]+[8\times Sin(25^o)]\\F_V = 13.9875\ N[/tex]
Resultant Force[tex]F_R = \sqrt{F_H^2+ F_V^2}[/tex]
[tex]F_R = \sqrt{3.3561^2+13.9875^2}\\F_R = 14.3844\ N[/tex]
Angle of Resultant Force[tex]Tan(\theta_R) = \dfrac{F_V}{F_H} = \dfrac{13.9875}{3.3561}= 4.167784\\\\(\theta_R) = Tan^{-1}(4.167784)\\(\theta_R) = 76.5^o[/tex]
As both the forces are positive the resultant force will be in the first quadrant, making an angle of 76.5° from the horizontal line.
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A space vehicle is traveling at 5000 km/h relative to the Earth when the exhausted rocket motor is disengaged and sent backward with a speed of 71 km/h relative to the command module. The mass of the motor is four times the mass of the module. What is the speed of the command module relative to Earth just after the separation?
Answer:
the speed of the command module relative to Earth just after the separation = 4943.2 Km/hr
Explanation:
Given:
speed of space vehicle =5000 km/hr
rocket motor speed = 71 km/hr relative to the command module
mass of module = m
mass of motor = 4m
By conservation of linear momentum
Pi = Pf
Pi= initial momentum
Pf= final momentum
Since, the motion is only in single direction
[tex]MV_i=4mV_{mE}+mV_{cE}[/tex]
Where M is the mass of the space vehicle which equals the sum of motor's mass and the command's mass, Vi its initial velocity, V_mE is velocity of motor relative to Earth, and V_cE is its velocity of the command relative to Earth.
The velocity of motor relative to Earth equals the velocity of motor relative to command plus the velocity of command relative to Earth.
V_mE = V_mc+V_cE
Where V_mc is the velocity of motor relative to command this yields
[tex]5mV_i = 4m(V_{mc}+V_{cE})+mV_{cE}[/tex]
[tex]5V_i = 4V_{mc}+5V_{cE}[/tex]
substituting the values we get
[tex]V_{cE} = \frac{5V_i-4V_{mc}}{5}[/tex]
[tex]V_{cE} = \frac{5(5000)-4(71)}{5}[/tex]
= 4943.2 Km/hr
the speed of the command module relative to Earth just after the separation = 4943.2 Km/hr
Answer:
The speed of the command module relative to earth just after the separation is 4985.8 km/h
Explanation:
Given that,
Velocity of vehicle = 5000 km/h
Relative velocity = 71 km/h
The mass of the motor is four times the mass of the module.
We need to calculate the velocity of motor
Using formula of relative velocity
[tex]v=v_{2}-v_{1}[/tex]
Put the value into the formula
[tex]71=v_{2}-v_{1}[/tex]
[tex]v_{2}=71+v_{1}[/tex]
We need to calculate the speed of the command module relative to Earth just after the separation
Using conservation of momentum
[tex]mu=m_{1}v_{1}+m_{2}v_{2}[/tex]
Put the value into the formula
[tex]5m\times5000=4m\times v_{1}+mv_{2}[/tex]
[tex]5\times5000=4\times v_{1}+(71+v_{1})[/tex]
[tex]25000=5v_{1}+71[/tex]
[tex]v_{1}=\dfrac{25000-71}{5}[/tex]
[tex]v_{1}=4985.8\ km/h[/tex]
Hence, The speed of the command module relative to earth just after the separation is 4985.8 km/h
One planet is three times farther from the Sun than another. Will the farther planet take more, less, or the same amount of time to orbit the Sun
The planet will take more time to orbit the Sun.
Explanation:
According to the Kepler's law of orbital motion, a planet which is far away from the Sun experiences a lower gravitational pull towards the Sun, thus it will move with a lower speed in its orbit. Thus the farther planet takes more time to orbit the Sun compared to the planets closer to the Sun.
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Assuming a Chevy Cruze may be operated indefinitely at 30 m/s with no stops for provisions,
and may be refueled in transit, how long (in years) would be required for the Chevy
vehicle to travel one light year?
A. 1,000 years
B. 100,000 years
C. 1,000,000 years
D. 10,000,000 years
To solve this problem we will apply the linear motion kinematic equations. In which the speed is defined as the distance traveled in a certain period of time. In turn we must emphasize that 1 light year is equivalent to [tex]9.461*10^{15}m[/tex]. Mathematically the speed is described as,
[tex]v = \frac{x}{t} \rightarrow t = \frac{x}{v}[/tex]
Replacing,
[tex]t = \frac{9.461*10^{15}}{30}[/tex]
[tex]t = 3.15366*10^{14}s (\frac{1 year}{3.154*10^7})[/tex]
[tex]t \approx 10,000,000 years[/tex]
Therefore the correct answer is D.
A 0.30 kg mass is sliding on a horizontal, frictionless air track with a speed of 4.0 m/s when it instantaneously hits and sticks to a 2.0 kg mass initially at rest on the track. The spring constant is 100 N/m . The other end of the spring is fixed.a. Determine the following for the 0.30 kg mass immediately before the impact:i. Its linear momentum ii. Its KEb. Determine the following for the combined masses immediately after the impact:i. The linear momentum in. The KEc. Besides the fact that the objects stick together, what other clue lets you know that this was an inelastic collision
Answer
given,
mass of block 1, m = 0.3 Kg
speed of block 1, v = 4 m/s
mass of second block,M = 2 Kg
initial speed of block = 0 m/s
spring constant, k = 100 N/m
a) for block 1
linear momentum before collision
P₁ = m v = 0.3 x 4 = 1.2 Kg.m/s
Kinetic energy
[tex]KE_1 = \dfrac{1}{2}mv^2[/tex]
[tex]KE_1 = \dfrac{1}{2}\times 0.3\times 4^2[/tex]
[tex]KE_1 =2.4\ J[/tex]
b) After impact
final velocity calculation
using conservation of momentum
m v = (m + M )v_f
0.3 x 4 = 2.3 x v_f
v_f = 0.522 m/s
Linear momentum
P₂ = (m+M) v_f
P₂ = 1.5 x 0.522
P₂ = 0.783 kg.m/s
Kinetic energy
[tex]KE_2= \dfrac{1}{2}(M+m)v^2[/tex]
[tex]KE_2= \dfrac{1}{2}\times 2.3\times 0.522^2[/tex]
[tex]KE_2=0.313\ J[/tex]
A box is moving along the x-axis and its position varies in time according to the expression:
x = (6.00 m/s^2)t^2,
where x is in meters and t is in seconds.
(a) Determine its position (in m) at t = 3.20 s.
(b) Determine its position (in m) at t = (3.20 s) + Δt.
(c) Evaluate the limit of Δx Δt as Δt approaches zero to find the velocity (in m/s) at t = 3.20 s.
Answer:
38.4 m/s
Explanation:
a) at t = 3.2s. [tex]x = 6 * 3.2^2 = 61.44 m[/tex]
b) at t = 3.2 + Δt. [tex]x = 6*(3.2 + \Delta t)^2[/tex]
c) As Δt approaches 0. We can find the velocity by the ratio of Δx/Δt
[tex]v = \frac{\Delta x}{\Delta t} = \frac{x_2 - x_1}{\Delta t}[/tex]
[tex]v = \frac{6*(3.2 + \Delta t)^2 - 61.44}{\Delta t}[/tex]
[tex]v = \frac{6(3.2^2 + 6.4\Delta t + \Delta t^2) - 61.44}{\Delta t}[/tex]
[tex]v = \frac{61.44 + 38.4\Delta t + \Delta t^2 - 61.44}{\Delta t}[/tex]
[tex]v = \frac{\Delta t(38.4 + \Delta t)}{\Delta t}[/tex]
[tex] v = 38.4 + \Delta t[/tex]
As Δt approaches 0, v = 38.4 + 0 = 38.4 m/s
a. The position of box at t = 3.20s is 61.44 meters
b. The position at t = 3.20+ Δt is, [tex]x(3.20+ \Delta t)=6*(3.20+ \Delta t)^{2}[/tex]
c. The velocity is 38.4 meter per second.
The position of box is given by as a function shown below,
[tex]x(t)=6t^{2}[/tex]
where x is in meters and t is in seconds.
a. The position of box at t = 3.20s is,
[tex]x(3.2)=6*(3.2)^{2}\\ \\x(3.2)=61.44m[/tex]
b. The position at t = 3.20+ Δt,
[tex]x(3.20+ \Delta t)=6*(3.20+ \Delta t)^{2}[/tex]
c. We have to find
[tex]\frac{\Delta x}{\Delta t}=\frac{x(3.20+\Delta t)-x(3.2)}{\Delta t}\\\\\frac{\Delta x}{\Delta t}=\frac{6[(3.2)^{2}+(\Delta t)^{2}+6.4\Delta t - (3.2)^{2} ]}{\Delta t} \\\\\frac{\Delta x}{\Delta t}=6 \Delta t +38.4[/tex]
When [tex]\Delta t[/tex] approaches to zero.
Velocity = [tex]\frac{\Delta x}{\Delta t}=38.4 m/s[/tex]
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A horse and a squirrel participate in a race over a 1.20 km long course. The horse travels at a speed of 19.0 m/s and the squirrel can do 3.50 m/s. The horse runs for 0.960 km and then stops to tease the slow-moving squirrel, which eventually passes by. The horse waits for a while after the squirrel passes and then runs toward the finish line. Both animals cross the finish line at the exact same instant. Assume both animals, when moving, move steadily at their respective speeds.
(a)
How far (in m) is the squirrel from the finish line when the horse resumes the race?
m
(b)
For how long in time (in s) was the horse stationary?
s
Answer:
44.21053 m
279.69925 seconds
Explanation:
Time taken by the squirrel to reach the finish line
[tex]\dfrac{1200}{3.5}=342.85714\ seconds[/tex]
Time taken by the horse to cover 0.96 km
[tex]\dfrac{960}{19}=50.52631\ seconds[/tex]
Time taken by the horse to cover 1.2 km
[tex]\dfrac{1200}{19}=63.15789\ seconds[/tex]
Let the distance that the squirrel is from the finish line when the horse resumes the race be x.
During this time the horse also reaches the finish line.
We deduce that
The time taken by the horse from one stop = Time taken by squirrel before x distance from finish line.
[tex]3.5\times (63.15789-50.52631)=x\\\Rightarrow x=44.21053\ m[/tex]
The squirrel is 44.21053 m from the finish line when the horse resumes the race.
Duration is given by
[tex](342.85714-63.15789)=279.69925\ s[/tex]
The duration is 279.69925 seconds
When the horse resumed the race, the squirrel was 240m from the finish line. The horse was stationary for approximately 68.57 seconds.
Explanation:The subject of this question is Physics, specifically it deals with the concept of speed and distance. The grade level is high school as it involves basic kinematics.
(a) The horse has traveled 0.96 km, which is 960 m. It leaves 240 m for the rest of the journey. When the horse stopped, the squirrel kept moving. By the time the horse started again, the squirrel must have traveled more than 240 m. We know the horse and the squirrel finished the race at the same time, hence, they must have started the remaining journey at the same time. So, the distance from the finish line where the squirrel was when the horse began to run again equals to the rest of the horse's journey, which is 240m.
(b) To figure out how long the horse was stationary we need to calculate how long it took the squirrel to cover the distance. This is done by dividing the distance the squirrel traveled (240m) by its speed (3.5 m/s). Hence, time = distance / speed = 240m / 3.5 m/s = 68.57 seconds.
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Find a unit vector that has the same direction as the given vector. −3i + 2j − k
The unit vector that has the same direction as the given vector [tex]\(-3i + 2j - k\) is \(\frac{-3}{\sqrt{14}}i + \frac{2}{\sqrt{14}}j - \frac{1}{\sqrt{14}}k\).[/tex]
To find a unit vector in the same direction as the given vector [tex]\(-3i + 2j - k\)[/tex], we first need to calculate the magnitude of the given vector and then divide each component by that magnitude.
The magnitude of the given vector is given by the formula:
[tex]\[|v| = \sqrt{(-3)^2 + (2)^2 + (-1)^2} = \sqrt{14}\][/tex]
Now, we can find the unit vector [tex]\(\hat{u}\)[/tex] by dividing each component of the given vector by its magnitude:
[tex]\[\hat{u} = \frac{-3}{\sqrt{14}}i + \frac{2}{\sqrt{14}}j - \frac{1}{\sqrt{14}}k\]\[\hat{u} = \frac{-3}{\sqrt{14}}i + \frac{2}{\sqrt{14}}j - \frac{1}{\sqrt{14}}k\][/tex]
So, the unit vector that has the same direction as the given vector [tex]\(-3i + 2j - k\) is \(\frac{-3}{\sqrt{14}}i + \frac{2}{\sqrt{14}}j - \frac{1}{\sqrt{14}}k\).[/tex]
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To find a unit vector with the same direction as the given vector -3i + 2j - k, divide the given vector by its magnitude.
Explanation:To find a unit vector with the same direction as the given vector -3i + 2j - k, we need to divide the given vector by its magnitude. The magnitude of the given vector is the square root of the sum of the squares of its components, which is sqrt((-3)^2 + 2^2 + (-1)^2) = sqrt(14).
So, the unit vector with the same direction as the given vector is (-3i + 2j - k) / sqrt(14).
In Michael Johnson's world-record 400 m sprint, he ran the first 100 m in 11.20 s; then he reached the 200 m mark after a total time of 21.32 s had elapsed, reached the 300 m mark after 31.76 s,and finished in 43.18 s.
During what 100 m segment was his speed the highest?
a) Between the start and the 100 m mark.
b) Between the 100 m mark and the 200 m mark.
c) Between the 200 m mark and the 300 m mark.
d)Between the 300 m mark and the finish.
Answer:
b)
Explanation:
Assuming that we are talking about average speed during any segment, we can apply the definition of average speed, as follows:
v(avg) = Δx / Δt = (xf-x₀) / (tfi-t₀)
Using this definition for the 4 segments, we have:
1) v(0-100m) = 100 m / 11.20 sec = 8.93 m/s
2) v(100m-200m) = 100 m / (21.32 s - 11.2 s) = 100 m / 10.12 s = 9.88 m/s
3) v(200m -300m) = 100 m / (31.76 s- 21.32s) = 100 m / 10.44 s = 9.58 m/s
4) v(300m-400m) = 100 m / (43.18 s - 31.76 s) = 100 m / 11.42 s = 8.76 m/s
As we can see, the highest speed was reached between the 100m mark and the 200m mark, so the statement b) is the one that results to be true.
Final answer:
The highest speed in Michael Johnson's world-record 400 m sprint was achieved during the segment between the 100 m mark and the 200 m mark with a speed of 9.88 m/s.
Explanation:
To determine which 100 m segment Michael Johnson had the highest speed, we should calculate his speed for each segment separately. Speed is calculated by dividing the distance by the time taken to cover that distance. We are given the total time at each 100 m interval, so we need to calculate the time for each interval separately and then calculate the speed.
For the first 100 m: Speed = 100 m / 11.20 s = 8.93 m/sFor the second 100 m (100 m to 200 m): Time taken = 21.32 s - 11.20 s = 10.12 s, so Speed = 100 m / 10.12 s = 9.88 m/sFor the third 100 m (200 m to 300 m): Time taken = 31.76 s - 21.32 s = 10.44 s, so Speed = 100 m / 10.44 s = 9.58 m/sFor the fourth 100 m (300 m to 400 m): Time taken = 43.18 s - 31.76 s = 11.42 s, so Speed = 100 m / 11.42 s= 8.76 m/sComparing these speeds, we can see that the highest speed was achieved during the segment between the 100 m mark and the 200 m mark.
Select the options that best complete the statement.
Positively charged particle trajectories
(always, never, the same as)
follow electric field lines, because
(electric field lines are defined by the path positive test charges travel.,
the particle velocities may or may not be in the same direction as the electric field lines.,
positive charges repel each other and, therefore, are repelled by electric field lines.,
the electric force on a positively charged particle is in the same direction as the electric field.)
Positively charged particles always follow the path set by electric field lines. This is because electric field lines are defined by the path that the positive test charges would naturally travel.
Explanation:The options that best complete the statement are 'always' and 'electric field lines are defined by the path positive test charges travel.' This is because the trajectory of positively charged particle trajectories always follow the direction of electric field lines. The electric field lines are essentially mapped out pathways that positive test charges would naturally follow due to the electric force acting upon them. Hence, a positively charged particle will always move along these field lines, from a region of higher potential to a region of lower potential.
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An electromagnetic wave in a vacuum traveling in the +x direction generated by a variable source initially has a
wavelength ? of 225 ?mand a maximum electric field Emax in the +ydirection of 7.40�10^-3 V/m .
If the period of the wave is then increased by a factor of 1.50, what is the equation of the
resulting magnetic field component of the wave?
What are the coefficients in the equation for the magnetic field component
of the wave after the period increases by a factor of 1.50?
Enter the maximum magnetic field, wavenumber,
and angular frequency in teslas, inverse meters, and radians per second separated by commas.
Answer:
B (z) = 2,467 10⁻¹¹ cos (1,486 10⁸ x - 5.571 10¹⁵ t )
Explanation:
An electromagnetic wave is a wave that is sustained in the perpendicular fluctuations of the electric and magnetic fields, the equation of the wave is
E (y) = Eo cos (kx –wt)
B (z) = Bo cos (kx-wt)
Let's look for the terms to build these equations. The speed of the wave is given by
c = λ f
The frequency and period are related
f = 1 / T
Let's start by applying this equation our case
f = c /λ
f = 3 10⁸/225 10⁻⁹
f = 1.33 10¹⁵ Hz
The angular velocity and the wave number are
w = 2π f
k = 2π /λ
w = 2π 1.33 10¹⁵ = 8.38 10¹⁵ rad / s
k = 2π / 225 10⁻⁹ = 2.79 10⁷ m⁻¹
It indicates that the period increases by a factor of 1.5, let's look for the new frequency
T = 1.5 T₀
f = 1 / T
f = 1 / 1.5T₀
f = 1 / 1.5 f₀
f = 1 / 1.5 1.33 10¹⁵ = 8.87 10¹⁴ Hz
c = λ f
λ = c / f
λ = 3 10⁸ / 8.87 10¹⁴ = 4,229 10⁻⁸ m
Let's find the new w and k
w = 2π f
w = 2π 8.87 10¹⁴ = 5.571 10¹⁵ rad/s
k = 2π / λ
k = 2π / 4,229 10⁻⁸ = 1,486 10⁸ cm⁻¹
We use the relationship that the fields are in phase
c = E₀ / B₀
B₀ = E₀ / c
B₀ = 7.4 10⁻³ / 3 10 ⁸ = 2.467 10⁻¹¹ T
With these values we can build the equation of the magnetic field
B (z) = 2,467 10⁻¹¹ cos (1,486 10⁸ x - 5.571 10¹⁵ t )
Two cars, one of mass 1300 kg, and the second of mass 2400 kg, are moving at right angles to each other when they collide and stick together. The initial velocity of the first car is 12 m/s in the positive x direction and that of the second car is 18 m/s in the positive y direction.
What is the magnitude of the velocity of the wreckage of the two cars immediately after the collision?
To solve this problem we will apply the momentum conservation theorem, that is, the initial momentum of the bodies must be the same final momentum of the bodies. The value that will be obtained will be a vector value of the final speed of which the magnitude will be found later. Our values are given as,
[tex]m_1 = 1300kg[/tex]
[tex]m_2 = 2400kg[/tex]
[tex]u_1 = 12m/s i[/tex]
[tex]u_2 = 18m/s j[/tex]
Using conservation of momentum,
[tex]m_1u_1+m_2u_2 = (m_1+m_2)v_f[/tex]
[tex]1300*12i-2400*18j = (1300+2400)v_f[/tex]
Solving for [tex]v_f[/tex]
[tex]v_f = 4.2162i-11.6756j[/tex]
Using the properties of vectors to find the magnitude we have,
[tex]|v| = \sqrt{(4.2162^2)+(-11.6756)^2}[/tex]
[tex]|v| = 12.4135m/s[/tex]
Therefore the magnitude of the velocity of the wreckage of the two cars immediately after the collision is 12.4135m/s
For the two cars the final velocity of the wreckage of the two cars immediately after the collision is 12.4135 m/s.
What is conservation of momentum?Momentum of a object is the force of speed of it in motion. Momentum of a moving body is the product of mass times velocity.
When the two objects collides, then the initial collision of the two body is equal to the final collision of two bodies by the law of conservation of momentum.
The mass of the car one is 1300 kg and the mass of the second car is 2400 kg.
The initial velocity of the first car is 12 m/s in the positive x direction, and that of the second car is 18 m/s in the positive y direction.
Final velocity of the two cars after the collision is equal. Thus, the final velocity of the wreckage of the two cars immediately after the collision by the law of conservation can be given as,
[tex](1300)12\hat i-(2400)18\hat j=(1300+2400)v\\v=4.2162\hat i-11.6756\hat j[/tex]
Solve the above equation further using the property of vectors as,
[tex]v=\sqrt{(4.2162)^2+(-11.6756)^2}\\v=12.4135\rm m/s[/tex]
Hence, For the two cars the final velocity of the wreckage of the two cars immediately after the collision is 12.4135 m/s.
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Particle A of charge 3.06 10-4 C is at the origin, particle B of charge -5.70 10-4 C is at (4.00 m, 0), and particle C of charge 1.08 10-4 C is at (0, 3.00 m). We wish to find the net electric force on C.
Answer:
F_net = 26.512 N
Explanation:
Given:
Q_a = 3.06 * 10^(-4 ) C
Q_b = -5.7 * 10^(-4 ) C
Q_c = 1.08 * 10^(-4 ) C
R_ac = 3 m
R_bc = sqrt (3^2 + 4^2) = 5m
k = 8.99 * 10^9
Coulomb's Law:
F_i = k * Q_i * Q_j / R_ij^2
Compute F_ac and F_bc :
F_ac = k * Q_a * Q_c / R^2_ac
F_ac = 8.99 * 10^9* ( 3.06 * 10^(-4 ))* (1.08 * 10^(-4 )) / 3^2
F_ac = 33.01128 N
F_bc = k * Q_b * Q_c / R^2_bc
F_bc = 8.99 * 10^9* ( 5.7 * 10^(-4 ))* (1.08 * 10^(-4 )) / 5^2
F_bc = - 22.137 N
Angle a is subtended between F_bc and y axis @ C
cos(a) = 3 / 5
sin (a) = 4 / 5
Compute F_net:
F_net = sqrt (F_x ^2 + F_y ^2)
F_x = sum of forces in x direction:
F_x = F_bc*sin(a) = 22.137*(4/5) = 17.71 N
F_y = sum of forces in y direction:
F_y = - F_bc*cos(a) + F_ac = - 22.137*(3/5) + 33.01128 = 19.72908 N
F_net = sqrt (17.71 ^2 + 19.72908 ^2) = 26.5119 N
Answer: F_net = 26.512 N
Final answer:
The question requires applying Coulomb's law to calculate the net electric force on particle C due to particles A and B, and summing the vector forces to find the resultant.
Explanation:
The question involves finding the net electric force on particle C, which is at the position (0, 3.00 m) in the presence of particles A and B with given charges and positions. The situation can be analyzed using Coulomb's law, which states that the force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them.
We would sum the vector forces from both charges on C to find the resultant force. This is a classical physics problem typically addressed in high school or introductory college physics courses.
A parallel-plate capacitor is made from two aluminum-foil sheets, each 3.70 cm wide and 12.0 m long. Between the sheets is a mica strip of the same width and length that is 0.0225 mm thick.
What is the maximum charge that can be stored in this capacitor? (The dielectric constant of mica is 5.4, and its dielectric strength is 1.00 \times 10^8 \;{\rm{V}}/{{\rm{m}}})
Qmax= ??mC
Answer:
2.121876 mC
Explanation:
[tex]\epsilon_0[/tex] = Permittivity of free space = [tex]8.85\times 10^{-12}\ F/m[/tex]
A = Area = [tex]0.037\times 12\ m^2[/tex]
d = Thickness = 0.0225 mm
E = Dielectric strength = [tex]1\times 10^8\ V/m[/tex]
k = Dielectric constant = 5.4
Capacitance is given by
[tex]C=\dfrac{k\epsilon_0A}{d}\\\Rightarrow C=\dfrac{5.4\times 8.85\times 10^{-12}\times 0.037\times 12}{0.0225\times 10^{-3}}\\\Rightarrow C=9.43056\times 10^{-7}\ F[/tex]
Maximum voltage is given by
[tex]V_m=E_md\\\Rightarrow V_m=1\times 10^8\times 0.0225\times 10^{-3}\\\Rightarrow V_m=2250\ V[/tex]
Maximum charge is given by
[tex]Q_m=CV_m\\\Rightarrow Q_m=9.43056\times 10^{-7}\times 2250\\\Rightarrow Q_m=0.002121876\ C=2.121876\ mC[/tex]
The maximum charge that can be stored in this capacitor is 2.121876 mC
The maximum charge that can be stored in this capacitor is 16.016 μC.
Explanation:To find the maximum charge that can be stored in the capacitor, we need to use the formula Q=CV, where Q is the charge, C is the capacitance, and V is the voltage. The capacitance can be calculated using the formula C = (εrε0A)/d, where εr is the relative permittivity (dielectric constant) of mica, ε0 is the permittivity of free space, A is the area of the plates, and d is the distance between the plates.
First, let's calculate the area of the plates using the given dimensions: A = (0.037 m) * (12.0 m) = 0.444 m².Next, let's calculate the capacitance using the formula: C = (5.4 * 8.854 × 10^-12 F/m) * (0.444 m²) / (0.0000225 m) = 1.6016 × 10^-8 F.Finally, we can calculate the maximum charge using the formula Q = CV: Q = (1.6016 × 10^-8 F) * (1 V) = 1.6016 × 10^-8 C = 16.016 μC.Learn more about Maximum charge in a capacitor here:https://brainly.com/question/32606892
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Suppose you are an observer standing on the Moon, looking back at Earth.
If the moon is in the waxingquarter phase, what phase of the Earth would you see?
A. waxing quarter
B. full
C. new
D. waning quarter
E. none; the Earth doesn't have phases
Answer:
D. waning quarter
Explanation:
The Moon is a natural satellite of Earth and reflects the light of Sun to become visible from the Earth. It shows various phases during its revolution around the Earth depending on how much part of the lit up portion of the Moon is towards Earth. Just like Moon shows phases, if we go on Moon the Earth will show various phases.
When the Moon will be in waxing quarter phase i.e. as seen from Earth, we will see a semicircular Moon with its right side lit up. At the same time if someone from the Moon sees Earth, the Earth will show a waning quarter phase. It will be seen as semicircular but left side will be lit up.
If you are an observer standing on the Moon looking back at Earth during the waxing quarter phase of the Moon, you would see a waning quarter phase of the Earth. Thus, the correct option is D.
Explanation:If you are an observer standing on the Moon looking back at Earth, the phase of the Earth that you would see depends on the relative positions of the Moon, Earth, and Sun. In the waxing quarter phase of the Moon, you would see a waning quarter phase of the Earth. This is because when the Moon is in the waxing quarter phase, the illuminated portion of the Moon is on the right side, meaning the Earth would be on the left side of the Moon and in the waning quarter phase.
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In a salt crystal, the distance between adjacent sodium and chloride ions is 2.82 Angstroms. What is the force of attraction between the two singly charged ions
The force of attraction between sodium and chloride ions in a salt crystal can be calculated using Coulomb's law, considering isotropic forces and applying the Madelung constant.
Explanation:
The force of attraction between the sodium and chloride ions in a salt crystal can be calculated using Coulomb's law, which involves the product of the charges divided by the distance between them squared. The force is proportional to the ionic charges and inversely proportional to the square of the distance between the ions. In a sodium chloride (NaCl) crystal, the resulting ions (Na+ and Cl¯) produce isotropic forces, which means the force is the same in all directions.
Given the distance mentioned in your question (2.82 Angstroms), you can use this along with the known charges of sodium and chloride ions to calculate the force. However, please note that this calculation needs to take into account the Madelung constant, which is a factor for considering the interaction of a sodium ion with all the nearby chloride and sodium ions in a tightly arranged three-dimensional lattice structure. For a NaCl crystal, the Madelung constant is approximately 1.75.
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A soft-drink bottler purchases glass bottles from a vendor. The bottles are required to have internal pressure strength of at least 150 pounds per square inch (psi). A prospective bottle vendor claims that its production process yields bottles with mean internal pressure strength of 157 psi and standard deviation of 3 psi. The bottler strikes an agreement with the vendor that permits the bottler to sample from the vendor's production process to verify the vendor's claim. The bottler obtains a random sample of 64 bottles. If the mean internal pressure strength of the sample falls below K, the bottler will conclude the vendor's claim about the mean internal pressure strength to be false. Suppose the bottler is willing to risk a 2% chance of concluding the vendor's claim to be false even if the claim is true. Find the value of K.
Answer: K =24 psi
Explanation:
Given: Standard deviation =3psi
Internal pressure strength =157psi
Number of random bottle =n=64
K= 3 × square root of 64
K= 3×8=24 psi
If mean internal pressure K fall below K,
157-1.3=155.7psi
At 2%:
0.16×64 = 10.24
Final answer:
The value of K is 150.835 psi.
Explanation:
To find the value of K, we need to calculate the critical value of the sample mean that separates the lower 2% of the distribution from the upper 98%. First, we need to determine the z-score corresponding to the desired probability of 2%. We can find this value using a standard normal distribution table or a calculator. The z-score for a 2% probability is approximately -2.055. Next, we can calculate the value of K by multiplying the z-score by the standard deviation and adding it to the mean:
K = 157 + (-2.055 * 3) = 150.835 psi
Therefore, if the sample mean falls below 150.835 psi, the bottler will conclude the vendor's claim about the mean internal pressure strength to be false.
When a second student joins the first, the height difference between the liquid levels in the right and left pistons is 40 cm . What is the second student's mass?
To find the second student's mass, we can use the principle of Pascal's Law which states that when a fluid is at rest in a container, the pressure is transmitted equally in all directions. The height difference between the liquid levels in the right and left pistons can be used to determine the mass.
Explanation:To find the second student's mass, we can use the principle of Pascal's Law which states that when a fluid is at rest in a container, the pressure is transmitted equally in all directions. In this case, the height difference between the liquid levels in the right and left pistons can be used to determine the mass. The height difference is 40 cm, so the pressure difference is equal to the weight of the second student:
P = ρgh, where P is the pressure difference, ρ is the density of the liquid, g is the acceleration due to gravity, and h is the height difference.
Using the formula, we can solve for the mass of the second student:
m = P/(g*h), where m is the mass of the second student.
Let's plug in the values:
m = (ρgh)/(g*h) = ρ. So, the mass of the second student is equal to the density of the liquid.
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To find the mass of the second student using the difference in liquid levels in a hydraulic system, we can rearrange the pressure difference equation, which takes into account the height difference, gravity, and fluid density.
Explanation:The subject of the question delves into the physical principles of hydraulics. If the height difference between the liquid levels in the right and left pistons is 40 cm when a second student joins the first, we can apply the principle of hydraulic machines which is essentially the conservation of energy principle.
In hydraulics, pressure applied on one piston gets transmitted to the other without being diminished. Hence, in this context, let's denote the mass of the second student as m2. The difference in heights or pressure difference ΔP can be measured by subtracting fluid devices and can be expressed using the formula ΔP = m2g
In this equation, 'g' is the acceleration due to gravity. Given the height difference Δh as 40cm, we can deduce the density (ρ) of the fluid. Remember, pressure difference ΔP = ρg Δh. Hence, we can obtain the mass of the second student by rearranging the equation as follows: m2 = (ρg Δh) / g. Please note: for this scenario, the value of 'g' needs to be considered as 9.8 m/s² (earth's acceleration) and ρ is the fluid density (e.g., for water, it is 1000 kg/m³).
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Sketch the function Ex versus x for both positive and negative values of x. Remember that Ex is negative when E points in the negative x direction.
Answer:
Here is the complete question.
A charge of 3 μC is at the origin. Sketch the function E x versus x for both positive and negative values of x. (Remember that Ex is negative when E points in the negative x direction.)
The graph is sketched in the attachment.
Explanation:
Since our electric field Eₓ = q/4πε₀x² (electric field for a point charge)
where q = electric charge = 3μC and x = distance from the origin of the charge. 1/4πε₀= 9 × 10⁹ Nm²/C².
Substituting the values into Eₓ = 3 × 10⁻⁶×9 × 10⁹ /x²= 0.027/x². We plot values of x on the x- axis ranging from -3 to 3 into the equation to give us the graph for Eₓ. Note that the sign of Eₓ changes when x crosses the origin because of the direction of the electric field due to the new position of the charge. The graph is in the attachment below. Note that as x tends to zero, Eₓ tends to infinity and x tends to infinity, Eₓ tends to zero.
A motor produces a torque of 22 Nm. It is used to accelerate a wheel of radius 10cm and moment of inertia 2 kg.m2 which is initially at rest. Calculate
a)
the number of revolutions made by the wheel in the first 5s
b)
the angular velocity after 5s
c)
the acceleration of a point on the rim of the wheel after 5s
Answer
given,
torque produced, τ = 22 N.m
Radius of the wheel. r = 10 cm
Moment of inertial = 2 kg.m²
initial angular speed = 0 rad/s
time, t = 5 s
a) we know,
τ = I α
22 = 2 x α
α = 11 rad/s²
using equation of rotation motion
[tex]\theta = \omega_o t + \dfrac{1}{2}\alpha t^2[/tex]
[tex]\theta =\dfrac{1}{2}\times 11 \times 5^2[/tex]
θ = 137.5 rad
θ = 137.5/2π = 22 revolution.
b) angular velocity of the motor
[tex]\omega_f = \omega_i + \alpha t[/tex]
[tex]\omega_f = 0 + 11 x 5[/tex]
[tex]\omega_f = 55\ rad/s[/tex]
c) acceleration of a point on the rim of the wheel
radial acceleration
[tex]a_r = \omega^2 r[/tex]
[tex]a_r = 55^2\times 0.1[/tex]
[tex]a_r =302.5 \ m/s^2[/tex]
tangential acceleration of the point on the rim
[tex]a_t = \alpha r[/tex]
[tex]a_t = 11\times 0.1[/tex]
[tex]a_t = 1.1\ m/s^2[/tex]
now, acceleration of the point
[tex]a = \sqrt{a_r^2+a_t^2}[/tex]
[tex]a = \sqrt{302.5^2+1.1^2}[/tex]
[tex]a = 302.5\ m/s^2[/tex]
3. Find the de Broglie wavelength of the following particles: (i) An electron in a semiconductor having average thermal velocity at T = 300 K
Answer:
The wavelength of the electron is [tex]\dfrac{4.4097\times10^{-9}}{\sqrt{a}}\ m[/tex]
Explanation:
Given that,
Temperature = 300 K
We know that,
The energy of free electron is
[tex]E=\dfrac{(\hbar)^2k^2}{2m}[/tex]
[tex]k=\dfrac{\sqrt{2mE}}{\hbar}[/tex]
Where, k = wave number
The momentum of the electron is
[tex]p=\hbar k[/tex]
Th effective mass is
[tex]m=am_{0}[/tex]
We need to calculate the wavelength of the electron
Using formula of wave number
[tex]k=\dfrac{2\pi}{\lambda}[/tex]
[tex]\lambda=\dfrac{2\pi}{k}[/tex]
Put the value of k
[tex]\lambda=\dfrac{2\pi}{\dfrac{\sqrt{2mE}}{\hbar}}[/tex]
[tex]\lambda=\dfrac{h}{\sqrt{2am_{0}E}}[/tex]
We know that,
Thermal energy of electron
[tex]E=3kT[/tex]
The de Broglie wavelength of the electron is
[tex]\lambda=\dfrac{h}{\sqrt{2am_{0}\times3kT}}[/tex]
Put the value into the formula
[tex]\lambda=\dfrac{6.63\times10^{-34}}{\sqrt{2\times9.1\times10^{-31}\times3\times300\times1.380\times10^{-23}\times a}}[/tex]
[tex]\lambda=\dfrac{4.4097\times10^{-9}}{\sqrt{a}}\ m[/tex]
Hence, The wavelength of the electron is [tex]\dfrac{4.4097\times10^{-9}}{\sqrt{a}}\ m[/tex]
The de Broglie wavelength of an electron can be calculated using de Broglie's equation, λ = h/p. To find the wavelength for an electron in a semiconductor at 300 K, one needs to know the electron's velocity, which is related to its kinetic energy at that temperature. Precise calculation would require details about the effective electron mass in the semiconductor.
Explanation:The de Broglie wavelength λ of a particle can be determined using de Broglie's equation, λ = h/p, where 'h' is Planck's constant (6.626 × 10-34 m2 kg/s) and 'p' is the momentum of the particle. The momentum of an electron moving at thermal velocities in a semiconductor at room temperature (T = 300 K) can be approximated using the formula p = mv, where 'm' is the mass of an electron (9.11 × 10-31 kg) and 'v' is the velocity. The average thermal velocity can be obtained from the kinetic theory of gases, which states that the average kinetic energy (KE) of a particle is ⅓kT, where 'k' is the Boltzmann constant (1.38 × 10-23 J/K) and 'T' is the temperature in kelvins. Therefore, v can be calculated as √(3kT/m). Substituting this velocity into the momentum formula and then into de Broglie's equation yields the de Broglie wavelength of the electron.
However, a precise value for the de Broglie wavelength of an electron in a semiconductor at T = 300 K would require additional information not provided in the question, such as the effective mass of the electron in the semiconductor, which can differ from the free electron mass due to the crystal structure's influence.
At t=0 a ball, initially at rest, starts to roll down a ramp with constant acceleration. You notice it moves 1 foot between t=0 seconds and t = 1 second. 1)How far does it move between t = 1 second and t = 2 seconds?
Answer: 3 ft
Explanation:
Given:
- the ball is initially at rest before it start moving at constant acceleration.
a = constant
u = initial speed = 0
Using the formula for displacement.
d = ut + 0.5at^2 .....1
Since u = 0, equation 1 becomes;
d = 0.5at^2 .....3
making a the subject of formula
a = 2d/t^2 ....2
Where; a = acceleration, d = distance travelled, t = time taken, u = initial speed
d = 1 ft, t =1 s
Substituting into equation 2 above.
a = (2×1)/1^2 = 2ft/s^2
Since the acceleration is 2ft/s^2, at t= 2 sec the distance it would have covered will be:
Using equation 3;
d = 0.5(2 × 2^2) = 0.5(8) = 4ft
therefore, the distance travelled between t = 1 to t= 2
Is :
d2 = d - d1
Where d = total distance (t=0 to t= 2) =4 ft
d1 = distance covered between (t=0 to t=1) = 1 ft
d2 = 4 - 1 = 3ft
The distance traveled between time 1s and time 2s is 3 ft.
The given parameters:Initial velocity of the ball, u = 0Acceleration = constantThe distance traveled by the ball at time t is calculated as follows;
[tex]s = ut + \frac{1}{2} at^2\\\\s = 0 + \frac{1}{2} at^2\\\\s = \frac{1}{2} at^2\\\\a = \frac{2s}{t^2}[/tex]
when the distance, s = 1 ft and time, t = 1 s
[tex]a = \frac{2(1)}{(1)^2} \\\\a = 2 \ ft/s^2[/tex]
When the time of motion, t = 2s, the distance traveled is calculated as;
[tex]s = \frac{1}{2} (2) (2)^2\\\\s = 4 \ ft[/tex]
The distance traveled between t =1 and t = 2 is calculated as follows;
[tex]s = s_2 - s_1\\\\s = 4 \ ft - 1\ ft\\\\s = 3 \ ft[/tex]
Learn more about change in distance here: https://brainly.com/question/2854969
One simple model for a person running the 100m dash is to assume the sprinter runs with constant acceleration until reaching top speed, then maintains that speed through the finish line.
If a sprinter reaches his top speed of 11.3m/s in 2.44s , what will be his total time?
Answer:
Explanation:
Given
Race length [tex]L=100\ m[/tex]
Runner reaches maximum velocity in [tex]t_1=2.4\ s[/tex]
[tex]v_{max}=11.3\ m/s[/tex]
distance traveled during this time
[tex]v=u+at[/tex]
[tex]11.3=0+a\times 2.4[/tex]
[tex]a=\frac{11.3}{2.4}=4.708\ m/s^2[/tex]
distance traveled in this time
[tex]s=ut+\frac{1}{2}at^2[/tex]
[tex]s=0+0.5\times 4.708\times (2.44)^2[/tex]
[tex]s=14.014\ m[/tex]
Remaining distance will be traveled with [tex]v_{max}[/tex]
remaining distance [tex]d=100-14.014=85.98\ m[/tex]
time taken [tex]t_2=\frac{d}{v_{max}}=\frac{85.98}{11.3}=7.609\ s[/tex]
total time [tex]t=t_1+t_2[/tex]
[tex]t=2.44+7.609=10.049\ s[/tex]