Answer:
T = 6.0 N
Explanation:
given,
mass of the cord = 0.46 Kg
length of the supports = 7.2 m
time taken to travel = 0.74 s
tension in the chord = ?
using formula for tension calculation
[tex]T = \dfrac{v^2.m}{l}[/tex]
[tex]v = \dfrac{l}{s}[/tex]
[tex]v = \dfrac{7.2}{0.74}[/tex]
v = 9.73 m/s
now, calculation of tension
[tex]T = \dfrac{9.73^2\times 0.46}{7.2}[/tex]
T = 6.0 N
The tension in the cord is equal to 6.0 N.
You are pulling a box so it moves at increasing speed. Compare the work (W1) you need to do to accelerate it from 0 m/s to speed v to the work (W2) needed to accelerate it from speed v to the speed of 2v in the presence of friction. Compare the work () you need to do to accelerate it from 0 to speed to the work () needed to accelerate it from speed to the speed of 2 in the presence of friction. W2 > W1 W1 > W2 W1
Answer:
Explanation:
In absence of friction
Work done by force is given by change in kinetic energy of box
According work Energy theorem change in kinetic Energy of object is equal to work done by all the forces
[tex]W_1=\frac{1}{2}mv^2-0[/tex]
[tex]W_1=\frac{1}{2}mv^2[/tex]
where m=mass of object
[tex]W_2=\frac{1}{2}m(2v)^2-\frac{1}{2}mv^2[/tex]
[tex]W_2=\frac{3}{2}mv^2[/tex]
[tex]\frac{W_2}{W_1}=\frac{\frac{3}{2}mv^2}{\frac{1}{2}mv^2}[/tex]
[tex]W_2>W_1[/tex]
so work done in second case is three times of first case
(b)In Presence of friction
[tex]W_f+W_1=\frac{1}{2}mv^2[/tex]
suppose [tex]f_r[/tex] is the friction force and d is the displacement
so [tex]W_f=f_r\cdot d\cos 180[/tex]
[tex]W_f=-f_r\cdot d[/tex]
[tex]W_1=\frac{1}{2}mv^2+f_r\cdot d[/tex]
for second case when speed increases v to 2v
[tex]W_2=\frac{3}{2}mv^2+f_r\cdot d[/tex]
thus [tex]W_2>W_1[/tex] when friction is present
The work done to accelerate the box from zero is lesser than the work done to accelerate the box from velocity [tex]v[/tex].
From Work- energy theorem,Work can be defined as the change in the kinetic energy of the object.
So,
In the first Scenario,
[tex]W_1 = \dfrac 12 mv^2- 0\\\\W_1 = \dfrac 12 mv^2[/tex]
In the second scenario,
[tex]W_2 = \dfrac 12 m(2v^2) -\dfrac 12 mv^2\\\\W_2 = \dfrac 32 mv^2[/tex]
Now compare both scenarios,
[tex]\dfrac {W_1}{W_2}=\dfrac{ \dfrac 32 mv^2}{\dfrac12 mv^2}[/tex]
So, [tex]W_1 <W_2[/tex]
Therefore, the work done to accelerate the box from zero is lesser than the work done to accelerate the box from velocity [tex]v[/tex].
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A farmer lifts his hay bales into the top loft of his barn by walking his horse forward with a constant velocity of 1 ft/s. Determine the velocity and acceleration of the hay bale when the horse is 10 ft away from the__________.
When an object moves with constant velocity, its acceleration is zero. So, the hay bale's velocity is 1 ft/s, and its acceleration is 0 ft/s2. These values persist irrespective of the horse's distance from the barn.
Explanation:In the case of the farmer's horse and the hay bale, the horse is moving with a constant velocity of 1 ft/s. When an object moves with constant velocity, its acceleration is zero. This principle stems from the fundamental definition of acceleration as the rate of change of velocity over time. Since the horse's velocity is not changing, acceleration is zero.
Moving on to the velocity of the hay bale: if we suppose that the horse's movement directly influences the lifting of the hay bale, the bale's velocity would also be 1 ft/s. Provided that the system of lifting the bale is suitable for the task, it does not matter how far the horse is away from the barn; the velocity and acceleration values persist. Therefore, regardless of whether the horse is 10 ft away or 100 ft away, the velocity of the hay bale remains 1 ft/s and acceleration 0 ft/s2.
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The hay bale being lifted by the farmer's horse, which is walking at a constant velocity of 1 ft/s, would have the sameconstant velocity with an acceleration of 0 m/s², because there is no change in velocity. Velocity vectors for the horse's movement would remain consistent, indicating the absence of acceleration.
The question involves determining the velocity and acceleration of a hay bale being lifted into a barn loft by a horse walking at a constant velocity. When dealing with such problems in physics, we typically look for changes in velocity over time to find acceleration. In this scenario, since the horse is moving with a constant velocity of 1 ft/s and no information is provided about forces acting on the hay bale or if there is any change in the speed of the horse, we can infer that the hay bale being lifted is moving at the same constant velocity of 1 ft/s, with an acceleration of 0 m/s² (or ft/s²) since there is no change in speed. This is because acceleration is defined as the rate of change of velocity over time, and a constant velocity means no change in velocity.
To conceptualize this, imagine sketching out a series of velocity vectors for the horse's path from one point to the next, indicating consistent motion with no alterations above or below the horizontal axis, representing that there is no acceleration occurring. An example in the given information refers to a racehorse accelerating from rest to 15.0 m/s, indicating that its acceleration can be calculated by dividing the change in velocity by the change in time, giving an average acceleration. But for the farmer's horse walking forward at a constant pace, the situation is quite different since there is no increase in speed.
To reiterate, if the horse maintains a constant velocity, the hay bale will have the same velocity (1 ft/s) and zero acceleration, provided that the system remains unaltered. Calculating acceleration in such cases is straightforward when there's a change in velocity; for constant velocity, the acceleration is simply zero.
Waves on a string are described by the following general equation
y(x,t)=Acos(kx−ωt).
A transverse wave on a string is traveling in the +xdirection with a wave speed of 8.75 m/s , an amplitude of 6.50×10−2 m , and a wavelength of 0.540 m . At time t=0, the x=0 end of the string has its maximum upward displacement. Find the transverse displacement y of a particle at x = 1.59 m and t = 0.150 s .
Q. In general, the cosine function has maximum displacements, either positive or negative, when its argument is equal to an integer multiple of π. When t = 0.150 s , k = 11.6 rad/m , and ω = 102 rad/s use the wave equation to select all of the x positions that correspond to points of maximum displacement.
Check all that apply considering only positive arguments of the cosine function.
0.795 m
1.59 m
1.73 m
1.86 m
2.00 m
2.13 m
Answer:
The displacement is at x=1.59m and t=0.150s is [tex]-6.50\cdot10^{-2}\text{ m}.[/tex]
Out of the given points, the argument of the cosine is an integer multiple of [tex]\pi[/tex] for x=1.59m, 1.86m, 2.13m.
Explanation:
The displacement at x and t is given by y(x,t). We now need to find [tex]k[/tex] and [tex]\omega[/tex]. The speed of the wave is given by
[tex]c=\frac{\omega}{k}[/tex]
while the wavelength satisfies
[tex]k=\frac{2\pi}{\lambda}=\frac{2\pi}{0.540\text{ m}}=11.6\text{ m}^{-1}.[/tex]
Substituting this into the previous equation we find
[tex]\omega=ck=8.75\text{ m/s}\cdot 11.6\text{ m}^{-1}=102\text{ rad/s}.[/tex]
Now we have
[tex]y(1.59\text{ m},0.150\text{ s})=6.50\cdot10^{-2}\text{ m}\cos(11.6\cdot1.59-102\cdot0.150)=-6.50\cdot10^{-2}\text{ m}.[/tex]
Now we calculate [tex]kx-\omega t[/tex] at [tex]t=0.150\text{ s}[/tex] at each given x and check whether it is integer multiple of [tex]\pi[/tex].
[tex]11.6\text{ m}^{-1}\cdot 0.795\text{ m}-102\text{ rad/s}\cdot0.150\text{ s}=-6.08=-1.93\pi\text{ not an integer multiple of }\pi;[/tex]
[tex]11.6\text{ m}^{-1}\cdot 1.59\text{ m}-102\text{ rad/s}\cdot0.150\text{ s}=3.14=\pi\text{ it is an integer multiple of }\pi;[/tex]
[tex]11.6\text{ m}^{-1}\cdot 1.73\text{ m}-102\text{ rad/s}\cdot0.150\text{ s}=4.77=1.52\pi\text{ not an integer multiple of }\pi;[/tex]
[tex]11.6\text{ m}^{-1}\cdot 1.86\text{ m}-102\text{ rad/s}\cdot0.150\text{ s}=6.28=2\pi\text{ it is an integer multiple of }\pi;[/tex]
[tex]11.6\text{ m}^{-1}\cdot 2.00\text{ m}-102\text{ rad/s}\cdot0.150\text{ s}=7.90=2.51\pi\text{ not an integer multiple of }\pi;[/tex]
[tex]11.6\text{ m}^{-1}\cdot 2.13\text{ m}-102\text{ rad/s}\cdot0.150\text{ s}=9.41=3\pi\text{ it is an integer multiple of }\pi;[/tex]
Final answer:
The wave function y(x, t) = Acos(kx−ωt) can be used to find x positions of maximum displacement for a transverse wave on a string.
Explanation:
The wave function y(x, t) = Acos(kx−ωt) represents a transverse wave on a string. Given the values of k = 11.6 rad/m and ω = 102 rad/s, we can find the x positions that correspond to points of maximum displacement by setting the argument of the cosine function to be an integer multiple of π.
Using the wave equation, the x positions that correspond to points of maximum displacement are:
0.795 m
1.59 m
2.00 m
2.13 m
A 0.23 kg mass at the end of a spring oscillates 2.0 times per second with an amplitude of 0.15 m
Part A
Determine the speed when it passes the equilibrium point.
Part B
Determine the speed when it is 0.10 m from equilibrium.
Part C
Determine the total energy of the system.
Part D
Determine the equation describing the motion of the mass, assuming that at t=0, x was a maximum.
Determine the equation describing the motion of the mass, assuming that at , was a maximum.
x(t)=(0.075m)cos[2π(2.0Hz)t]
x(t)=(0.15m)sin[2π(2.0Hz)t]
x(t)=(0.15m)cos[2π(2.0Hz)t]
x(t)=(0.15m)cos[(2.0Hz)t]
Answer:
A) v = 1.885 m/s
B) v = 0.39 m/s
C) E = 0.03 J
D) [tex]x(t) = (0.15m)\cos(2\pi (2.0Hz)t)[/tex]
Explanation:
Part A
We will use the conservation of energy to find the speed at equilibrium.
[tex]K_{eq} + U_{eq} = K_A + U_A\\
\frac{1}{2}mv^2 + 0 = 0 + \frac{1}{2}kA^2\\
v = \sqrt{\frac{k}{m}}A[/tex]
where [tex]\omega = \sqrt{k/m}[/tex] and [tex]\omega = 2\pi f[/tex]
Therefore,
[tex] v = 2\pi f A = 2(3.14)(2)(0.15) = 1.885~m/s[/tex]
Part B
The conservation of energy will be used again.
[tex]K_1 + U_1 = K_2 + U_2\\
\frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \frac{1}{2}kA^2\\
mv^2 + kx^2 = kA^2\\
(0.23)v^2 + k(0.10)^2 = k(0.15)^2\\
v^2 = \frac{k(0.15)^2-(0.10)^2}{0.23}\\
v = \sqrt{0.054k}[/tex]
where [tex]k = \omega^2 m = (2\pi f)^2 m = 2(3.14)(2)(0.23) = 2.89[/tex]
Therefore, v = 0.39 m/s.
Part C
Total energy of the system is equal to the potential energy at amplitude.
[tex]E = \frac{1}{2}kA^2 = \frac{1}{2}(2.89)(0.15)^2 = 0.03~J[/tex]
Part D
The general equation of motion in simple harmonic motion is
[tex]x(t) = A\cos(\omega t + \phi)\\
x(t) = (0.15m)\cos(2\pi (2.0Hz)t + \phi)[/tex]
where [tex]\phi[/tex] is the phase angle to be determined by the initial conditions. In this case, the initial condition is that at t = 0, x is maximum. Therefore,
[tex]x(t) = (0.15m)\cos(2\pi (2.0Hz)t)[/tex]
The speed of the mass when passing the equilibrium point is approximately 1.88 m/s, and when it is 0.10 m from equilibrium, the speed is approximately 1.26 m/s. The equation describing the motion of the mass when x was a maximum at t=0 is x(t) = (0.15 m)cos[2π(2.0 Hz)t].
Explanation:Part A: To determine the speed when the mass passes the equilibrium point, we can use the formula for the velocity of an object in simple harmonic motion, which is v = ωA, where ω is the angular frequency and A is the amplitude. In this case, the angular frequency is given by ω = 2πf, where f is the frequency. So, ω = 2π(2.0 Hz) = 4π rad/s. The amplitude is 0.15 m. Substituting these values into the formula, we get v = (4π rad/s)(0.15 m) ≈ 1.88 m/s.
Part B: To determine the speed when the mass is 0.10 m from equilibrium, we can again use the formula for velocity. Using the same angular frequency ω, we find that v = (4π rad/s)(0.10 m) ≈ 1.26 m/s.
Part C: The total energy of the system can be found using the formula for the total energy of an object in simple harmonic motion, which is E = (1/2)kA², where k is the spring constant and A is the amplitude. In this case, k is not given, so we cannot determine the total energy.
Part D: The equation describing the motion of the mass, assuming that at t=0, x was a maximum, is given by x(t) = (0.15 m)cos[2π(2.0 Hz)t].
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A toy train rolls around a horizontal 1.0-m-diameter track.The coefficient of rolling friction is 0.10.a) What is the magnitude of the trains angular acceleration afterit is released?b)How long does it take the train to stop if it's released with anangular speed of 30 rpm?In step one of the question, the solution supplied states thattangential acceleration is the coefficient of rolling frictionmultiplied by gravity. Why is this so?
Answer:
1.962 rad / s², 1.6 s
Explanation:
Radius of the part = 1.0m / 2 = 0.5 m
angular speed = 30 rpm = 30 rpm × (2πrad / rev) × 1 minutes / 60 seconds = 3.142 rads⁻¹
μk = frictional force / normal ( mg )
normal is the force acting upward against the force of gravity
frictional force = - μk mg
since the body came to rest then
Fnet + Ff = 0
Fnet = - Ff
Fnet = ma
ma = - μk mg
a = - μkg where g = 9.81
a = - 0.1 × 9.81 = 0.981 m/s²
magnitude of angular acceleration = tangential acceleration / radius = 0.981 / 0.5 = 1.962 rad / s²
b) time for the train to come to rest = angular velocity / angular angular acceleration = 3.142/ 1.962 = 1.6 s
The equation earlier derived answer this question
Fnet + Ff = 0 since the body came to a rest
Fnet = - Ff and Ff = - μk mg, Fnet = ma
ma = - μk mg
m cancel m on both side
a = - μkg since it magnitude
a = μkg
The coefficient of rolling friction multiplied by gravity gives the tangential acceleration in rolling motion due to the relationship between frictional force and the object's acceleration.
The reason why the coefficient of rolling friction multiplied by gravity gives the tangential acceleration is due to the relationship between the frictional force and the force responsible for the acceleration of the object. In the case of rolling motion, the frictional force opposes the rotation and contributes to the overall acceleration.
A rocket has a mass 280(103) slugs on earth. Specify its mass in SI units, and its weight in SI units.
To solve this problem we will apply the concepts related to the conversion of units for which we will have that 1 slug is equal to 14.59kg. At the same time we will use Newton's second law for which weight is defined as the product between mass and acceleration (Due to gravity). This is then
A: Using the conversion ratio of slug to kilogram we have to,
[tex]1 slug = 14.59kg[/tex]
Then
[tex]m = 280*10^3 slugs (\frac{14.59kg}{1slug})[/tex]
[tex]m = 4.09*10^6kg[/tex]
B: Using Newton's second law we have to,
[tex]W = mg[/tex]
[tex]W = (4.09*10^6)(9.8)[/tex]
[tex]W = 40034960N\approx 4*10^7N[/tex]
A strong man is compressing a lightweight spring between two weights. One weight has a mass of 2.3 kg , the other a mass of 5.3 kg . He is holding the weights stationary, but then he loses his grip and the weights fly off in opposite directions. The lighter of the two is shot out at a speed of 6.0 m/s .
What is the speed of the heavier weight?
To solve this problem we will apply the concepts related to the conservation of momentum. For this purpose we have that the initial momentum must be equivalent to the final momentum of the system. Mathematically this can be expressed as
[tex]m_1v_1 = m_2v_2[/tex]
Here,
[tex]m_{1,2}[/tex] = Mass of each object
[tex]v_{1,2}[/tex]= Velocity of each object
Rearranging to find the speed of the heavier weight,
[tex]v_2 = \frac{m_1v_1}{m_2}[/tex]
[tex]v_2 = \frac{(2.3)(6)}{5.3}[/tex]
[tex]v_2 = 2.6m/s[/tex]
Therefore the speed of the heavier weight is 2.6m/s
How do the properties of an electromagnetic wave change as a result of increasing the period of the wave?
As the period of an electromagnetic wave increases, the wavelength of the wave also increases. However, the frequency of the wave remains the same. The energy carried by the wave is inversely proportional to its wavelength, so as the wavelength increases, the energy of the wave decreases.
Explanation:Electromagnetic waves have properties such as wavelength and frequency. As the period of an electromagnetic wave increases, the wavelength of the wave also increases. This means that the distance between successive crests or troughs of the wave becomes larger. However, the frequency of the wave remains the same. The energy carried by the wave is inversely proportional to its wavelength, so as the wavelength increases, the energy of the wave decreases.
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Assume the equation x 5 At3 1 Bt describes the motion of a particular object, with x having the dimension of length and t having the dimension of time. Determine the dimen- sions of the constants A and B. (b) Determine the dimen- sions of the derivative dx/dt 5 3At2 1 B.
Answer:
(a) A = m/s^3, B = m/s.
(b) dx/dt = m/s.
Explanation:
(a)
[tex]x = At^3 + Bt\\m = As^3 + Bs\\m = (\frac{m}{s^3})s^3 + (\frac{m}{s})s[/tex]
Therefore, the dimension of A is m/s^3, and of B is m/s in order to satisfy the above equation.
(b) [tex]\frac{dx}{dt} = 3At^2 + B = 3(\frac{m}{s^3})s^2 + \frac{m}{s} = m/s[/tex]
This makes sense, because the position function has a unit of 'm'. The derivative of the position function is velocity, and its unit is m/s.
determine the required values of the tension in ac and ad so that the resultant of the three forces applied at A is verticalFigure:A wire is coupled from three sides
Answer:
Fac = 21 KN
Fad = 64.29 KN
Explanation:
Step 1: Find the unit vectors in BA , CA, and DA directions
BA = -16 i + 48 j - 12 k
CA = -16 i + 48 j + 24 k
DA = 14 i + 48 j
The following magnitudes
mag (BA) = sqrt ( 16 ^2 + 48^2 + 12^2 ) = 52
mag (CA) = sqrt ( 16 ^2 + 48^2 + 24^2 ) = 56
mag (DA) = sqrt ( 14 ^2 + 48^2 ) = 50
Now compute unit vectors
unit (BA) = (-4 / 13) i + (12 / 13) j - (3 / 13) k
unit (CA) = (-2 / 7) i + (6 / 7) j - (3 / 7) k
unit (DA) = (7 / 25) i + (24 / 25) j
Step 2: Compute Force Vectors F(ml) dot unit( ML )
Fab = (-4*Fab / 13) i + (12*Fab / 13) j - (3*Fab / 13) k
Fac = (-2*Fac / 7) i + (6*Fac / 7) j - (3*Fac / 7) k
Fad = (7*Fad / 25) i + (24*Fad / 25) j
Step 3: Equilibrium force equations in x and z direction
Fr,x = 0 = 39*( -4 / 13 ) - Fac*( 2 / 7 ) + Fad*( 7 / 25 )
Fr,z = 0 = 39*( -3 / 13 ) + Fad*( 3 / 7 )
Step 4: Solve the above equations for Fca and Fda
Fca = 21 KN
Fda = 64.29 KN
Two point charges are released at rest in otherwise empty space. As time goes on, the magnitude of the force exerted on each of the particles decreases. Which of the following must be true?a. Both particles are negatively chargedb. Not enough infoc. The particles may be either both positively charged or both negatively chargedd. One particle is positively charged and the other is negatively chargede. Both particles are positively charged
Answer:
c.)
Explanation:
Assuming no other forces acting on the charges, if as time goes on, the magnitude of the force exerted on each of the particles decreases, as the electric force is inversely proportional to the square of the distance between charges, the only explanation is that this distance is increasing, so the force on both charges is repulsive.
When the force is repulsive, all we can say, is that both charges are of the same type, so they can be both positively charged or both negatively charged, as stated by the choice c).
A kicker punts a football from the very center of the field to the sideline 39 yards downfield.
What is the magnitude of the net displacement of the ball? (a,football field is 53 yards wide)
What is the angle between the direction of the net displacement of the ball and the central line of the feild.
Answer:
L = 47.15 yards
∅ = 55.80°
Explanation:
Given
y = 39 yards
x = 53/2 yards
L = ?
We apply the formula
L = √(x²+y²)
L = √((53/2 yards)²+(39 yards)²)
L = 47.15 yards
The angle between the direction of the net displacement of the ball and the central line of the feild is obtained as follows
∅ = tg⁻¹(y/x)
∅ = tg⁻¹(39/(53/2)) = 55.80°
A block with mass m=1.50 kg is initially at rest on a horizontal frictionless surface at x =0 , where x is the horizontal coordinate. A horizontally directed force is then applied to the block. The force is not constant: instead, its magnitude as a function of position is described by the relationship F(x)=(α-βx 2 )i , where x is given in units of meters, α = 2.50 N , β = 1.00 N/m2 , and i is the unit vector in the x direction.
a) What is the kinetic energy of the block as it passes through x=2.00 m?
b) What is the maximum speed of the block in the interval during which it moves from its initial position to x=2.00 m?
Answer:
(a) 2.33 J.
(b) 1.87 m/s.
Explanation:
(a)
[tex]F(x) = (2.5 - x^2)\^i[/tex]
The force as a function of position is given above. Since the force is a function of position, we can assume that we will use work-energy theorem.
[tex]W = \Delta K = K_2 - K_1\\\int\limits^2_0 {F(x)} \, dx = \frac{1}{2}mv_2 - 0\\\int\limits^2_0 {(2.5-x^2)} \, dx = (2.5x - \frac{x^3}{3})\left \{ {{x=2} \atop {x=0}} \right. = (5 - 8/3) = 7/3[/tex]
Therefore, the kinetic energy of the block is 2.33 J.
(b) In order to find the maximum speed in this interval, we need to investigate the acceleration of the block. Since acceleration is the derivative of velocity, velocity is at its maximum when acceleration is zero.
From Newton's Second Law:
[tex]F = ma\\a(x) = F(x)/m = 1.66 - 0.66x^2[/tex]
In order this to be zero:
[tex]1.66 - 0.66x^2 = 0\\x = 1.58~m[/tex]
The velocity of the block at x = 1.58 m can be found by work-energy theorem.
[tex]\int\limits^{1.58}_0 {(2.5-x^2)} \, dx = \frac{1}{2}(1.5)v^2\\ (2.5x-\frac{x^3}{3})\left \{ {{x=1.58} \atop {x=0}} \right. = 2.63 J\\2.63 = \frac{1}{2}(1.5)v^2\\v = 1.87~m/s[/tex]
1. Suppose you have a pipe producing standing sound waves. Two adjacent harmonics of standing waves (i.e., no standing waves in between these) have wavelengths 2.000 meters and 1.500 meters.
Which wavelength corresponds to a higher mode?
Answer:
1.5 m
Explanation:
given,
wavelength of of the standing waves
λ₁ = 2 m
λ₂ = 1.5 m
to find wavelength corresponding to higher mode.
we know,
[tex]wavelength\ \alpha\ \dfrac{1}{n}[/tex]
where n is the mode number.
From the above expression we can say that wavelength is inversely proportional to mode.
Hence , the wavelength corresponding to higher mode is equal to 1.5 m
Similar to what you see in your textbook, you can generally omit the multiplication symbol as you answer questions online, except when the symbol is needed to make your meaning clear. For example, 1⋅105 is not the same as 1105. When you need to be explicit, type * (Shift + 8) to insert the multiplication operator. You will see a multiplication dot (⋅) appear in the answer box. Do not use the symbol ×. For example, for the expression ma, typing m ⋅ a would be correct, but mxa would be incorrect.Enter the expression ma .
Final answer:
Type m · a to correctly enter the multiplication of m and a, ensuring clarity in expressions, especially when dealing with scientific notation which simplifies multiplication and division by manipulating coefficients and powers of ten separately.
Explanation:
When entering the expression m times a, you should use proper notation to differentiate between multiplication and adjacent variables or numbers. Instead of using 'x' as the multiplication symbol, as you might on paper, use the multiplication dot, represented by typing m · a with your keyboard. This avoids confusion, especially in contexts like scientific notation, where clarity is crucial. For example, 1·105 is not the same as 1 followed by 105. So, it is important to insert the multiplication dot to clarify that you are multiplying 1 by 10 raised to the 5th power.
In scientific notation, multiplication and division are simplified by separately manipulating the coefficients and the powers of ten. For instance, to multiply two numbers in scientific notation, such as (3 × 105) × (2 × 10°), you would multiply 3 by 2 to get 6, and then add the exponents (5 + 0) to get 5, resulting in an answer of 6 · 105. This simplifies calculations incredibly, especially with large numbers.
How strong is the electric field between two parallel plates 4.8 mm apart if the potential difference between them is 220 V?
Answer:
Electric field, E = 45833.33 N/C
Explanation:
Given that,
Separation between the plates, d = 4.8 mm = 0.0048 m
The potential difference between the plates, V = 220 volts
We need to find the electric field between two parallel plates. The relation between the electric field and electric potential is given by :
[tex]V=E\times d[/tex]
[tex]E=\dfrac{V}{d}[/tex]
[tex]E=\dfrac{220}{0.0048}[/tex]
E = 45833.33 N/C
So, the electric field between two parallel plates is 45833.33 N/C. Hence, this is the required solution.
The electric field between the plates is 45833.33 N/C.
Electric Field
Given that the separation d between the plates is 4.8 mm and the potential difference V between the plates is 220 V.
The electric field E between the plates can be calculated by the formula given below.
[tex]E =\dfrac {V}{d}[/tex]
Substituting the values in the above equation, we get
[tex]E = \dfrac {220}{0.0048}[/tex]
[tex]E = 45833.33 \;\rm N/C[/tex]
Hence we can conclude that the electric field between the plates is 45833.33 N/C.
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A crowbar 2727 in. long is pivoted 66 in. from the end. What force must be applied at the long end in order to lift a 600600 lb object at the short end?
To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.
[tex]\sum \tau = F*d[/tex]
If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,
[tex]F*(27-6)= 6*600[/tex]
[tex]F = \frac{6*600}{21}[/tex]
[tex]F= 171.42 lb[/tex]
So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb
Air is contained in a rigid well-insulated tank with a volume of 0.6 m3. The tank is fitted with a paddle wheel that transfers energy to the air at a constant rate of 4 W for 1 h. The initial density of the air is 1.2 kg/m3.
If no changes in kinetic or potential energy occur, determine:
(a) the specific volume at the final state, in m^3/kg
(b) the energy transfer by work, in kJ.
(c) the change in specific internal energy of the air, in kJ/kg.
Answer:
0.833 m³/kg
-14.4 kJ
20 kJ/kg
Explanation:
[tex]\rho[/tex] = Density = 1.2 kg/m³
[tex]V[/tex] = Volume = 0.6 m³
t = Time taken = 1 hour
P = Power = 4 W
Mass is given by
[tex]m=\rho V\\\Rightarrow m=1.2\times 0.6\\\Rightarrow m=0.72\ kg[/tex]
As there is no change in kinetic and potential energy, the specific volume of the tank will be unaffected
[tex]V_{s}=\dfrac{0.6}{0.72}\\\Rightarrow V_s=0.833\ m^3/kg[/tex]
The specific volume at the final state is 0.833 m³/kg
Energy is given by
[tex]E=Pt\\\Rightarrow E=-4\times 1\times 3600\\\Rightarrow E=-14400\ J[/tex]
The energy transfer by work is -14.4 kJ
Change in specific internal energy is given by
[tex]E=-m\Delta u\\\Rightarrow \Delta u=-\dfrac{E}{m}\\\Rightarrow \Delta u=-\dfrac{-14.4}{0.72}\\\Rightarrow \Delta u=20\ kJ/kg[/tex]
The change in specific internal energy is 20 kJ/kg
Answers:
0.833 m³/kg-14.4 kJ20 kJ/kgWater is pumped from a lake to a storage tank 15 m above at a rate of 90 L/s while consuming 15.4 kW of electric power. Disregarding any frictional losses in the pipes and any changes in kinetic energy, answer the questions that follow.
a) the overall efficiency of the pump motor unit.
b) the pressure difference between the inlet and the exit of the pump.
Answer:
B
Explanation:
Total lung capacity of a typical adult is approximately 5.0L. Approximately 20% of the air is oxygen, as air is 20% oxygen. At sea level and at anaverage body temperature of 37°C. How many moles ofoxygen do the lungs contain at the end of an inflation?
Answer:
n=0.03928 moles
Moles of oxygen do the lungs contain at the end of an inflation are 0.03928 moles
Explanation:
The amount of oxygen which lung can have is 20% of 5 L which is the capacity of lungs
Volume of oxygen in lungs =V=5*20%= 1 L=[tex]1*10^{-3} m^3[/tex]
Temperature=T=[tex]37^oC=273+37=310K[/tex]
Pressure at sea level = P= 1 atm=[tex]1.0125*10^5 Pa[/tex]
R is universal Gas Constant =8.314 J/mol.K
Formula:
[tex]n=\frac{PV}{RT}\\n=\frac{(1.0125*10^5) *(1*10^{-3})}{(8.314)*310} \\n=0.03928 mol[/tex]
Moles of oxygen do the lungs contain at the end of an inflation are 0.03928 moles
The specific volume of 5 kg of water vapor at 1.5 MPa, 440°C is 0.2160 m3 /kg. Determine (a) the volume, in m3 , occupied by the water vapor, (b) the amount of water vapor present, in gram moles, and (c) the number of molecules.
(a) The volume occupied by the water vapor is V = 1.08 m³.
(b) The amount of water vapor present is 277.78 gram-moles.
(c) The number of molecules is [tex]\rm\(1.67305 \times 10^{26}\)[/tex] molecules.
Given:
Specific volume (v) = 0.2160 m³/kg
Mass of water vapor (m) = 5 kg
One gram-mole of water vapor = 18 g
Avogadro's number [tex]\rm (\(N_A\))[/tex] = [tex]\rm \(6.023 \times 10^{23}\)[/tex]
(a) The volume occupied by the water vapor (V) can be calculated using the formula: [tex]\rm \[ V = m \times v \][/tex]
Substitute the values:
[tex]\rm \[ V = 5 \, \text{kg} \times 0.2160 \, \text{m³/kg} \]\\\rm V = 1.08 \, \text m\³[/tex]
(b) The amount of water vapour present in gram moles can be calculated using the formula:
[tex]\rm \[ \text{Amount of water vapor} = \frac{m}{\text{One gram-mole of water vapor}} \][/tex]
Substitute the value of one gram-mole of water vapor [tex](\(18 \, \text{g}\))[/tex]:
[tex]\[ \text{Amount of water vapor} = \frac{5 \, \text{kg} \times 1000}{18 \, \text{g}} \]\\\\\ \text{Amount of water vapor} = 277.78 \, \text{gram-moles} \][/tex]
(c) The number of molecules can be calculated using Avogadro's number:
[tex]\rm \[ \text{Number of molecules in 1 gram-mole} = N_A \\= 6.023 \times 10^{23} \, \text{molecules} \][/tex]
So, the number of molecules in 1 gram of water vapor is [tex]\rm \(\frac{N_A}{18}\)[/tex] molecules.
The number of molecules in [tex]\rm \(5 \, \text{kg}\)[/tex] of water vapor is:
[tex]\rm \[ \text{Number of molecules} = 5 \times 1000 \times \frac{N_A}{18} \, \text{molecules} \][/tex]
Now, calculate the number of molecules:
[tex]\rm \[ \text{Number of molecules} = 1.67305 \times 10^{26} \, \text{molecules} \][/tex]
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The sled is pulled up a steeperhill of the sameheightas the hill described above.
How will the velocity of the sled at the bottom of the hill (after it has slid down) compare to that of the sled at the bottom of the original hill?
Choose the best answer below.
A. The speed at the bottom is greater for the steeper hill.
B. The speed at the bottom is the same for both hills.
C. The speed at the bottom is greater for the original hill because the sled travels further.
D. There is not enough information given to say which speed at the bottom is faster.
E. None of these descriptions is correct.
Answer: B. The speed at the bottom is the same for both hills.
Explanation A sled which can also be called a sledge is a vehicle built with a smooth body on the side touching the ground underside which makes it easy for it to be able to slide towards the ground when on a slant hill. The velocity of the sled after it has slid down the Hill and the velocity at the bottom of the Hill will be the same because both point have the same level of steepness. Velocity of a material is directly proportional to the mass of a body and inversely proportional to the time traveled.
The speed of the sled at the bottom of both hills will be the same because the total energy, determined by the height of the hill and not its steepness, remains constant.
Explanation:The best answer to this question is B. The speed at the bottom is the same for both hills. This is because the total energy of the sled (kinetic plus gravitational potential energy) must remain constant if we assume no friction or air resistance. The height of the hill, not its steepness, determines the total energy. The sled will convert all its potential energy (which depends on the height) into kinetic energy (which affects the speed) as it travels downhill. Hence, if the two hills are of the same height, the sled will have the same speed at the bottom of both hills.
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The electric force between two identical positively charged ions is 33 × 10 − 9 N when they are 0.50 nm apart. How many electrons are missing from each ion?
Answer:
There are 5.98 electrons are missing from each ions.
Explanation:
Given that,
The electric force between two identical charges is, [tex]F=33\times 10^{-9}\ N[/tex]
The distance between the charges, [tex]d=0.5\ nm=0.5\times 10^{-9}\ m[/tex]
The electric force between charges is given by :
[tex]F=\dfrac{kq^2}{d^2}[/tex]
[tex]q=\sqrt{\dfrac{Fd^2}{k}}[/tex]
[tex]q=\sqrt{\dfrac{33\times 10^{-9}\times (0.5\times 10^{-9})^2}{9\times 10^9}}[/tex]
[tex]q=9.57\times 10^{-19}\ C[/tex]
Let there are n number of electrons are missing from each ions. It is given by :
[tex]n=\dfrac{q}{e}[/tex]
[tex]n=\dfrac{9.57\times 10^{-19}}{1.6\times 10^{-19}}[/tex]
n = 5.98 electrons
So, there are 5.98 electrons are missing from each ions. Hence, this is the required solution.
If the x-component of a force vector is 2.16 newtons and its y-component is 4.11 newtons, then what is its magnitude? F = newtons
Answer:
Magnitude of resultant vector will be 4.643 N
Explanation:
We have given x-component of the force vector is given [tex]F_{X}=2.16N[/tex]
And y component of the force vector is given [tex]F_{y}=4.11N[/tex]
We have to find the magnitude of resultant force F
Resultant force will be equal to vector sum of magnitude of x competent and y component of force vector
So F = [tex]\sqrt{F_X^2+F_Y^2}=\sqrt{2.16^2+4.11^2}=4.643N[/tex]
So magnitude of resultant vector will be 4.643 N
A 20 cm tall object is placed in front of a concave mirror with a radius of 31 cm. The distance of the object to the mirror is 94 cm. Calculate the focal length of the mirror.
Answer:
The focal length of the concave mirror is -15.5 cm
Explanation:
Given that,
Height of the object, h = 20 cm
Radius of curvature of the mirror, R = -31 cm (direction is opposite)
Object distance, u = -94 cm
We need to find the focal length of the mirror. The relation between the focal length and the radius of curvature of the mirror is as follows :
R = 2f
f is the focal length
[tex]f=\dfrac{R}{2}[/tex]
[tex]f=\dfrac{-31}{2}[/tex]
f = -15.5 cm
So, the focal length of the concave mirror is -15.5 cm. Hence, this is the required solution.
Starting with an initial value of P(0)equals30, the population of a prairie dog community grows at a rate of Pprime(t)equals30minusStartFraction t Over 2 EndFraction (in units of prairie dogs/month), for 0less than or equalstless than or equals60.
a. What is the population 9 months later?
b. Find the population P(t) for 0less than or equalstless than or equals60.
Answer:
a) The population of prairie dogs after nine months is 280.
b) P(t) = 30 + 30 · t - t²/4 for 0 ≤ t ≤ 60
Explanation:
Hi there!
We have the following information:
The initial population is P(0) = 30.
The rate of growth of the population is the following:
P´(t) = 30 - t/2 where
a) Let´s find the function of the population of prairie dogs P(t). For that, let´s integrate the P´(t) function between t = 0 and t and between P = 30 and P
P(t) = ∫P´(t)
P´(t) = dP/dt = 30 - t/2
Separating variables:
dP = (30 - t/2) dt
∫dP = ∫(30 - t/2) dt
P - 30 = 30 · t - t²/4
P(t) = 30 + 30 · t - t²/4
The population of prairie dogs at t = 9 months will be equal to P(9):
P(9) = 30 + 30(9) - (9)²/ 4
P(9) = 280 prairie dogs
The population of prairie dogs after nine months is 280.
b) P(t) = 30 + 30 · t - t²/4 (it was obtained in part a).
The population of the prairie dog community 9 months later would be 294 prairie dogs, and the population equation for the entire prairie dog community for t months is P(t) = 30t - t^2 / 4 + 30 for 0 ≤ t ≤ 60.
Explanation:The equation given is P'(t) = 30 - t / 2. This is a differential equation, and to get P(t), we need to find the anti-derivative or integrate P'(t) with respect to time, t. The integral of P'(t) = ∫(30 - t / 2) dt = 30t - t^2/4.
Adding the initial condition, P(0) = 30, we find P(t) = 30t - t^2 / 4 + 30 for 0 ≤ t ≤ 60. Therefore, a. The population of the prairie dog community 9 months later would be P(9) = 30*9 - (9^2) / 4 + 30 = 294 prairie dogs. b. P(t) = 30t - t^2 / 4 + 30 for 0 ≤ t ≤ 60 is the population equation for the entire prairie dog community for t months.
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Wilma I. Ball walks at a constant speed of 5.93 m/s along a straight line from point A to point B and then back from B to A at a constant speed of 3.15 m/s.
(a)
What is Wilma's average speed over the entire trip?Wilma I. Ball walks at a constant speed of 4.51 m/s along a straight line from point A to point B and then back from B to A at a constant speed of 3.15 m/s.
(a)
What is Wilma's average speed over the entire trip?
(b)
What is Wilma's average velocity over the entire trip?
Wilma's average speed over the entire trip is approximately 3.71 m/s, calculated by dividing the total distance by the total time. However, as she returned to her starting point, her total displacement is zero, and thus, her average velocity is 0 m/s.
Explanation:The average speed is given by the total distance divided by the total time. Since Wilma I. Ball travels the same distance twice (A to B and back), we can say the total distance is 2d. Suppose the distance between A to B is 'd' m. The time taken for the first trip is d/4.51 s, and the time taken for the return trip is d/3.15 s. Therefore, the average speed is given by (2d) / ((d/4.51) +(d/3.15)). This simplifies to approximately 3.71 m/s.
However, average velocity is the total displacement divided by total time. Since she ends at her starting point, her total displacement is zero. Thus, her average velocity over her round trip is 0 m/s.
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When the balls are brought close together, without touching, the following observations are made: Balls B, C, and D are attracted to ball A. Balls B and D have no effect on each other. Balls B is attracted to ball C. What are the charge states (positive, negative, or neutral) of balls A, B, C, and D
Answer:
B and D is neutral. A and C have opposite charges.
Explanation:
Opposite charges attract each other and similar charges repel each other. Furthermore, a neutral object is attracted to a charged object whether negative or positive.
- If the balls B, C, and D is attracted to A, then their charge is either opposite to that of A or neutral.
- If the balls B and D have no effect on each other, then both balls are neutral.
- If B is neutral and attracted to C, then ball C have opposite charges than A.
Finally, B and D is neutral. A and C have opposite charges. Whether they are positive or negative cannot be determined by the information given in the question.
If Star A is magnitude 1.0 and Star B is magnitude 9.6 , which is brighter and by what factor?
Answer:
Star A is brighter than Star B by a factor of 2754.22
Explanation:
Lets assume,
the magnitude of star A = m₁ = 1
the magnitude of star B = m₂ = 9.6
the apparent brightness of star A and star B are b₁ and b₂ respectively
Then, relation between the difference of magnitudes and apparent brightness of two stars are related as give below: [tex](m_{2} - m_{1}) = 2.5\log_{10}(b_{1}/b_{2})[/tex]
The current magnitude scale followed was formalized by Sir Norman Pogson in 1856. On this scale a magnitude 1 star is 2.512 times brighter than magnitude 2 star. A magnitude 2 star is 2.512 time brighter than a magnitude 3 star. That means a magnitude 1 star is (2.512x2.512) brighter than magnitude 3 bright star.
We need to find the factor by which star A is brighter than star B. Using the equation given above,
[tex](9.6 - 1) = 2.5\log_{10}(b_{1}/b_{2})[/tex]
[tex]\frac{8.6}{2.5} = \log_{10}(b_{1}/b_{2})[/tex]
[tex]\log_{10}(b_{1}/b_{2}) = 3.44[/tex]
Thus,
[tex](b_{1}/b_{2}) = 2754.22[/tex]
It means star A is 2754.22 time brighter than Star B.
Star A is brighter than Star B by approximately 2512 times.
Explanation:The brightness of stars is measured using the magnitude scale. The smaller the magnitude, the brighter the star. In this case, Star A has a magnitude of 1.0 and Star B has a magnitude of 9.6. Since Star A has a smaller magnitude, it is brighter than Star B. The difference in magnitude between the two stars can be calculated using the formula:
difference = 2.512^(m₁ - m₂)
where m₁ is the magnitude of Star A and m₂ is the magnitude of Star B. Plugging in the values, we get:
difference = 2.512^(1.0 - 9.6) ≈ 2512
Therefore, Star A is approximately 2512 times brighter than Star B.
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Suppose that a particular artillery piece has a range R = 5580 yards . Find its range in miles. Use the facts that 1mile=5280ft and 3ft=1yard. Express your answer in miles to three significant figures..
Answer:
3.170 miles
Explanation:
Conversion from yards to feet:
5580 yards = 5580 yard * 3 ft/yard = 16740 feet
Conversion from feet to miles:
16740 feet = 16740*(1/5280) mile/ft = 3.170 miles.
Therefore, that particular artillery piece has a range R = 3.170 miles