Answer:
Choice d. The set of vectors: [tex]\displaystyle \left\{\begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix},\; a,\, b\in \mathbb{R} \right\}[/tex] isn't a vector space over [tex]\mathbb{R}[/tex].
Explanation:
Let a set of vectors [tex]V[/tex] to be a vector field over some field [tex]\mathbb{F}[/tex] (for this question, that "field" is the set of all real number.) The following must be true:
The set of vectors [tex]V[/tex] includes the identity element [tex]\mathbf{0}[/tex]. In other words, there exists a vector [tex]\mathbf{0} \in V[/tex] such that for all [tex]\mathbf{v} \in V[/tex], [tex]\mathbf{v} + \mathbf{0} = \mathbf{v}[/tex].[tex]V[/tex] should be closed under vector addition. In other words, for all [tex]\mathbf{u},\, \mathbf{v} \in V[/tex], [tex]\mathbf{u} + \mathbf{v} \in V[/tex].[tex]V[/tex] should also be closed under scalar multiplication. In other words, for all [tex]\mathbf{v} \in V[/tex] and all "scalar" [tex]m \in \mathbb{F}[/tex] (in this question, the "field" is the set of all real numbers, so [tex]m[/tex] can be any real number,) [tex]a\,\mathbf{v} \in V[/tex].Note that in the general form of a vector in [tex]V[/tex], the second component is a always non-zero. Because of that non-zero component,
Assume by contradiction that [tex]V[/tex] is indeed a vector field. Therefore, it should contain a zero vector. Let [tex]\mathbf{0}[/tex] denote that zero vector. For all [tex]\mathbf{v} \in V[/tex], [tex]\mathbf{v} + \mathbf{0} = \mathbf{v}[/tex].
Using the definition of set [tex]V[/tex]: [tex]\displaystyle \left\{\begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix},\; a,\, b\in \mathbb{R} \right\}[/tex], there exist real numbers [tex]a[/tex] and [tex]b[/tex], such that:
[tex]\displaystyle \mathbf{v} = \begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix}[/tex].
Hence, [tex]\mathbf{v} + \mathbf{0} = \mathbf{v}[/tex] is equivalent to:
[tex]\displaystyle \begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix} + \mathbf{0} = \begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix}[/tex].
Apply the third property that [tex]V[/tex] is closed under scalar multiplication. [tex]-1[/tex] is indeed a real number. Therefore, if [tex]\mathbf{v}[/tex] is in
Therefore:
[tex]\displaystyle -\begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix} \in V[/tex].
Apply the second property and add [tex]\displaystyle - \mathbf{v} = -\begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix}[/tex] to both sides of [tex]\mathbf{v} + \mathbf{0} = \mathbf{v}[/tex]. The left-hand side becomes:
[tex]\mathbf{v} - \mathbf{v} + \mathbf{0} = \mathbf{0}[/tex].
The right-hand side becomes:
[tex]\displaystyle \begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix} -\begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix} = \begin{bmatrix}a - 4\, b - (a - 4\, b) \\ 5 - 5 \\ 4\, a+ b-(4\, a+ b)\\ -a -b - (-a -b)\end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ 0 \\0\end{bmatrix}[/tex].
Therefore:
[tex]\displaystyle \mathbf{0} = \begin{bmatrix}0 \\ 0 \\ 0 \\0\end{bmatrix}[/tex].
However, [tex]\mathbf{0} = \displaystyle \begin{bmatrix}0 \\ 0 \\ 0 \\0\end{bmatrix}[/tex] isn't a member of the set [tex]\displaystyle V = \left\{\begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix},\; a,\, b\in \mathbb{R} \right\}[/tex]. That's a contradiction, because [tex]\mathbf{0}[/tex] was supposed to be part of [tex]V[/tex].
Hence, [tex]V[/tex] isn't a vector space by contradiction.
Consider a skateboarder who starts from rest at the top of ramp that is inclined at an angle of 18.0 ∘ to the horizontal.
Assuming that the skateboarder's acceleration is gsin 18.0 ∘, find his speed when he reaches the bottom of the ramp in 3.50 s .
Answer:
Explanation:
v= u + at
v is final velocity , u is initial velocity . a is acceleration and t is time
Initial velocity u = 0 . Putting the given values in the equation
v = 0 + g sin 18 x 3.5
= 10.6 m /s
For a skateboarder who starts from the rest, the speed when he reaches the bottom of the ramp will be 10.6 m/s.
What are Velocity and Acceleration?The term "velocity" refers to a vector measurement of the rate and direction of motion. Velocity is the rate of movement in a single direction, to put it simply. Velocity can be used to determine how fast a rocket is heading into space and how fast a car is moving north on a congested motorway.
There are several types of velocity :
Instantaneous velocityAverage VelocityUniform VelocityNon-Uniform VelocityThe pace at which a person's velocity changes is known as acceleration. This implies that an object is accelerating if its velocity is rising or falling. An object that is accelerating won't have a steady change in location every second like an item moving at a constant speed does.
According to the question, the given values are :
Time, t = 3.50 sec
Initial Velocity, u = 0 m/s
Use equation of motion :
v = u+at
v = 0+ g sin 18 × 3.5
v = 10.6 m/s.
So, the final velocity will be 10.6 m/s.
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A force of 30 N is required to hold a spring that has been stretched from its natural length of 20 cm to a length of 35 cm. How much work is done in stretching the spring from 35 cm to 40 cm
Answer:
0.25 J
Explanation:
Given that
Force on the spring, F = 30 N
Natural length of the spring, l1 = 20 cm = 0.2 m
Final length of the spring, l2 = 35 cm = 0.35 m
Extension of the spring, x = 0.35 - 0.2 = 0.15 m
The other extension is 40 - 35 cm = 5 cm = 0.05 m
Work done = ?
Considering Hooke's Law of Elasticity.
We're told that the spring stretches through 15 cm, and thereafter asked to find the work done in stretching it through 5 cm
The question is solved in the attachment below
Consider a sound wave modeled with the equation s(x, t) = 3.00 nm cos(3.50 m−1x − 1,800 s−1t). What is the maximum displacement (in nm), the wavelength (in m), the frequency (in Hz), and the speed (in m/s) of the sound wave?
Answer:
- maximum displacement = 3.00nm
- λ = 1.79m
- f = 286.47 s^-1
Explanation:
You have the following equation for a sound wave:
[tex]s(x,t)=3.00nm\ cos(3.50m^{-1}x- 1,800s^{-1} t)[/tex] (1)
The general form of the equation of a sound wave can be expressed as the following formula:
[tex]s(x,t)=Acos(kx-\omega t)[/tex] (2)
A: amplitude of the wave = 3.00nm
k: wave number = 3.50m^-1
w: angular frequency = 1,800s^-1
- The maximum displacement of the wave is given by the amplitude of the wave, then you have:
maximum displacement = A = 3.00nm
- The wavelength is given by :
[tex]\lambda=\frac{2\pi}{k}=\frac{2\pi}{3.50m^{-1}}=1.79m[/tex]
The values for the wavelength is 1.79m
- The frequency is:
[tex]f=\frac{\omega}{2\pi}=\frac{1,800s^{-1}}{2\pi}=286.47s^{-1}[/tex]
The frequency is 286.47s-1
If a pickup is placed 16.25 cm from one of the fixed ends of a 65.00-cm-long string, which of the harmonics from n=1 to n=12 will not be "picked up" by this pickup?
Answer:
The answer to this question can be defined as follows:
Explanation:
Therefore the 4th harmonicas its node is right and over the pickup so, can not be captured from 16.25, which is 1:4 out of 65. Normally, it's only conceptual for the certain harmonic, this will be low, would still be heard by the catcher.
Instead, every harmonic node has maximum fractions along its string; the very first node is the complete string length and the second node is half a mile to the third node, which is one-third up and so on.
Answer:
b
Explanation:
because:/
A particle with kinetic energy equal to 282 J has a momentum of magnitude 26.4 kg · m/s. Calculate the speed (in m/s) and the mass (in kg) of the particle.
Answer:
[tex]v=21.36\,\,\frac{m}{s}\\[/tex]
[tex]m=1.2357\,\,kg[/tex]
Explanation:
Recall the formula for linear momentum (p):
[tex]p = m\,v[/tex] which in our case equals 26.4 kg m/s
and notice that the kinetic energy can be written in terms of the linear momentum (p) as shown below:
[tex]K=\frac{1}{2} m\,v^2=\frac{1}{2} \frac{m^2\,v^2}{m} =\frac{1}{2}\frac{(m\,v)^2}{m} =\frac{p^2}{2\,m}[/tex]
Then, we can solve for the mass (m) given the information we have on the kinetic energy and momentum of the particle:
[tex]K=\frac{p^2}{2\,m}\\282=\frac{26.4^2}{2\,m}\\m=\frac{26.4^2}{2\,(282)}\,kg\\m=1.2357\,\,kg[/tex]
Now by knowing the particle's mass, we use the momentum formula to find its speed:
[tex]p=m\,v\\26.4=1.2357\,v\\v=\frac{26.4}{1.2357} \,\frac{m}{s} \\v=21.36\,\,\frac{m}{s}[/tex]
An automobile accelerates from zero to 30 m/s in 6 s. The wheels have a diameter of 0.4 m. What is the angular acceleration of each wheel
Answer:
12.5 rad/s²
Explanation:
Angular Acceleration: This can be defined as the ratio of linear acceleration and radius. The S.I unit is rad/s²
From the question,
a = αr................... Equation 1
Where a = linear acceleration, α = angular acceleration, r = radius.
But,
a = (v-u)/t.............. Equation 2
Where v = final velocity, u = initial velocity, t = time.
Substitute equation 2 into equation 1
(v-u)/t = αr
make α the subject of the equation
α = (v-u)/tr................. Equation 3
Given: v = 30 m/s, u = 0 m/s, t = 6 s, r = 0.4 m
Substitute into equation 3
α = (30-0)/(0.4×6)
α = 30/2.4
α = 12.5 rad/s²
What is the force of gravity acting on a 1-kg m mass? (g = 9.8 m/s ^ 2)
Answer: Use this F=Ma.
Explanation: So your answer will be
F=1 Kg+9.8 ms-2
So the answer will be
F=9.8N
How'd I do this?
I just used Newton's second law of motion.
I'll also put the derivation just in case.
Applied force α (Not its alpha, proportionality symbol) change in momentum
Δp α p final- p initial
Δp α mv-mu (v=final velocity, u=initial velocity and p=v*m)
or then
F α m(v-u)/t
So, as we know v=final velocity & u= initial velocity and v-u/t =a.
So F α ma, we now remove the proportionality symbol so we'll add a proportionality constant to make the RHS & LHS equal.
So, F=kma (where k is the proportionality constant)
k is 1 so you can ignore it.
So, our equation becomes F=ma
A 5000 kg railcar hits a bumper (a spring) at 1 m/s, and the spring compresses 0.1 meters. Assume no damping. a) Find the spring constant k.
Answer:
k = 0.5 MN/m
Explanation:
Mass of the railcar, m = 5000 kg
Speed of the rail car, v = 1 m/s
The Kinetic energy(KE) of the railcar is given by the equation:
KE = 0.5 mv²
KE = 0.5 * 5000 * 1²
KE = 2500 J
The spring's compression, x = 0.1 m
The potential energy(PE) stored in the spring is given by the equation:
PE = 0.5kx²
PE = 0.5 * k * 0.1²
PE = 0.005k
According to the principle of energy conservation, Kinetic energy of the railcar equals the potential energy stored in the spring
KE = PE
2500 = 0.005k
k = 2500/0.005
k = 500000 N/m
k = 0.5 MN/m
what is the difference between a good conductor and a good insulator?
Answer:
Explanation:
In a conductor, electric current can flow freely, in an insulator it cannot.
Metals such as copper typify conductors, while most non-metallic solids are said to be good insulators, having extremely high resistance to the flow of charge through them.
Most atoms hold on to their electrons tightly and are insulators.
What is the change in internal energy of an engine if you put 15 gallon of gasoline into its tank? The energy content of gasoline is 1.5 x 106 J/gallon. All other factors, such as the engine’s temperature, are constant. How many hours the engine can work if the power of the engine’s motor is 600 W? (8 marks)
Answer:
ΔU = 2.25 x 10⁸ J
t = 104.17 s
Explanation:
The change in internal energy of the engine can be given by the following formula:
ΔU = (Mass of Gasoline)(Energy Content of Gasoline)
ΔU = (1.5 x 10⁶ J/gallon)(15 gallon)
ΔU = 2.25 x 10⁸ J
Now, for the time of operation, we use the following formula of power.
P = W/t = ΔU/t
t = ΔU/P
where,
t = time of operation = ?
ΔU = Change in internal energy = 2.25 x 10⁸ J
P = Power of motor = 600 W
Therefore,
t = (2.25 x 10⁸ J)/(600 W)
t = (375000 s)(1 h/3600 s)
t = 104.17 s
Water flowing through a garden hose of diameter 2.76 cm fills a 20.0-L bucket in 1.45 min. (a) What is the speed of the water leaving the end of the hose
Answer:
v = 31.84 cm/s or 0.318 m/s
the speed of the water leaving the end of the hose is 31.84 cm/s or 0.318 m/s
Explanation:
Given;
Diameter of hose d = 2.76 cm
Volume filled V = 20.0 L = 20,000 cm^3
Time t = 1.45 min = 105 seconds
The volumetric flow rate of water is;
F = V/t = 20,000cm^3 ÷ 105 seconds
F = 190.48 cm^3/s
The volumetric flow rate is equal the cross sectional area of pipe multiply by the speed of flow.
F = Av
v = F/A
Area A = πd^2/4
Speed v = F/(πd^2/4)
v = 4F/πd^2 ......1
Substituting the given values;
v = (4×190.48)/(π×2.76^2)
v = 31.83767439628 cm/s
v = 31.84 cm/s or 0.318 m/s
the speed of the water leaving the end of the hose is 31.84 cm/s or 0.318 m/s
A tightly wound toroid of inner radius 1.2 cm and outer radius 2.4 cm has 960 turns of wire and carries a current of 2.5 A.
Requried:
a. What is the magnetic field at a distance of 0.9 cm from the center?
b. What is the field 1.2 cm from the center?
Answer:
a
[tex]B = 0.0533 \ T[/tex]
b
[tex]B = 0.04 \ T[/tex]
Explanation:
From the question we are told that
The inner radius is [tex]r = 1.2 \ cm = 0.012 \ m[/tex]
The outer radius is [tex]r_o = 2.4 \ cm = \frac{2.4}{100} = 0.024 \ m[/tex]
The nu umber of turns is [tex]N = 960[/tex]
The current it is carrying is [tex]I = 2. 5 A[/tex]
Generally the magnetic field is mathematically represented as
[tex]B = \frac{\mu_o * N* I }{2 * \pi * r }[/tex]
Where [tex]\mu_o[/tex] is the permeability of free space with a constant value
[tex]\mu = 4\pi * 10^{-7} N/A^2[/tex]
And the given distance where the magnetic field is felt is r = 0.9 cm = 0.009 m
Now substituting values
[tex]B = \frac{ 4\pi * 10^{-7} * 960* 2.5 }{2 * 3.142 * 0.009 }[/tex]
[tex]B = 0.0533 \ T[/tex]
Fro the second question the distance of the position considered from the center is r = 1.2 cm = 0.012 m
So the magnetic field is
[tex]B = \frac{ 4\pi * 10^{-7} * 960* 2.5 }{2 * 3.142 * 0.012 }[/tex]
[tex]B = 0.04 \ T[/tex]
The magnetic field at a distance of 0.9 cm from the center of the toroid is 0.053 T.
The magnetic field at a distance of 1.2 cm from the center of the toroid is 0.04 T.
The given parameters;
radius of the toroid, r = 1.2 cm = 0.012 mouter radius of the toroid, R = 2.4 cm = 0.024 mnumber of turns, N = 960 turnscurrent in wire, I = 2.5 AThe magnetic field at a distance of 0.9 cm from the center of the toroid is calculated as follows;
[tex]B = \frac{\mu_o NI}{2\pi r} \\\\B = \frac{(4\pi \times 10^{-7})\times (960) \times (2.5)}{2\pi \times 0.009} \\\\B = 0.053 \ T[/tex]
The magnetic field at a distance of 1.2 cm from the center of the toroid is calculated as follows;
[tex]B = \frac{\mu_o NI}{2\pi r} \\\\B = \frac{(4\pi \times 10^{-7})\times (960) \times (2.5)}{2\pi \times 0.012} \\\\B = 0.04 \ T[/tex]
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Suppose the current in a conductor decreases exponentially with time according to the equation I(t) = I0e-t/τ, where I0 is the initial current (at t = 0), and τ is a constant having dimensions of time. Consider a fixed observation point within the conductor.
Required:
a. How much charge passes this point between t = 0 and t = τ?
b. How much charge passes this point between t = 0 and t = 10 τ?
c. What If ? How much charge passes this point between t = 0 and t = [infinity]?
Answer:
Pls see attached file
Explanation:
How fast is the spaceship traveling towards the Sun? The radius of the orbit of Jupiter is 43.2 light-minutes, and that of the orbit of Mars is 12.6 light-minutes.
Question:
A spaceship enters the solar system moving toward the Sun at a constant speed relative to the Sun. By its own clock, the time elapsed between the time it crosses the orbit of Jupiter and the time it crosses the orbit of Mars is 35.0 minutes
How fast is the spaceship traveling towards the Sun? The radius of the orbit of Jupiter is 43.2 light-minutes, and that of the orbit of Mars is 12.6 light-minutes.
Answer:
S = 5.508 × 10¹¹m
V = 2.62 × 10⁸ m/s
Explanation:
The radius of the orbit of Jupiter, Rj is 43.2 light-minutes
radius of the orbit of Mars, Rm is 12.6 light-minutes
Distance travelled S = (Rj - Rm)
= 43.2 - 12.6 = 30.6 light- minutes
= 30.6 × (3 ×10⁸m/s) × 60 s
= 5.508 × 10¹¹m
time = 35mins = (35 × 60 secs)
= 2100 secs
speed = distance/time
V = 5.508 × 10¹¹m / 2100 s
V = 2.62 × 10⁸ m/s
find the value of k for which the given pair of vectors are not equal
2ki +3j and 8i + 4kj
Answer:
5
Explanation:
Which observation have scientists used to support Einstein's general theory of relativity?
The orbital path of Mercury around the Sun has changed.
O GPS clocks function at the same rate on both Earth and in space.
O The Sun has gotten more massive over time.
Objects act differently in a gravity field than in an accelerating reference frame.
Answer:
Objects act differently in a gravity field than in an accelerating reference frame.
Explanation:
The main thrust of the theory general relativity as proposed by Albert Einstein boarders on space and time as the two fundamental aspects of spacetime. Spacetime is curved in the presence of gravity, matter, energy, and momentum. The theory of general relativity explains gravity based on the way space can 'curve', that is, it seeks to relate gravitational force to the changing geometry of space-time.
The Einstein general theory of relativity has replaced Newton's ideas proposed in earlier centuries as a means of predicting gravitational interactions. This concept is quite helpful but cannot be fitted into the context of quantum mechanics due to obvious incompatibilities.
Answer:
A - The orbital path of mercury around the sun has changed.
Explanation:
got right on edg.
A bicycle tire pump has a piston with area 0.43 In2. If a person exerts a force of 16 lb on the piston while Inflating a tire, what pressure does this produce on the air in the pump?
Answer:
The pressure produced on the air in the pump is 37.209 pounds per square inch.
Explanation:
By definition, the pressure is the force exerted on the piston divided by its area. Given that, force is distributed uniformly on the piston area, the formula to determine the pressure is:
[tex]p = \frac{F}{A}[/tex]
Where:
[tex]p[/tex] - Pressure, measured in pounds per square inch.
[tex]F[/tex] - Force exerted on the piston, measured in pounds.
[tex]A[/tex] - Piston area, measured in square inches.
If [tex]F = 16\,lb[/tex] and [tex]A = 0.43\,in^{2}[/tex], the pressure produced on the air in the pump is:
[tex]p = \frac{16\,lb}{0.43\,in^{2}}[/tex]
[tex]p = 37.209\,psi[/tex]
The pressure produced on the air in the pump is 37.209 pounds per square inch.
6. Two forces of 50 N and 30 N, respectively, are acting on an object. Find the net force (in
N) on the object if
the forces are acting in the same direction
b. the forces are acting in opposite directions.
Answer:
same direction = 80 (n)
opposite direction = 20 (n) going one direction
Explanation:
same direction means they are added to each other
and opposite means acting on eachother
In an experiment different wavelengths of light, all able to eject photoelectrons, shine on a freshly prepared (oxide-free) zinc surface. Which statement is true
Answer:
the energy of the photons is greater than the work function of the zinc oxide.
h f> = Ф
Explanation:
In this experiment on the photoelectric effect, it is explained by the Einstein relation that considers the light beam formed by discrete energy packages.
K_max = h f - Ф
in the exercise phase, they indicate that different wavelengths can inject electrons, so the energy of the photons is greater than the work function of the zinc oxide.
h f > = Ф
According to Newton, when the distance between two interacting objects doubles, the gravitational force is
Answer:
1/4 of its original value
Explanation:
Newton's law of universal gravitation states that when two bodies of masses M₁ and M₂ interact, the force of attraction (F) between these bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance (r) between these bodies. i.e
F ∝ [tex]\frac{M_1 M_2}{r^2}[/tex] ------------(i)
From the equation above, it can be deduced that;
F ∝ [tex]\frac{1}{r^2}[/tex]
=> F = G [tex]\frac{1}{r^2}[/tex] -----------(ii)
Where;
G = constant of proportionality called the gravitational constant
Equation (ii) can be re-written as
Fr² = G
=> F₁r₁² = F₂r₂² -----------(iii)
Where;
F₁ and r₁ are the initial values of the force and distance respectively
F₂ and r₂ are the final values of the force and distance respectively
From the question, if the distance doubles i.e;
r₂ = 2r₁,
Then the final value of the gravitational force F₂ is calculated as follows;
Substitute the value of r₂ = 2r₁ into equation (iii) as follows;
F₁r₁² = F₂(2r₁)²
F₁r₁² = 4F₂r₁² [Divide through by r₁²]
F₁ = 4F₂ [Make F₂ subject of the formula]
F₂ = F₁ / 4 [Re-write this]
F₂ = [tex]\frac{1}{4} F_1[/tex]
Therefore the gravitational force will be 1/4 of its original value when the distance between the bodies doubles.
When the pivot point of a balance is not at the center of mass of the balance, how is the net torque on the balance calculated
Answer:
It is calculated as Force × perpendicular distance.
Explanation:
Torque is a rotational force and twisting force that can cause an object to rotate in it's axis. This cause angular rotation.
The torque due to gravity on a body about its centre of mass is zero because the centre of mass is the that point of the body at which the force acts by the gravity that is mg.
But if the pivot point of a balance is not at the centre of mass of the balance, it will be FORCE × PERPENDICULAR DISTANCE because at that point, there is no centre in which the force act on a body by gravity. The distance and force will be use to calculate.
An object is launched at a velocity of 20 m/s in a direction making an angle of 25° upward with the horizontal.
a) What is the maximum height reached by the object?
b) What is the total flight time (between launch and touching the ground) of the object?
c) What is the horizontal range (maximum x above ground) of the object?
d) What is the magnitude of the velocity of the object just before it hits the ground?
Answer:
(a) max. height = 3.641 m
(b) flight time = 1.723 s
(c) horizontal range = 31.235 m
(d) impact velocity = 20 m/s
Above values have been given to third decimal. Adjust significant figures to suit accuracy required.
Explanation:
This problem requires the use of kinematics equations
v1^2-v0^2=2aS .............(1)
v1.t + at^2/2 = S ............(2)
where
v0=initial velocity
v1=final velocity
a=acceleration
S=distance travelled
SI units and degrees will be used throughout
Let
theta = angle of elevation = 25 degrees above horizontal
v=initial velocity at 25 degrees elevation in m/s
a = g = -9.81 = acceleration due to gravity (downwards)
(a) Maximum height
Consider vertical direction,
v0 = v sin(theta) = 8.452 m/s
To find maximum height, we find the distance travelled when vertical velocity = 0, i.e. v1=0,
solve for S in equation (1)
v1^2 - v0^2 = 2aS
S = (v1^2-v0^2)/2g = (0-8.452^2)/(2*(-9.81)) = 3.641 m/s
(b) total flight time
We solve for the time t when the vertical height of the object is AGAIN = 0.
Using equation (2) for vertical direction,
v0*t + at^2/2 = S substitute values
8.452*t + (-9.81)t^2 = 3.641
Solve for t in the above quadratic equation to get t=0, or t=1.723 s.
So time for the flight = 1.723 s
(c) Horiontal range
We know the horizontal velocity is constant (neglect air resistance) at
vh = v*cos(theta) = 25*cos(25) = 18.126 m/s
Time of flight = 1.723 s
Horizontal range = 18.126 m/s * 1.723 s = 31.235 m
(d) Magnitude of object on hitting ground, Vfinal
By symmetry of the trajectory, Vfinal = v = 20, or
Vfinal = sqrt(v0^2+vh^2) = sqrt(8.452^2+18.126^2) = 20 m/s
a heat engine with an efficiency of 30.0% performs 2500 j of work. how much heat is discharged to the lower temperature reservoir
Answer:
Q₂ = 5833.33 J
Explanation:
First we need to find the energy supplied to the heat engine. The formula for the efficiency of the heat engine is given as:
η = W/Q₁
where,
η = efficiency of engine = 30% = 0.3
W = Work done by engine = 2500 J
Q₁ = Heat supplied to the engine = ?
Therefore,
0.3 = 2500 J/Q₁
Q₁ = 2500 J/0.3
Q₁ = 8333.33 J
Now, we find the heat discharged to lower temperature reservoir by using the formula of work:
W = Q₁ - Q₂
Q₂ = Q₁ - W
where,
Q₂ = Heat discharged to the lower temperature reservoir = ?
Therefore,
Q₂ = 8333.33 J - 2500 J
Q₂ = 5833.33 J
Two children are balanced on a seesaw, but one child weighs twice as much as the other child. The heavier child is sitting half as far from the pivot as is the lighter child. Since the seesaw is balanced, the heavier child is exerting on the seesaw:_______.
a. a force that is less than the force the lighter child is exerting.
b. a force that is equal in amount but oppositely directed to the force the lighter child is exerting.
Answer:
B. A force that is equal in amount but oppositely directed to the force the lighter child is exerting.
Explanation:
If they are sitting at the same distance away from the pivot yet the seesaw is balanced, the only conclusion is the heavier child is exerting a lower force. This causes the pivot exertion and balances to be equal. The equilibrium of the pivot-seesaw is not affected by the weight because of force exertion.
The compressor of an air conditioner draws an electric current of 16.2 A when it starts up. If the start-up time is 1.45 s long, then how much electric charge passes through the circuit during this period
Answer:
Q = 23.49 C
Explanation:
We have,
Electric current drawn by the air conditioner is 16.2 A
Time, t = 1.45 s
It is required to find the electric charge passes through the circuit during this period. We know that electric current is defined as the electric charge flowing per unit time. So,
[tex]I=\dfrac{q}{t}\\\\q=It\\\\q=16.2\times 1.45\\\\q=23.49\ C[/tex]
So, the charge of 23.49 C is passing through the circuit during this period.
Consider a heat engine that inputs 10 kJ of heat and outputs 5 kJ of work. What are the signs on the total heat transfer and total work transfer
Answer:
Total heat transfer is positive
Total work transfer is positive
Explanation:
The first law of thermodynamics states that when a system interacts with its surrounding, the amount of energy gained by the system must be equal to the amount of energy lost by the surrounding. In a closed system, exchange of energy with the surrounding can be done through heat and work transfer.
Heat transfer to a system is positive and that transferred from the system is negative.
Also, work done by a system is positive while the work done on the system is negative.
Therefore, from the question, since the heat engine inputs 10kJ of heat, then heat is being transferred to the system. Hence, the sign of the total heat transfer is positive (+ve)
Also, since the heat engine outputs 5kJ of work, it implies that work is being done by the system. Hence the sign of the total work transfer is also positive (+ve).
A bucket of water with total mass 23 kg is attached to a rope, which in turn is wound around a 0.050-m radius pulley at the top of a well. The bucket is raised to the top of the well and released from rest. The bucket is falling for 2 s and has a speed of 8.0 m/s upon hitting the water surface in the well. What is the moment of inertia of the pulley?
Answer:
[tex]I = 0.083 kg m^2[/tex]
Explanation:
Mass of the bucket, m = 23 kg
Radius of the pulley, r = 0.050 m
The bucket is released from rest, u = 0 m/s
The time taken to fall, t = 2 s
Speed, v = 8.0 m/s
Moment of Inertia of the pulley, I = ?
Using the equation of motion:
v = u + at
8 = 0 + 2a
a = 8/2
a = 4 m/s²
The relationship between the linear and angular accelerations is given by the equation:
[tex]a = \alpha r[/tex]
Angular acceleration, [tex]\alpha = a/r[/tex]
[tex]\alpha = 4/0.050\\\alpha = 80 rad/s^2[/tex]
Since the bucket is falling, it can be modeled by the equation:
mg - T = ma
T = mg - ma = m(g-a)
T = 23(9.8 - 4)
The tension, T = 133.4 N
The equation for the pulley can be modeled by:
[tex]T* r = I * \alpha\\133.4 * 0.050 = I * 80\\6.67 = 80 I\\I = 6.67/80\\I = 0.083 kg m^2[/tex]
A particle is released as part of an experiment. Its speed t seconds after release is given by v (t )equalsnegative 0.4 t squared plus 2 t, where v (t )is in meters per second. a) How far does the particle travel during the first 2 sec? b) How far does it travel during the second 2 sec?
Answer:
a) 2.933 m
b) 4.534 m
Explanation:
We're given the equation
v(t) = -0.4t² + 2t
If we're to find the distance, then we'd have to integrate the velocity, since integration of velocity gives distance, just as differentiation of distance gives velocity.
See attachment for the calculations
The conclusion of the attachment will be
7.467 - 2.933 and that is 4.534 m
Thus, The distance it travels in the second 2 sec is 4.534 m
What force is required so that a particle of mass m has the position function r(t) = t3 i + 7t2 j + t3 k?
Answer:
[tex]F(t)=m\,\,a(t)=6\,m\,t\,\hat i+14\,m\,\hat j+6\,m\,t\,\hat k\\F(t)=\,(6\,m\,t,14\,m,6\,m\,t)[/tex]
Explanation:
Recall that force is defined as mass times acceleration, and acceleration is the second derivative with respect to time of the position. Since the position comes in terms of time, and with separate functions for each component in the three dimensional space, we first calculate the velocity (with the first derivative, and then the acceleration as the second derivative:
[tex]r(t)=t^3\,\hat i+7\,t^2\,\hat j+t^3\,\hat k\\v(t)=3\,t^2\,\hat i+14\,t\,\hat j+3\,t^2\,\hat k\\a(t)=6\,t\,\hat i+14\,\hat j+6\,t\,\hat k[/tex]
Therefore, the force will be given by the product of this acceleration times the mass "m":
[tex]F(t)=m\,\,a(t)=6\,m\,t\,\hat i+14\,m\,\hat j+6\,m\,t\,\hat k[/tex]
A length of organ pipe is closed at one end. If the speed of sound is 344 m/s, what length of pipe (in cm) is needed to obtain a fundamental frequency of 50 Hz
Answer:
The length = 27.52m
Explanation:
v=f x wavelength