Answer:
[tex]T(1)=21 [/tex]
Explanation:
The equation of the position in kinematics is given:
[tex]x(t)=x_{0}+v_{0}t+0.5at^{2}[/tex]
x(0) is the initial position, in this it is 0 v(0) is the initial velocity (20 m/s)a is the acceleration (2 m/s²)So the equation will be:
[tex]x(t)=20t+0.5*2*t^{2}[/tex]
[tex]x(t)=20t+t^{2}[/tex]
Now, the Taylor polynomial equation is:
[tex]f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^{2}+...[/tex]
Using our position equation we can find f'(t)=v(t) and f''(x)=a(t). In our case a=0, so let's find each derivative.
[tex]f(t)=x(t)=20t+t^{2}[/tex]
[tex]f'(t)=\frac{dx(t)}{dt}=v(t)=20+2t[/tex]
[tex]f''(t)=\frac{dv(t)}{dt}=a(t)=2[/tex]
Using the Taylor polynomial with a = 0 and take just the second order of the derivative.
[tex]f(0)+\frac{f'(0)}{1!}(x)+\frac{f''(0)}{2!}(x)^{2}[/tex]
[tex]f(0)=x(0)=0[/tex]
[tex]f'(0)=v(0)=20[/tex]
[tex]f''(0)=a(0)=2[/tex]
[tex]T(t)=f(0)+\frac{f'(0)}{1!}(t)+\frac{f''(0)}{2!}(t)^{2}[/tex]
[tex]T(t)=\frac{20}{1!}(t)+\frac{2}{2!}(t)^{2}[/tex]
[tex]T(t)=20t+t^{2}[/tex]
Let's put t=1 so find the how far the car moves in the next second:
[tex]T(1)=20*1+1^{2}[/tex]
[tex]T(1)=21 [/tex]
Therefore, the position in the next second is 21 m.
We need to know if the acceleration remains at this value to use this polynomial in the next minute, so I suggest that it would be reasonable to use this method just under this condition.
I hope it helps you!
How might a nitrogen-oxygen atmosphere on the moon affect the range of temperatures on the moon?
A) An atmosphere might hold in heat, making the moon very hot
B) An atmosphere might block heat radiating from the sun, making the moon very cold
C) An atmosphere might moderate temperatures, making more or even, as on Earth
D) An atmosphere would have no effect on the range of temperatures on the moon
Answer:
C) An atmosphere might moderate temperatures, making more or even, as on Earth
Explanation:
Atmosphere has moderating effect on a planet . It not only protects the planet from getting heated excessively during day time but it also protects it from getting too much cooled during night . In day time , it filters excessive heat out and during night it traps the heat coming out . Both ways , it tries to keep the temperature of the planet evenly balanced in day and night.
The Answer Is C) An atmosphere might moderate temperatures, making more or even, as on Earth
A zone reconnaissance involves a directed effort to obtain detailed information on all routes, obstacles, terrain, enemy forces, or specific civil considerations within a zone defined by boundaries to include obstacles (existing and reinforcing) as well as areas of CBRN contamination. True or false?
Answer:
It is True
Explanation:.
A commander assigns a zone reconnaissance mission when he seeks additional information on a zone before committing other forces in the zone. It is appropriate when the enemy situation is vague, existing knowledge of the terrain is limited, or combat operations have altered the terrain. A zone reconnaissance could include several route or area reconnaissance missions assigned to subordinate units.
At the top of the hill, the
ball has 100 J of potential energy. Explain what happens
to Mechanical Energy, Kinetic Energy and Potential
Energy as the ball rolls down and reaches the bottom.
Answer:
The answer to your question is below
Explanation:
To explain what happens with the ball we must remember the Law of Conservation of Energy.
This law states that the energy can be neither created nor destroyed only converted from one form of energy to another.
Then,
At the top of the hill, the potential energy is maximum and the kinetic energy equals to zero.
When the ball starts to roll down the potential energy will be lower and the kinetic energy will have a low value.
At the middle of the hill, both energies have the same values.
At the end of the hill, the potential energy will be equal to zero and the kinetic energy will be maximum.
According to this law, energy cannot be generated or annihilated; it can only be transformed from one type of energy to another.
The gravitational potential is highest at the summit of the slope, whereas the kinetic energy is negative.The potential energy of the ball will be reduced when it begins to roll down, and the velocity will be low.Both energy have had the same value at the center of the slope.The gravitational potential will be minimal and the velocity will be highest at the top of the slope.Learn more:https://brainly.com/question/17858145?referrer=searchResults
To study the properties of various particles, you can accelerate the particles with electric fields. A positron is a particle with the same mass as an electron but the opposite charge ( e). If a positron is accelerated by a constant electric field of magnitude 326 N/C, find the following.(a) Find the acceleration of the positron. m/s2 (b) Find the positron's speed after 8.70 Ã 10-9 s. Assume that the positron started from rest. m/s
Answer:
(a) 5.73 * 10¹³ m/s²
(b) 4.99 * 10⁵ m/s
Explanation:
(a) We know that Electrical force is the product of electric charge and electric field:
F = qE
We also know that Force is the product of mass and acceleration:
F = ma
Therefore, equating both of them:
ma = qE
Acceleration, a, will be:
a = (qE) / m
Electric charge of a positron = 1.602 * 10^(-19) C
Mass of a positron = 9.11 * 10^(-31) kg
Acceleration will be:
a = (1.602 * 10^(-19)) * 9.11 * 10^(-31)) / 326
a = 5.73 * 10¹³ m/s²
(b) Acceleration is the time rate of change of velocity. It is given as:
a = v/t
Velocity, v, will then be:
v = a * t
Time, t = 8.7 * 10^(-9) s
Therefore, velocity will be:
v = 5.73 * 10^(13) * 8.7 * 10^(-9)
v = 49.85 * 10⁴ m/s = 4.99 * 10⁵ m/s
All of the following are functions of the skeletal system EXCEPT:
A.
producing red blood cells
B.
providing support for the body
C.
producing white blood cells
D.
producing ligaments and tendons
Answer:
D. Producing ligaments and tendons.
just to the quiz on edge
The skeletal system provides structure, assists in movement, protects internal organs, and produces both red and white blood cells. However, it does not produce ligaments and tendons, which are part of the muscular system and connective tissues. Thus, the correct option is D.
Explanation:The skeletal system has multiple functions within the body. These include providing structure, facilitating movement, protecting internal organs, producing red blood cells, and producing white blood cells. These cells are crucial as they are responsible for oxygen transport around the body and defense against disease. However, producing ligaments and tendons is not a function of the skeletal system. Ligaments and tendons are part of the muscular system and connective tissues. Ligaments connect bones to each other, and tendons connect muscle to bone.
Therefore, the correct answer would be D. producing ligaments and tendons.
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A circuit consists of a series combination of 5.50 âkΩ and 5.00 âkΩ resistors connected across a 50.0-V battery having negligible internal resistance. You want to measure the true potential difference (that is, the potential difference without the meter present) across the 5.00 âkΩ resistor using a voltmeter having an internal resistance of 10.0 kΩ.1. What potential difference does the voltmeter measure across the 5.00 âkΩ resistor?2. What is the true potential difference across this resistor when the meter is not present?3. By what percentage is the voltmeter reading in error from the true potential difference?
Answer:
(a) 18.87 V
(b) 23.81 V
(c) 20.75%
Explanation:
The answers are given in the pictures. I have attached the pictures because circuits were needed to be drawn which are easier to understand when done on page. The page are numbered on top left corner.
A block of unknown mass is attached to a spring of spring constant 6.1 N/m and undergoes simple harmonic motion with an amplitude of 10.3 cm. When the mass is halfway between its equilibrium position and the endpoint, its speed is measured to be 31.3 cm/s. Calculate the mass of the block. Answer in units of kg
Answer:
Explanation:
Given that, .
Spring constant.
K = 6.1 N/m
Amplitude of oscillation
A = 10.3cm = 0.103m
Half wave, it speed is
V = 31.3cm/s = 0.313m/s
Mass of block
M =?
The is half of the distance between the equilibrium position and endpoint
Then,
x = A/2
Where,
A is the amplitude
x is the position of the block
x = A / 2 = 0.103/2
x = 0.0515m
The velocity if the block at any given position is given as
v = ω√(A²—x²)
Then,
ω = v / √(A²—x²)
Where
ω is angular frequency
v Is the velocity of the block
The force constant is given as
k = mω²
Where,
K is spring constant
ω is angular frequency
Substitute ω into k
Then,
k = m (v / √(A²—x²))²
k = mv² / (A²—x²)
Make m subject of formula since we want to find m
m = k(A²—x²) / v²
m = 6.1 (0.103²—0.0515²) / 0.313²
m = 6.1 × 7.96 × 10^-3 / 0.313²
m = 0.495 kg
The mass of the block is approximately 0.5kg
Answer:
The mass of the block is 496 g
Explanation:
Here we have by the principle of conservation of energy;
Energy in spring, E = 0.5·k·A²
Where:
k = Spring constant = 6.1 N/m
A = Amplitude of motion = 10.3 cm = 0.103 m
E = 0.5×6.1×0.103² = 0.03235745 J= 3.24 × 10⁻² J
At half way, we have
[tex]E = \frac{1}{2} k(\frac{A}{2} )^{2} + \frac{1}{2} mv^{2} = \frac{E}{4} + \frac{1}{2} mv^{2}[/tex]
Where:
m = Mass of the block
v = Velocity of block at the instant (Halfway between its equilibrium position and the endpoint)
Therefore,
[tex]\frac{3}{4}E = \frac{1}{2} mv^{2}[/tex] or
m = [tex]\frac{3}{2v^2}E[/tex] = [tex]\frac{3}{2\times 0.313^2} \times 3.24 \times 10^{-2}[/tex]= 0.496 kg = 496 g.
Your little sister is building a radio from scratch. Plans call for a 500 μH inductor wound on a cardboard tube. She brings you the tube from a toilet-paper roll (12 cm long, 4.0 cm diameter), and asks how many turns she should wind on the full length of the tube.
Answer:
N = 195 turns
Explanation:
The inductance of the inductor, L = 500 μH = 500 * 10⁻⁶H
The length of the tube, l = 12 cm = 0.12 m
The diameter of the tube, d = 4 cm = 0.04 m
Radius, r = 0.04/2 = 0.02 m
Area of the tube, A = πr² = 0.02²π = 0.0004π m²
[tex]\mu_{0} = 4\pi * 10^{-7}[/tex]
The inductance of a solenoid is given by:
[tex]L = \frac{\mu_{0}N^{2} A }{l}[/tex]
[tex]500 * 10^{-6} = \frac{4\pi *10^{-7} N^{2} *4\pi *10^{-4} }{0.12}\\500 * 10^{-6} = 0.00000001316N^{2} \\N^{2} = \frac{500 * 10^{-6}}{0.00000001316}\\N^{2} = 37995.44\\N = \sqrt{37995.44} \\N = 194.92 turns[/tex]
When convection currents flow in the mantle they also move the
Answer:
Crust
Explanation:
When the convection currents flow in the mantle they also move the crust. This is best explained as that the crust gets a free ride with the convection currents in the mantle; just as a conveyor belt in a factory moves boxes.
When a constant force acts upon an object, the acceleration of the object varies inversely with its mass 2kg. When a certain constant force acts upon an object with mass , the acceleration of the object is 26m/s^2 . If the same force acts upon another object whose mass is 13 , what is this object's acceleration
If the same force acts upon another object whose mass is 13 kg the acceleration is [tex]4m/s^2[/tex].
Force is a fundamental concept in physics that describes the interaction between objects that can cause a change in their motion or shape. It's a vector quantity, meaning it has both magnitude and direction. In simpler terms, force is what can make things move, stop, change direction, or deform.
Given:
Masses, [tex]m_1 = 2\ kg[/tex] and [tex]m_2 = 13\ kg[/tex]
Acceleration, [tex]a = 26\ m/s^2[/tex]
The force is computed as:
[tex]F = m_1a\\F = 2\times26\\F = 52\ N[/tex]
The acceleration for mass 52 kilograms is:
[tex]F = m_2a\\a = F/m_2\\a = 52/ 13\\a = 4\ m/s^2[/tex]
Hence, if the same force acts upon another object whose mass is 13 kg the acceleration is [tex]4m/s^2[/tex].
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Applying Newton's second law of motion, the acceleration of an object is directly proportional to the force acting on it, but inversely proportional to its mass. The acceleration of a 13kg object, when acted upon by the same force that gives a 2kg object an acceleration of 26m/s^2, would be 4 m/s^2.
Explanation:The question basically asks us to apply Newton's second law of motion, which states that the acceleration of a system is directly proportional to the net external force acting on the system, but inversely proportional to its mass. In equation form, this can be written as F=ma, where F is the force, m is the mass, and a is the acceleration. This is also referred to as the principle of constant acceleration, in which the force applied to an object will either accelerate or decelerate it at a constant rate, provided the mass of the object remains unchanged.
In this case, we know that a 2kg object accelerates at 26m/s^2 under the influence of a constant force. Therefore, the force (F) acting on it can be computed as F = m * a = 2kg * 26m/s^2 = 52N. Now, if the same force acts on another object with a mass of 13kg, the acceleration (a) of this object can be computed by rearranging the equation as a = F/m = 52N / 13kg = 4m/s^2. So, the acceleration of the second object, under the same force, would be 4 m/s^2.
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The linear momentum of a truck of mass 5000 kg that is moving at a velocity of +30 m/a is ___ kg m/s
Linear momentum of a truck is 1,50,000 kg.m/s
Explanation:
Linear momentum is the product of the mass and velocity of an object. It is a vector quantity, which have a magnitude and a direction.
Linear momentum is a property of an object which is in motion with respect to a reference point (i.e. any object changing its position with respect to the reference point).
It's SI units are kg.m/s
Linear momentum is a vector quantity.
Linear momentum formula (p) = mass × velocity
Given data mass = 5000 kg ; velocity = 30 m/s
P = 5000 × 30
Linear momentum p= 1,50,000 kg.m/s
The linear momentum of the truck is 150,000 kgm/s.
The given parameters:
Mass of truck, m = 5000 kgVelocity of the truck, v = 30 m/sThe linear momentum of the truck is calculated as the product of the mass and velocity of the truck.
The linear momentum of the truck is calculated as follows;
P = mv
P = 5000 x 30
P = 150,000 kgm/s
Thus, the linear momentum of the truck is 150,000 kgm/s.
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Which statement best describes pseudoscience? Pseudoscience is
A) not a branch of science and involves beliefs and opinions
B) a method that scientists use to study the living world
C) a trustworthy branch of science that involves theories and laws
D) a way of testing a hypothesis to prove or disprove your theory
(FLVS Physical Science 03.09)
Answer:
Pseudoscience is not science - it involves beliefs and opinions.
Explanation:
i got it right on the test
Answer:
Pseudoscience is not science - it involves beliefs and opinions.
Explanation:
A 6.4 kg block with a speed of 5.4 m/s collides with a 12.8 kg block that has a speed of 3.6 m/s in the same direction. After the collision, the 12.8 kg block is observed to be traveling in the original direction with a speed of 4.5 m/s. (a) What is the velocity of the 6.4 kg block immediately after the collision
Answer:
The velocity of the 6.4 kg block immediately after the collision is 3.6 m/s in the original direction.
Explanation:
Given;
mass of first block, m₁ = 6.4 kg
initial speed of first block, u₁ = 5.4 m/s
mass of second block, m₂ = 12.8 kg
initial speed of second block, u₂ = 3.6 m/s
final speed of second block, v₂ = 4.5 m/s
To determine the final speed of 6.4 kg block immediately after the collision, we apply principle of conservation linear momentum;
Total momentum before collision = Total momentum after collision
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
where;
v₁ is the final speed of 6.4 kg block
v₂ is the final speed of 12.8 kg block
Substitute the given values of m₁, u₁, m₂, u₂ , v₂ and calculate v₁
6.4 x 5.4 + 12.8 x 3.6 = 6.4v₁ + 12.8 x 4.5
34.56 + 46.08 = 6.4v₁ + 57.6
80.64 = 6.4v₁ + 57.6
80.64 - 57.6 = 6.4v₁
23.04 = 6.4v₁
v₁ = 23.04 / 6.4
v₁ = 3.6 m/s
Therefore, the velocity of the 6.4 kg block immediately after the collision is 3.6 m/s in the original direction.
Determine whether each of the following is exothermic or endothermic and indicate the sign of δh. A. Natural gas burning on a stove b. Isopropyl alcohol evaporating from skin c. Water condensing from steam
Natural gas burning on a stove is exothermic (δh < 0), isopropyl alcohol evaporating from the skin is endothermic (δh > 0), and water condensing from steam is exothermic (δh < 0).
Explanation:(a) Natural gas burning on a stove: This is an exothermic process. The combustion of natural gas on a stove releases a huge amount of heat which cooks the food. The sign of δh (enthalpy change) in this process is negative (δh < 0) as heat is released.
(b) Isopropyl alcohol evaporating from the skin: This is an endothermic process. When isopropyl alcohol evaporates from our skin, it absorbs heat from our body which causes a cooling effect. Hence, the sign of δh is positive (δh > 0) as heat is absorbed.
(c) Water condensing from steam: This is an exothermic process. The condensation process releases heat and hence, the sign of δh is negative (δh < 0).
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A fisherman fishing from a pier observes that the float on his line bobs up and down, taking 2.4 s to move from its highest to its lowest point. He also estimates that the distance between adjacent wave crests is 48 m. What is the speed of the waves going past the pier?
(A) 20 m/s
(B) 1.0 m/s
(C) 10 m/s
(D) 5.0 m/s
(E) 115 m/s
Answer:
(c) 10m/s
Explanation:
to find the speed of the waves you can use the following formula:
[tex]v=\frac{\lambda}{T}[/tex]
λ: wavelength of the wave
T: period
the wavelength is the distance between crests = 48m
the period is the time of a complete oscillation of the wave. In this case you have that the float takes 2.4 s to go from its highest to the lowest point. The period will be twice that time:
T = 2(2.4s)=4.8s
by replacing you obtain:
[tex]v=\frac{48m}{4.8s}=10\frac{m}{s}[/tex]
the answer is (c) 10m/s
A researcher measures the thickness of a layer of benzene (n = 1.50) floating on water by shining monochromatic light onto the film and varying the wavelength of the light. she finds that light of wavelength 600 nm is reflected most strongly from the film. part a what does she calculate for the minimum thickness of the film? express your answer with the appropriate units.
Answer:
Explanation:
This problem relates to interference of light in thin films .
The condition of bright fringe in thin films which is sandwitched by two layers of medium having lesser refractive index is as follows.
2nt = (2n+1) λ / 2 , n is refractive index of thin layer , t is its thickness , λ is wavelength of light .
2 x 1.5 t = λ / 2 , if n = 0 for minimum thickness.
2 x 1.5 t = 600 / 2 nm
t = 100 nm .
An oscillator consists of a block attached to a spring (k = 427 N/m). At some time t, the position (measured from the system's equilibrium location), velocity, and acceleration of the block are x = 0.134 m, v = -12.1 m/s, and a = -107 m/s2. Calculate (a) the frequency of oscillation, (b) the mass of the block, and (c) the amplitude of the motion.
Answer:
a) 4.49Hz
b) 0.536kg
c) 2.57s
Explanation:
This problem can be solved by using the equation for he position and velocity of an object in a mass-string system:
[tex]x=Acos(\omega t)\\\\v=-\omega Asin(\omega t)\\\\a=-\omega^2Acos(\omega t)[/tex]
for some time t you have:
x=0.134m
v=-12.1m/s
a=-107m/s^2
If you divide the first equation and the third equation, you can calculate w:
[tex]\frac{x}{a}=\frac{Acos(\omega t)}{-\omega^2 Acos(\omega t)}\\\\\omega=\sqrt{-\frac{a}{x}}=\sqrt{-\frac{-107m/s^2}{0.134m}}=28.25\frac{rad}{s}[/tex]
with this value you can compute the frequency:
a)
[tex]f=\frac{\omega}{2\pi}=\frac{28.25rad/s}{2\pi}=4.49Hz[/tex]
b)
the mass of the block is given by the formula:
[tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\\\\m=\frac{k}{4\pi^2f^2}=\frac{427N/m}{(4\pi^2)(4.49Hz)^2}=0.536kg[/tex]
c) to find the amplitude of the motion you need to know the time t. This can computed by dividing the equation for v with the equation for x and taking the arctan:
[tex]\frac{v}{x}=-\omega tan(\omega t)\\\\t=\frac{1}{\omega}arctan(-\frac{v}{x\omega })=\frac{1}{28.25rad/s}arctan(-\frac{-12.1m/s}{(0.134m)(28.25rad/s)})=2.57s[/tex]
Finally, the amplitude is:
[tex]x=Acos(\omega t)\\\\A=\frac{0.134m}{cos(28.25rad/s*2.57s )}=0.45m[/tex]
A 0.100-kg ball traveling horizontally on a frictionless surface approaches a very massive stone at 20.0 m/s perpendicular to wall and rebounds with 70.0% of its initial kinetic energy. What is the magnitude of the change in momentum of the stone
Answer:
Change in momentum of the stone is 3.673 kg.m/s.
Explanation:
Given:
Mass of the ball on the horizontal the surface, m = 0.10 kg
Velocity of the ball with which it hits the stone, v = 20 m/s
According to the question it rebounds with 70% of the initial kinetic energy.
We have to find the change in momentum i.e Δp
Before that:
We have to calculate the rebound velocity with which the object rebounds.
Lets say that the rebound velocity be "v1" and KE remaining after the object rebounds be "KE1".
⇒ [tex]KE_1=0.7\times \frac{mv^2}{2}[/tex]
⇒ [tex]KE_1=0.7\times \frac{0.10\times (20)^2}{2}[/tex]
⇒ [tex]KE_1=0.7\times \frac{0.10\times 400}{2}[/tex]
⇒ [tex]KE_1=14[/tex] Joules (J).
Rebound velocity "v1".
⇒ [tex]KE_1=\frac{m(v_1)^2}{2}[/tex]
⇒ [tex]v_1 = \sqrt{\frac{2KE_1}{m} }[/tex]
⇒ [tex]v_1 = \sqrt{\frac{2\times 14}{0.10} }[/tex]
⇒ [tex]v_1=16.73[/tex]
⇒ [tex]v_1=-16.73[/tex] m/s ...as it rebounds.
Change in momentum Δp.
⇒ [tex]\triangle p= m\triangle v[/tex]
⇒ [tex]\triangle p= 0.10\times (20-(-16.73)[/tex]
⇒ [tex]\triangle p= 0.10\times (20+16.73)[/tex]
⇒ [tex]\triangle p= 0.10\times (36.73)[/tex]
⇒ [tex]\triangle p = 3.673[/tex] Kg.m/s
The magnitude of the change in momentum of the stone is 3.673 kg.m/s.
In physics how much energy is required to raise the temp of 9.78kg of water from 40.82c to 52.07c? The specific heat of water in this state is 4186j/kg-c.
Answer:
469.6KJ
Explanation:
Heat energy required can be calculated using the formula
H = mc∆t where
m is the mass of the water
c is the specific heat capacity of the water
∆t is the change in temperature of the water
Given m = 9.78kg
c = 4186j/kg-c.
∆t = 52.07°C - 40.82°C
∆t = 11.25°C
H = 9.78 × 4186 × 11.25
H = 460,564.65Joules
= 460.6KJ
Answer:
the amount of energy required to raise the temperature of the water is 460564.65 J
Explanation:
The energy required to raise the temperature of water can be calculated as follows;
Q = mcΔθ
where;
Q is the quantity of heat or energy required to raise the temperature of water
m is mass of water
c is specific heat capacity of water
Δθ is change in temperature = T₂ - T₁
Given;
m = 9.78kg
c = 4186j/kg-c
Δθ = T₂ - T₁ = 52.07°C - 40.82°C = 11.25°C
Q = mcΔθ
Q = (9.78)(4186)(11.25)
Q = 460564.65 J
Therefore, the amount of energy required to raise the temperature of the water is 460564.65 J
A passenger compartment of a rotating amusement park ride contains a bench on which a book of mass
mb is
placed, as indicated by the dot in the left figure above. The compartment moves with a constant angular speed
about the center of the ride along a circular path of radius R. The bench remains horizontal throughout the
compartment’s motion. The right figure above shows a magnified view of the compartment.
The graph below shows the horizontal ( ) x component of the book’s position as a function of time, where the
+x-direction is to the right.
(a)
i. Determine the period of revolution of the book.
ii. Calculate the tangential speed vb (not the angular speed) of the book.
a) 120 s
b) v = 0.052R [m/s]
Explanation:
a)
The period of a revolution in a simple harmonic motion is the time taken for the object in motion to complete one cycle (in this case, the time taken to complete one revolution).
The graph of the problem is missing, find it in attachment.
To find the period of revolution of the book, we have to find the time between two consecutive points of the graph that have exactly the same shape, which correspond to two points in which the book is located at the same position.
The first point we take is t = 0, when the position of the book is x = 0.
Then, the next point with same shape is at t = 120 s, where the book returns at x = 0 m.
Therefore, the period is
T = 120 s - 0 s = 120 s
b)
The tangential speed of the book is given by the ratio between the distance covered during one revolution, which is the perimeter of the wheel, and the time taken, which is the period.
The perimeter of the wheel is:
[tex]L=2\pi R[/tex]
where R is the radius of the wheel.
The period of revolution is:
[tex]T=120 s[/tex]
Therefore, the tangential speed of the book is:
[tex]v=\frac{L}{T}=\frac{2\pi R}{120}=0.052R[/tex]
The diagrams below each show the motion of a distant star relative to Earth (not to scale). The red arrows indicate the speed and direction of the star’s motion: Longer arrows mean faster speed. Rank the stars based on the Doppler shift that we would detect on Earth, from largest blueshift, through no shift, to largest redshift.Note:a picture in resemble to solar system is shown in the figure
Answer:
The ranking of the stars are 5 → 2 → 4 → 3 → 1
Explanation:
To solve the question, we note that an approaching star has a blue-shift, a regressing star has a red-shift and a crossing star will have no shift.
The magnitude of the blue-shift or red-shift is dependent on the speed of the stars.
Therefore we have the rank of the stars based on the Doppler shift that would be detected on earth from largest blue-shift to largest red-shift given as
5 → 2 → 4 → 3 → 1.
The Doppler shift affects the wavelengths of light emitted or reflected by objects in motion. Larger blueshifts indicate faster speeds towards us, while larger redshifts indicate faster speeds away from us. Stars with no shift do not have a detected Doppler shift.
The Doppler shift is a phenomenon that occurs when an object's motion causes a change in the wavelengths of light that it emits or reflects. In the case of stars, if a star is moving away from us, its light will be redshifted, meaning the wavelengths of the light will be longer. On the other hand, if a star is moving towards us, its light will be blueshifted, meaning the wavelengths of the light will be shorter.
To rank the stars based on the Doppler shift that we would detect on Earth, we need to understand that larger blueshifts indicate faster speeds towards us, and larger redshifts indicate faster speeds away from us. So, the star with the largest blueshift will have the highest ranking for blueshift, while the star with the largest redshift will have the highest ranking for redshift. Stars with no shift will have no Doppler shift detected.
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Which statement describes Newton's law of universal gravitation?
Mass has hittle effect on gravity between objects.
Gravity pushes objects away from Earth's center,
Gravity does not act between Earth and the moon.
Every object in the universe attracts every other object
Final answer:
Newton's law of universal gravitation states that all objects in the universe attract each other with a force proportional to the product of their masses and inversely proportional to the square of the distance between them.
Explanation:
Newton's Law of Universal Gravitation:
Sir Isaac Newton's law of universal gravitation states that every object in the universe attracts every other object with a force. This force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The gravitational force is universally attractive and only depends on mass and distance, following the equation F = G × (m1 × m2) / r², where F is the force of gravity, G is the gravitational constant, m1 and m2 are the masses of the objects, and r is the distance between the centers of the two masses.
A technician is checking refrigerant system pressures. Both high- and low-side service ports are located on the A/C compressor. When the compressor is engaged, the high-side pressure instantly goes over 375 psig. What is the most likely cause of this condition
Answer:
Two major causes are outline bellow
1. The presence of air in the system
2. Clogged condenser
Explanation:
1. The presence of air in the system
One of the causes that have been established in relation to high compressor discharge pressure is the presence of air in the system. When this takes place, your best solution is to recharge the system.
2. Clogged condenser
Another is a clogged condenser in which case you will need to clean the condenser so that it will function properly. When you happen to spot that the discharge valve is closed and it is causing high discharge pressure on the compressor, you can solve that easily by opening the valve
The most likely cause of a refrigerant system's high-side pressure exceeding 375 psig is a blockage, overcharging, or a malfunctioning expansion valve. These issues prevent proper refrigerant flow and pressure regulation within the vapor compression system.
If a technician observes that the high-side pressure of a refrigerant system instantly exceeds 375 psig once the compressor is engaged, the most likely cause for this condition could be a blockage in the condenser or the liquid line, overcharging of the refrigerant, or a malfunctioning expansion valve that is not allowing the liquefied refrigerant to flow properly into the evaporator. The blockage could result in an excessively high pressure build-up because the refrigerant cannot circulate properly. Overcharging of the system adds too much refrigerant, which also leads to high pressure. Additionally, if the expansion valve is stuck closed or not functioning correctly, it would prevent the refrigerant from moving into the evaporator, causing the pressure to build up on the high side.
In a typical vapor compression system, as the electrically driven compressor introduces work (W), it raises the temperature and pressure of the refrigerant, forcing it into the condenser coils. Here, heat transfer occurs as the gas inside the coils is higher in temperature than the room, leading to condensation of the gas back into a liquid. The cooled pressurized liquid passes through the expansion, or pressure-reducing valve, and goes to the evaporator coils located outdoors where it is further cooled by expansion.
On a good night, the front row of the Twisted Sister concert would surely result in a 120 dB sound level. An IPod produces 100 dB. How many IPods would be needed to produce the same intensity as the front row of the Twisted Sister concert?
Answer:
100 iPods
Explanation:
The deciBell scale is logarithmic, and thus, it turns multiplying into adding.
Initially, it was the Bell scale, purely logarithmic, where "times 10" is translated into "plus 1" (just like normal logs). However, the steps became too big and so they divided the Bell in 10 parts, the deciBell.
The levels above could well have been called 10B and 12B.
Usually, we define the dB scale for intensity as:
I(dB) = 10•log(I)
Thus,
I = 10^(I(dB)/10)
Now 120 dB gives us units of I = 10^(120/10) = 10^12 Pa (assume the dB are measured to 1 Pa) and 100 dB is 10^10 Pa.
Thus, we would need 100 ipods to get the same intensity
A stone is dropped at t = 0. A second stone, with 3 times the mass of the first, is dropped from the same point at t = 55 ms. (a) How far below the release point is the center of mass of the two stones at t = 470 ms? (Neither stone has yet reached the ground.) (b) How fast is the center of mass of the two-stone system moving at that time?
Answer:
Part(a): At [tex]\bf{t = 470~ms}[/tex] the center of mass will travel [tex]\bf{0.903~m}[/tex].
Part(b): The velocity of the first stone is [tex]\bf{4.606~m.s^{-1}}[/tex] and the velocity of the second stone is [tex]\bf{4.067~m.s^{-1}}[/tex].
Explanation:
Given:
The first stone is dropped at [tex]t_{1}=0~s[/tex].
The second stone is dropped at, [tex]t_{2}=55~ms=0.055~s[/tex]
The mass of the second stone is 3 times the mass of the first.
Both the stones are dropped from the same point.
Consider the mass of the first stone be [tex]m[/tex]. So the mass of the second stone is [tex]3m[/tex].
(a)
The formula to calculate the distance traveled by each stone is given by
[tex]y = \dfrac{1}{2}gt^{2}~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)[/tex]
where [tex]g[/tex] is the acceleration due to gravity and [tex]t[/tex] is the time taken by each stone.
Substituting [tex]9.8~m.s^{-2}[/tex] for [tex]g[/tex], [tex]y_{1}[/tex] for [tex]y[/tex] and [tex]470~ms[/tex] or [tex]0.47~s[/tex] for first stone in equation (1), we have
[tex]y_{1}&=& \dfrac{1}{2}(9.8~m.s^{-2})(0.47~s)^{2}\\~~~~&=& 1.08~m[/tex]
where [tex]y_{1}[/tex] is the distance traveled by the first stone.
Substituting [tex]9.8~m.s^{-2}[/tex] for [tex]g[/tex], [tex]y_{2}[/tex] for [tex]y[/tex] and [tex](470-55)~ms = 415~ms[/tex] or [tex]0.415~s[/tex] for second stone in equation (1), we have
[tex]y_{2}&=& \dfrac{1}{2}(9.8~m.s^{-2})(0.415~s)^{2}\\~~~~&=& 0.844~m[/tex]
The formula to calculate the distance traveled by the center of mass is given by
[tex]y_{c} &=& \dfrac{my_{1}+3my_{2}}{m+3m} \\~~~~&=& \dfrac{1.08m + 0.844(3m)}{4m}\\~~~~&=& 0.903~m[/tex]
(b)
The formula to calculate the velocity of each stone is given by
[tex]v=gt~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)[/tex]
Substituting [tex]v_{1}[/tex] for [tex]v[/tex], [tex]9.8~m.s^{-2}[/tex] for [tex]g[/tex] and [tex]0.47~s[/tex] for [tex]t[/tex] in equation (2), we have
[tex]v_{1} &=& (9.8~m.s^{-2})(0.47~s)\\~~~~&=& 4.606~m.s^{-1}[/tex]
where [tex]v_{1}[/tex] is the velocity of the first stone after [tex]470~ms[/tex].
Substituting [tex]v_{2}[/tex] for [tex]v[/tex], [tex]9.8~m.s^{-2}[/tex] for [tex]g[/tex] and [tex]0.415~s[/tex] for [tex]t[/tex] in equation (2), we have
[tex]v_{2} &=& (9.8~m.s^{-2})(0.415~s)\\~~~~&=& 4.067~m.s^{-1}[/tex]
where [tex]v_{2}[/tex] is the velocity of the second stone after [tex]470~ms[/tex].
Strontium has density of 2.64 g/cm3 and crystallizes with the face-centered cubic unit cell. Calculate the radius of a strontium atom in units of picometers.
Answer:
The radius of strontium atom is
2.14 E-8cm
Explanation:
Go through the attached file for a comprehensive detailed explanation.
Calculate the total resistance in a series circuit made up of resistances of 3Ω, 4Ω, and 5Ω.
The resistance means resist, means to resist the current, so we are here given that, three resistors are connected in series, so as we know that, if n resistors are connected in series combination, then the total resistance is the individual sum of all n Resistors, so if we apply the same formula here, we will be having :
[tex]{:\implies \quad \sf R_{Total}=3+4+5}[/tex]
[tex]{:\implies \quad \boxed{\bf{R_{Total}=12\Omega}}}[/tex]
For more information related to Resistances, voltage and current see the similar question from here :
https://brainly.com/question/27101567the difference between an electric motor and an electric generator is that a motor converts _____ energy into ______ energy, while a generator converts _______ energy into ________ energy
Answer:electrical,kinetic,kinetic,electrical
Explanation:
Answer:
Electrical, kinetic, kinetic, electrical
Explanation:
Just did it
What type of air mass would form over the northern atlantic ocean?
Answer:
Maritime polar air masses form over the northern Atlantic ocean
Explanation:
Maritime polar (mP) air masses are cool, moist, and also unstable. Some maritime polar air masses arise as continental polar air masses over Asia and move westward over the Pacific, gathering warmth and moisture from the ocean. .
Answer:
Maritime Polar
Explanation:
Calculate the mag-netic field (magnitude and direc-tion) at a point p due to a current i=12.0 a in the wire shown in fig. p28.68. segment bc is an arc of a circle with radius 30.0 cm, and point p is at the center of cur- vature of the arc. segment dais an arc of a circle with radius 20.0 cm, and point p is at its cen- ter of curvature. segments cd and ab are straight lines of length 10.0 cm each
Complete Question
The diagram for this question is shown on the first uploaded image
Answer:
The magnitude is [tex]B= 4.2 *10^ {-6}T[/tex] , the direction is into the page
Explanation:
From the question we are told that
The current is [tex]i = 12.0 A[/tex]
The radius of arc bc is [tex]r_{bc} = 30.0 \ cm =\frac{30}{100} = 0.3m[/tex]
The radius of arc da is [tex]r_{da} = 20.0 \ cm = \frac{20}{100} = 0.20 \ m[/tex]
The length of segment cd and ab is = [tex]l = 10cm = \frac{10}{100} = 0.10 m[/tex]
The objective of the solution is to obtain the magnetic field
Generally magnetic due to the current flowing in the arc is mathematically represented as
[tex]B = \frac{\mu_o I}{4 \pi r}[/tex]
Here I is the current
[tex]\mu_o[/tex] is the permeability of free space with a value of [tex]4\pi *10^{-7}T \cdot m/A[/tex]
r is the distance
Considering Arc da
[tex]B_{da} = \frac{\mu_o I}{4 \pi r_{da}} \theta[/tex]
Where [tex]\theta[/tex] is the angle the arc da makes with the center from the diagram its value is [tex]\theta = 120^o = 120^o * \frac{\pi}{180} = \frac{2\pi}{3} rad[/tex]
Now substituting values into formula for magnetic field for da
[tex]B_{da} = \frac{4\p *10^{-7} * 12}{4 \pi (0.20)}[\frac{2 \pi}{3} ][/tex]
[tex]= \frac{10^{-7} * 12}{0.20} * [\frac{2 \pi}{3} ][/tex]
[tex]B_{da}= 12.56*10^{-6} T[/tex]
Looking at the diagram to obtain the direction of the current and using right hand rule then we would obtain the the direction of magnetic field due to da is into the pages of the paper
Considering Arc bc
[tex]B_{bc} = \frac{\mu_o I}{4 \pi r_{bc}} \theta[/tex]
Substituting value
[tex]B_{bc} = \frac{4 \pi *10^{-7} * 12}{4 \pi (0.30)} [\frac{2 \pi}{3} ][/tex]
[tex]B_{bc}= 8.37*10^{-6}T[/tex]
Looking at the diagram to obtain the direction of the current and using right hand rule then we would obtain the the direction of magnetic field due to bc is out of the pages of the paper
Since the line joining P to segment bc and da makes angle = 0°
Then the net magnetic field would be
[tex]B = B_{da} - B{bc}[/tex]
[tex]= 12.56*10^{-6} - 8.37*10^{-6}[/tex]
[tex]= 4.2 *10^ {-6}T[/tex]
Since [tex]B_{da} > B_{bc}[/tex] then the direction of the net charge would be into the page
Applying the Biot-Savart law and Ampere's law, the straight segments AB and CD contribute nothing to the magnetic field at point P, while the arcs BC and DA contribute a magnetic field given by μ0I/2R with direction as indicated by the right-hand rule.
Explanation:To calculate the magnetic field at a point due to a current in a wire, we can use the Biot-Savart law. Each section of the wire will contribute differently to the total magnetic field due to its geometry and position relative to point P.
For the straight segments AB and CD, they will likely contribute nothing to the magnetic field at point P, because of their symmetry around P and the linear nature of the wires. In each case, the magnetic field they create at P points in opposite directions, cancelling each other.
For the circular segments BC and DA, we can use Ampere's law and the right-hand rule to calculate their effects. The magnetic field B at the centre of a current-carrying loop is given by B = μ0I/2R, where I is the current, R is the radius of the loop, and μ0 is the permeability of free space.
Using this formula, we can calculate the magnetic field due to each arc, taking into account the direction of current (as given by the right-hand rule) and the radius of curvature. Add these two results to get the total magnetic field at point P, while keeping in mind the sign conventions of the vectors.
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