Answer:
The constant of proportionality is the ratio between two directly proportional quantities. Two quantities are directly proportional when they increase and decrease at the same rate. The constant of proportionality k is given by k=y/x where y and x are two quantities that are directly proportional to each other.
Explanation:
A 1.10 kg block is attached to a spring with spring constant 17 N/m. While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 46 cm/s.
A) What is the amplitude of the subsequent oscillations?
B) What is the block's speed at the point where x = 0.25 A?
Answer:
Explanation:
The kinetic energy of block will be converted into potential energy of spring .
If A be the amplitude of oscillations
1 /2 k A² = 1/2 m v²
17 A² = 1.1 x .46²
A² = .0137
A= 11.7 cm
B )
when x = .25 A = .25 x 11.7 = 2.9 cm
potential energy = 1/2 k x²
= .5 x 17 x ( .029 )² = .00715 J
kinetic energy = 1/2 m v²
1/2 m v² + .00715 = .5 1.1 x .46²
1/2 m v² + .00715 = .1164
1/2 m v² = .10925
.5 x 1.1 x v²= .10925
v² = .1986
v = .4456 m /s
= 44.56 cm /s
A skater spins with an angular speed of 5.9 rad/s with her arms outstretched. She lowers her arms, decreasing her moment of inertia by a factor of 1.7. Ignoring friction on the skates, determine the ratio of her final kinetic energy to her initial kinetic energy.
Answer:
the ratio of her final kinetic energy to her initial kinetic energy is 1.7.
Explanation:
Given;
initial angular speed, ω₁ = 5.9 rad/s
let her initial moment of inertia = I₁
her final moment of inertia [tex]I_2 = \frac{I_1}{1.7}[/tex]
Apply the principle of conservation of angular momentum to determine the final angular speed of the girl;
[tex]\omega_1I_1 = \omega_f I_2\\\\\omega_f = \frac{\omega _1 I_1}{I_2} \\\\\omega_f = \frac{5.9 \times I_1}{I_1/1.7} \\\\\omega = 5.9 \times 1.7 \\\\\omega_f = 10.03 \ rad/s[/tex]
The initial rotational kinetic energy is given as;
[tex]K.E_I = \frac{1}{2}I_1 \omega_I ^2[/tex]
The final rotational kinetic energy is given as;
[tex]K.E_f = \frac{1}{2}I_2 \omega_f ^2[/tex]
The ratio of her final kinetic energy to her initial kinetic energy is given as;
[tex]\frac{K.E_f}{K.E_I}= \frac{\frac{1}{2}I_2 \omega_f^2 }{\frac{1}{2} I_1\omega _1^2} \\\\\frac{K.E_f}{K.E_I}= \frac{I_2 \omega_f^2}{ I_1\omega _1^2} \\\\\frac{K.E_f}{K.E_I}= \frac{I_1/1.7 \times \omega_f^2}{ I_1 \times \omega _1^2} \\\\\frac{K.E_f}{K.E_I}= \frac{ \omega_f^2}{ 1.7 \omega _1^2} \\\\\frac{K.E_f}{K.E_I}= \frac{ (10.03)^2}{ 1.7(5.9)^2} = \frac{17}{10} = 1.7[/tex]
Therefore, the ratio of her final kinetic energy to her initial kinetic energy is 1.7.
what type of image does
a dilated and a Constricted
pupil produce?
Answer:blue
Explanation:
I read it
All the questions are in the photos above. Thanks guys!
Answer:
right
Explanation:
In a little league baseball game, the 145 grams ball enters the strike zone with a speed of 11.0 meters per second. The batter hits the ball and it leaves his bat with a speed of 25.0 meters per second in exactly the opposite direction. If the bat is in contact with the ball for 1.0 m/s, what is the magnitude of the average force exerted by the bat on the ball?
Answer:
Force = 5.22 N
Explanation:
According to Newton's Second Law of motion:
[tex]Force = Rate\ of\ Change\ of\ Momentum\\\\Force = \frac{mv_f-mv_i}{t}\\[/tex]
where,
m = mass of ball = 145 g = 0.145 kg
vf = final speed of ball after hit = 25 m/s
vi = initial speed of ball before hit = - 11 m/s (negative sign due to opposite direction)
t = time of contact = 1 s
Therefore,
[tex]Force = \frac{(0.145\ kg)(25\ m/s)-(0.145\ kg)(-11\ m/s)}{1\ s} \\\\[/tex]
Force = 5.22 N
A 2:2 kg toy train is con ned to roll along a straight, frictionless track parallel to the x-axis. The train starts at the origin moving at a speed of 1:6m=s in the +x direction, and continues until it reaches a position 7:5m down the track from where it started. During its journey, it experiences a force pointing in the same direction as the vector 0:6 +0:8 , with magnitude initially 2:8N and decreasing linearly with its x-position to 0N when the train has finished its journey.
Required:
a. Calculate the work done by this force over the entire journey of the train.
b. Find the speed of the train at the end of its journey.
Answer:
a) 10.51 J
b) 3.48 m/s
Explanation:
Given data :
mass of train ( M ) = 2.2 kg
Given initial velocity ( u ) = 1.6 m/s
a) calculating work done by the force over the journey of the train
F = mx + b ------ ( 1 )
m = slope = ( Δ f / Δ x ) = 2.8 / -7.5 = - 0.373 N/m
x = distance travelled on the x axis by the train = 7.5 m
F = force experienced by the train = 2.8 N
x = 0
∴ b = 2.8
hence equation 1 can be written as
F = ( -0.373) x + 2.8 ----- ( 2 )
hence to determine the work done by the force
W = [tex]\int\limits^7_0 { ( -0.373) x + 2.8 )} \, dx[/tex] Note: the limits are actually 7.5 and 0
∴ W ( work done ) = -10.49 + 21 = 10.51 J
b) calculate the speed of the train at the end of its journey
we will apply the work energy theorem
W = 1/2 m*v^2 - 1/2 m*u^2
∴ V^2 = 2 / M ( W + 1/2 M*u^2 ) ( input values into equation )
V^2 = 12.11
hence V = 3.48 m/s
Police radar guns measure the speed of moving vehicles by transmitting electromagnetic waves at a vehicle and detecting a Doppler shift in the reflected wave. Suppose police radar transmits at a frequency of 24.0 GHz and receives a wave reflected from a car moving toward the radar at 65.0 mph. Find the frequency shift Df 5 fO 2 fS between the observed (received) and source (transmitted) frequencies.
Answer:
[tex]4649.20\ \text{Hz}[/tex]
Explanation:
u = Speed of car = 65 mph
[tex]f_s[/tex] = Frequency of source = 24 GHz
c = Speed of light = [tex]3\times 10^8\ \text{m/s}[/tex]
Beat frequency or frequency shift between the observed (received) and source (transmitted) frequencies is given by
[tex]f_o-f_s=\dfrac{2uf_s}{c}\\\Rightarrow \Delta f=\dfrac{2\times 65\times \dfrac{1609.34}{3600}\times 24\times 10^9}{3\times 10^8}\\\Rightarrow \Delta f=4649.20\ \text{Hz}[/tex]
The difference between te required frequencies is [tex]4649.20\ \text{Hz}[/tex].
A truck is traveling on a level road. The driver suddenly applies the brakes, causing the truck to decelerate by an amount g/2. This causes a box in the rear of the truck to slide forward. If the coefficient of sliding friction between the box and the truckbed is 2/5, find the acceleration of the box relative to the truck and relative to the road.
Answer:
Truck [tex]\dfrac{g}{10}[/tex]
Road [tex]-\dfrac{g}{10}[/tex]
Explanation:
[tex]a_1[/tex] = Acceleration of truck = [tex]-\dfrac{g}{2}[/tex]
[tex]\mu[/tex] = Coefficient of friction = [tex]\dfrac{2}{5}[/tex]
Frictional force is given by
[tex]f=-\mu mg\\\Rightarrow f=-\dfrac{2}{5}mg\\\Rightarrow ma_2=-\dfrac{2}{5}mg\\\Rightarrow a_2=-\dfrac{2}{5}g[/tex]
Net acceleration is given by
[tex]a=a_2-a_1\\\Rightarrow a=-\dfrac{2}{5}g+\dfrac{g}{2}\\\Rightarrow a=\dfrac{g}{10}[/tex]
The acceleration of the box relative to the truck is [tex]\dfrac{g}{10}[/tex] and [tex]-\dfrac{g}{10}[/tex] relative to the road.
what is the meaning of love ? ...What is love?
Answer:
love is a feeling that you get for people that you care about.
Explanation:
Answer:
Love is a feeling of affection for someone, when you adore somebody.
I wish true love and soul mates were real ( •_•)
A car is travelling at 27m/s and decelerates at a=5m/s2 for a distance of 10m. Calculate its final velocity. (Hint does deceleration imply that the acceleration is positive or negative?)[
Answer:
use the formula to calculate acceleration and you'll get the answers
If you were on a spaceship traveling at 0.50c away from a star, what speed
would the starlight pass you?
Explanation:
The speed of light in vacuum is the same by all observers (2nd Principle). You would find that the starlight passes you at c. This is more than a thousand times faster than existing space craft.
According to all observers, the velocity of light in a vacuum will be the same (2nd Principle). You'd discover that the starlight is passing you at c. Compared to current spacecraft, this really is a thousand times faster.
What is Speed?The amount of the shift in approach per unit of time or the size of the displacement over time for an object can be used to describe speed, which would be a scalar quantity in everyday language and kinematics.
The maximum speed that can be maintained when a time period grows closer to zero is the starting speed.
By dividing the object's distance traveled by the duration of the interval, the mean pace of the object for the given period of time is calculated. Speed and velocity are not always the same thing.
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Identical satellites X and Y of mass m are in circular orbits around a planet of mass M. The radius of the planet is R. Satellite X has an orbital radius of 3R, and satellite Y has an orbital radius of 4R. The kinetic energy of satellite X is Kx . Satellite X is moved to the same orbit as satellite Y by a force doing work on the satellite. In terms of Kx , the work done on satellite X by the force is
Answer:
The work down on satellite X by the force in terms of Kx is [tex]\dfrac{-K_x}{4}[/tex].
Explanation:
The work done is given as in terms of
[tex]W=\Delta TE[/tex]
Where ΔTE is the change in total energy.
This is given as
[tex]W=\Delta TE\\W=TE_y-TE_x\\W=\dfrac{-GMm}{2(4R)}-\dfrac{-GMm}{2(3R)}\\W=\dfrac{-GMm}{8R}+\dfrac{GMm}{6R}\\W=\dfrac{-6GMm+8GMm}{48R}\\W=\dfrac{2GMm}{48R}\\W=\dfrac{GMm}{24R}[/tex]
Rearranging it in terms of K_x gives
[tex]W=\dfrac{GMm}{24R}\\W=\dfrac{GMm}{-4\times -6R}\\W=\dfrac{1}{-4}\dfrac{-GMm}{6R}\\W=\dfrac{1}{-4}\dfrac{-GMm}{2(3R)}\\W=\dfrac{1}{-4}K_x\\W=\dfrac{-K_x}{4}[/tex]
Think of a hydropower dam . How is electrical energy produced from potential and kinetic energy ?
hydroelectric dam converts the potential energy stored in a water reservoir behind a dam to mechanical energy—mechanical energy is also known as kinetic energy. ... The generator converts the turbine's mechanical energy into electricity.
Hope this helps!
Answer:
Potential energy and kinetic energy are constituents of mechanical energy.
When a turbine is switched on, it rotates with mechanical energy.
Since a motor runs the turbine, it converts this mechanical energy to electrical energy.
What is perihelion and aphelion in short.
I dont know you could start by asking your teacher, parent, and or siblings.
Ps the last part was a joke they never help.
Answer:
I think that perihelion is the point of the Earth's course that is close to the Sun. And i'm sure that aphelion in short is astronomy or otherwise the point in the orbit of a planet.
Explanation:
Hope this helps :)
In medieval times it was believed that projectiles were pushed through the air until they reached their impetus.
a. True
b. Fals
Answer:
false
Explanation:
1.What is the Kinetic energy of a 3 kg object moving at 4 m/s?
Plz help I’ll give points
Answer:
24 J
Explanation:
[tex]K = \frac{1}{2} mv^{2} = \frac{1}{2} (3kg)(4m/s)^{2} = 24 J[/tex]
g A high-speed flywheel in a motor is spinning at 500 rpm when a power failure suddenly occurs. The flywheel has mass 39.0kg and diameter 78.0cm. The power is off for 34.0s, and during this time the flywheel slows due to friction in its axle bearings. During the time the power is off, the flywheel makes 170 complete revolutions.At what rate is the flywheel spinning when the power comes back on?
Answer:
[tex]10.54\ \text{rad/s}[/tex]
Explanation:
[tex]\omega_i[/tex] = Initial angular velocity = 500 rpm = [tex]500\times \dfrac{2\pi}{60}\ \text{rad/s}[/tex]
[tex]\omega_f[/tex] = Final angular velocity
t = Time = 34 s
[tex]\theta[/tex] = Angular displacement = 170 revs = [tex]170\times 2\pi\ \text{rad}[/tex]
[tex]\alpha[/tex] = Angulr acceleration
From the kinematic equations of angular motion we have
[tex]\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\Rightarrow \alpha=\dfrac{\theta-\omega_it}{\dfrac{1}{2}t^2}\\\Rightarrow \alpha=\dfrac{170\times 2\pi-500\times \dfrac{2\pi}{60}\times 34}{\dfrac{1}{2}\times 34^2}\\\Rightarrow \alpha=-1.23\ \text{rad/s}^2[/tex]
[tex]\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=500\times \dfrac{2\pi}{60}+(-1.23)\times 34\\\Rightarrow \omega_f=10.54\ \text{rad/s}[/tex]
The rate at which the wheel is spinning when the power comes back on is [tex]10.54\ \text{rad/s}[/tex].
If F = force, which equation illustrates the Law of Conservation of Momentum?
A) F1 = F2
B) F1 = - F2
C) - F1 = -F 2
D) F1 + - F2 = F3
Answer:
b
Explanation:
f1=-f2 that could be thank u
what kind of charge does an object have if it has extra positive charges
How does the presence of a nucleus provide a method of basic cell
classification? *
Answer:
The nucleus-containing cells are called eukaryotic cells. Eukaryote means having membrane-bound organelles.
Explanation:
I hope this is what you were looking for?!
Hope this helps!
Have a great day!
-Hailey!
The cells which have a nucleus are called Eukaryotic cells. The cells which do not have a nucleus are called prokaryotic cells.
What is a nucleus?
The nucleus is the controlling center of the body. It performs all the major activities in the cell. The cell which possesses a nucleus is called a eukaryote. The cell which does not have a nucleus is called prokaryote.
A prokaryotic cell is a straightforward, one-celled (unicellular) organism that is devoid of a nucleus or any other organelle that is membrane-bound. The nucleoid, a dark area in the center of the cell, is where prokaryotic DNA is located.
Therefore, The cells which have a nucleus are called Eukaryotic cells. The cells which do not have a nucleus are called prokaryotic cells.
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Plutonium-238 has a half life of 87.7 years. What percentage of a 5 kilogram (kg) sample remains after 50 years?
Answer:
i dont know but i should know try g o o g l e
Explanation:
Urgent!!!!! A student heated 235 g of water in a beaker until the water reached 100°C. The student removed the beaker from the heat and placed the beaker on a counter in a 23°C room. The student recorded the temperature of the water every 4 minutes for 20 minutes. The data are shown in the table. Estimate the average temperature of the air in the room at 20 min. explain your answer.
The average temperature of the air in the room at 20 min is 23°C.
What is temperature?Temperature is the degree of hotness or coldness of the object.
A student heated 235 g of water in a beaker until the water reached 100°C. The student removed the beaker from the heat and placed the beaker on a counter in a 23°C room. The student recorded the temperature of the water every 4 minutes for 20 minutes.
The surrounding is vast, its temperature does not get affected by small amount of water. So, the temperature of air remains constant.
Thus, the average temperature of the air in the room at 20 min is 23°C.
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A wheel has eight spokes and a radius of 30 cm. It is mounted on a fixed axle and is spinning at 2.5 rev/s. You want to shoot a 24- cm arrow parallel to this axle and through the wheel without hitting any of the spokes Assume that the arrow and the spokes hitting any of the spokes. Assume that the arrow and the spokes are very thin. (a) What minimum speed must the arrow have to pass through (a) What minimum speed must the arrow have to pass through without contact
Answer:
4.8 m/s
Explanation:
Given: angular velocity of wheel ω = 2.5 rev/sec
radius r = 30 cm
length of arrow = 24 cm
For arrow to pass through spinning ring it has to pass between any two spokes of the wheel.
angle between two spokes = π/4
time taken by a spook to reach the position of adjacent spoke t =θ/ω
= π/4/(2.5×2π) = 1/20 sec
for the arrow to pass through the spokes of the wheel it should take time t <1/20 sec to pass through the wheel
a) therefore, minimum speed = (24/100)/(1/20) = 4.8 m/s
A winch is capable of hauling a ton of bricks vertically two stories (6.35 m ) in 24.5 s .
If the winch’s motor is rated at 5.80 hp , determine its efficiency during raising the load.
Answer: 84 %
Explanation:
stored energy is _________ ___________
kinetic energy
energy in motion
potential energy
Answer:
Potential energy
Explanation:
Potential energy is stored energy
Extra CreditA particle is directed along the axis of the instrument in the gure. Aparallel plate capacitor sets up an electric eld E, which is orientedperpendicular to a uniform magnetic eld B. If the plates are separated byd= 2:0 mm and the value of the magnetic eld isB= 0:60T. Calculatethe potential di erence, between the capacitor plates, required to allow aparticle
This question is incomplete, the complete question is;
A particle is directed along the axis of the instrument in the figure below. A parallel plate capacitor sets up an electric field E, which is oriented perpendicular to a uniform magnetic field B. If the plates are separated by d = 2.0 mm and the value of the magnetic field is B = 0.60T.
Calculate the potential difference, between the capacitor plates, required to allow a particle with speed v = 5.0 × 10⁵ m/s to pass straight through without deflection.
Hint : ΔV = Ed
Answer:
the required potential difference, between the capacitor plates is 600 V
Explanation:
Given the data in the question;
B = 0.60 T
d = 2.0 mm = 0.002 m
v = 5.0 × 10⁵ m/s.
since particle pass straight through without deflection.
F[tex]_{net[/tex] = 0
so, F[tex]_E[/tex] = F[tex]_B[/tex]
qE = qvB
divide both sides by q
E = vB
we substitute
E = (5.0 × 10⁵) × 0.6
E = 300000 N/C
given that; potential difference ΔV = Ed
we substitute
ΔV = 300000 × 0.002
ΔV = 600 V
Therefore, the required potential difference, between the capacitor plates is 600 V
The owner of a van installs a rear-window lens that has a focal length of -0.304 m. When the owner looks out through the lens at a person standing directly behind the van, the person appears to be just 0.237 m from the back of the van, and appears to be 0.343 m tall. (a) How far from the van is the person actually standing
Answer:
[tex]p =-1.03[/tex]
Explanation:
From the question we are told that:
Focal length of lens [tex]f=-0.304 m[/tex]
Image distance [tex]q=0.237 m[/tex]
Height of image [tex]H_i=0.343[/tex]
Generally the lens equation is mathematically given by
[tex]\frac{1}{f} = \frac{1}{q} - \frac{1}{p}[/tex]
Where [tex]p[/tex] is Subject
[tex]p = \frac{(qf) }{(f - q)}[/tex]
[tex]p = \frac{(-0.237)(-0.304)) }{((-0.304) - (0.237))}[/tex]
[tex]p =-1.03[/tex]
Therefore the distance between the person and the car is
[tex]p =-1.03[/tex]
what does loudness of a sound depend on?
Answer:
Amplitude
Explanation:
The loudness of a sound depends on the amplitude of vibration producing the sound
A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surface of the water. It takes a time of 2.00 s for the boat to travel from its highest point to its lowest, a total distance of 0.600 m . The fisherman sees that the wave crests are spaced a horizontal distance of 6.40 m apart.
Required:
a. How fast are the waves traveling?
b. What is the amplitude of each wave?
c. If the total vertical distance traveled by the boat were 0.30 m but the other data remained the same, how would the answers to parts (a) and (b) be affected?
Answer:
a. Speed = 1.6 m/s
b. Amplitude = 0.3 m
c. Speed = 1.6 m/s
Amplitude = 0.15 m
Explanation:
a.
The frequency of the wave must be equal to the reciprocal of the time taken by the boat to move from the highest point to the highest point again. This time will be twice the value of the time taken to travel from the highest point to the lowest point:
frequency = [tex]\frac{1}{2(2\ s)}[/tex] = 0.25 Hz
The wavelength of the wave is the distance between consecutive crests of wave. Therefore,
Wavelength = 6.4 m
Now, the speed of the wave is given as:
Speed = (Frequency)(Wavelength)
Speed = (0.25 Hz)(6.4 m)
Speed = 1.6 m/s
b.
Amplitude is the distance between the mean position of the wave and the extreme position. Hence, it will be half the distance between the highest and lowest point:
Amplitude = (0.5)(0.6 m)
Amplitude = 0.3 m
c.
frequency = [tex]\frac{1}{2(2\ s)}[/tex] = 0.25 Hz
Speed = (Frequency)(Wavelength)
Speed = (0.25 Hz)(6.4 m)
Speed = 1.6 m/s
Amplitude = (0.5)(0.3 m)
Amplitude = 0.15 m
Saved Which of the following is NOT an important function of facial display? Multinio Choic
A. emotion
B. attractiveness
c. Primacy
d. identity
Answer:
C
Explanation:
Primacy means being first or important so thats not an important facial display as the others.
