Answer:
Bob's angular speed is the same as that of lily
Explanation:
Because for a carousel the angular speed remains the same since velocity at center and edge are the same
An object of mass 2 kg has a speed of 6 m/s and moves a distance of 8 m. What is its kinetic energy in joules?
Answer:
36 JoulesExplanation:
Mass ( m ) = 2 kg
Speed of the object (v) = 6 metre per second
Kinetic energy =?
Now,
We have,
Kinetic Energy = [tex] \frac{1}{2} \times m \times {v}^{2} [/tex]
Plugging the values,
[tex] = \frac{1}{2} \times 2 \times {(6)}^{2} [/tex]
Reduce the numbers with Greatest Common Factor 2
[tex] = {(6)}^{2} [/tex]
Calculate
[tex] = 36 \: joule[/tex]
Hope this helps...
Good luck on your assignment...
The Kinetic energy of the object will be "36 joules".
Kinetic energyThe excess energy of moving can be observed as that of the movement of an object, component, as well as the group of components. There would never be a negative (-) amount of kinetic energy.
According to the question,
Mass of object, m = 2 kg
Speed of object, v = 6 m/s
As we know the formula,
→ Kinetic energy (K.E),
= [tex]\frac{1}{2}[/tex] × m × v²
By substituting the values, we get
= [tex]\frac{1}{2}[/tex] × 2 × (6)²
= [tex]\frac{1}{2}[/tex] × 2 × 36
= 36 joule
Thus the above answer is appropriate.
Find out more information about Kinetic energy here:
https://brainly.com/question/25959744
A car has a mass of 1200 kg and an acceleration of 4 m/s^2. If the friction on the car is 200 N, how much force is the thrust providing?
Answer:
5000N
Explanation:
According to Newton's second law of motion, the net force (∑F) acting on a body is the product of the mass (m) of the body and the acceleration (a) of the body caused by the force. i.e
∑F = m x a -------------(i)
From the question, the net force is the combined effect of the thrust (F) and the friction force (Fₓ). i.e
∑F = F + Fₓ -------------(ii)
Where;
Fₓ = -200N [negative sign because the friction force opposes motion]
Combine equations(i) and (ii) together to get;
F + Fₓ = m x a
F = ma - Fₓ -------------(iii)
Where;
m = mass of car = 1200kg
a = acceleration of the car = 4m/s²
Now substitute the values of m, a and Fₓ into equation (iii) as follows;
F = (1200 x 4) - (-200)
F = 4800 + 200
F = 5000N
Therefore, the force the thrust is providing is 5000N
The only factor connecting horizontal and vertical components of projectile motion is _____.
Answer:
VelocityExplanation:
When a body is launched in air and allowed to fall freely under the influence of gravity, the motion experienced by the body is known as a projectile motion. The body is launched at a particular velocity and at an angle theta to the horizontal. The velocity of the body ca be resolved towards the horizontal component and the vertical component.
Along the horizontal Ux = Ucos(theta)
Along the vertical Uy = Ucos(theta)
Ux and Uy are the velocities of the body along the horizontal and vertical components respectively.
This means that the only factor connecting horizontal and vertical components of projectile motion is its velocity since we are able to calculate the velocity of the body along both components irrespective of its initial velocity.
Inside a 30.2 cm internal diameter stainless steel pan on a gas stove water is being boiled at 1 atm pressure. If the water level in the pan drops by 1.45 cm in 18.6 min, determine the rate of heat transfer to the pan in watts. (Give your answer in 3 significant digits.)
Answer:
Q = 20.22 x 10³ W = 20.22 KW
Explanation:
First we need to find the volume of water dropped.
Volume = V = πr²h
where,
r = radius of pan = 30.2 cm/2 = 15.1 cm = 0.151 m
h = height drop = 1.45 cm = 0.0145 m
Therefore,
V = π(0.151 m)²(0.0145 m)
V = 1.038 x 10⁻³ m³
Now, we find the mass of the water that is vaporized.
m = ρV
where,
m = mass = ?
ρ = density of water = 1000 kg/m³
Therefore,
m = (1000 kg/m³)(1.038 x 10⁻³ m³)
m = 1.038 kg
Now, we calculate the heat required to vaporize this amount of water.
q = mH
where,
H = Heat of vaporization of water = 22.6 x 10⁵ J/kg
Therefore,
q = (1.038 kg)(22.6 x 10⁵ J/kg)
q = 23.46 x 10⁵ J
Now, for the rate of heat transfer:
Rate of Heat Transfer = Q = q/t
where,
t = time = (18.6 min)(60 s/1 min) = 1116 s
Therefore,
Q = (23.46 x 10⁵ J)/1116 s
Q = 20.22 x 10³ W = 20.22 KW
A motorcyclist changes his speed from 20 km / h to 100 km / h in 3 seconds, maintaining a constant acceleration in that time interval. If the mass of the motorcycle is 200 kg and that of its rider is 80 kg, what is the value of the net force to accelerate the motorcycle? Help!
Answer:
2000 N
Explanation:
20 km/h = 5.56 m/s
100 km/h = 27.78 m/s
F = ma
F = m Δv/Δt
F = (200 kg + 80 kg) (27.78 m/s − 5.56 m/s) / (3 s)
F = 2074 N
Rounded to one significant figure, the force is 2000 N.
The gravitational energy of a swimmer on a driving board at different heights is shown in the table below. What is the driver's gravitational energy at 5m high? (A) 5500 J (B) 2750 J (C) 8800 J (D) 3300 J
Answer:
E = 2750 J at h = 5 m
Explanation:
The gravitational potential energy is given by :
[tex]E=mgh[/tex]
In this case, m is the mass of swimmer is constant at every heights. So,
At h = 1 m, E = 550 J
[tex]550=m\times 10\times 1\\\\m=55\ kg[/tex]
So, at h = 5 m, gravitational potential energy is given by :
[tex]E=55\times 10\times 5\\\\E=2750\ J[/tex]
So, the correct option is (B).
A child is sitting on the seat of a swing with ropes 10 m long. Their father pulls the swing back until the ropes make a 37o angle with the vertical and then releases the swing. If air resistance is neglected, what is the speed of the child at the bottom of the arc of the swing when the ropes are vertical
Answer:
Explanation:
We shall apply conservation of mechanical energy law to solve the problem .
loss of height = L ( 1 - cos 37 ) where L is length of rope
loss of potential energy at the bottom = gain of kinetic energy .
mg L ( 1 - cos 37 ) = 1/2 m v² where v is velocity at the bottom
v² = 2 L g ( 1 - cos 37 )
= 2 x 10 x 9.8 ( 1 - cos 37 )
= 39.46
v = 6.28 m /s
Think about a time you heard a sound come from another room in your house. How did those sound waves get to you if there were obstacles like walls in the way?
Answer:
this sound wave get to the other by either undergoing either reflection,absorption or transmission
Explanation:
When a sound wave meets an obstacle, some of the sound is reflected back from the front surface and some of the sound passes into the obstacle material, where it is absorbed or transmitted through the material like the building walls
Thus, Reflection and absorption are dependent on the wavelength of the sound. Hence percentage of the sound transmitted through an obstacle depends on how much sound is reflected and how much is absorbed.
A hungry 177 kg lion running northward at 81.8 km/hr attacks and holds onto a 32.0 kg Thomson's gazelle running eastward at 59.0 km/hr. Find the final speed of the lion–gazelle system immediately after the attack.
Answer:
The final speed of the lion-gazelle system immediately after the attack is 69.862 kilometers per hour.
Explanation:
Let suppose that lion and Thomson's gazelle are running at constant speed before and after collision and that collision is entirely inelastic. Given the absence of external force, the Principle of Momentum Conservation is applied such that:
[tex]\vec p_{L} + \vec p_{G} = \vec p_{F}[/tex]
Where:
[tex]\vec p_{L}[/tex] - Linear momentum of the lion, measured in kilograms-meters per second.
[tex]\vec p_{G}[/tex] - Linear momentum of the Thomson's gazelle, measured in kilograms-meters per second.
[tex]\vec p_{F}[/tex] - Linear momentum of the lion-Thomson's gazelle, measured in kilograms-meters per second.
After using the definition of momentum, the system is expanded:
[tex]m_{L}\cdot \vec v_{L} + m_{G}\cdot \vec v_{G} = (m_{L} + m_{G})\cdot \vec v_{F}[/tex]
Vectorially speaking, the final velocity of the lion-gazelle system is:
[tex]\vec v_{F} = \frac{m_{L}}{m_{L}+m_{G}}\cdot \vec v_{L} + \frac{m_{G}}{m_{L}+m_{G}}\cdot \vec v_{G}[/tex]
Where:
[tex]m_{L}[/tex], [tex]m_{G}[/tex] - Masses of the lion and the Thomson's gazelle, respectively. Measured in kilograms.
[tex]\vec v_{L}[/tex], [tex]\vec v_{G}[/tex], [tex]\vec v_{F}[/tex] - Velocities of the lion, Thomson's gazelle and the lion-gazelle system. respectively. Measured in meters per second.
If [tex]m_{L} = 177\,kg[/tex], [tex]m_{G} = 32\,kg[/tex], [tex]\vec v_{L} = 81.8\cdot j\,\left[\frac{km}{h} \right][/tex] and [tex]\vec v_{G} = 59.0\cdot i\,\left[\frac{km}{h} \right][/tex], the final velocity of the lion-gazelle system is:
[tex]\vec v_{F} = \frac{177\,kg}{177\,kg+32\,kg}\cdot \left(81.8\cdot j\right)\,\left[\frac{km}{h} \right] + \frac{32\,kg}{177\,kg+32\,kg}\cdot \left(59.0\cdot i\right)\,\left[\frac{km}{h} \right][/tex]
[tex]\vec v_{F} = 9.033\cdot i + 69.276\cdot j\,\left[\frac{km}{h} \right][/tex]
The speed of the system is the magnitude of the velocity vector, which can be found by means of the Pythagorean theorem:
[tex]\|\vec v_{F}\| = \sqrt{\left(9.033\frac{km}{h} \right)^{2}+\left(69.276\frac{km}{h} \right)^{2}}[/tex]
[tex]\|\vec v_{F}\| \approx 69.862\,\frac{km}{h}[/tex]
The final speed of the lion-gazelle system immediately after the attack is 69.862 kilometers per hour.
A parallel-plate air capacitor is made from two plates 0.200 m square, spaced 0.900 cm apart. It is connected to a 140 V battery.
A. What is the capacitance?
B. What is the charge on each plate?
C. What is the electric field between the plates?
D. What is the energy stored in the capacitor?
E. If the battery is disconnected and then the plates are pulled apart to a separation of 1.40 cm , what are the answers to parts A, B, C, and D?
Enter your answer as four numbers corresponding to C, Q, E, U. Please enter the answer in the given order and in the same units as in parts A, B, C, and D.
Answer:
See detailed solution below
Explanation:
a) From C= εoεrA/d
Where;
C= capacitance of the capacitor
εo= permittivity of free space
εr= relative permittivity
A= cross sectional area
d= distance between the plates
Since the relative permittivity of air=1 and permittivity of free space = 8.85 × 10^−12 Fm−1
Then;
C= 8.85 × 10^−12 Fm−1 × 0.2m^2/0.009 m
C= 196.67 × 10^-12 F or 1.967 ×10^-10 F
b) Q= CV = 1.967 ×10^-10 F × 140 V = 2.75 × 10^-8 C
c) E= V/d = 140 V/0.009m = 15.56 Vm-1
d) W= 1/2 CV^2 = 1/2 × 1.967 ×10^-10 F × (140)^2 =1.93×10^-6J
Part II
When the distance is now 0.014 m
a) C= 8.85 × 10^−12 Fm−1 × 0.2m^2/0.014 m = 1.26×10^-10 F
b) W= 1/2 Q^2/C = 1/2 × ( 2.75 × 10^-8 C)^2 / 1.26×10^-10= 3×10^-6 J
Note that the voltage changes when the distance is changed but the charge remains the same
How many components do vectors have, and what are they?
Answer:
There are two components for a two-dimensional coordinate system/vector.
Explanation:
For two-dimensional vectors, such as velocity, acceleraton, etc, there are two components, the x- and y-components.
These components could be rotated or translated, depending on the coordinate system.
Instead of rectangular cartesian system, the components could also be in the form of polar coordinates, such as radius and theta (angle).
For three-dimensional vectors, such as velocity in space, there are three components, in various coordinate systems.
Two parallel plates 0.800 cm apart are equally and oppositely charged. An electron is released from rest at the surface of the negative plate and simultaneously a proton is released from rest at the surface of the positive plate.
How far from the negative plate is the point at which the electron and proton pass each other?
Express your answer with the appropriate units.
Answer:
0.79 cm
Explanation:
The computation is shown below:-
Particle acceleration is
[tex]a = \frac{qE}{m}[/tex]
We will take d which indicates distance as from the negative plate, so the travel by proton is 0.800 cm - d at the same time
[tex]d = \frac{1}{2} a_et^2\\\\0.800 cm - d = \frac{1}{2} a_pt^2\\\\\frac{d}{0.800 cm - d} = \frac{a_e}{a_p} \\\\\frac{d}{0.800 cm - d} = \frac{m_p}{m_e} \\\\\frac{d}{0.800 cm - d} = \frac{1836m_e}{m_e}[/tex]
After solving the equation we will get 0.79 cm from the negative plate.
Therefore it is 0.79 cm far from the negative pate i.e the point at which the electron and proton pass each other
The point at which the electron and proton pass each other will be 0.79 cm.
What is the charge?When the matter is put in an electromagnetic field, it has an electric charge, which causes it to experience a force. A positive or negative electric charge can exist.
The given data in the problem is;
d' is the distance between the two parallel plates= 0.800 cm
The acceleration is given as;
[tex]\rm a= \frac{qE}{m} \\\\[/tex]
The distance from Newton's law is found as;
[tex]d = ut+\frac{1}{2} at^2 \\\\ u=0 \\\\ d= \frac{1}{2} at^2 \\\\ d-d' = \frac{1}{2} a_pt^2 \\\\ 0.800-d= \frac{1}{2} a_pt^2 \\\\\ \frac{d}{0.800-d} =\frac{a}{a_p} \\\\ \frac{d}{0.800-d} =\frac{m_p}{m} \\\\ \frac{d}{0.800-d} =\frac{1836m_e}{m_e} \\\\ d=0.79 \ cm[/tex]
Hence the point at which the electron and proton pass each other will be 0.79 cm.
To learn more about the charge refer to the link;
https://brainly.com/question/24391667
The time constant of a simple RL circuit is defined as _______. We say that R is the resistance of the circuit and L is the inductance of the circuit.
Answer:
The correct answer will be "[tex]\tau =\frac{L}{R}[/tex]".
Explanation:
The time it would take again for current or electricity flows throughout the circuit including its LR modules can be connected its full steady-state condition is equal to approximately 5[tex]\tau[/tex] as well as five-time constants.
It would be calculated in seconds by:
⇒ [tex]\tau=\frac{L}{R}[/tex]
, where
R seems to be the resistor function in ohms.L seems to be the inductor function in Henries.A scientist claims to be a solid-state physicist. She is working on electromagnetics and trying to see if she can create an electromagnetic field by causing different solutions to flow through a tube. Is this scientist a solid-state physicist?
no, because she is working in electromagnetics, which is not a solid-state field
no, because she is not working on the atomic structures of a solid
yes, because she is working in electromagnetics, which is a solid-state field
yes, because she is causing solutions to flow through a tube
The correct answer is B. No, because she is not working on the atomic structures of a solid
Explanation:
Solid-state physics is a sub-discipline of physics that focuses on studying solids, this includes analyzing solids structures, features, and other phenomena that occur in substances in this state of the matter. This means a solid-state physics will not study or gases.
In this context, the fact the scientist is trying to create an electromagnetic field by using solutions and the flow of these show the scientists is not working with solids but with liquids or gases as solids do not flow. Also, her focus is not solids, and therefore she is not a solid-state physicist. Thus, it can be concluded she is not a solid-state physicists because she is not working on the structures of solids.
The relationship between the Period (T) caused by the oscillation of the mass on the end of a hanging spring and the mass (m) is:
Answer:
T= 2p√m/k
Explanation:
This is because the period of oscillation of the mass of spring system is directly proportional to the square root of the mass and it is inversely proportional to the square root of the spring constant.
The period of a mass on a spring is given by the equation
T=2π√m/k.
Where T is the period,
M is mass
K is spring constant.
An increase in mass in a spring increases the period of oscillation and decrease in mass decrease period of oscillation.
When there is the relationship between the Period (T) caused by the oscillation of the mass should be considered as the T= 2p√m/k.
Oscillation of the mass:The mass of the spring system with respect to period of oscillation should be directly proportional to the square root of the mass and it is inversely proportional to the square root of the spring constant.
So the following equation should be considered
T=2π√m/k.
Here,
T is the period,
M is mass
K is spring constant.
An increase in mass in a spring rises the period of oscillation and reduce in mass decrease period of oscillation.
Learn more about mass here: https://brainly.com/question/21860379
A wave traveling in a string in the positive x direction has a wavelength of 35 cm, an amplitude of 8.4 cm, and a period of 1.2 s. What is the wave equation (in base SI units) that correctly describes this wave
Explanation:
It is given that,
Wavelength of a wave, [tex]\lambda=35\ cm=0.35\ m[/tex]
Amplitude, [tex]A=8.4\ cm[/tex]
Period of the wavelength, t = 1.2 s
The wave is traveling in a string in the positive x direction. We need to write the wave equation for this conditions.
The general equation of the wave when it is traveling in +x direction is given by :
[tex]y=A\sin(kx-\omega t)[/tex]
A is amplitude
k is propagation constant
[tex]k=\dfrac{2\pi}{\lambda}\\\\k=\dfrac{2\pi}{0.35}\\\\k=17.95\ m^{-1}\approx 18\ m^{-1}[/tex]
[tex]\omega[/tex] is angular frequency
[tex]\omega=\dfrac{2\pi}{T}\\\\\omega=\dfrac{2\pi}{1.2}\\\\\omega=5.23\ s^{-1}[/tex]
So, the wave equation is given by :
[tex]y(x,t)=(0.084) \sin (18 x - 5.2 t)[/tex]
Hence, this is the required solution.
Zuckerman’s test for sensation seeking measures which of the following characteristics?
dangerousness, antisocial traits, “letting loose,’ and intolerance for boredom
thrill and adventure seeking, experience seeking, disinhibition, and susceptibility to boredom
adventurousness, physical prowess, creative morality, and charisma
dangerousness, adventurousness, creativity, and thrill and adventure seeking
The correct answer is B. thrill and adventure seeking, experience seeking, disinhibition, and susceptibility to boredom
Explanation:
Marvin Zuckerman was an important American Psychologists mainly known for his research about personality and the creation of a model to study this aspect of human psychology. This model purposes five factors define personality, these are the thrill and adventure-seeking that involves seeking for adventures and danger; experience seeking that implies a strong interest in participating in new activities; disinhibition that implies being open and extrovert; and susceptibility to boredom that implies avoiding boredom or repetition. Thus, option B correctly describes the characteristics used in Zuckerman's test.
g If the interaction of a particle with its environment restricts the particle to a finite region of space, the result is the quantization of ____ of the particle.
Answer:
the result is the quantization of __Energy__ of the particle
Explanation:
While running, a person dissipates about 0.60 J of mechanical energy per step per kilogram of body mass. If a 52-kg person develops a power of 80 W during a race, how fast is the person running
Answer:
The person is running at a speed of 2.564 m/s
Explanation:
Given;
mechanical energy dissipated per kilogram per step, E/kg/S = 0.6 J/kg/S
mass of the person, m = 52 kg
power developed by the person, P = 80 W
mechanical energy of the person per step, E = 0.6 J/kg x 52 kg
[tex]E_{step}[/tex] = 31.2 J
mechanical energy for the total step, [tex]E_{total}[/tex] = 31.2 J x S
P = E / t
[tex]P_{avg} = \frac{E_{total}}{t} \\\\P_{avg} = \frac{E_{step}*S}{t}\\\\\frac{P_{avg}}{E_{step}} = \frac{S}{t} \\\\\frac{S}{t} = \frac{80}{31.2} \\\\\frac{S}{t} = 2.564 \ m/s[/tex]
Therefore, the person is running at a speed of 2.564 m/s
A speed skater moving to the left across frictionless ice at 8.6 m/s hits a 4.7-m-wide patch of rough ice. She slows steadily, then continues on at 5.3 m/s.
What is her acceleration on the rough ice?
Answer:
a = 4.88 m / s²
Explanation:
We can solve this exercise using the expressions of kinematics in one dimension
v² = v₀² - 2a x
where v is the velocity, v₀ is the initial velocity, at acceleration and ax is the distance, the negative sign is because the velocity decreases because
a = (v₀² - v²) / 2x
let's calculate
a = (8.6² - 5.3²) / (2 4.7)
a = 4.88 m / s²
3. A particle of charge +7.5 µC is released from rest at the point x = 60 cm on an x-axis. The particle begins to move due to the presence of a charge ???? that remains fixed at the origin. What is the kinetic energy of the particle at the instant it has moved 40 cm if a) ???? = +20 µC and b) ???? = −20 µC?
Answer:
HSBC keen vs kg get it yyyyyuuy
Explanation:
hgccccxfcffgbbbbbbbbbbghhyhhhgdghcjyddhhyfdghhhfdgbxbbndgnncvbhcxgnjffccggshgdggjhddh
nnnbvvvvvggfxrugdfutdfjhyfggigftffghhjjhhjyhrdffddfvvvvvvvvvvvbbbbbbbbbvvcxccghhyhhhjjjhjnnnnnnnnnnnnnbhbfgjgfhhccccccvvjjfdbngxvncnccbnxcvbchvxxghfdgvvhhihbvhbbhhvxcgbbbcxzxvbjhcxvvbnnxvnn
B. A car moving at an initial speed vi applies its brakes and skids for some distance until coming to a complete stop. If the coefficient of kinetic friction between tires and the road is µk what distance did the car skid?
Complete Question
The complete question is shown on the first uploaded image
Answer:
The distance which the car skid is [tex]l = \frac{v_i^2 }{2 * \mu_k * g }[/tex]
Explanation:
From the question we are told that
The initial velocity of the car is [tex]v_i[/tex]
The coefficient of kinetic friction is [tex]\mu_k[/tex]
According to the law of energy conservation
The initial Mechanical Energy = The final Mechanical Energy
[tex]M_i = M_f[/tex]
The initial mechanical energy is mathematically represented as
[tex]M_i = KE _o + PE_e[/tex]
where KE is the initial kinetic energy which is mathematically represented as
[tex]KE = \frac{1}{2} m v_i^2[/tex]
And PE is the initial potential energy which is zero given that the car is on the ground
Now
[tex]M_f = W_{\mu}[/tex]
Where [tex]W_{\mu}[/tex] is the work which friction exerted on the car which is mathematically represented as
[tex]W_{\mu} = m* \mu_k * g * l[/tex]
Where [tex]l[/tex] is the distance covered by the car before it slowed down
[tex]\frac{1}{2} m v_i^2 = m* \mu_k * g * l[/tex]
=> [tex]l = \frac{v_i^2 }{2 * \mu_k * g }[/tex]
Two small charged spheres are 7.59 cmcm apart. They are moved, and the force on each of them is found to have been tripled. How far apart are they now?
Answer:
The two small charged spheres are now 4.382 cm apart
Explanation:
Given;
distance between the two small charged sphere, r = 7.59 cm
The force on each of the charged sphere can be calculated by applying Coulomb's law;
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
where;
F is the force on each sphere
q₁ and q₂ are the charges of the spheres
r is the distance between the spheres
[tex]F = \frac{kq_1q_2}{r^2} \\\\kq_1q_2 = Fr^2 \ \ (keep \ kq_1q_2 \ constant)\\\\F_1r_1^2 = F_2r_2^2\\\\r_2^2 = \frac{F_1r_1^2}{F_2} \\\\r_2 = \sqrt{\frac{F_1r_1^2}{F_2}} \\\\r_2 = r_1\sqrt{\frac{F_1}{F_2}}\\\\(r_1 = 7.59 \ cm, \ F_2 = 3F_1)\\\\r_2 = 7.59cm\sqrt{\frac{F_1}{3F_1}}\\\\r_2 = 7.59cm\sqrt{\frac{1}{3}}\\\\r_2 = 7.59cm *0.5773\\\\r_2 = 4.382 \ cm[/tex]
Therefore, the two small charged spheres are now 4.382 cm apart.
Earth orbiting the Sun The Earth is 1.5 ⋅ 10 8 km from the Sun (on average). How fast is the Earth orbiting the Sun in kilometers per second (on average)? You can assume the orbit of the Earth is a circle and that the circumference of a circle is equal to C = 2 π R where R is the radius of a circle (the distance between the center and the edge. Note that for our purposes, it is perfectly fine to assume π = 3 which allows for a pretty good approximation C = 6 R . Your answer does not need to be put into scientific notation, but if you choose to do so it will be marked correct! kilometers per second
Answer:
1 yr = 24 * 3600 * 365 = 3.2 * 10E7 sec
C = 6 R = 1.5 * 10E8 * 6 = 9 * 10E8 km circumference of orbit
v = C / t = 9 * 10E8 km / 3 * 10E7 sec = 30 km / sec = 18 mi/sec
A car is traveling down a highway. It was moving with a velocity of 50m/s when the driver reads the speed limit and has to decelerate with an acceleration of -5m/s for 2 seconds. What is the momentum of this 500kg car after it decelerates?
Answer:
20,000 kg m/s
Explanation:
Given:
v₀ = 50 m/s
a = -5 m/s²
t = 2 s
Find: v
v = at + v₀
v = (-5 m/s²) (2 s) + (50 m/s)
v = 40 m/s
p = mv
p = (500 kg) (40 m/s)
p = 20,000 kg m/s
What value of D is required to make vt = 42.7 m/s the terminal velocity of a skydiver of mass 85.0 kg . Express your answer using two significant figures.
Complete Question
For a human body falling through air in a spread edge position , the numerical value of the constant D is about [tex]D = 0.2500 kg/m[/tex]
What value of D is required to make vt = 42.7 m/s the terminal velocity of a skydiver of mass 85.0 kg . Express your answer using two significant figures?
Answer:
The value of D is [tex]D = 0.457 \ kg/m[/tex]
Explanation:
From the question we are told that
The terminal velocity is [tex]v_t = 42.7 \ m/s[/tex]
The mass of the skydiver is [tex]m = 85.0 \ kg[/tex]
The numerical value of D is [tex]D = 0.2500 kg/m[/tex]
From the unit of D in the question we can evaluate D as
[tex]D = \frac{m * g }{v^2}[/tex]
substituting values
[tex]D = \frac{85 * 9.8 }{(42.7)^2}[/tex]
[tex]D = 0.457 \ kg/m[/tex]
A wave is described by the equation y(x, t) = 35 sin (2 π x - 3 t +1.5), where all numerical values and variables have the appropriate units to produce displacement in cm, speed in cm/s, and time in seconds. What is the speed of the wave?
Answer:
v = 0.477m/s
Explanation:
You have the following wave function:
[tex]y(x,t)=35sin(2\pix-3t+1.5)[/tex] (1)
where y is the vertical displacement of the wave for the position x.
The general form of a wave function can be written as follow:
[tex]y(x,t)=Asin(kx-\omega t+\phi)[/tex] (2)
by comparing the equation (2) and (1) you have:
A: amplitude of the wave = 35
k: wave number = 2π
w: angular frequency of the wave = 3
φ: phase of the wave = 1.5
The speed of the wave is given by the following formula:
[tex]v=\frac{\omega}{k}[/tex]
you replace the values of the parameters in the previous formula:
[tex]v=\frac{3}{2\pi}=0.477\frac{m}{s}[/tex]
The speed of the wave is 0.477m/s
Match each term to the best description.
1. Composed of numerous narrowly spaced parallel slits or grooves
2. Having the same wavelength, frequency, and in-phase Interaction of waves where they meet in space
3. The bending of waves near a boundary or as a wave passes through an opening
4. The zeroth order direct reflection fringe
a. Coherent
b. Diffraction
c. Grating
d. Specular dot
e. Interference
Answer:
1. Grating
2. Interference
3. Diffraction
4. Specular dot
Explanation:
1. Composed of numerous narrowly spaced slits and grooves ........ Grating
2. Having the same wavelength, frequency, and in-phase Interaction of waves where they meet in space ....... Interference
3. The bending of waves near a boundary or as a wave passes through an opening ...... Diffraction
4. The zeroth order direct reflection fringe ...... Specular dot
A particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. Find the velocity and the speed when t = 4. f(t) = 12 + 35 t + 1
A particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. Find the velocity and the speed when t = 4. f(t) = 12t² + 35 t + 1
Answer:
Velocity = 131 m/s
Speed = 131 m/s
Explanation:
Equation of motion, s = f(t) = 12t² + 35 t + 1
To get velocity of the particle, let us find the first derivative of s
v (t) = ds/dt = 24t + 35
At t = 4
v(4) = 24(4) + 35
v(4) = 131 m/s
Speed is the magnitude of velocity. Since the velocity is already positive, speed is also 131 m/s
A street light is at the top of a pole that has a height of 17 ft . A woman 5 ft tall walks away from the pole with a speed of 8 ft/s along a straight path. How fast is the tip of his shadow moving when he is 40ft from the pole?
Answer:
8 ft/s
Explanation:
This is a straight forward question without much ado.
It is given from the question that she walks with a speed of 8 ft/s