Answers
b). The power depends on the RATE at which work is done.
Power = (Work or Energy) / (time)
So to calculate it, you have to know how much work is done AND how much time that takes.
In part (a), you calculated the amount of work it takes to lift the car from the ground to Point-A. But the question doesn't tell us anywhere how much time that takes. So there's NO WAY to calculate the power needed to do it.
The more power is used, the faster the car is lifted. The less power is used, the slower the car creeps up the first hill. If the people in the car have a lot of time to sit and wait, the car can be dragged from the ground up to Point-A with a very very very small power ... you could do it with a hamster on a treadmill. That would just take a long time, but it could be done if the power is small enough.
Without knowing the time, we can't calculate the power.
...
d). Kinetic energy = (1/2) · (mass) · (speed squared)
On the way up, the car stops when it reaches point-A.
On the way down, the car leaves point-A from "rest".
WHILE it's at point-A, it has no speed. So it has no (zero) kinetic energy.
Related Questions
Sam heaves a 16lb shot straight upward, giving it a constant upward acceleration from rest of 35 m/s^2 for 64.0 cm. He releases it 2.20m above the ground. You may ignore air resistance.
(a) What is the speed of the shot when Sam releases it?
(b) How high above the ground does it go?
(c ) How much time does he have to get out of its way before it returns to the height of the top of his head, 1.83 m above the ground?
Answers
Answer:
6.69 m/s
4.483 m
1.42s
Explanation:
Given that:
Initial Velocity, u = 0
Final velocity, v =?
Acceleration, a = 35m/s²
1.) using the relation :
v² = u² + 2as
v² = 0 + 2(35) * 64*10^-2m
v² = 70 * 0.64
v = sqrt(44.8)
v = 6.693
v = 6.69 m/s
B.) height from the ground, h0 = 2.2
How high ball went , h:
Using :
v² = u² + 2as
Upward motion, g = - ve
0 = 6.69² + 2(-9.8)*(h - 2.2)
0= 6.69² - 19.6(h - 2.2)
44.7561 + 43.12 - 19.6h = 0
19.6h = 44.7561 - 43.12
h = 87.8761 / 19.6
h = 4.483 m
C.)
vt - 0.5gt² = h - h0
6.69t - 0.5(9.8)t²
6.69t - 4.9t² = 1.83 - 2.2
-4.9t² + 6.69t + 0.37 = 0
Using the quadratic equation solver :
Taking the positive root:
1.4185 = 1.42s
state four law of photoelectric effect
Answers
Answer:
LAW 1 : For a given metal and frequency, the number of photoelectrons emitted is directly proportional to the intensity of the incident radiation.
---------------------------------------------
LAW 2: For a given metal, there exists a certain frequency below which the photoelectric emission does not take place. This frequency is called threshold frequency.
-----------------------------------------------
LAW 3: For a frequency greater than the threshold frequency, the kinetic energy of photoelectrons is dependent upon frequency or wavelength but not on the intensity of light.
-----------------------------------------------
LAW 4: Photoelectric emission is an instantaneous process. The time lag between incidence of radiations and emission of electron is 10^-9 seconds.
Explanation:
Answer:
LAW 1 : For a given metal and frequency, the number of photoelectrons emitted is directly proportional to the intensity of the incident radiation. ... LAW 4: Photoelectric emission is an instantaneous process.
I don’t even understand anyone help please.
Answers
Answer:
a) A:170572.5 J
C: 55794.9J
b) 170572.5 J
c) 41.4413265306m
d) 2.7455874717m/s
Explanation:
a) Kinetic energy = 0.5*m*v²
KE at A = 0.5*420*28.5² = 170572.5 J
KE at C = 0.5*420*16.3² = 55794.9 J
b) Mechanical energy is the total kinetic energy plus potential energy at a point on the system. There is no potential energy at A.
ANSWER: 170572.5 J
c) v²=u²+2as
28.5²=2(9.8)s
812.25/19.6=s
s=41.4413265306m
d) h=height from part c, r=radius of loop
v²=u²+2as
v²=gr or a=v²/r
Ei=Ef
mgh=0.5mv²+mg(2r)
gh=0.5v²+2gr
h=0.5r+2r
h=5/2r
r=2/5h=(2/5)(41.4413265306)=16.5765306122
F=ma
mg=m(v²/r)
g=v²/r
v²=(9.8)(16.5765306122)
v=√162.45
=12.7455874717m/s
A daredevil is attempting to jump his motorcycle over a line of buses parked end to end by driving up a ramp of 64.8 degrees at a speed of 25.4 m/s. What would be the largest number of buses he can clear if the top of the takeoff ramp is at the same height as the bus tops and the buses are 10.0 m long
Answers
Answer: he can only make it over 5 buses
Explanation:
Given the data in the question;
we know that range is expressed as;
R = (V₀²sin2∅₀)/g
V₀ is the initial velocity( 25.4 m/s), ∅₀ is the angle of projection( 64.8°), g is acceleration due to gravity( 9.8 m/s²),
so we substitute
R = ((25.4)²sin2(64.8))/9.8
R = 50.7 m
now, them number of buses will be;
n = R / bus length
given that bus length is 10.0 m
we substitute
n = 50.7 m / 10.0
n = 5.07 ≈ 5
Therefore, he can only make it over 5 buses
What is average acceleration due to gravity on Earth for a 2000 kg boulder, in proper SI units?
Answers
Answer:
9.8m/s²
Explanation:
The average acceleration due to gravity on Earth for a 2000kg boulder is 9.8m/s².
Every object on earth is accelerated towards the center by a rate of change of velocity with time value of 9.8m/s².
The acceleration due to gravity on earth is a constant value from places to places.
For other planetary bodies, the value varies and it differs.
But on earth every object is accelerated at 9.8m/s².
An electric range has a constant current of 10 A entering the positive voltage terminal with a voltage of 110 V. The range is operated for two hours, (a) Find the charge in coulombs that passes through the range, (b) Find the power absorbed by the range, (c) If electric energy costs 12 cents per kilowatt-hour, determine the cost of operating the range for two hours.
Answers
Answer:
A. 72000 C
B. 1100 W
C. 26.4 cents.
Explanation:
From the question given above, the following data were obtained:
Current (I) = 10 A
Voltage (V) = 110 V
Time (t) = 2 h
A. Determination of the charge.
We'll begin by converting 2 h to seconds. This can be obtained as follow:
1 h = 3600 s
Therefore,
2 h = 2 h × 3600 s / 1 h
2 h = 7200 s
Thus, 2 h is equivalent to 7200 s.
Finally, we shall determine the charge. This can be obtained as follow:
Current (I) = 10 A
Time (t) = 7200 s
Charge (Q) =?
Q = It
Q = 10 × 7200
Q = 72000 C
B. Determination of the power.
Current (I) = 10 A
Voltage (V) = 110 V
Power (P) =?
P = IV
P = 10 × 110
P = 1100 W
C. Determination of the cost of operation.
We'll begin by converting 1100 W to KW. This can be obtained as follow:
1000 W = 1 KW
Therefore,
1100 W = 1100 W × 1 KW / 1000 W
1100 W = 1.1 KW
Thus, 1100 W is equivalent to 1.1 KW
Next, we shall determine the energy consumption of the range. This can be obtained as follow:
Power (P) = 1.1 KW
Time (t) = 2 h
Energy (E) =?
E = Pt
E = 1.1 × 2
E = 2.2 KWh
Finally, we shall determine the cost of operation. This can be obtained as follow:
1 KWh cost 12 cents.
Therefore, 2.2 KWh will cost = 2.2 × 12
= 26.4 cents.
Thus, the cost of operating the range for 2 h is 26.4 cents.
7. If the impact of the golf club on the ball in the previous question occurs over a time of 2 x 10 seconds, what
force does the ball experience to accelerate from rest to 73 m/s?
Answers
Answer:
3.65 x mass
Explanation:
Given parameters:
Time = 20s
Initial velocity = 0m/s
Final velocity = 73m/s
Unknown:
Force the ball experience = ?
Solution:
To solve this problem, we apply the equation from newton's second law of motion:
F = m [tex]\frac{v - u}{t}[/tex]
m is the mass
v is the final velocity
u is the initial velocity
t is the time taken
So;
F = m ([tex]\frac{73 - 0}{20}[/tex] ) = 3.65 x mass
How do you calculate area when pressure and force are given to you
Answers
Answer:
This is my answer
Explanation:
First convert 150 kPa to Pa:
150 × 1,000 = 150,000.
Next substitute the values into the equation:
force normal to a surface area = pressure × area of that surface.
force = 150,000 × 180.
force = 27,000,000 N.
1.First convert 150kPato Pa:
2.150 x 1,000 + 150,000
3.next substitute the values into the equations:
4.force normal to a surface area =pressure x area of that surface.
5.force=150,000 x 180.
6.force = 27,000,000N.
can i have brainliest please
What is displacement?
a. The distance an object travels.
b. The distance between the starting point and the ending point of an object's
journey.
C. The amount of time it takes an object to travel to a destination.
d. The path in which an object travels.
Answers
Answer:
displacement is the distance between the starting point and the ending point of an object's journey
2.19 The drag characteristics of a blimp traveling at 4 m/s are to be studied by experiments in a water tunnel. The prototype is 20 m in diameter and 110 m long. The model is one-twentieth scale. What velocity must the model have for dynamic similarity
Answers
Answer:
[tex]Vm=0.894m/s[/tex]
Explanation:
From the question we are told that
Velocity if travel [tex]v=4m/s[/tex]
Diameter of prototype [tex]d_1=20m[/tex] and [tex]d_2=110m[/tex]
Scale ratio=[tex]\frac{1}{20}[/tex]
Generally Velocity of of the model using Froud's model is mathematically given as
[tex]Fm=Fp[/tex]
[tex]\frac{Vm}{\sqrt{Lmg}} =\frac{Vp}{\sqrt{Lpg}}[/tex]
[tex]Vm=Vp*\frac{Vp}{\sqrt{Lpg} }[/tex]
[tex]Vm=4*\frac{1}{\sqrt{20}}[/tex]
[tex]Vm=0.894m/s[/tex]
As a laudably skeptical physics student, you want to test Coulomb's law. For this purpose, you set up a measurement in which a proton and an electron are situated 865 nm from each other and you study the forces that the particles exert on each other. As expected, the predictions of Coulomb's law are well confirmed. You find that the forces are attractive and the magnitude of each force is:______
Answers
Answer:
force F = 1.66 × [tex]10^{-13}[/tex] N
Explanation:
given data
proton and an electron = 865 nm
solution
we get here force that is express as
force F = k q1 q2 ÷ r² ......................1
put here value and we get
force F = 9 × [tex]10^{9}[/tex] × [tex]\frac{1.6\times (10^{-19})^{2}}{865 \times (10^{-9})^{2}}[/tex]
force F = 1.66 × [tex]10^{-13}[/tex] N
The pickup truck has a changing velocity because the pickup truck
A.can accelerate faster than the other two vehicles
B.is traveling in the opposite direction from the other two vehicles
C.is traveling on a curve in the road
D.needs a large amount of force to move
please get right i need awnser today
Answers
Answer:
C. Is traveling on a curve in the road
Hope this helps :3
The pick up truck has a changing velocity because, it is travelling on a curve in the road. A change in direction results in its change in velocity because, velocity is a vector quantity.
What is velocity ?Velocity is a physical quantity that measures the distance covered by an object per unit time. It is a vector quantity, thus having magnitude as well direction.
The rate of change in velocity is called acceleration of the object. Like velocity, acceleration also is a vector quantity. Thus, a change in magnitude or direction or change in both for velocity make the object to accelerate.
Here, all the three vehicles are travelling with the same velocity. But, the truck is moving to a curve on the road. The curvature in the path will make a change in its velocity.
Find more on velocity:
https://brainly.com/question/16379705
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The image related with this question is attached below:
A circus tightrope walker weighing 800 N is standing in the middle of a 15 meter long cable stretched between two posts. The cable was originally horizontal. The lowest point of the cable is now at his feet and is 30 cm below the horizontal. Assume the cable is massless. What is the tension in the cable
Answers
Answer:
T = 10010 N
Explanation:
To solve this problem we must use the translational equilibrium relation, let's set a reference frame
X axis
Fₓ-Fₓ = 0
Fₓ = Fₓ
whereby the horizontal components of the tension in the cable cancel
Y Axis
[tex]F_{y} + F_{y} - W =0[/tex]
2[tex]F_{y}[/tex] = W
let's use trigonometry to find the angles
tan θ = y / x
θ = tan⁻¹ (0.30 / 0.50 L)
θ = tan⁻¹ (0.30 / 0.50 15)
θ = 2.29º
the components of stress are
F_{y} = T sin θ
we substitute
2 T sin θ = W
T = W / 2sin θ
T = [tex]\frac{ 800}{ 2sin 2.29}[/tex]
T = 10010 N
While investigating Kirchhoff's Laws, you begin observing a blackbody, such as a star, from Earth using advanced technology that can analyze spectra. While pointing it at the star with nothing between you and the star, you observe a full spectrum. You come back and repeat this same experiment a year later using the same star, except this time you observe an absorption spectrum. What is the most likely explanation for this
Answers
Answer:
the second time there is a gas between you and the star,
Explanation:
When you observe the star for the first time you do not have a given between you and the star, therefore you observe the emission spectrum of the same that is formed by lines of different intensity and position that indicate the type and percentage of the atoms that make up the star.
When you observe the same phenomenon for the second time there is a gas between you and the star, this gas absorbs the wavelengths of the star that has the same energies and the atomisms and molecular gas, therefore these lines are not observed by seeing a series of dark bands,
The information obtained from the two spectra is the same, the type of atoms that make up the star
calculate ine gravitational potential energy of the ball using pe=m×g×h.(use g=9.8 n/kg)
A 4.0-kilogram ball held 1.5 meters above the floor has ________ joules of potential energy
Answers
Answer:
58.8J
Explanation:
Given parameters;
Mass of ball = 4kg
Height above the floor = 1.5m
g = 9.8n/kg
Unknown:
Potential energy = ?
Solution:
The potential energy of a body is the energy due to the position of the body.
It is mathematically expressed as:
Potential energy = mass x acceleration due to gravity x height
Potential energy = 4 x 9.8 x 1.5 = 58.8J
On a low-friction track, a 0.66-kg cart initially going at 1.85 m/s to the right collides with a cart of unknown inertia initially going at 2.17 m/s to the left. After the collision, the 0.66-kg cart is going at 1.32 m/s to the left, and the cart of unknown inertia is going at 3.22 m/s to the right. The collision takes 0.010 s.
What is the unknown inertia?
What is the average acceleration of the heavier cart?
What is the average acceleration of the lighter cart?
Answers
Answer:
(a) the unknown inertia is 0.388 kg
(b) the average acceleration of the heavier cart is 317 m/s²
(c) the average acceleration of the lighter cart is 539 m/s²
Explanation:
Given;
mass of the first cart, m₁ = 0.66 kg
initial speed of the first cart, u₁ = 1.85 m/s
let the mass of the cart with unknown inertia be m₂
initial velocity of the second cart, u₂ = 2.17 m/s to the left
velocity of the first cart after collision, v₁ = 1.32 m/s to the left
velocity of the second cart after collision, v₂ = 3.22 m/s
time of collision, t = 0.010 s
(a) What is the unknown inertia?
Apply the principle of conservation of linear momentum, to determine the unknown inertia.
let leftward direction be negative direction
let rightward direction be positive direction
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
0.66(1.85) + m₂(-2.17) = 0.66(-1.32) + m₂(3.22)
1.221 - 2.17m₂ = -0.8712 + 3.22m₂
1.221 + 0.8712 = 3.22m₂ + 2.17m₂
2.0922 = 5.39m₂
m₂ = 2.0922 / 5.39
m₂ = 0.388 kg
The unknown inertia is 0.388 kg
(b) the average acceleration of the heavier cart
the heavier cart has a mass of 0.66 kg
[tex]a = \frac{v_1 - u_1}{t} \\\\a = \frac{-1.32 - 1.85}{0.01} \\\\a = -317 \ m/s^2\\\\|a| = 317 \ m/s^2[/tex]
(c) the average acceleration of the lighter cart;
the lighter cart has a mass of 0.388 kg
[tex]a = \frac{v_2 - u_2}{t} \\\\a = \frac{3.22 - (-2.17)}{0.01} \\\\a =\frac{3.22 \ +\ 2.17}{0.01} \\\\a= 539\ m/s^2[/tex]
What energy store is in the torch
BEFORE it gets switched on?
Answers
Answer:
Chemical energy
Explanation:
The energy in the torch is stored as chemical energy before the torch gets switch on.
The chemical energy energy in the battery of cell will power the cell and allows it to produce light.
Chemical energy is a form of potential energy. The electrolytes within the battery are capable of producing electric current. So the chemical energy is transformed into electrical energy which is used to produce the light of the torch.Required
Momentum
The magnitude of the momentum of an object is 64 kg*m/s. If the velocity of the
object is doubled, what will be the magnitude of the momentum of the object? *
32 kg*m/s
64 kg*m/s
128 kg*m/s
256 kg*m/s
Answers
Answer:
C) 128 kg*m/s
Explanation:
When you double something you multiply it by 2 most of the time. 64*2=128 or you can add it 64+64=128. Hope this helps.
A car enters a 105-m radius flat curve on a rainy day when the coefficient of static friction between its tires and the road is 0.4. What is
the maximum speed which the car can travel around the curve without sliding
Answers
Static friction (magnitude Fs) keeps the car on the road, and is the only force acting on it parallel to the road. By Newton's second law,
Fs = m a = W a / g
(a = centripetal acceleration, m = mass, g = acceleration due to gravity)
We have
a = v ² / R
(v = tangential speed, R = radius of the curve)
so that
Fs = W v ² / (g R)
Solving for v gives
v = √(Fs g R / W)
Perpendicular to the road, the car is in equilibrium, so Newton's second law gives
N - W = 0
(N = normal force, W = weight)
so that
N = W
We're given a coefficient of static friction µ = 0.4, so
Fs = µ N = 0.4 W
Substitute this into the equation for v. The factors of W cancel, so we get
v = √((0.4 W) g R / W) = √(0.4 g R) = √(0.4 (9.80 m/s²) (105 m)) ≈ 20.3 m/s
Which of the following is NOT a step used to perform a scientific inquiry
Answers
Answer:
b. Designing an uncontrolled experiment.
Explanation:
They always have it controlled.
Answer:
B. Designing an uncontrolled experiment.
Explanation:
Correct Answer!!!!!!
A particle with charge Q and mass M has instantaneous speed uy when it is at a position where the electric potential is V. At a later time, the particle has moved a distance R away to a position where the electric potential is V2 ) Which of the following equations can be used to find the speed uz of the particle at the new position?
a. 1/2M(μ2^2-μ1^2)=Q (v1-v2)
b. 1/2M(μ2^2-μ1^2)^2=Q(v1-v2)
c. 1/2Mμ2^2=Qv1
d. 1/2Mμ2^2=1/4πx0 (Q^2/R)
Answers
Answer:
A
Explanation:
Ke = 1/2 MV^2
Give an example of mass making a difference in the amount of gravitational energy. Tell how you know the gravitational energy is different and your example
Please help due today!!
Answers
Answer:
The Gravitational potential energy at large distances is directly proportional to the masses and inversely proportional to the distance between them. The gravitational potential energy increases as r increases.
Examples of Gravitational Energy
A raised weight.
Water that is behind a dam.
A car that is parked at the top of a hill.
A yoyo before it is released.
Surface currents are on the
of the Earth's oceans
Answers
Fill in the blank with the correct word below (from the reading_):
helps you track your progress once you have made a lifestyle
change.
Self-monitoring
Healthy food
Regular xxercise
Goals
Answers
Answer:I think it’s self monitoring sorry if wrong
Explanation:
Answer:
It self monitoring
Explanation:
I took the test
When monochromatic light passes through the interface between two unknown materials at an angle θ where 0∘<θ<90∘, no changes in the direction of propagation of light are observed. What can be said about the two materials? Check all that apply. View Available Hint(s) Check all that apply. The two materials have matching indexes of refraction. The second material through which light propagates has a lower index of refraction. The second material through which light propagates has a higher index of refraction. The two materials are identical.
Answers
Answer:
the correct one is the first, the refractive index of the two materials must be the same
Explanation:
When a beam of light passes through two materials, it must comply with the law of refraction
n₁ sin θ₁ = n₂ sin θ₂
where n₁ and n₂ are the refractive indices of each medium.
In this case, it indicates that the light does not change direction, so the input and output angle of the interface must be the same,
θ₁ = θ₂ = θ
substituting
n₁ = n₂
therefore the refractive index of the two materials must be the same
When reviewing the answers, the correct one is the first
What are the major properties of the sun?
PLEASE HELP!!
Answers
Answer:
Explanation:
Properties of the sun
Introduction to Stars A. The Sun 1. ...
Layers of the sun (Atmosphere) a. Photosphere: Layer that emits the radiation we see. ...
Layers of the sun (Interior) a. ...
Solar Wind: a. ...
The sun in X-Rays: a. ...
Luminosity: The measure of the energy put out by the sun. ...
Luminosity continued: c. ...
Luminosity continued: e.
The heat of the Sun's surface is so large that neither solid nor liquid can reside there. The heat is the physical property of the sun.
What is the sun?The Sun is a billion star at the core of our universe, a hot. It is an incandescent ball of hydrogen and helium.
The Sun is millions of kilometers away from the land, and life is not possible without its energy.
The major properties of the sun are as follows;
1. The mass of the sun is 1.98892 x 10³⁰ kg.
2 The density of the sun is 1.622 x 10⁵ kg/m³
3. The surface gravity of the sun is 27.94 g.
4. The diameter is 1,391,000 kilometers.
5. The volume of the Sun is 1.412 x 1018 km³
The heat of the Sun's surface is so hot that neither solid nor liquid can reside there instead, the component materials are mostly gaseous atoms with a few molecules. As a result, no permanent surface exists.
Hence heat, density, and mass are the major properties of the sun.
To learn more about the sun refer to the link;
https://brainly.com/question/2526507
a graduated cylinder.measures 15.3 mL. Convert this measurement to DaL
Answers:
A. 0.0153
B. 0.00153
C. 0.000153
D. 0.153
Answers
Answer:
0.000153DaL
Explanation:
We have been given:
15.3mL to convert to DaL
DaL is a unit of volume which indicates a decaliter.
This implies that;
1 Da L = 1 x 10²L
So:
1 mL = 1 x 10⁻³L
So 15.3mL will give 15.3 x 10⁻³L
So;
1 x 10²L = 1 DaL
15.3 x 10⁻³L will give [tex]\frac{15.3 x 10^{-3} }{1 x 10^{2} }[/tex] = 15.3 x 10⁻⁵DaL
Therefore, this is 0.000153DaL
6 A test of a driver's perception/reaction time is being conducted on a special testing track with level, wet pavement and a driving speed of 50 mi/h. When the driver is sober, a stop can be made just in time to avoid hitting an object that is first visible 385 ft ahead. After a few drinks under exactly the same conditions, the driver fails to stop in time and strikes the object at a speed of 30 mi/h. Determine the driver's perception/reaction time before and after drinking. (Assume practical stopping distance.)
Answers
Answer:
a. 10.5 s b. 6.6 s
Explanation:
a. The driver's perception/reaction time before drinking.
To find the driver's perception time before drinking, we first find his deceleration from
v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m
So, a = v² - u²/2s
substituting the values of the variables into the equation, we have
a = v² - u²/2s
a = (0 m/s)² - (22.35 m/s)²/2(117.35 m)
a = - 499.52 m²/s²/234.7 m
a = -2.13 m/s²
Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver = -2.13 m/s² and t = reaction time
So, t = (v - u)/a
Substituting the values of the variables into the equation, we have
t = (0 m/s - 22.35 m/s)/-2.13 m/s²
t = - 22.35 m/s/-2.13 m/s²
t = 10.5 s
b. The driver's perception/reaction time after drinking.
To find the driver's perception time after drinking, we first find his deceleration from
v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m
So, a = v² - u²/2s
substituting the values of the variables into the equation, we have
a = v² - u²/2s
a = (13.41 m/s)² - (22.35 m/s)²/2(117.35 m)
a = 179.83 m²/s² - 499.52 m²/s²/234.7 m
a = -319.69 m²/s² ÷ 234.7 m
a = -1.36 m/s²
Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver = -1.36 m/s² and t = reaction time
So, t = (v - u)/a
Substituting the values of the variables into the equation, we have
t = (13.41 m/s - 22.35 m/s)/-1.36 m/s²
t = - 8.94 m/s/-1.36 m/s²
t = 6.6 s
12. A bag weighing 20 N CARRIED horizontally a distance of 35 m, How much
work is done on the bag in Joules? (Do not put units with your answer.) W=Fd *
Your answer
13. A child performs 10J of work in lifting a box 1 m in 2 seconds. How much
power did the child apply to the box? (Do not include units with your answer.)
P=W/t *
Your answer
Answers
Answer:
Explanation:
Well they told you the exact formula to use. Work is the force multiplied by the distance through which its applied.
W = (20N)(35m)
= 700 Joules
13.) Power is the amount of work done over the time through which the work is being done.
P = W/t
= 10J/2s
= 5J/s
The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 4.0 rev/s in 9.0 s. At this point, the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub slows to rest in 15.0 s. Through how many revolutions does the tub turn during this 24 s interval
Answers
Answer:
48 rev
Explanation:
a) we can calculate the distance covered by the tube using the formula:
θ = (ω + ωo)t/2
where ω is the final angular speed, θ is the distance covered, t is the time taken, ωo is the initial angular speed.
Firstly, we calculate the distance covered from 0 to 9 s then from 9s to 24 s.
within 9s, the tub runs from rest (0) to 4 rev/s, hence:
t = 9s, wo = 0, w = 4 rev/s = (4 * 2π) rad/s = 8π rad/s. Hence:
θ = (ω + ωo)t/2 = (0 + 8π)9 / 2 = 36π rad
θ = 36π rad = (36π)/2π rev = 18 rev
Also, within 15 s, the tub runs from 4 rev/s to rest, hence:
t = 15 s, wo = 4 rev/s = 8π rad/s, w = 0 rad/s. Hence:
θ = (ω + ωo)t/2 = (8π + 0)15 / 2 = 60π rad
θ = 60π rad = (60π)/2π rev = 30 rev
Therefore the total revolutions by the tube during 24 s interval = 30 rev + 18 rev = 48 rev
