Answer:
The final speed of the electron is 6.626 x 10⁷ m/s
Explanation:
force applied to the electron, f = 4 x 10⁻¹⁵ N
distance traveled by the electron, d = 50 cm = 0.5 m
The work done on the electron;
W = F x d
W = (4 x 10⁻¹⁵ N) x (0.5 m)
W = 2 x 10⁻¹⁵ J
The kinetic energy of the electron is given by;
[tex]K.E = \frac{1}{2}m_ev^2[/tex]
Apply work - energy theorem;
K.E = W
[tex]\frac{1}{2}m_ev^2 = 2 *10^{-15}\\\\ v^2 = \frac{2( 2 *10^{-15})}{m_e}\\\\v= \sqrt{\frac{2( 2 *10^{-15})}{m_e}}\\\\ v = \sqrt{\frac{2( 2 *10^{-15})}{9.11*10^{-31}}}\\\\v = 6.626*10^7 \ m/s[/tex]
Therefore, the final speed of the electron is 6.626 x 10⁷ m/s
what vehicle are runaway ramps designed for
Answer:truck
Explanation:
truck
What is the minimum compression of the spring necessary for the block to make it to the top of the hill on the other side?
Answer:
sciencephysicsphysics questions and answersA 12-kg Block Is Pressed Against A Spring (spring Constant 870 N/m ), Compressing It Some Distance. ...
Question: A 12-kg Block Is Pressed Against A Spring (spring Constant 870 N/m ), Compressing It Some Distance. The Block Is Released From Rest And Slides Across A Track As Shown In (Figure 1). While Most Of The Track Is Frictionless, There Is A 55-cm Section Of Track That Has A Coefficient Of Friction With The Block Of 0.4. A Bit Further On, The Track Ascends ...
A 12-kg block is pressed against a spring (spring constant 870 N/m ), compressing it some distance. The block is released from rest and slides across a track as shown in (Figure 1). While most of the track is frictionless, there is a 55-cm section of track that has a coefficient of friction with the block of 0.4. A bit further on, the track ascends into a hill that is 40-cm tall.
What is the minimum compression of the spring necessary for the block to slide past the section with friction?
What is the minimum compression of the spring necessary for the block to make it to the top of the hill on the other side? 40cm 12kg Ww.w.wwwww 55cm
Explanation:
Based on these photographs, how do Mars's surface and atmosphere differ from Earth's?
Answer:
Explanation:
Mars does not appear to have any water, while Earth is mostly water. Mars also has yellow-brown sky (pink-brown at sunrise and sunset), which suggests that it has a different atmospheric makeup.
Answer:
Mars does not appear to have any water, while Earth is mostly water. Mars also has yellow-brown sky (pink-brown at sunrise and sunset), which suggests that it has a different atmospheric makeup.
Explanation:
a force of 193 pounds makes an angle of 86.27 with a second force. The resultant of the two forces makes an angle of 31.4 to the first force. Find the magnitudes of the second force and of the resultant.
Answer:
Explanation:
Let force P = 193 pounds
second force = Q
Angle that resultant makes with force P is θ
Tanθ = Q sin86.27 / (P + Q cos86.27 )
Tan 31.4 = Q sin86.27 / (193 + Q cos86.27 )
.61 = .99 Q / (193 + .065Q )
117.73 + .04 Q = .99Q
117.73 = .95 Q
Q = 123.93 pounds .
Which is one way that ultrasound technology may be used?
O repairing fractured bones
O locating misplaced electrical equipment
O cleaning dirty surgical tools
O canceling out unwanted noise
Answer:
Locating misplaced electrical equipment.
Explanation:
Ultrasound is primarily used by doctors to locate fractured bones or damaged organs, so I assume it can be used to locate other things too.
Which is the right one?
(Show work)
A scout hikes 2.35KM east of a campsite he takes a break for lunch and then hikes another 1.25 KM north of the location where he ate lunch. what distance is the scout from the campsite
Answer:
mk7000000.0000000000
I HAVE A QUESTION ABOVE CAN ANYONE PLEASE HELP ME I WOULD REALLY APPRECIATE THE HELP I HAVE BEEN ON THIS QUESTION FOREVER i will give you the brainlist !!!!!
Answer:
answer is C
Explanation:
Answer: Monetary policy used during times of recession
Explanation:
All living creatures have some form of sleep that they engage in at regular periods, making sleep a(n) __________ function. A. environmental B. biological C. species-specific D. psychological
The answer is B. biological
Answer:
ans: a biological function
Answer:
B
Explanation:
EDGE2020
A 3.05 kg block is sliding across level ground and feels a 5.27 N friction force. What is the coefficient of friction?
Answer:
0.176
Explanation:
(5.27)/(3.05*9.8)
It is indeed 0.176 for the answer.
A segment of wire of total length 3 m carries a 10 A current and is formed into a semicircle. Determine the magnitude of the magnetic field (in micro-tesla) at the center of the circle along which the wire is placed.
Answer:
[tex]B=3.29\ \mu T[/tex]
Explanation:
Given that,
Length of a wire is 3 m
Current in the wore, I = 10 A
The wire formed a semicircle.
We need to find the magnitude of the magnetic field (in micro-tesla) at the center of the circle along which the wire is placed.
The magnetic field at the center of semicircle is given by :
[tex]B=\dfrac{\mu_o I}{4R}[/tex]
R is radius of semicircle,
[tex]R=\dfrac{3}{\pi}=0.954\ m[/tex]
So,
[tex]B=\dfrac{4\pi \times 10^{-7}\times 10I}{4(0.954)}\\\\B=3.29\times 10^{-6}\ T\\\\B=3.29\ \mu T[/tex]
So, the magnitude of the magnetic field is [tex]3.29\ \mu T[/tex].
A block with mass 0.50 kg is forced against a horizontal spring of negligible mass, compressing the spring a distance of 0.20 m . When released, the block moves on a horizontal tabletop for 1.00 m before coming to rest. The spring constant k is 100 N/m. What is the coefficient of kinetic friction between the block and the tabletop
Answer:
μ = 0.41
Explanation:
[tex]\frac{1}{2} kx^2 = \mu mgd[/tex]
[tex]\mu = (\frac{1}{2} kx^2)/mgd[/tex]
[tex]\mu = (\frac{1}{2} (100)(0.2)^2)/(0.5)(9.81)(1)[/tex]
[tex]\mu = 0.41[/tex]
The coefficient of kinetic friction between the block and the tabletop is 0.41.
The given parameters;
mass of the block, m = 0.5 kgextension of the spring, x = 0.2 mspring constant, K = 100 N/mThe coefficient of kinetic friction between the block and the tabletop by applying the principle of conservation of energy as shown below.
Fd = Uₓ
μmgd = ¹/₂kx²
where;
μ is the coefficient of kinetic frictionThe coefficient of kinetic friction between the block and the tabletop;
[tex]\mu_k = \frac{kx^2}{mgd} \\\\\mu_k = \frac{100\times 0.2^2}{2\times 0.5\times9.8 \times 1 } \\\\\mu_k = 0.41[/tex]
Thus, the coefficient of kinetic friction between the block and the tabletop is 0.41.
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a mass of .4 kg is raised by a vertical distance of .450 m in the earth's gravitational field. what is the change in its gravitational potential energy
Answer:
E = 1.76 J
Explanation:
Given that,
Mass of an object, m = 0.4 kg
It moves by a vertical distance of 0.45 m in the Earth's gravitational field.
We need to find the change in its gravitational potential energy. It can be given by the formula as follow :
[tex]E=mgh\\\\E=0.4\times 9.8\times 0.45\\\\E=1.76\ J[/tex]
So, the change in its gravitational potential energy is 1.76 J.
The gravity on the moon is 1/6 as strong as that of earth.An object on earth weighs 1 Newton another object on the moon also weighs 1 Newton.Which of the two objects has the greater mass
Answer:
The object on the moon has greater mass
Explanation:
Given;
weight of the object on Earth, W₁ = 1 N
wight of the object on Moon, W₂ = 1 N
gravity on Earth, g₁ = 9.8 m/s²
gravity on the Moon, g₂ = g₁/6 = 1.633 m/s²
Apply Newton's second law of motion;
Weight of the object on Earth is given by;
W₁ = m₁g₁
m₁ = W₁/g₁
m₁ = 1 / 9.8
m₁ = 0.102 kg
Weight of the object on Moon is given by;
W₂ = m₂g₂
m₂ = W₂ / g₂
m₂ = 1 / 1.633
m₂ = 0.612 kg
Therefore, the object on the moon has greater mass.
Additional Practice on Speed
1. A group of campers and one group leader left a campsite in a canoe.
They traveled at an average rate of 10 km/h. Two hours later, the other
group leader left the campsite in a motorboat. He traveled at an average
rate of 22 km/h. How long had each boat traveled when the motorboat
caught up with the canoe?
a.
canoe: 2.2 h; motorboat: 1 h
b. canoe: 2.2 h; motorboat: 10 h
c. O canoe: 3 2 h; motorboat: 12 h
3
3
d. O canoe: 4.4 h; motorboat: 2 h
Answer:
canoe = 3 h 40 min. motorboat = 1 h 40 min
Explanation:
As both boats travel at a constant speed, we can apply the definition of average velocity to both, so we can get:[tex]v_{b} = \frac{\Delta x_{b} }{\Delta t_{b} } (1)[/tex]
[tex]v_{mb} = \frac{\Delta x_{mb} }{\Delta t_{mb} } (2)[/tex]
Now, when the motorboat caught up with the canoe, distance travelled by both boats were the same, so, solving (1) and (2) for Δx, we get:[tex]v_{b} *\Delta t_{b} = v_{mb} *\Delta t_{mb} (3)[/tex]
if we take the initial time for the canoe to be t₀ =0, the initial time for the motorboat will be tmb₀ = 2 Hs.So, replacing in (3) we have:[tex]v_{b} *t_{b} = v_{mb} *(t_{b} - 2 hs.) = 10 km/h* t_{b} = 22 km/h* (t_{b} - 2 hs.)[/tex]
Solving for tb:[tex]t_{b} = \frac{-44 }{-12} hs = 3h 40 min.[/tex]
tmb = tb - 2 hs. = 1h 40 min.
Two identical objects of the same size and shape are dropped. One is dropped from a table, while the other is dropped from the top of a tall building. Which will hit the ground with more impact?
Answer:
These two objects hit the ground at the same time. But their impact velocity will be different
because they are dropped from different heights.
The one that is dropped from the building will have more impact when it hits the ground because gravity causes this onject to fall toward the ground at a faster and faster velocity the longer the object falls. In fact, its velocity increases by 9.8 m/s2.
Explanation:
Initial velocity of the object u=0
Final velocity of the object when it reaches the ground v=A m/s
Using v^2−u^2=2gS where g=10m/s^2
A^2 −0^2=2(10)H
A^2=20*H H- heigh from the table
Object falling from the building will have the same initial velocity but different heigh.
T>H
so
B^2=20*T
20*T>20*H
so B^2>A^2
it equal to B>A so object falling from the building will have more impact than falling from the table.
A golf ball rolls up a hill toward a miniature-golf hole. Assume the direction toward the hole is positive. If the golf ball starts with a speed of 2.0 m/s and slows at a constant rate of 0.50 m/s2. What is the golf ball’s velocity if the constant acceleration continues for 6.0 s? *
Answer:
0.2533333334
Explanation:
It is simple, just do the math
A truck is traveling at an initial velocity
of +39m/s and it starts slowing down smoothly for 3.3s. It covers a distance of 45m while slowing down. What is the acceleration?
Answer:
Approximately [tex]-15.37m/s^2[/tex]
Explanation:
The acceleration can be found using the formula:
[tex]x_{f} = x_{i} + v_{i} (t)+\frac{1}{2} (a)(t^{2} )[/tex]
The work is as shown:
[tex]45m = 0 + 39 m/s (3.3s)+\frac{1}{2} a(3.3s)^2[/tex]
[tex]45m=128.7m + 5.445s^2 a[/tex]
[tex]-83.7m=5.445s^2a[/tex]
[tex]a = -15.37190083m/s^2[/tex]
a is about -15.37[tex]m/s^2[/tex]
A dart is loaded into a compressed spring. Starting from rest the spring is released and shoots the dart up into the air. From the instant the dart is released (before it starts to move) until it reaches its maximum height, what is the Net work done on the dart
Answer:
Wnet = 0
Explanation:
The work-energy theorem says that the total net work done on an object, is equal to the change in the kinetic energy of that object.In this case, the total change in kinetic energy of the dart is zero, due to the initial status is at rest (loaded into a compressed spring) and the final is also at rest (when it reaches to its maximum height).The total work done on the dart can be written as follows:[tex]W_{net} = \frac{1}{2} k* \Delta x^{2} - m*g*h_{max} (1)[/tex]
Now, in absence of friction, all the elastic potential energy becomes gravitational potential energy, so the following is true:[tex]\frac{1}{2} k* \Delta x^{2} = m*g*h_{max} (2)[/tex]
From (1) and (2) we conclude that Wnet = 0.What would be the luminosity of the Sun if its surface temperature were 4000 K and its radius were 2.0 AU ?
Answer:
0.04558
Explanation:
Luminosity is defined as the total amount of energy a star can produce in a second. Luminosity is given by the formula
σ * A * T⁴
where;
σ = Stefan-Boltzmann constant
= 5.670367 * 10⁻⁸
A= Surface Area
=4πr²
T= Temperature in Kelvin
In the question,
Temperature= 4000K
Radius= 2.0 A.U
Substituting the variables into the equation,
0.000000056704 * 4(3.14)(2.0²) * 4000⁴
= 0.000000056704 * 50.24 * 16000
= 0.04558
The speed of light in vacuum is exactly 299,792,458 m/s. A beam of light has a wavelength of 621 nm in vacuum. This light propagates in a liquid whose index of refraction at this wavelength is 1.18.What is the speed of this light in the liquid
Answer:
2.54*10^8 m/s
Explanation:
Given that
Speed of light in space, c = 299792458 m/s
Wavelength of the beam of light, λ = 621 nm or 621*10^-9 m
Index of refraction of the liquid, n = 1.18
Speed of light in the liquid, v = ?
The index of refraction of an optical material, n is given by the following formula.
n = c/v, and if we substitute directly, we have
1.18 = 299792458 / v, if we make v subject of formula, we have
v = 299792458 / 1.18
v = 2.54*10^8 m/s
Calculate the peak voltage (in Volts) of a generator that rotates its 200-turn, 0.100 m diameter coil at 3.5 x 103 rpm in a 0.7 T field. You should round your answer to the nearest integer, do not include unit.
Angular velocity of the coil :
[tex]\omega=\dfrac{2\pi n}{60}[/tex]
[tex]\omega=\dfrac{2\times \pi\times 3500}{60}\\\\\omega =366.52\ rad/s[/tex]
Now,
Peak voltage is given by :
[tex]V=NAB\omega\\\\V= 200\times \dfrac{\pi (0.1)^2}{4}\times 0.7\times 366.52\\\\V=403.01\ V[/tex]
Hence, this is the required solution.
A half-eaten apple turns brown which of these is evidence a chemical change took place A.th3e apple is softer b.the apple changed color C.half the apple was eaten D.the apple is warmer \
Answer:
b the apple changed color
A mass that has a force of the 10N to the right, 7N up, 3N down, and 4N to the left. What is the net external force on that mass? If the mass is 2kg find the acceleration.
What is the mass of an object if a force of 17N causes it to accelerate at 1.5 m/sec/sec
A. 11.38
B. 11.3 kg
C. 12.3 kg
D. 14.3 kg
Answer:
B.11.3 kg
Explanation :
F=ma here force is given as 17N and acceleration as 1.5 m/sec/sec using the formula mass can be calculated
∴ F=ma
17N =m × 1.5 m[tex]s^{-2}[/tex]
[tex]m=\frac{F}{a}[/tex]
[tex]m= \frac{17 kg m/sec/sec}{1.5 m/sec/sec}[/tex]
∴ m=11.3 kg
What is the average velocity of the cart? Is the velocity increasing decreasing or constant as the cart moves along the air track?
Answer:
It is increasing .60 m/s
Explanation:
Find μsμsmu_s, the coefficient of static friction between the rod and the rails. Express the coefficient of static friction in terms of variables given in the introduction.
Answer:
The question is incomplete. However the complete question is :
A Rail Gun uses electromagnetic forces to accelerate a projectile to very high velocities. The basic mechanism of acceleration is relatively simple and can be illustrated in the following example. A metal rod of mass m and electrical resistance R rests on parallel horizontal rails (that have negligible electric resistance), which are a distance L apart. The rails are also connected to a voltage source V, so a current loop is formed.
The vertical magnetic field, initially zero, is slowly increased. When the field strength reaches the value [tex]$B_0$[/tex], the rod, which was initially at rest, begins to move. Assume that the rod has a slightly flattened bottom so that it slides instead of rolling. Use g for the magnitude of the acceleration due to gravity.
Find μs, the coefficient of static friction between the rod and the rails.
Express the coefficient of static friction in terms of variables given in the introduction.
So the answer is : [tex]$\mu_s=\frac{\Delta V LB_0}{Rmg}$[/tex]
Explanation:
It is given : B = [tex]$B_0$[/tex]
Therefore, using force on the current carrying wire is:
[tex]$\mu_s mg = B_0IL$[/tex]
[tex]$\mu_s= \frac{ILB_0}{mg}$[/tex]
Therefore,
[tex]$\mu_s=\frac{\Delta V LB_0}{Rmg}$[/tex]
The coefficient of static friction in terms of variables will be [tex]\rm \mu_s =\frac{ \triangle VLB_0}{Rmg} \\\\[/tex].
What is the cofficient of static friction?The ratio of the greatest static friction force (F) between the surfaces in contact before movement begins to the normal (N) force is the coefficient of static friction.
The given data in the problem is;
[tex]\rm B=B_0[/tex]
The force on a current-carrying wire is balanced by the friction force.
The force on a current-carrying wire is given by;
[tex]\rm \mu_s mg= B_0IL \\\\ \rm \mu_s=\frac{ILB_0}{mg}\\\\ \rm \mu_s= \frac{\triangle VLB_0}{Rmg}[/tex]
Hence the coefficient of static friction in terms of variables will be [tex]\rm \mu_s =\frac{ \triangle VLB_0}{Rmg} \\\\[/tex].
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You are given two vectors A⃗ =−3.00i^+7.00j^ and B⃗ =5.00i^+2.00j^. Let counterclockwise angles be positive.
1- What angle θA, where 0∘≤θA<360∘, does A⃗ make with the +x-axis?
2-What angle θB, where 0∘≤θB<360∘, does B⃗ make with the +x-axis?
Express your answer in degrees.
3-Vector C⃗ is the sum of A⃗ and B⃗ , so C⃗ =A⃗ +B⃗ . What angle θC, where 0∘≤θC<360∘, does C⃗ make with the +x-axis?
Express your answer in degrees
Answer:
Ax = -3 By= 7 tan phi = -7 / 3 phi = -66.8
Since theta is in the second quadrant theta = 180 - 66.8 = 113.2 deg
Bx = 5 By = 2 tan theta = .4 theta = 21.8 deg
Cx = Ax + Bx = 2 Cy = Ay + By = 9
tan theta = 9 / 2 = 4.5 theta = 77.5 deg
1) The solution is [tex]\theta_{1} \approx 113.141^{\circ}[/tex].
2) The solution is [tex]\theta_{1} \approx 21.874^{\circ}[/tex].
3) The solution is [tex]\theta_{1} \approx 77.467^{\circ}[/tex].
1) In this case, we can determine the angle, measured in sexagesimal degrees, with respect to the x-axis by definition of dot point:
[tex]\cos \theta = \frac{\vec A\,\bullet \, \vec u}{\|\vec A\|\cdot \|\vec u\|}[/tex] (1)
Where [tex]\|\vec A\|[/tex] and [tex]\|\vec u\|[/tex] are the norms of [tex]\vec A[/tex] and [tex]\vec u[/tex].
If we know that [tex]\vec A = (-3, 7)[/tex] and [tex]\vec u = (1, 0)[/tex], then the angle that [tex]\vec A[/tex] make with the x-axis is:
[tex]\|\vec A\| = \sqrt{(-3)^{2}+7^{2}}[/tex]
[tex]\|\vec A\| = \sqrt{58}[/tex]
[tex]\|\vec u\| = 1[/tex]
[tex]\cos \theta = \frac{(-3)\cdot (1)}{\sqrt{58} \cdot 1}[/tex]
[tex]\cos \theta \approx -0.393[/tex]
The solution is [tex]\theta_{1} \approx 113.141^{\circ}[/tex].
2) In this case, we can determine the angle, measured in sexagesimal degrees, with respect to the x-axis by definition of dot point:
[tex]\cos \theta = \frac{\vec B\,\bullet \, \vec u}{\|\vec B\|\cdot \|\vec u\|}[/tex] (1)
Where [tex]\|\vec B\|[/tex] and [tex]\|\vec u\|[/tex] are the norms of [tex]\vec B[/tex] and [tex]\vec u[/tex].
If we know that [tex]\vec B = (5, 2)[/tex] and [tex]\vec u = (1, 0)[/tex], then the angle that [tex]\vec A[/tex] make with the x-axis is:
[tex]\|\vec B\| = \sqrt{5^{2}+2^{2}}[/tex]
[tex]\|\vec B\| = \sqrt{29}[/tex]
[tex]\|\vec u\| = 1[/tex]
[tex]\cos \theta = \frac{(5)\cdot (1)}{\sqrt{29} \cdot 1}[/tex]
[tex]\cos \theta \approx 0.928[/tex]
The solution is [tex]\theta_{1} \approx 21.874^{\circ}[/tex].
3) In this case, we need to find the value of [tex]\vec C[/tex] by vectorial sum before calculating the angles with respect to x-axis:
[tex]\vec C = \vec A + \vec B = (-3, 7) + (5, 2)[/tex]
[tex]\vec C = (2, 9)[/tex]
In this case, we can determine the angle, measured in sexagesimal degrees, with respect to the x-axis by definition of dot point:
[tex]\cos \theta = \frac{\vec C\,\bullet \, \vec u}{\|\vec C\|\cdot \|\vec u\|}[/tex] (1)
Where [tex]\|\vec C\|[/tex] and [tex]\|\vec u\|[/tex] are the norms of [tex]\vec C[/tex] and [tex]\vec u[/tex].
If we know that [tex]\vec B = (2, 9)[/tex] and [tex]\vec u = (1, 0)[/tex], then the angle that [tex]\vec A[/tex] make with the x-axis is:
[tex]\|\vec C\| = \sqrt{2^{2}+9^{2}}[/tex]
[tex]\|\vec C\| = \sqrt{85}[/tex]
[tex]\|\vec u\| = 1[/tex]
[tex]\cos \theta = \frac{(2)\cdot (1)}{\sqrt{85} \cdot 1}[/tex]
[tex]\cos \theta \approx 0.217[/tex]
The solution is [tex]\theta_{1} \approx 77.467^{\circ}[/tex].
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Two points are on a disk that rotates about an axis perpendicular to the plane of the disk at its center. Point B is 3 times as far from the axis as point A. If the linear speed of point B is V, then the linear speed of point A is:_______
a. 9V.
b. 3V.
c. V.
d. V/3.
e. V/9.
Answer:
D. V/3.
Explanation:
V = rw
At point a
Linear velocity
= Va = rw
At point b
We have,
V = 3rw
So linear velocity of a
= V/Va = 3rw/re
I cross multiples and after which I got Va = v/3
Please view attachment
Answer:
The linear speed of point A is [tex]\frac{V}{3}[/tex] (Option D)Explanation:
The relation between linear and angular velocity,
[tex]V = rw[/tex]
Now, for point A the linear velocity is
[tex]V_A = rw[/tex]
And for point B
[tex]V = 3rw[/tex]
Therefore, the linear velocity of the point A
[tex]\frac{V}{V_A} = \frac{3rw}{rw}\\\\V_A = \frac{V}{3}[/tex]
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A 55kg skateboarder wants to just make it to the upper edge of a quarter pipe, a pipe that is one-quarter of a circle with a radius of 3.10m. What speed does he need at the bottom?
Answer:
Explanation:
A 52.0 kg skateboarder wants to just make it to the upper edge of a "quarter pipe," a track that is one-quarter of a circle with a radius of 2.60 m .
What speed does he need at the bottom
Let the lowest point of the circle = (0,0)
Center of circle = (0, 2.6)
The height of the circle = 2 * radius = 2 * 2.60 = 5.20 m
Highest point = (0, 5.2)
As the skateboarder moves around ¼ of a vertical circle, the skateboarder moves from the lowest position to a position that is ½ the way up to the highest position. This is the point that is 2.6 meters directly to the right of left of the center = (2.6, 2.6)
As the skateboarder has moved 2.6 m upward, the potential energy increase = m * g * ∆h = 52.0 * 9.8 * 2.6
During this same time, the kinetic energy decreased from the maximum to 0. The decrease of KE = the increase of PE
½ * m * ∆v^2 = m * g * ∆h
½ * 52 * ∆v^2 = 52.0 * 9.8 * 2.6
½ * ∆v^2 = 9.8 * 2.6
∆v = (2 * 9.8 * 2.6)^0.5 = 7.14 m/s
The velocity at highest point, (2.6,2.6) is 0 m/s
So the velocity at the lowest point must be 7.14 m/s
Let’s see if the skateboarder has enough KE to move upward 2.6 m.
KE at bottom = ½ * 52 * 7.14^2 = 1325.5
PE = 52 * 9.8 * h
52 * 9.8 * h = 1325.5
h = 2.6 m
The answer is OK
The speed needed by the skateboarder at the bottom is 7.8 m/s.
The given parameters;
mass of the skateboarder, m = 55 kgradius of the quarter pipe, r = 3.1 mApply the principle of conservation of mechanical energy to determine the sped of the skateboarder at the bottom;
[tex]P.E_{top} = K.E_{bottom}\\\\mgh = \frac{1}{2}mv^2\\\\gh = \frac{1}{2} v^2\\\\v^2 = 2gh\\\\v = \sqrt{2gh} \\\\v = \sqrt{2\times 9.8 \times 3.1} \\\\v = 7.8 \ m/s[/tex]
Thus, the speed needed by the skateboarder at the bottom is 7.8 m/s.
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