Answer with Explanation:
We are given that
Length of pole(l0)=20 m
Length of garage(l)=10 m
a. We have to find the speed by which the Patrick is running.
We know that
[tex]l=l_0\sqrt{1-\frac{v^2}{c^2}}[/tex]
Substitute the values
[tex]10=20\sqrt{1-\frac{v^2}{c^2}}[/tex]
[tex]\frac{10}{20}=\sqrt{1-\frac{v^2}{c^2}}[/tex]
[tex]\frac{1}{2}=\sqrt{1-\frac{v^2}{c^2}}[/tex]
[tex]\frac{1}{4}=1-\frac{v^2}{c^2}[/tex]
[tex]\frac{v^2}{c^2}=1-\frac{1}{4}=\frac{3}{4}[/tex]
[tex]v^2=0.75c^2[/tex]
[tex]v=0.87c[/tex]
b.
l0=10 m
[tex]l=10\sqrt{1-\frac{3}{4}}[/tex]
[tex]l=5m[/tex]
Hence, the length of garage appear to him=5 m long.
What is the approximate heat of water in kj/kg k?
Answer:
Specific heat (Cp) water (at 15°C/60°F): 4.187 kJ/kgK = 1.001 Btu(IT)/(lbm °F) or kcal/(kg K)
Arrange the objects in order from greatst to least of potential energy assume that gravity is constant
Answer:
Water > Box of books > Stone > Ball
Explanation:
We'll begin by calculating the potential energy of each object. This can be obtained as follow:
For stone:
Mass (m) = 15 Kg
Acceleration due to gravity (g) = 10 m/s²
Height (h) = 3 m
Potential energy (PE) =?
PE = mgh
PE = 15 × 10 × 3
PE = 450 J
For water:
Mass (m) = 10 Kg
Acceleration due to gravity (g) = 10 m/s²
Height (h) = 9 m
Potential energy (PE) =?
PE = mgh
PE = 10 × 10 × 9
PE = 900 J
For ball:
Mass (m) = 1 Kg
Acceleration due to gravity (g) = 10 m/s²
Height (h) = 20 m
Potential energy (PE) =?
PE = mgh
PE = 1 × 10 × 20
PE = 200 J
For box of books:
Mass (m) = 25 Kg
Acceleration due to gravity (g) = 10 m/s²
Height (h) = 2 m
Potential energy (PE) =?
PE = mgh
PE = 25 × 10 × 2
PE = 500 J
Summary:
Object >>>>>>>> Potential energy
Stone >>>>>>>>> 450 J
Water >>>>>>>>> 900 J
Ball >>>>>>>>>>> 200 J
Box of books >>> 500 J
Arranging from greatest to least, we have:
Object >>>>>>>> Potential energy
Water >>>>>>>>> 900 J
Box of books >>> 500 J
Stone >>>>>>>>> 450 J
Ball >>>>>>>>>>> 200 J
Water > Box of books > Stone > Ball
A bottle of water at a room temperature of 21.0 C is placed into a refrigerator
with an air temperature of 4.5C. The thermal energy will move — *
A. from the cooler air to lower the temperature of the water to 4.5 C
B. in both directions until the temperature is equal in the water and the air
C. from the water to the air until the water temperature is zero degrees Celsius
O D. from the water to the air until the temperature is equal in both
Answer:
B. in both directions until the temperature is equal in the water and the air
Explanation:
When a warm body is in contact with a cool body , there is exchange of heat energy in both sides until there is attainment of equilibrium temperature . At this temperature both the body attains equal temperature . Initially rate of heat radiated by warm body is more than that from cool body , but after attainment of equilibrium , the rate becomes equal to each other . This is called dynamic equilibrium .
Hence option B is correct .
2. What is the cheetah's speed for the first four seconds. She
Explanation:
Cheetahs can go from 0 to 60 miles per hour in just 3.4 seconds and reach a top speed of 70 miles per hour. While they are the fastest land animal in the world, they can only maintain their speed for only 20 to 30 seconds.
what do you call these sound waves whose frequency is above 20000 hertz
Answer:
Untrasound
Explanation:
Your welcome :)
A runner starts from rest and stops in 12 seconds. He covers
100m distance. Using this information you can clain the
maximum absolute value of his acceleration was not less than:
a 0.69 m/s 2
b 1.39 m/s 2
c 2.78 m/s 2
d 3.47 m/s 2
Answer:
b 1.39 m/s²
Explanation:
Given the following data;
Time = 12 seconds
Distance, S = 100 m
Since it's starting from rest, the initial velocity is equal to 0m/s.
To find the acceleration, we would use the second equation of motion;
[tex] S = ut + \frac {1}{2}at^{2}[/tex]
Where;
S represents the displacement or height measured in meters.
u represents the initial velocity measured in meters per seconds.
t represents the time measured in seconds.
a represents acceleration measured in meters per seconds square.
Substituting into the equation, we have;
100 = 0(12) + ½*a*12²
100 = 0 + 72
100 = 72a
Acceleration, a = 100/72
Acceleration, a = 1.389 ≈ 1.39 m/s²
A 20-turn coil of area 0.32 m2 is placed in a uniform magnetic field of 0.055 T so that the perpendicular to the plane of the coil makes an angle of 30∘ with respect to the magnetic field.
The flux through the coil is
Answer:
1.5 * 10^-2 Tm^2
Explanation:
Electric Flux = B.A cos(theta)
B = 0.055 T
A = 0.32 m^2
theta = 30
Electric Flux = (0.055 T).(0.32 m^2).Cos(30) = 0.0152 = 1.5 * 10^-2 Tm^2
In ionic compounds which group from the periodic table usually provide anion?
Determine mass flow rate and velocity of efflux from circular hole of 0.1 diameter at the bottom of water tank at this instant .
The tank is open to atmosphere and H=4m
Answer:
Mixed in a smoothie it like it licked
Explanation:
2. Am 80.0 kg astronaut is training for accelerations that he will experience upon re-entry to earth’s gravity from space. He is placed in a centrifuge (r = 25.0 m) and spun at a constant angular velocity of 10.0 rpm (revolutions per minute).
a. Find the linear velocity of the centrifuge in m/s. Show your work
b. Find the magnitude and direction of the centripetal acceleration when he is spinning at this constant velocity.
c. How many g’s is the astronaut experiencing? (at constant velocity)
d. Find the linear deceleration and torque required to bring the centrifuge (5000.0 kg) to a stop over a 5 minute time period.
Answer:
a) v = 26.2 m / s, b) acceleration is radial, a = 27.4 m / s², c) a = 2.8 g and
d) a = - 8.73 10⁻² m / s², τ = 1.09 10⁴ N m
Explanation:
a) For this exercise we can use the relationships between rotational and linear motion
v = w r
let's reduce the magnitudes to the SI system
w = 10 rpm (2pi rad / 1 rev) 1 min / 60s) = 1,047 rad / s
r = 25.0 m
let's calculate
v = 1.047 25.0
v = 26.2 m / s
b) When the body is rotating at constant speed, the relationship must be perpendicular to the speed, therefore the direction of acceleration is radial, that is, towards the center of the circle and its magnitude is
a = v² / r
a = 26.2²/25
a = 27.4 m / s²
c) Let's look for the relationship between the centripetal acceleration and the acceleration due to gravity
a / g = 27.4 / 9.8
a / g = 2.8
a = 2.8 g
d) let's find the deceleration and torque to stop the centripette in 5 min
t = 5 min (60 s / 1min) = 300 s
let's use the rotational kinematics relations
w = w₀ + α t
initial angular velocity is wo = 1,047 rad / s and the final as is stop do w = 0
α = - w₀ / t
α = - 1,047 / 300
α = -3.49 10⁻³ rad / s²
angular and linear are related
a = α r
a = -3.49 10⁻³ 25
a = - 8.73 10⁻² m / s²
the negative sign indicates that the acceleration is stopping the movement
torque is
τ = F r
The force can be found with Newton's second law
F = m a
we substitute
τ = m a r
τ = 5000.0 8.73 10⁻² 25
τ = 1.09 10⁴ N m
6th grade science I mark as brainliest
Answer:
7 would be C, a cell.
Explanation:
Hi.
7 would be C, a cell.
A cell is the basic unit of structure and function in all living things.
If it is living, it is made of cells.
Hope this helps.
Answer:
7. Cell
8. Organelle
A dog runs at 35 m/s at 45 degrees N of E. What are its x and y components (all answers
are in m/s)?
Answer:
its x and y component is 24.749m/s
Explanation:
Given
Speed of the dog = 35m/s
x component of the speed = xcos theta
y component of the speed = ycos theta
Given theta =45 degrees
x-component = 35cos45
x-component = 35(0.7071)
x-component = 24.749m/s
y-component = 35sin45
y-component = 35(0.7071)
y-component = 24.749m/s
Hence its x and y component is 24.749m/s
A 0.11 kg bullet traveling at speed hits a 18.3 kg block of wood and stays in the wood. The block with the bullet imbedded in it moves forward with a velocity of 8.8 m/s. What was the velocity (speed) of the bullet immediately before it hit the block (in m/s)?
Explanation:
The energy of the system before the collision must equal the energy after the collision.
After the collision the bullet and the block have a total mass of 18.41 kg and they move at a speed of 8.8 m/s. The kinetic energy after the collision is
[tex]\frac{18.41 kg (8.8 m/s)^2}{2} = 713 J[/tex]
Before the collision only the bullet has kinetic energy.
So we can now determine the speed of the bullet using
[tex]\frac{0.11kg (v^2)}{2} = 713 J\\v = 114 m/s[/tex]
You are sitting in your car at rest at a traffic light with a bicyclist at rest next to you in the adjoining bicycle lane. As soon as the traffic light turns green, your car speeds up from rest to 49.0 mi/h with constant acceleration 8.00 mi/h/s and thereafter moves with a constant speed of 49.0 mi/h. At the same time, the cyclist speeds up from rest to 21.0 mi/h with constant acceleration 14.00 mi/h/s and thereafter moves with a constant speed of 21.0 mi/h.
A. For what time interval (in s) after the light turned green is the bicycle ahead of your car?
B. By what maximum distance does the bicycle lead the car?
Answer:
A. 2.63 s B. 12.38 m
Explanation:
A. For what time interval (in s) after the light turned green is the bicycle ahead of your car?
The time interval at which the bicycle is ahead of the car is the time it takes for the car to reach the bicycle's speed of 21.0 mi/h.
So, using v = u + at where u = initial speed of car = 0 mi/h, v = final speed of car = 21.0 mi/h, a = acceleration of car = 8.00 mi/h/s and t = time taken for acceleration.
So, v = u + at
t = (v - u)/a
substituting the values of the variables into the equation, we have
t = (21.0 mi/h - 0 mi/h)/8.00 mi/h/s
= 21.0 mi/h ÷ 8.00 mi/h/s
= 2.63 s
B. By what maximum distance does the bicycle lead the car?
To find this distance, we find the distance moved by both the car in this time of t = 2.63 s
So, using s = ut + 1/2at² where u = initial speed of car = 0 mi/h = 0 m/s, t = time = 2.63 s, a = acceleration of car = 8.00 mi/h/s = 8.00 × 1609 m/3600 s = 3.58 m/s/s = 3.58 m/s² and s = distance moved by car.
So, substituting the values of the variables into the equation, we have
s = ut + 1/2at²
s = 0 m/s × 2.63 s + 1/2 × 3.58 m/s² × (2.63 s)²
s = 0 m + 1/2 × 3.58 m/s² × (2.63 s)²
s = 1.79 m/s² × 6.9169 s²
s = 12.38 m
which is also the maximum distance with which the bicycle leads the car.
A 107 gram apple falls from a branch that is 2 meters above the ground. (a) How much time elapses before the apple hits the ground
Answer:
The time of motion is 0.64 s.
Explanation:
Given;
mass of the apple, m = 107 g
height of fall, h = 2 m
The velocity of the apple when it hits the ground is calculated from the law of conservation of energy;
[tex]P.E = K.E\\\\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 2} \\\\v = 6.261 \ m / s[/tex]
The time of motion is calculated;
v = u + gt
6.261 = 0 + 9.8t
6.261 = 9.8t
t = 6.261 / 9.8
t = 0.64 s
Therefore, the time of motion is 0.64 s
The time taken for the apple to hit the ground is 0.64 s.
The time taken for the apple to hit the ground can be calculated using the formula below.
Formula:
s = ut+gt²/2............ Equation 1Where:
s = heightt = timeu = initial velocityg = acceleration due to gravity.
From the question,
Given:
s = 2 mu = 0 m/s (fall from a height)g = 9.8 m/s²Substitute these values into equation 1
2 = 0(t)+9.8(t²)/2Solve for t.
9.8t² = 4t² = 4/9.8t² = 0.4081t = √0.4081t = 0.64 s.Hence, The time taken for the apple to hit the ground is 0.64 s
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in a football game, the kicker kicks a football a horizontal distance of 43 yards if the ball lands 3.9 seconds later, what is the balls horizontal velocity
Answer:
10s
Explanation:
Horizontal velocity is the velocity of an object in an horizontal direction
The ball's horizontal velocity is approximately 33.078 ft./s
Reason:
The known parameter are;
The horizontal distance the footballer kicks the ball, d = 43 yards
The time after which the ball lands, Δt = 3.9 seconds
Required:
To find the velocity of the ball
Solution:
[tex]Velocity = \dfrac{Distance}{Time} = \dfrac{d}{\Delta t}[/tex]
Therefore;
[tex]Horizontal \ velocity \ of \ the \ ball, \ v_x= \dfrac{43 \ yard}{3.9 \ seconds} \approx 11.026 \ yd/s[/tex]
The ball's horizontal velocity, vₓ ≈ 11.026 yd/s
1 yard = 3 feet
[tex]11.026 \ yard = 11.026 \ yard \times \dfrac{3 \ feet}{yard} = 22.078 \ feet[/tex]
The ball's horizontal velocity, vₓ ≈ 33.078 ft./s
Learn more about horizontal velocity here:
https://brainly.com/question/14898646
The "problem of perception" is best characterized as?
Answer:
making sense of a 3-d world from 2-d data
Explanation:
Emma is working in a shoe test lab measuring the coefficient of friction for tennis shoes on a variety of surfaces. The shoes are pushed against the surface with a force of 400 N, and a sample of the surface material is then pulled out from under the shoe by a machine. The machine pulls with a force of 300 N before the material begins to slide. When the material is sliding, the machine has to pull with a force of only 200 N to keep the material moving.
a. What is the coefficient of static friction between the shoe and the material?
b. What is the coefficient of dynamic friction between the shoe and the material?
c. Draw a Free Body Diagram for the above.
Answer:
Explanation:
Force of friction = μ N , where μ is coefficient of friction , N is normal force on the body .
a )
Given,
Normal force N = 400 N
Force of friction = 300 N
μ = coefficient of static friction = ?
Putting the values ,
300 = 400 μ
μ = .75
b )
Normal force N = 400 N
Force of friction = 200 N
μ = coefficient of kinetic friction = ?
Putting the values ,
200 = 400 μ
μ = .50
c ) see attached file .
what is air resistance means explain it with free falling body
Calculate the density of a substance that has mass 10g and volume 2mL
You're supposed to divide the mass by the volume, which is going to equal to 5
Grace drives her car 168 km in 2 hours. What is her average speed in kilometers per hour?
Answer:
84kliometers
Explanation:
divide one hundred and sixty eight kilo meters by two hours
A 0.1 kg arrow with an initial velocity of 30 m/s hits a 4.0 kg melon initially at rest on a friction-less surface. The arrow emerges out the other side of the melon with a speed of 20 m/s. What is the speed of the melon? Why would we normally not expect to see the melon move with the is speed after being hit by the arrow?
Answer:
Speed of the melon = 0.25 m/s
we would normally don't see the melon moving due to friction with the resting surface.
Explanation:
We use conservation of momentum:
Pi = Pf
with Pi = 0.1 kg * 30 m/s = 3 kg m/s
and Pf = 0.1 kg * 20 m/s + 4.0 kg * V = 2 kg m/s + 4 * V
Then using the equality above, we solve for V (velocity of the melon)
3 kg m/s = 2 kg m/s + 4 V
1 kg m/s = 4 kg * V
Then V = 1 / 4 M/s = 0.25 m/s
So we would normally don't see the melon moving due to friction with the resting surface.
A student swings a 0.5kg rubber ball attached to a string over her head in a horizontal, circular
path. The string is 1.5 meters long and in 60 seconds the ball makes 120 complete circles.
What is the velocity of the ball?
What is the ball’s centripetal acceleration?
What is the ball's centripetal force?
Answer:
The balls velocity is 1 divided by 3
The velocity of the ball is 18.85 m/s.
The ball’s centripetal acceleration is 236.87 m/s².
The ball's centripetal force is 118.44 Newton.
What is centripetal acceleration?Centripetal acceleration is a characteristic of an object's motion along a circular path. Centripetal acceleration applies to any item travelling in a circle with an acceleration vector pointing in the direction of the circle's center.
Given parameters:
length of the string: l = 1.5 meters.
Time interval = 60 seconds.
Total number of complete rotation = 120.
Hence, the velocity of the ball = 120×2π×1.5/60 m/s
= 18.85 m/s.
The ball’s centripetal acceleration = (velocity)²/ radius
= (18.85)²/1.5 m/s²
= 236.87 m/s²
The ball's centripetal force = mass × centripetal acceleration
= 0.5 × 236.87 Newton
= 118.44 Newton
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A student throws a stone upward at an angle of
45° above the horizontal. Which statement best
describes the stone at the highest point that it
reaches?
(1) Its acceleration is zero.
(2) Its acceleration is a minimum, but not zero.
(3) Its gravitational potential energy is a
minimum
(4) Its kinetic energy is a minimum
Answer:
(4) Its kinetic energy is a minimum.
Explanation:
Stone experiments a parabolic motion, which is a combination of horizontal motion at constant speed and vertical uniform accelerated motion due to gravity, where effects from air viscosity and Earth's rotation are neglected. Meaning that stone represents a conservative system.
When stone reachest highest point, horizontal velocity remains unchanged and vertical velocity is zero. Acceleration remains constant and different of zero. Hence, gravitational potential energy is a maximum and kinetic energy is a minimum.
Correct answer is: (4) Its kinetic energy is a minimum.
The highest point, that stone reaches its kinetic energy is a minimum. therefore the option 4 is correct.
Projectile motion -The motion of an object, thrown (projected) into the air is known as projectile motion.
(1) Its acceleration is zero-
When an object is at its highest point the acceleration will be equal to the gravitational force (9.8 m/sec squared). If we take air resistance into the account it will be slightly greater than the 9.8 meters per second squared but not equal to zero in any case. Hence, statement 1 is incorrect.
(2) Its acceleration is a minimum, but not zero-
At the highest point, the object will be at the one place where air resistance does not affect the object, and thus acceleration is exactly equal to the acceleration due to gravity and at this position, it will be the maximum. Hence, statement 2 is incorrect.
(3) Its gravitational potential energy is a minimum-
At the highest point, the object will stop for the moment and have zero velocity. Thus it will have zero kinetic energy. Therefore total energy will have in the form of gravitational potential energy and which is maximum at this point. Hence, statement 3 is incorrect.
(4) Its kinetic energy is a minimum-
At the highest point the object will stop for the moment and have zero velocity. Thus it will have zero kinetic energy. Hence, statement 4 is correct.
At the highest point, that stone reaches its kinetic energy is a minimum. therefore the option 4 is correct.
For more about the projectile motion, follow the link below-
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Imagine two cases: Block N is pushed by a hand, which exerts a constant force F_o. AND moves a distance d_ 0. In case 1, it takes a time T to move this distance. In a case 2, it takes time 2T to move this distance. The work done by the hand on N in case 1 is ____________ the work done by the hand in case 2.
a. greater than
b. less than
c, equal to
Answer:
C: Equal to
Explanation:
In calculating workdone, it is pertinent to know that it doesn't depend on time. The only relationship between work and time is when we want to calcite power where workdone/time taken = power.
Now, even if it took 2T time to love the same distance, it just means lesser force was used but still the workdone doesn't change.
Thus, the workdone in the first case will be equal to the workdone in the second case.
A 3.50 kg box that is several hundred meters above the surface of the earth is suspended from the end of a short vertical rope of negligible mass. A time-dependent upward force is applied to the upper end of the rope, and this results in a tension in the rope of T(t) = (36.0 N/s)t. The box is at rest at t = 0. The only forces on the box are the tension in the rope and gravity.
Required:
a. What is the velocity of the box at (i) t = 1.00 s and (ii) t = 3.00 s?
b. What is the maximum distance that the box descends below its initial position?
c. At what value of t does the box return to its initial position?
Answer:
a. i. -4.65 m/s ii. -13.95 m/s b. 5.89 m c. 2.85 s
Explanation:
a. What is the velocity of the box at (i) t = 1.00 s and (ii) t = 3.00 s?
We write the equation of the forces acting on the mass.
So, T - mg = ma where T = tension in vertical rope = (36.0 N/s)t, m = mass of box = 3.50 kg, g = acceleration due to gravity = 9.8 m/s² and a = acceleration of box = dv/dt where v = velocity of box and t = time.
So, T - mg = ma
T/m - g = a
dv/dt = T/m - g
dv/dt = (36.0 N/s)t/3.50 kg - 9.8 m/s²
dv/dt = (10.3 m/s²)t - 9.8 m/s²
dv = [(10.3 m/s²)t - 9.8 m/s²]dt
Integrating, we have
∫dv = ∫[(10.3 m/s³)t - 9.8 m/s²]dt
∫dv = ∫(10.3 m/s³)tdt - ∫(9.8 m/s²)dt
v = (10.3 m/s³)t²/2 - (9.8 m/s²)t + C
v = (5.15 m/s³)t² - (9.8 m/s²)t + C
when t = 0, v = 0 (since at t = 0, box is at rest)
So,
0 = (5.15 m/s³)(0)² - (9.8 m/s²)(0) + C
0 = 0 + 0 + C
C = 0
So, v = (5.15 m/s³)t² - (9.8 m/s²)t
i. What is the velocity of the box at t = 1.00 s,
v = (5.15 m/s³)(1.00 s)² - (9.8 m/s²)(1.00 s)
v = 5.15 m/s - 9.8 m/s
v = -4.65 m/s
ii. What is the velocity of the box at t = 3.00 s,
v = (5.15 m/s³)(3.00 s)² - (9.8 m/s²)(3.00 s)
v = 15.45 m/s - 29.4 m/s
v = -13.95 m/s
b. What is the maximum distance that the box descends below its initial position?
Since v = (5.15 m/s³)t² - (9.8 m/s²)t and dy/dt = v where y = vertical distance moved by mass and t = time, we need to find y.
dy/dt = (5.15 m/s³)t² - (9.8 m/s²)t
dy = [(5.15 m/s³)t² - (9.8 m/s²)t]dt
Integrating, we have
∫dy = ∫[(5.15 m/s³)t² - (9.8 m/s²)t]dt
∫dy = ∫(5.15 m/s³)t²dt - ∫(9.8 m/s²)tdt
∫dy = ∫(5.15 m/s³)t³/3dt - ∫(9.8 m/s²)t²/2dt
y = (1.72 m/s³)t³ - (4.9 m/s²)t² + C'
when t = 0, y = 0.
So,
0 = (1.72 m/s³)(0)³ - (4.9 m/s²)(0)² + C'
0 = 0 + 0 + C'
C' = 0
y = (1.72 m/s³)t³ - (4.9 m/s²)t²
The maximum distance is obtained at the time when v = dy/dt = 0.
So,
dy/dt = (5.15 m/s³)t² - (9.8 m/s²)t = 0
(5.15 m/s³)t² - (9.8 m/s²)t = 0
t[(5.15 m/s³)t - (9.8 m/s²)] = 0
t = 0 or [(5.15 m/s³)t - (9.8 m/s²)] = 0
t = 0 or (5.15 m/s³)t = (9.8 m/s²)
t = 0 or t = (9.8 m/s²)/(5.15 m/s³)
t = 0 or t = 1.9 s
Substituting t = 1.9 s into y, we have
y = (1.72 m/s³)t³ - (4.9 m/s²)t²
y = (1.72 m/s³)(1.9 s)³ - (4.9 m/s²)(1.9 s)²
y = (1.72 m/s³)(6.859 s³) - (4.9 m/s²)(3.61 s²)
y = 11.798 m - 17.689 m
y = -5.891 m
y ≅ - 5.89 m
So, the maximum distance that the box descends below its initial position is 5.89 m
c. At what value of t does the box return to its initial position?
The box returns to its original position when y = 0. So
y = (1.72 m/s³)t³ - (4.9 m/s²)t²
0 = (1.72 m/s³)t³ - (4.9 m/s²)t²
(1.72 m/s³)t³ - (4.9 m/s²)t² = 0
t²[(1.72 m/s³)t - (4.9 m/s²)] = 0
t² = 0 or (1.72 m/s³)t - (4.9 m/s²) = 0
t = √0 or (1.72 m/s³)t = (4.9 m/s²)
t = 0 or t = (4.9 m/s²)/(1.72 m/s³)
t = 0 or t = 2.85 s
So, the box returns to its original position when t = 2.85 s
What three factors determine the amount of potential energy in a object are ______,______,and ______.
Answer:
It should be Mass, Gravity and Height
Explanation:
1. State the law of conservation of energy and what it means for you as a human considering how energy works.
2. Explain how different forms of energy are related.
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Answer: 1. The law of consevation of energy sates that energy can neither be created nor destroyed. It can only be transformed or transfered from one form to another. The law of conservation of energy is found everywhere for example, Water falls from the sky, converting potential energy to kinetic energy.
2. Different forms of energy are related because energy cannot be created or destroyed. they can all be transformed into from one form to another.
Explanation:
Express the speed of the electron in the Bohr model in terms of the fundamental constants (me, e, h, e0), the nuclear charge Z, and the quantum number n. Evaluate the speed of an electron in the ground states of He1 ion and U911. Compare these speeds with the speed of light c. As the speed of an object approaches the speed of light, relativistic effects become important. In which kinds of atoms do you expect relativistic effects to be greatest
Answer:
a) v = 4.37 10⁶ m / s, speed is much less than c
b) v = 2.01 10⁸ m / s, this value is 67% of the speed of light, , for which relativistic corrections should be used
Explanation:
The bohr model for the hydrogen atom and dendroids is a classical model with a quantization of the angular momentum
let's start by using Newton's second law with the electric force
F = m a
Coulomb's law electric force
F = [tex]k \frac{q_1q_2}{r^2}[/tex]
in this case in an atom the number of protons is equal to the atomic number and there is only one electron
q₁ = Ze
q₂ = e
acceleration is centripetal
a = v² / r
we substitute
[tex]k \frac{Z e^2}{r^2} = m \frac{v^2}{r}[/tex]
v² = [tex]k \frac{Ze^2}{m r}[/tex]
quantization is imposed without justification in this model,
L = p x r = n [tex]\hbar[/tex]
\hbar= h /2π
if we consider circular orbits, the speed and position are perpendicular
m v r = n \hbar
r = [tex]\frac{n \hbar}{m v}[/tex]
we substitute
v² = [tex]k \frac{Z e^2}{m} \frac{m v}{n \hbar}[/tex]
v = [tex]k \frac{Z e^2 }{ n \hbar}[/tex]
let's apply this equation
\hbar= h / 2π
\hbar= 6.626 10-34 / 2π
\hbar= 1.05456 10⁻³⁴ J s
a) He1 ion, the atomic number of helium is 2
v = [tex]\frac{9 \ 10^9 \ 2 ( 1.6 \ 10^{-19})^2 }{n \ 1.0546 \ 10^{-34}}[/tex]
v =4.3695 10⁶ / n m / s
the ground state occurs for N = 1
v = 4.37 10⁶ m / s
the relationship of this value to the speed of light is
v / c = 4.37 10⁶/3 10⁸
v / c = 1.46 10⁻²
speed is much less than c
b) the uranium ion with atomic number Z = 92
v = [tex]\frac{9 \ 10^9 \ 92 ( 1.6 \ 10^{-19})^2 }{n \ 1.054 \ 10^{-34} }[/tex]
v = 2.01 10⁸ m / s
v/c = [tex]\frac{2.01 \ 10^8 }{3 \ 10^8}[/tex]
v/c = 0.67
this value is 67% of the speed of light, for atoms with a higher atomic number the effects are increasingly important, for which relativistic corrections should be used
A car is sitting still. It accelerates to a constant speed then it decelerates again to zero speed. While the car is accelerating how do the directions of the angular acceleration and angular velocity of one of the wheels compare
Answer:
in the acceleration process the quantity α and w must increase
the deceleration process the alpha quantity must constant a direction opposite to the angular velocity
Explanation:
Acceleration and angular velocity are related to linear
v = w xr
a = αx r
The bold letters indicate vectors and the cross is a vector product, therefore if
we can see that the relationship between linear and angular variables is direct
therefore in the acceleration process the quantity α and w must increase as well as their linear counterparts
in the deceleration process the alpha quantity must constant as the linear acceleration and must have a direction opposite to the angular velocity