Answer: [tex]p_{x}=[/tex] 0.0135814 kg.m/s
[tex]p_{y}=-0.000219[/tex] kg.m/s
Explanation: When an object with mass is in motion, we say the object has momentum (p). Momentum is dependent on mass and velocity:
p = m.v
The total momentum of Bryce's mice is calculated as
x-axis
[tex]p_{x}=\Sigma m.v_{x}[/tex]
[tex]p_{x}=[(0.0213)(0.675)+(0.0165)(-0.249)+(0.0185)(0.395)+(0.0193)(-0.207)][/tex]
[tex]p_{x}=[/tex] 0.0135814
At the x-axis, total momentum of Bryce's mice is 0.0135814 kg.m/s.
y-axis
[tex]p_{y}=\Sigma m.v_{y}[/tex]
[tex]p_{y}=[(0.0213)(-0.417)+(0.0165)(-0.809)+(0.0185)(0.953)+(0.0193)(0.227)][/tex]
[tex]p_{y}=-0.000219[/tex]
At the y-axis, total momentum of Bryce's mice is -0.000219 kg.m/s.
The membrane that surrounds a certain type of living cell has a surface area of 4.7 x 10-9 m2 and a thickness of 1.3 x 10-8 m. Assume that the membrane behaves like a parallel plate capacitor and has a dielectric constant of 4.7. (a) The potential on the outer surface of the membrane is 79.5 mV greater than that on the inside surface. How much charge resides on the outer surface
Answer:
Q = 1.2*10⁻¹² C
Explanation:
For any capacitor, by definition the capacitance C is equal to the relationship between the charge on one of the conductors and the potential difference between them, as follows:[tex]C = \frac{Q}{V} (1)[/tex]
For the special case of a parallel plate capacitor, just by application of Gauss' law to a rectangular surface half out of the outer surface, and half inside it, it can be showed that the value of the capacitance C is a parameter defined only by geometric constants, as follows:[tex]C = \frac{\epsilon_{0}*\epsilon _{r} * A}{d} (2)[/tex]
So, due to the left sides in (1) and (2) are equal each other, right sides must be equal too.Replacing ε₀, εr (dielectric constant), A, d and V by their values, we can solve for Q, as follows:[tex]Q =\frac{\epsilon_{0} * \epsilon_{r} *A* V}{d} = \frac{(8.85*(4.7)^{2}*79.5)e-24 (F/m*m2*V)}{1.3e-8m} = 1.2e-12 C = 1.2 pC (3)[/tex]
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]
A steel ball (mass = 50 grams) and a plastic ball of the same dimensions (mass = 10 grams) are dropped from the same height at the same time. Which of the following will occur? (ASSUME NO AIR RESISTANCE)
a. the plastic ball will hit the ground before the steel ball hits the ground
b. the steel ball will hit the ground before the plastic ball hits the ground
c. the steel ball and the plastic ball will hit the ground at the same time
What is a measure of how much matter an object is made of?
Answer:
grams
Explanation:
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 student holds a 3.0kg mass in each hand while sitting on a rotating stool. when his arms are extended horizontally, the masses are 1.0m from the axis of rotation and he rotates with an angular speed of 0.75 rad/s. if the student pulls the masses horizontally to 0.30m from the axis of rotation, what is his new angular speed
Answer:
Explanation:
Due to change in the position of 3 kg mass , the moment of inertia of the system changes , due to which angular speed changes . We shall apply conservation of angular momentum , because no external torque is acting .
Initial moment of inertia I₁ = M R² = 3 x 1 ² = 3 kg m²
Final moment of inertia I₂ = M R² = 3 x .3 ² = 0.27 kg m²
Applying law of conservation of angular momentum
I₁ ω₁ = I₂ ω₂
Putting the values ,
3 x .75 = .27 x ω₂
ω₂ = 8.33 rad / s
New angular speed = 8.33 rad /s .
A Chinook salmon can jump out of water with a speed of 7.20 m/s . How far horizontally can a Chinook salmon travel through the air if it leaves the water with an initial angle of =29.0° with respect to the horizontal? (Neglect any effects due to air resistance.)
d=
Explanation:
The vertical component of the salmon's velocity is 7.2 m/s x sin 29 = 3.49 m/s
If g = 9.81 m/s^2, the salmon takes
(3.49 m/s) / (9.81 m/s^2) = 0.356 s to reach the highest point of its trajectory.
It takes another 0.356 s to fall back into the water again.
So the salmon is out of the water for a total of 0.712 s.
In this time the salmon travels horizontally with a velocity of 7.2 m/s x cos 29 = 6.30 m/s
We can now calculate the horizontal distance tavelled by multiplying the horizontal velocity by the time spent out of water;
0.712 s x 6.30 m/s = 4.48 m
3.
A student swings a ball attached to the end of a string 0.5m in length in
a vertical circle. The speed of the ball is 2m/s at the highest point and
6m/s at its lowest point: Find the acceleration of the ball at (ii) its highest
point and (ii) its lowest point.
Answer:
I.72m/s²
II.8m/s²
Explanation:
acceleration equal velocity² divided by length
what occurred when the photosynthetic began to pump free oxygen into oceans?
when the photosynthetic began to pump free oxygen into oceans, the ocean had enough oxygen to support the life of non-photosynthetic organisms. So, non-photosynthetic organisms came into being.
What is the difference between elastic PE and gravitational PE?
A car is traveling 100 km/hr. How many hours will it take to cover a distance of 850 km?
Your answer:
.118 hours
8.5 hours
7.5 hours
23 hours
Answer:
8.5 hours
Explanation:
What is the power of 600j of work done in 4 seconds?
Explanation:
Power = change in work/change in time
P = 600 joules/ 4 seconds
P= 150 watts
hope this helps :)
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 10-meter-long, 150 kg beam extends horizontally from a wall.One end of the beam is fixed to the wall and the other end is attached to the same wall by a cable that makes an angle of 60° with the horizontal. A 75 kg sign is hung from the beam 2.50 meters from the wall.
Determine the magnitude of the tension, in [N] on the cable necessary to keep the system in equilibrium.
Answer:
the magnitude of the tension on the cable necessary to keep the system in equilibrium is 1060.9 N
Explanation:
Given that;
Length L = 10 m
mass of beam m_b = 150 kg; weight W_beam = 150×9.8
mass of sign m = 75 kg
distance of sign hung from the beam from the wall d = 2.50 m
angle ∅ = 60°
g = 9.8 m/s²
Now,
Torque acting at one end of the beam will be;
[tex]T_{net}[/tex] = Tsin∅ × L - mg(d)-W × (L/2)
for equilibrium, [tex]T_{net}[/tex] = 0
therefore, 0 = Tsin∅ × L - mg(d)-W × (L/2)
so we substitute
Tsin(60°) × 10 - 75×9.8(2.50) - 150 × 9.8× (10/2) = 0
Tsin(60°) × 10 - 1837.5 - 7350 = 0
Tsin(60°) × 10 - 9187.5 = 0
Tsin(60°) × 10 = 9187.5
divide both side by 10
Tsin(60°) = 918.75
T × 0.8660 = 918.75
T = 918.75 / 0.8660
T = 1060.9 N
Therefore, the magnitude of the tension on the cable necessary to keep the system in equilibrium is 1060.9 N
The magnitude of the tension on the cable that keep the system in equilibrium is 1060.9 N.
Torque acting at one end of the beam,
= Tsin∅ × L - mg(d)-W × (L/2)
When equilibrium = 0
Tsin∅ × L - mg(d)-W × (L/2) = 0
Where,
L - Length = 10 m
m - mass of sign bord= 75 kg
g- gravitational accelaration = 9.8 m/s²
W - weight of beam = 150×9.8 = 1470 kg
Put the values in the formula,
Tsin(60°) × 10 - 75×9.8(2.50) - 150 × 9.8× (10/2) = 0
Tsin(60°) × 10 - 1837.5 - 7350 = 0
Tsin(60°) × 10 - 9187.5 = 0
Tsin(60°) × 10 = 9187.5
Tsin(60°) = 918.75
T × 0.8660 = 918.75
T = 918.75 / 0.8660
T = 1060.9 N
Therefore, the magnitude of the tension on the cable that keep the system in equilibrium is 1060.9 N.
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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 .
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
1. An object that does not give off its own light is called ____
2. The tilt of the Earth's axis creates the
_______
3. Both of the movements of the Earth causes interesting changes on the _______
4. The word solar means of the ______
5. When the Sun, the Earth and the Moon are in line there is an/a _____
Answer:
1. non-luminous objects
sorry have to say the rest in the comments because brai.nly is doing the most for no reason
Explanation:
hope this helps sorry if it doesn't have a good rest of your day/afternoon :) ❤
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 police officer is parked by the side of the road, when a speeding car travelling at 50 mi/hrpasses. The police car immediately pursues it, accelerating at a rate of 10 mi/hr per second.The road is fairly busy, so the officer will not go faster than a top speed of 70 mi/hr. How longwill it take before the officer catches up to the speeding car, and how far will it have travelled inorder to do so
Answer:
a) time taken to catch up with speeding car is 12.25 secs
b) the police car will travel 273.8 m to catch up with the speeding car
Explanation:
Given that;
speed of car [tex]V_{c}[/tex] = 50 mi/hr = 22.352 m/s
acceleration of police car = 10 mi/hr = 4.47 m/s²
[tex]V_{f}[/tex] = 70 mi/hr = 31.29 m/s
Now time taken to reach maximum speed is t₁
so
[tex]V_{f}[/tex] = [tex]V_{i}[/tex] + at₁
we substitute
31.29 = 0 + 4.47t₁
t₁ = 31.29 / 4.47
t₁ = 7 sec
now
d₁ = 0 + 1/2 × at₁²
d₁ = 0 + 1/2 × 0 + 4.47×(7)²
d₁ = 109.5 m
so distance travelled by the speeding car in time t₁ will be
[tex]d_{c}[/tex] = [tex]V_{c}[/tex] × t₁
we substitute
[tex]d_{c}[/tex] = 22.352 × 7
[tex]d_{c}[/tex] = 156.46 m
now distance between polive car and speeding car
Δd = [tex]d_{c}[/tex] - d₁
Δd = 156.46 - 109.5
Δd = 46.96 m
time taken to cover Δd will be
t₂ = Δd / ( [tex]V_{f}[/tex] - [tex]V_{c}[/tex] )
t₂ = 46.96 / ( 31.29 - 22.352 )
t₂ = 46.96 / 8.938
t₂ = 5.25 sec
distance travelled by the police in time t₂ will be
d₂ = [tex]V_{f}[/tex] × t₂
d₂ = 31.29 × 5.25
d₂ = 164.3 m
a) How long will it take before the officer catches up to the speeding car;
time taken to catch up with speeding car;
t = t₁ + t₂
t = 7 + 5.25
t = 12.25 secs
Therefore, time taken to catch up with speeding car is 12.25 secs
b) how far will it have travelled in order to do so;
distance = d₁ + d₂
distance = 109.5 + 164.3
distance = 273.8 m
Therefore, the police car will travel 273.8 m to catch up with the speeding car
You serve a volleyball with a mass of 2.1 kg. The ball leaves your hand with a speed of 2.1 m/s. The
ball has ____
energy. Calculate it.
Answer:
4.6 Joules
Explanation:
K=1/2*MV^2
1/2 * 2.1kg * 2.1^2m/s
==4.6305 Joules
simplified to 4.6 Joules
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
how do i get the answer for keplers law 3
A spider accelerates from a standstill to 5m/s in 10s. What is its acceleration?
Answer:
Acceleration = 0.5m/s²
Explanation:
Given the following data;
Final velocity = 5m/s
Time = 10 seconds
Since the spider started from rest, its initial velocity is equal to 0m/s
To find the acceleration;
In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.
This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.
Mathematically, acceleration is given by the equation;
[tex]Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}[/tex]
[tex]a = \frac{v - u}{t}[/tex]
Where,
a is acceleration measured in [tex]ms^{-2}[/tex]
v and u is final and initial velocity respectively, measured in [tex]ms^{-1}[/tex]
t is time measured in seconds.
Substituting into the equation, we have;
[tex]a = \frac{5 - 0}{10}[/tex]
[tex]a = \frac{5}{10}[/tex]
Acceleration = 0.5m/s²
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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How do protons neutrons and electrons differ
Answer/Explanation:
They have a relatively small mass compared to Protons and Neutrons. Protons are electrochemically positive in charge and the Neutrons are electrochemically neutral in charge.
Answer:
They differ because
Proton means positive charge
Electrons are negatively charged
Neutrons are neutral
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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PLEASE I REALLY NEED HELP!
Question 6
If the car traveled a total of 1,200 meters during this test, what was the average speed of the car? Include the
correct units.
Answer:
[tex]v=\dfrac{1200}{t}\ m/s[/tex]
Explanation:
Given that,
The car traveled a total of 1,200 meters during this test.
We need to find the average speed of the car. The average speed of the car is given by total distance covered divided by the time taken. So,
[tex]v=\dfrac{1200}{t}\ m/s[/tex]
But putting the value of t we can find the average speed of the car.
A movie stunt double is supposed to run across the top of a train (in the opposite direction that the train is moving) and just barely jump off before reaching a tunnel, but after reaching the end of the train (starting from the front). If the train is moving at 150 km/hr, is 2 km long and the tunnel is 20 km away from the end (where the stunt double is going to jump from), how fast (in km/hr) will the stunt double need to run
Answer:
the required speed/velocity of the stunt double is 13.633 km/h
Explanation:
Given the data in the question;
velocity of train V = 150 km/h
distance = length of train + distance between the tunnel and the end
= 2 km + 20 km = 22 km
first we calculate time t taken by the train to reach the tunnel;
t = distance / velocity
we substitute
t = 22 km / 250 km/h
t = 0.1467 hr
so the velocity of the of the stunt double will be;
velocity = distance / time
we substitute
velocity = 2 km / 0.1467 hr
velocity = 13.633 km/h
Therefore, the required speed/velocity of the stunt double is 13.633 km/h
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
g A boat is anchored 2000 ft from shore anddirects its searchlight towards an automobile travelingdown the straight road. At the particular moment whenthe distance rfrom the searchlight to the automobile is3000 ft, the automobile has speed 80 ft/s and increasesits speed at a rate of 15 ft/s2 down the road.Find the required angular velocity and angularacceleration of the boat's searchlight to track the automobile at this instant.(Answers: 0.018 rad/s CCW, 0.004 rad/s2 CCW)
Answer:
Explanation:
The distance of searchlight will act as radius R and velocity of car may be supposed to be tangential velocity v . We are required to calculate angular velocity ω .
v = 80 ft /s
R = 3000 ft
ω = v / R
= 80 / 3000 = .027 rad / s
For angular acceleration the formula is
angular acceleration α = a / R
a is linear acceleration = 15 ft / s²
α = 15 / 3000 = .005 rad / s².