Answer:
(a) 508.37 m
(b) 47.53 s
(c) 21.165 m
(d) 19.365 m
Explanation:
initial velocity, u = 77 km/h = 21.39 m/s
acceleration, a = - 0.45 m/s^2
(a) final velocity, v = 0
Let the distance traveled is s.
Use third equation of motion
[tex]v^{2}=u^{2}+2as[/tex]
[tex]0^{2}=21.39^{2}-2 \times 0.45 \times s[/tex]
s = 508.37 m
(b) Let t be the time taken to stop.
Use first equation of motion
v = u + at
0 = 21.39 - 0.45 t
t = 47.53 s
(c) Use the formula for the distance traveled in nth second
[tex]s_{n^{th}=u+\frac{1}{2}a\left ( 2n-1 \right )}[/tex]
where n be the number of second, a be the acceleration, u be the initial velocity.
put n = 1, u = 21.39 m/s , a = - 0.45m/s^2
[tex]s_{n^{th}=21.39-\frac{1}{2}\times 0.45\left ( 2\times 1-1 \right )}[/tex]
[tex]s_{n^{th}=21.165m[/tex]
(d) Use the formula for the distance traveled in nth second
[tex]s_{n^{th}=u+\frac{1}{2}a\left ( 2n-1 \right )}[/tex]
where n be the number of second, a be the acceleration, u be the initial velocity.
put n = 5, u = 21.39 m/s , a = - 0.45m/s^2
[tex]s_{n^{th}=21.39-\frac{1}{2}\times 0.45\left ( 2\times 5-1 \right )}[/tex]
[tex]s_{n^{th}=19.365m[/tex]
1 microgram equals how many milligrams?
Answer: 1 microgram is equal to 0.001 miligrams
Explanation: The factor micro is equal 10^-3 while the factor mili is equal to 10^-3 so to converte the micro to mile we have to multiply by 0.001.
An object is originally moving at a constant velocity of 8 m/s in the -x direction. It moves at this constant velocity for 3 seconds. Then, a constant acceleration of 7 m/s2 is applied to it in the +x direction for 9 seconds. What is the total distance covered by this object in meters? Please give a detailed explanation.
Answer:
244.64m
Explanation:
First, we find the distance traveled with constant velocity. It's simply multiplying velocity time the time that elapsed:
[tex]x = V*t = -8\frac{m}{s} *3s = -24m[/tex]
After this, the ball will start traveling with a constant acceleration motion. Due to the fact that the acceleration is the opposite direction to the initial velocity, this motion will have 2 phases:
1. The velocity will start to decrease untill it reaches 0m/s.
2. Then, the velocity will start to increase at the rate of the acceleration.
The distance that the ball travels in the first phase can be found with the following expression:
[tex]v^2 = v_0^2 + 2a*d[/tex]
Where v is the final velocity (0m/s), v_0 is the initial velocity (-8m/s) and a is the acceleration (+9m/s^2). We solve for d:
[tex]d = \frac{v^2 - v_0^2}{2a} = \frac{(0m/s)^2 - (-8m/s)^2}{2*7m/s^2}= -4.57m[/tex]
Now, before finding the distance traveled in the second phase, we need to find the time that took for the velocity to reach 0:
[tex]t_1 = \frac{v}{a} = \frac{8m/s}{7m/s^2} = 1.143 s[/tex]
Then, the time of the second phase will be:
[tex]t_2 = 9s - t_1 = 9s - 1.143s = 7.857s[/tex]
Using this, we using the equations for constant acceleration motion in order to calculate the distance traveled in the second phase:
[tex]x = \frac{1}{2}a*t^2 + v_0*t + x_0[/tex]
V_0, the initial velocity of the second phase, will be 0 as previously mentioned. X_0, the initial position, will be 0, for simplicity:
[tex]x = \frac{1}{2}*7\frac{m}{s^2}*t^2 + 0m/s*t + 0m = 216.07m[/tex]
So, the total distance covered by this object in meters will be the sum of all the distances we found:
[tex]x_total = 24m + 4.57m + 216.07m = 244.64m[/tex]
A rocket accelerates upward from rest, due to the first stage, with a constant acceleration of a1 = 67 m/s2 for t1 = 39 s. The first stage then detaches and the second stage fires, providing a constant acceleration of a2 = 34 m/s2 for the time interval t2 = 49 s.
(a) Enter an expression for the rocket's speed, v1, at time t1 in terms of the variables provided.
(b) Enter an expression for the rocket's speed, v2, at the end of the second period of acceleration, in terms of the variables provided in the problem statement.
(c) Using your expressions for speeds v1 and v2, calculate the total distance traveled, in meters, by the rocket from launch until the end of the second period of acceleration.
Answer:
(a) [tex]v_{1}=0+a_{1}t_{1} = a_{1}t_{1}[/tex].
(b) [tex]v_{final}=v_{initial}+at\\ v_{2}=v_{1}+a_{2} t_{2}[/tex]
(c) [tex]219807.5m[/tex]
Explanation:
Part (a):To find an expression for the rocket's speed [tex]v_{1}[/tex] at time [tex]t_{1}[/tex], we use the constant acceleration model, which relates these variables with the expression:[tex]v_{final} =v_{intial}+at[/tex]. In this case, the initial velocity is null, because accelerates from rest. So, we take all values of the first interval, and we replace it to find the expression:
[tex]v_{1}=0+a_{1}t_{1} = a_{1}t_{1}[/tex]. (expression for the first interval).
Part (b):Then, we do the same process to find the expression of the second interval, we just replace the variables given:
[tex]v_{final}=v_{initial}+at\\ v_{2}=v_{1}+a_{2} t_{2}[/tex]
In this case, you can notice that the initial velocity used is the one we obtain from the first interval, because the end of the first period is the beginning of the second period.
Part (c):To calculate the total distance we have to sum the distance covered during the two intervals, that it's translated as: [tex]d_{total} = d_{1} + d_{2}[/tex].
Then, we use this equation to replace in each distance: [tex]v_{final}^{2} = v_{initial}^{2} +2ad[/tex].
Isolating d we have:
[tex]d=\frac{v_{final}^{2}-v_{initial}^{2}}{2a}[/tex].
Now, we apply the equation to each interval to obtain [tex]d_{1}[/tex] and [tex]d_{2}[/tex]:
[tex]d_{1}=\frac{v_{1}^{2}-0}{2a_{1}}[/tex].
[tex]d_{2}=\frac{v_{2}^{2}-v_{1}^{2}}{2a_{2}}[/tex].
Before calculating the total distance, we need to know the magnitude of each speed.
[tex]v_{1}=a_{1}t_{1}[/tex]
[tex]v_{1}=(67)(39)=[tex]v_{2}=v_{1}+a_{2} t_{2}= 2613\frac{m}{s^{2}}+34\frac{m}{s^{2}}(49sec)=4279\frac{m}{s}[/tex][/tex]
At last, we use all values known to calculate the total distance:
[tex]d_{total} = d_{1} + d_{2}[/tex]
[tex]d_{total} =\frac{v_{1}^{2}-0}{2a_{1}} + \frac{v_{2}^{2}-v_{1}^{2}}{2a_{2}}[/tex].
[tex]d_{total} =\frac{(2613)^{2} }{2(67)} +\frac{(4279)^{2}-(2613)^{2} }{2(34)}\\ d_{total}=50953.5+168854=219807.5m[/tex]
Therefore, the total distance traveled until the ends of the second period is [tex]219807.5m[/tex]. The rocket is in the Thermosphere.
On a hot summer day in the state of Washington while kayaking, I saw several swimmers jump from a railroad bridge into the Snohomish River below. The swimmers stepped off the bridge, and I estimated that they hit the water 1.50 s later. a.) How high was the bridge?
b.) How fast were the swimmers moving when they hit the water?
c.) What would the swimmer's drop time be if the bridge were twice as high?
The bridge was approximately 11.01 meters high. The swimmers were moving at a speed of around 14.72 m/s when they hit the water. If the bridge were twice as high, the drop time would be about 2.12 seconds.
Explanation:The subject in question involves principles of physics, specifically gravitational acceleration. When an object is dropped, it falls under the influence of gravity. This is usually around 9.81 m/s2 on Earth. Using the formula for motion d=1/2gt^2, where d is the distance or height, g is the acceleration due to gravity, and t is the time it takes for the object to fall.
a.) For calculation of the bridge's height, plug in the values of time (t=1.5s) and gravity (g=9.81m/s^2) into the equation, we get d=1/2*(9.81)*(1.5)^2 = 11.01 meters. The height of the bridge from which they jumped is approximately 11 meters.
b.) To calculate how fast the swimmers were moving when they hit the water, we can use the equation v=gt, where v is the velocity or speed. Plugging in the values, we get v=(9.81)*(1.5) = 14.72 m/s. So, the swimmers were moving at about 14.7 meters per second when they hit the water.
c.) If the bridge were twice as high, the time of drop would be found using the equation t= sqrt(2d/g), where d is now twice the original distance = 2*11.01 = 22.02 meters. Substituting the given values, we get t= sqrt(2(22.02)/9.81) = 2.12 seconds. The swimmer's drop time would be approximately 2.12 seconds.
Learn more about Gravitational Acceleration here:https://brainly.com/question/30429868
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Time dilation: A missile moves with speed 6.5-10 m/s with respect to an observer on the ground. How long will it take the missile's clock to fall behind the ground observer's clock by 1 millisecond? Hint: use the binomial formula:(1+x)a1+ ax.
Answer:
The time taken by missile's clock is [tex]4.6\times 10^{6} s[/tex]
Solution:
As per the question:
Speed of the missile, [tex]v_{m = 6.5\times 10^{3}} m/s[/tex]
Now,
If 'T' be the time of the frame at rest then the dilated time as per the question is given as:
T' = T + 1
Now, using the time dilation eqn:
[tex]T' = \frac{T}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}[/tex]
[tex]\frac{T'}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}[/tex]
[tex]\frac{T + 1}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}[/tex]
[tex]1 + \frac{1}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}[/tex]
[tex]1 + \frac{1}{T} = (1 + (\frac{v_{m}}{c})^{2})^{- \frac{1}{2}}[/tex] (1)
Using binomial theorem in the above eqn:
We know that:
[tex](1 + x)^{a} = 1 + ax[/tex]
Thus eqn (1) becomes:
[tex]1 + \frac{1}{T} = 1 - \frac{- 1}{2}.\frac{v_{m}^{2}}{c^{2}}[/tex]
[tex]T = \frac{2c^{2}}{v_{m}^{2}}[/tex]
Now, putting appropriate values in the above eqn:
[tex]T = \frac{2(3\times 10^{8})^{2}}{(6.5\times 10^{3})^{2}}[/tex]
[tex]T = 4.6\times 10^{6} s[/tex]
Consider two displacements, one of magnitude 15 m and another of magnitude 20 m. What angle between the directions of this two displacements give a resultant displacement of magnitude (a) 35 m, (b) 5 m, and (c) 25 m.
Answer:
a) 0°
b) 180°
c) 90°
Explanation:
Hello!
To solve this question let a be the vector whose length is 15 m and b the vector of length 20 m
So:
|a | = 15
|b | = 20
Since we are looking for the angle between the vectors we need to calculate the length of the sum of the two vectors, this is:
[tex]|a+b|^{2} = |a|^{2} + |b|^{2} + 2 |a||b|cos(\theta)[/tex]
Now we replace the value of the lengths:
[tex]|a+b|^{2} = 15^{2} + 20^{2} + 2*15*20*cos(\theta)[/tex]
[tex]|a+b|^{2} = 625 + 600*cos(\theta)[/tex] --- (1)
Now, if:
a) |a+b| = 35
First we can see that 20 + 15 = 35, so the angle must be 0, lets check this:
[tex]35^{2} = 625 + 600*cos(\theta)[/tex]
[tex]1225 = 625 + 600*cos(\theta)[/tex]
[tex]600 = 600*cos(\theta)[/tex]
[tex]1= cos(\theta)[/tex]
and :
[tex]\theta = arccos(1)[/tex]
θ = 0
b) |a+b|=5
From eq 1 we got:
[tex]\theta = arccos(\frac{|a+b|^{2}-625}{600})[/tex] --- (2)
[tex]\theta = arccos(\frac{|a+b|^{2}-625}{600})[/tex]
[tex]\theta = arccos(-1)[/tex]
θ = π or θ = 180°
c) |a+b|=25
[tex]\theta = arccos(\frac{|25|^{2}-625}{600})[/tex]
[tex]\theta = arccos(-1)[/tex]
θ = π/2 or θ = 90°
Final answer:
In vector addition, an angle of 0° between two vectors gives a resultant of 35 m, an angle of 180° gives a resultant of 5 m, and the angle for a resultant of 25 m can be found using the Law of Cosines.
Explanation:
The question involves the concept of vector addition and the use of trigonometry to determine the resultant displacement when two vectors are combined at various angles. The displacement vectors have magnitudes of 15 m and 20 m, and we are interested in finding the angles that would result in resultant displacements of 35 m, 5 m, and 25 m, respectively.
For (a) a resultant displacement of 35 m, the two vectors must be added in the same direction. This implies that the angle between them is 0°.
For (b) a resultant displacement of 5 m, the two vectors must be in exactly opposite directions. Since the difference in magnitudes is 5 m, this means that the larger vector (20 m) partially cancels out the smaller vector (15 m). Hence, the angle between them is 180°.
For (c) a resultant displacement of 25 m, we can use the Law of Cosines to determine the angle:
c2 = a2 + b2 - 2ab cos(θ)
Where a = 15 m, b = 20 m, and c = 25 m. Solving this equation will give us the value of θ.
An airplane in the process of taking off travels with a speed of 80 m/s at an angle of 15° above the horizontal. What is the ground speed of the airplane? O 80 m/s O 21 m/s O 77 m/s O 2.6 m/s
Answer:
Option C
Explanation:
given,
velocity of airplane = 80 m/s
angle with the horizontal = 15°
speed of the ground= ?
when the plane is taking off the horizontal component of the velocity is v cosθ
so,
ground speed of the airplane is = [tex]v\times cos\theta[/tex]
= [tex]80 \times cos 15^0[/tex]
v = 77.27 m/s
horizontal velocity of the air plane comes out to be 77.27 m/s ≅ 77 m/s
so, the correct option is Option C
What is the significance of each of the following in the study of astronomy:
(a) Dark Matter
(b) 21 cm Radiation
(5 pts) Describe the overall structure and main parameters of the Milky Way galaxy.
(10 pts) Describe the main characteristics of Einstein’s Theory of General Relativity.
(4 pts) Explain the major characteristics/properties of Pulsars.
Explanation:
a) Dark matter: a kind of matter we can't directly observe but that we can imply its presence in various observations including gravitational effects which can't be explained by accepting theories of gravity unless more matter is present. This is the reason why dark matter is thought to account for around 80% of the matter in the universe.
b)21cm Radiation: also called hydrogen line, is a spectral line emitted by neutral hydrogen, it has a frequency of 1420megahertz and 21 cm wavelength. In astronomy, this line is used to study the amount and velocity of hydrogen in the galaxy.
Milky way: it's a barred spiral galaxy with more than 200 billion stars, approximately 100000 light-years in diameter and the sun is located about 28000light years from the center.
From the outside its structure has the following characteristics:
Galactic disk: it is made out of old and young stars, as well as gas and dust, gravitational interactions between stars cause a circular motion with up and down motions, this disk is divided in three other parts, a nucleus (the center of the disk), a bulge (around the nucleus) and spiral arms (extended areas)Globular clusters: located above and below the disk, the stars in this zone are older and there's no gar or dust.Halo: large region surrounding the galaxy, it is made of hot gas and dark matter.Two fundamentals parameters of the milky Way are [tex]R_{0}[/tex] (the radial distance from de sun to the galactic center) and [tex]Θ_{0}[/tex], the galactic rotational velocity at [tex]R_{0}[/tex]
Einstein's Theory of General Relativity: general relativity is a metric theory of gravitation, that defies gravity as a geometric property of space and time, this means there's no gravitational force deflecting objects from their natural straight paths, but a change in properties of space and time that changes this straight path into a curve.
At weak gravitational fields and slow speed, this theory overlaps with Newton's.
Pulsars: they are rotating neutron stars that emit a focused beam of electromagnetic radiation its formation happens when a medium mass star dies and it maintains its angular momentum emitting a powerful blast of radiation along its magnetic field lines. They are useful to search for gravitational waves, and even to find extrasolar planets.
I hope you find this information interesting and useful! Good luck!
If the wavelength of an electron is 4.63 x 10^−7 m, how fast is it moving?
Answer:
it move with velocity 1571 m/s
Explanation:
given data
wavelength λ = 4.63 × [tex]10^{-7}[/tex] m
to find out
how fast is it moving
solution
we will use here de Broglie wavelength equation
that is
wavelength λ = [tex]\frac{h}{mv}[/tex] ..........1
here h is planck constant = 6.626068 × [tex]10^{-34}[/tex]
and m is mass of electron i.e = 9.10938188 × [tex]10^{-31}[/tex]
and v is velocity
put all value we find velocity in equation 1
wavelength λ = [tex]\frac{h}{mv}[/tex]
v = [tex]\frac{6.626068*10^{-34}}{9.10938188*10^{-31}*4.63*10^{-7}}[/tex]
v = 1571.035464
so it move with velocity 1571 m/s
Answer:
[tex]v=1.57*10^{3}\frac{m}{s}[/tex]
Explanation:
As DeBroglie equation proved by Davisson-Germer experiment says, the wavelength of an electron is related with its velocity with the equation:
λ = [tex]\frac{h}{mv}[/tex]
where m is the mass of the electron [tex]m=9.11*10^{-31}kg[/tex], h is the Planck´s constant [tex]h=6.626*10^{-34}J.s[/tex] and v its velocity.
Solving the equation for the velocity of the electron, we have:
v = h/mλ
And replacing the values:
[tex]v=\frac{6.626*10^{-34}J.s}{(9.11*10^{-31}Kg)*(4.63*10^{-7}m)}[/tex]
[tex]v=1570.9\frac{m}{s}[/tex]
[tex]v=1.57*10^{3}\frac{m}{s}[/tex]
A charge q1= 3nC and a charge q2 = 4nC are located 2m apart. Where on the line passing through these charges is the total electric field zero?
Answer:
Explanation:
Electric field due to a charge Q at a point d distance away is given by the expression
E = k Q / d , k is a constant equal to 9 x 10⁹
Field due to charge = 3 X 10⁻⁹ C
E = E = [tex]\frac{9\times 10^9\times3\times10^{-9}}{d^2}[/tex]
Field due to charge = 4 X 10⁻⁹ C
[tex]E = [tex]\frac{9\times 10^9\times4\times10^{-9}}{(2-d)^2}[/tex]
These two fields will be equal and opposite to make net field zero
[tex]\frac{9\times 10^9\times3\times10^{-9}}{d^2}[/tex] = [tex][tex]\frac{9\times 10^9\times4\times10^{-9}}{(2-d)^2}[/tex][/tex]
[tex]\frac{3}{d^2} =\frac{4}{(2-d)^2}[/tex]
[tex]\frac{2-d}{d} =\frac{2}{1.732}[/tex]
d = 0.928
It has been said that in his youth George Washington threw a silver dollar across a river. Assuming that the wide, (a) what minimum initial speed river was 75 m was necessary to get the coin across the river and b) how long was the coin in flight?
Answer:
(a) 27.1 m/s
(b) 3.9 second
Explanation:
Let the speed is u.
Maximum horizontal range, R = 75 m
The range is maximum when the angle of projection is 45°.
(a) Use the formula for the maximum horizontal range
[tex]R=\frac{u^{2}}{g}[/tex]
[tex]75=\frac{u^{2}}{9.8}[/tex]
u = 27.1 m/s
(b) Let the time of flight is T.
Use the formula for the time of flight
[tex]T=\frac{2uSin\theta}{g}[/tex]
[tex]T=\frac{2\times 27.1 \times Sin45}{9.8}[/tex]
T = 3.9 second
Answer:
A and B
Explanation:
A light year is defined as the distance that light can travel in 1 year. What is the value of 1 light year in meter? Show your complete work. [Note: light travels 3 x 10^8 m in 1s.]
Answer:
d=9.462×10^15 meters
Explanation:
Relation between distance, temps and velocity:
d=v*t
t=1year*(365days/1year)*/(24hours/1day)*(3600s/1h)=31536000s
So:
1 light year=d=3*10^8m/s*3.154*10^7s=9.462×10^15 meters
Suppose you are sitting on a rotating stool holding a 2 kgmass
in each outstretched hand. If you suddenly drop the masses,will
your angular velocity increase, decrease, or stay the same?Please
Explain.
Answer:Increase
Explanation:
Given
You are holding 2 kg mass in each outstreched hand
If the masses are dropped then Moment of inertia will decease by [tex]2mr^2[/tex]
Where m=2 kg
r=length of stretched arm
Since angular momentum is conserved therefore decrease in Moment of inertia will result in increase of angular velocity
as I[tex]\omega [/tex]=constant
I=Moment of inertia
[tex]\omega [/tex]=angular velocity
A basketball player is running at 4.80 m/s directly toward the basket when he jumps into the air to dunk the ball. He maintains his horizontal velocity. (a) What vertical velocity (in m/s) does he need to rise 0.650 meters above the floor? _______m/s (b) How far (in m) from the basket (measured in the horizontal direction) must he start his jump to reach his maximum height at the same time as he reaches the basket?_______m
Answer:
a) 3.56m/s
b) 1.73m
Explanation:
We have to treat this as a parabolic motion problem:
we will use the next formulas:
[tex]V=Vo+a*t\\Vy^2=Vyo^2+2*a*Y\\X=Vox*t[/tex]
we first have to calculate the initial velocity of the basketball player:
[tex]Vy^2=Vyo^2+2*a*Y\\0^2=Vyo^2+2*(-9.8)*(0.650)\\Vyo=\sqrt{2*9.8*0.650} \\Vyo=3.56 m/s[/tex]
the final velocity is zero when he reaches the maximun height.
To answer the second part we need to obtain the time to reach the maximun height, so:
[tex]V=Vo+a*t\\\\0=3.56+(-9.8)*t\\t=0.36 seconds[/tex]
Now having that time, let's find the distance on the X axis, the X axis behaves as constant velocity movement, so:
[tex]X=Vox*t\\X=4.80*0.36\\X=1.73m[/tex]
The force of attraction between a -130.0 C and +180.0 C charge is 8.00 N. What is the separation between these two charges in meter rounded to three decimal places? (k = 1/470 - 9.00 10°N.m2/C2 1uC = 106C)
Answer:
distance between the charges is 5.12 × 10⁶ m
Explanation:
charges q₁ = -130.0 C
q₂ = 180 C
force between the charges = 8 N
force between two charge
[tex]F = \dfrac{k q_1q_2}{r^2}[/tex]
value of K =8.975 × 10⁹ N.m²/C²
[tex]8 = \dfrac{8.975 \times 10^{9}\times 130 \times 180}{r^2}[/tex]
[tex]r^2 = \dfrac{8.975 \times 10^{9}\times 130 \times 180}{8}[/tex]
[tex]r^2 =2.625 \times 10^{13} [/tex]
r = 5.12 × 10⁶ m
hence, distance between the charges is 5.12 × 10⁶ m.
Assuming that a rabbit runs at a speed of 12.0 meters per second. How far would the rabbit travel in 11.5 seconds? A hiker takes a hiking trip around the camp. He first travels 5.10 km in one hour at a constant speed. He then travels 16.5 km in two hours at a different constant speed and reaches a ranch tower. What is the average speed of the hiker for the entire trip?
Answer:
(A). The rabbit travel the distance 138 m.
(B). The average speed of the hiker for the entire trip is 7.2 km/h.
Explanation:
Given that,
Speed = 12.0 m
Time = 11.5 second
(A). We need to calculate the distance
Using formula of distance
[tex]d=v\times t[/tex]
[tex]d =12.0\times11.5[/tex]
[tex]d=138\ m[/tex]
The rabbit travel the distance 138 m.
(B). Given that,
Distance = 5.10 km
Time = 1 hours
Distance 16.5 km
Time = 2 hours
We need to calculate the average speed of the hiker for the entire trip
Using formula of average speed
[tex]v=\dfrac{D}{T}[/tex]
Where, D = Total distance
T = Total time
Put the value into formula
[tex]v=\dfrac{5.10+16.5}{1+2}[/tex]
[tex]v=7.2\ km/h[/tex]
The average speed of the hiker for the entire trip is 7.2 km/h.
Hence, This is the required solution.
Which of the following statements is/are true? Select all correct answers. An orbital is the probability distribution function describing the volume in which we are most likely to find an electron. The emission spectrum of a particular element is always the same and can be used to identify the element. The uncertainty principle states that we can never know both the exact location and speed of an electron. Part of the Bohr model proposed that electrons in the hydrogen atom are located in "stationary states" or particular orbits around the nucleus.
Answer:
The emission spectrum is always the same and can be used to identify the element and part of the Bohr model proposed that electrons in the hydrogen are located in particular orbits around the nucleos are True.
Explanation:
The Niels Bohr and quantic mecanic theorys are both based on the study of atomics spectrums. The atomic spectrum is a characteristic pattern of a light wavelenght emited wich is unique to each element.
For example, if we put some low pressure hydrogen in a glass tube and in the tp of the glass we apply a voltage big enough to produce a electric current in the hydrogen gas, the tube its going to emit light wich have a color dependig of the gas element in the interior. If we observe this light with a spectrometer we are going to see shining lines and each one of this lines have a wavelenght and diferent colors. This lines are called emission spectrum and the wavelength of that spectrum are unique to eache element.
Summering up, we can identify elements using the emission spectrum because any element produces the same spectrum than other element.
According to Niels Bhor theory the electron only can be in especific discret ratios to the nucleus. Where this electron moves himself in circukar orbits under the influence of the Coulomb attraction force.
(This is a non-relativistic warm-up problem, to get you to think about reference frames.) A girl throws a baseball upwards at time t=0. She catches it at exactly t=2.0 seconds. A) Calculate the position of the ball as a function of time (0
Answer:
X(t) = 9.8 *t - 4.9 * t^2
Explanation:
We set a frame of reference with origin at the hand of the girl the moment she releases the ball. We assume her hand will be in the same position when she catches it again. The positive X axis point upwards.The ball will be subject to a constant gravitational acceleration of -9.81 m/s^2.
We use the equation for position under constant acceleration:
X(t) = X0 + V0 * t + 1/2 * a *t^2
X0 = 0 because it is at the origin of the coordinate system.
We know that at t = 2, the position will be zero.
X(2) = 0 = V0 * 2 + 1/2 * -9.81 * 2^2
0 = 2 * V0 - 4.9 * 4
2 * V0 = 19.6
V0 = 9.8 m/s
Then the position of the ball as a function of time is:
X(t) = 9.8 *t - 4.9 * t^2
Find the electric energy density between the plates of a 225-μF parallel-plate capacitor. The potential difference between the plates is 365 V , and the plate separation is 0.200 mm .
Answer:
Energy density will be 14.73 [tex]J/m^3[/tex]
Explanation:
We have given capacitance [tex]C=225\mu F=225\times 10^{-6}F[/tex]
Potential difference between the plates = 365 V
Plate separation d = 0.200 mm [tex]0.2\times 10^{-3}m[/tex]
We know that there is relation between electric field and potential
[tex]E=\frac{V}{d}[/tex], here E is electric field, V is potential and d is separation between the plates
So [tex]E=\frac{V}{d}=\frac{365}{0.2\times 10^{-3}}=1825000N/C[/tex]
Energy density is given by [tex]E=\frac{1}{2}\varepsilon _0E^2=\frac{1}{2}\times 8.85\times 10^{-12}\times (1.825\times 10^6)^2=14.73J/m^3[/tex]
A girl is helping her brother slide their dog m = 25kg along an icy frictionless sidewalk. The boy is pulling a rope T = 10N tied to the dogs collar and the girl is pushing with 7N of force. What is the Net force on the dog and it's acceleration?
Answer:
0.68 m/s²
Explanation:
Given:
Mass of the dog, m = 25 kg
Tension in the rope = 10 N
Force applied by the girl on the dog = 7 N
Now,
since the boy is pulling the dog and the girl is pushing the dog
Thus,
The net force on the dog = 10 N + 7 N = 17 N
also,
Net force on the dog = Mass × Acceleration
thus,
25 kg × Acceleration = 17 N
or
Acceleration = [tex]\frac{\textup{17 N}}{\textup{25 kg}}[/tex]
or
Acceleration = 0.68 m/s²
If the Earth’s crust contained twice as much iron as it does, would atmospheric oxygen buildup have been slowed down or sped up during the past 2.5 billion years?
What would the consequence of this have been for the evolution of aerobic respiration, the ozone layer, green plants, and animals?
500 elves currently live in the great elf kingdom. The elves are very prolific, with a birth rate of 20% per year. The elves also live for a long time, so the death rate per year is only 5% of the population. in Addition, 55 new elves per year are coming into the kingdom due to a volcanic eruption in the not-so-great elf kingdom. Use a mass balance to determine the accumulation rate.
Answer:
The yearly accumulation for the current year is 130
Explanation:
Accumulation is defined as
input - output = accumulation
The accumulation rate is accumulation per unit of time (in this case a year)
The accumulation rate will then be the amount of births, plus the amount of immigrants minus the amount of deaths.
ar = 500*0.2 + 55 - 0.05*500
ar = 100 + 55 - 25
ar = 130
Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.116 N when their center-to-center separation is 65.4 cm. The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres repel each other with an electrostatic force of 0.0273 N. Of the initial charges on the spheres, with a positive net charge, what was (a) the negative charge on one of them and (b) the positive charge on the other? (Assume the negative charge has smaller magnitude.)
Answer:
Part a)
[tex]q_1 = -1.47 \times 10^{-6} C[/tex]
Part b)
[tex]q_2 = 3.75 \times 10^{-6} C[/tex]
Explanation:
Let the charge on two spheres is q1 and q2
now the force between two charges are
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
[tex]0.116 = \frac{(9\times 10^9)(q_1)(q_2)}{0.654^2}[/tex]
[tex]q_1 q_2 = 5.51 \times 10^{-12}[/tex]
now when we connect then with conducting wire then both sphere will equally divide the charge
so we will have
[tex]q = \frac{q_1-q_2}{2}[/tex]
now we have
[tex]0.0273 = \frac{(9\times 10^9)(\frac{q_1- q_2}{2})^2}{0.654^2}[/tex]
[tex]q_1 - q_2 = 2.28\times 10^{-6} C[/tex]
now we will have
Now we can solve above two equations
Part a)
negative charge on the sphere is
[tex]q_1 = -1.47 \times 10^{-6} C[/tex]
Part b)
positive charge on the sphere is
[tex]q_2 = 3.75 \times 10^{-6} C[/tex]
A speed skater moving across frictionless ice at 8.4 m/s hits a 5.7 m -wide patch of rough ice. She slows steadily, then continues on at 6.5 m/s. What is her acceleration on the rough ice?
Answer:
Acceleration, [tex]a=-2.48\ m/s^2[/tex]
Explanation:
Initial speed of the skater, u = 8.4 m/s
Final speed of the skater, v = 6.5 m/s
It hits a 5.7 m wide patch of rough ice, s = 5.7 m
We need to find the acceleration on the rough ice. The third equation of motion gives the relationship between the speed and the distance covered. Mathematically, it is given by :
[tex]v^2-u^2=2as[/tex]
[tex]a=\dfrac{v^2-u^2}{2s}[/tex]
[tex]a=\dfrac{(6.5)^2-(8.4)^2}{2\times 5.7}[/tex]
[tex]a=-2.48\ m/s^2[/tex]
So, the acceleration on the rough ice [tex]-2.48\ m/s^2[/tex] and negative sign shows deceleration.
A man strikes one end of a thin rod with a hammer. The speed of sound in the rod is 15 times the speed of sound in air. A woman, at the other end with her ear close to the rod, hears the sound of the blow twice with a 0.12 s interval between; one sound comes through the rod and the other comes through the air alongside the rod. If the speed of sound in air is 343 m/s, what is the length of the rod?
Answer:
44.1 m
Explanation:
Given:
[tex]V_a[/tex] = speed of sound in air = 343 m/s[tex]V_r[/tex] = speed of sound in the rod = [tex]15V_a[/tex][tex]\Delta t[/tex] = times interval between the hearing the sound twice = 0.12 sAssumptions:
[tex]l[/tex] = length of the rod[tex]t[/tex] = time taken by the sound to travel through the rod [tex]T[/tex] = time taken by the sound to travel to through air to the same point = [tex]t+\Delta t = t+0.12\ s[/tex]We know that the distance traveled by the sound in a particular medium is equal to the product of the speed of sound in that medium and the time taken.
For traveling sound through the rod, we have
[tex]l=V_r t\\\Rightarrow t = \dfrac{l}{V_r}[/tex]..........eqn(1)
For traveling sound through the air to the women ear for traveling the same distance, we have
[tex]l=V_aT\\\Rightarrow l=V_a(t+0.12)\\\Rightarrow l=V_a(\dfrac{l}{V_r}+0.12)\,\,\,\,\,\,(\textrm{From eqn (1)})\\\Rightarrow l=V_a(\dfrac{l}{15V_a}+0.12)\\\Rightarrow l=\dfrac{l}{15}+0.12V_a\\\Rightarrow l-\dfrac{l}{15}=0.12V_a\\\Rightarrow \dfrac{14l}{15}=0.12V_a\\\Rightarrow l = \dfrac{15}{14}\times 0.12V_a\\\Rightarrow l = \dfrac{15}{14}\times 0.12\times 343\\\Rightarrow l = \dfrac{15}{14}\times 0.12\times 343\\\Rightarrow l = 44.1\ m[/tex]
Hence, the length of the rod is 44.1 m.
The length of the rod can be calculated using the difference in hearing times and the speed of sound in the rod and air. Using the formula for distance (speed x time), and given that the speed of sound in the rod is 15 times the speed of sound in air, the length of the rod is found to be approximately 41.16 meters.
Explanation:In this problem, we know that the speed of sound in the rod is 15 times the speed of sound in the air, and that the woman hears the sound of the strike twice with a 0.12 second gap. The first sound is transmitted through the rod and the second, through the air. Therefore, we can use this information to conclude that the difference in time is the amount of time it takes for the sound to travel the length of the rod in air after it already traveled through the rod.
The speed of sound in the rod is 15 times the speed of sound in air, which is given as 343 m/s. So, the speed of sound in the rod is 15 * 343 = 5145 m/s.
We are looking for the distance travelled, which is the length of the rod. We can find the distance by using the formula distance = speed x time. In this case we are calculating distance as time taken for sound to travel through air minus the time taken to travel through the rod. Therefore, the length of the rod can be calculated to be 343 m/s * 0.12 s = 41.16 meters.
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You are designing a delivery ramp for crates containing exercise equipment. The 1890 N crates will move at 1.8 m/s at the top of a ramp that slopes downward at 22.0◦. The ramp exerts a 515 N kinetic friction force on each crate, and the maximum static friction force also has this value. Each crate will compress a spring at the bottom of the ramp and will come to rest after traveling a total distance of 5.0 m along the ramp. Once stopped, a crate must not rebound back up the ramp. Calculate the largest force constant of the spring that will be needed to meet the design criteria.
The largest force constant of the spring can be found by balancing the kinetic energy of the crate at the top of the ramp with the elastic potential energy of the compressed spring at the bottom of the ramp and taking into consideration the static friction.
Explanation:Calculating the force constantTo calculate the largest force constant of the spring at the bottom of the ramp, we would use Hooke's law which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. This force can be represented by -kx, where k is the force constant and x is the displacement. The displacement in our scenario is the movement of the crate down the ramp. We know that at the base of the ramp, kinetic energy of the crate will be transformed into elastic potential energy as it compresses the spring.
The kinetic energy of the crate at the top of the ramp will be 1/2*m*v^2 (mass * velocity squared) and at the bottom it will be converted into the potential energy of the compressed spring represented by 1/2*k*x^2. Here, given mass of the crate as 1890/9.81 kg, its velocity as 1.8 m/s, and it travels 5m along the ramp.
Setting these two energies equal, we can find the value of spring constant k.
Also, based on the problem, the static friction must be overcome at the bottom of the ramp to avoid the crate rebound. This scenario involves the interplay of several fundamental concepts in physics including kinetic and potential energy, the work-energy principle, and concepts of kinetic and static friction. Different parts of the given information and reference text refer to these aspects of physics.
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A basketball center holds a basketball straight out, 2.0 m above the floor, and releases it. It bounces off the floor and rises to a height of 1.5 m. a) What is the ball's velocity just before it hits the floor? b) What is the ball's velocity just after it leaves the floor? c) If the ball is in contact with the floor for 0.02 seconds, what are the magnitude and direction of the ball's average acceleration while in contact with the floor?
The basketball's velocity just before hitting the floor is approximately -6.26 m/s (downward), while just after it leaves the floor, it reaches around 5.42 m/s (upward). During the brief 0.02 seconds it's in contact with the floor, its average acceleration is enormous, about 584 m/s², and directed upward.
Explanation:This question involves the physics concept of velocity and acceleration related to a bouncing basketball. Let's dig into the details one by one:
(a) The ball's velocity just before hitting the floor:
Using Physics, the velocity of an object just before it hits the ground can be calculated using the equation for motion that involves falling from a height namely v² = u² + 2gh, where u is the initial velocity, g is acceleration due to gravity, v is the final velocity and h is the height. Since the ball was released, the initial velocity (u) is 0, g is approximately 9.8 m/s² (negative because the ball is falling downwards), and h is -2.0 m (negative because it is below the release point). Solving for v, the equation transforms to v = sqrt(u² + 2gh) = sqrt(0 + 2*(-9.8)*(-2)) = sqrt(39.2) which is approximately 6.26 m/s (negative, indicating downward).
(b) The ball's velocity just after it leaves the floor:
Assuming the ball reaches the height of 1.5 m with uniform acceleration, we can use the same equation, but this time treating it as a ground-to-air motion with initial velocity 0, g = 9.8 m/s² (positive because the motion is upward), and h = 1.5 m. Solving for v, we get v = sqrt(u² + 2gh) = sqrt(0 + 2*9.8*1.5) = sqrt(29.4) which is approximately 5.42 m/s (positive, indicating upward).
(c) Average acceleration while the ball is in contact with the floor:
Acceleration can be calculated using the formula a = (v_final - v_initial) / t, where v_final is the final velocity, v_initial is the initial velocity, and t is the time. The change in velocity here is the difference between the velocity just after the ball leaves the floor and the velocity just before it hits the floor, i.e., (5.42 m/s - -6.26 m/s) = 11.68 m/s. Given that the ball is in contact with the floor for 0.02 seconds, the average acceleration is therefore a = (11.68 m/s) / 0.02 s = 584 m/s². This is considerably higher than g because while in contact with the floor, the ball is being rapidly decelerated and then accelerated in the opposite direction due to the impact force. The direction is upward, same as the final velocity.
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A test charge of +4 µC is placed halfway between a charge of +6 µC and another of +2 µC separated by 20 cm. (a) What is the magnitude of the force (in N) on the test charge?
(b) What is the direction of this force (away from or toward the +6 µC charge)?
a. away from the +6 µC charge or
b. toward the +6 µC charge
Answer:
(a) Magnitude: 14.4 N
(b) Away from the +6 µC charge
Explanation:
As the test charge has the same sign, the force that the other charges exert on it will be a repulsive force. The magnitude of each of the forces will be:
[tex]F_e = K\frac{qq_{test}}{r^2}[/tex]
K is the Coulomb constant equal to 9*10^9 N*m^2/C^2, q and qtest is the charge of the particles, and r is the distance between the particles.
Let's say that a force that goes toward the +6 µC charge is positive, then:
[tex]F_e_1 = K\frac{q_1q_{test}}{r^2}=-9*10^9 \frac{Nm^2}{C^2} \frac{6*10^{-6}C*4*10^{-6}C}{(0.1m)^2} =-21.6 N[/tex]
[tex]F_e_2 = K\frac{q_2q_{test}}{r^2}=9*10^9 \frac{Nm^2}{C^2} \frac{2*10^{-6}C*4*10^{-6}C}{(0.1m)^2} =7.2 N[/tex]
The magnitude will be:
[tex]F_e = -21.6 + 7.2 = -14.4 N[/tex], away from the +6 µC charge
A parachutist descending at a speed of 15.1 m/s loses a shoe at an altitude of 41.2 m. What is the speed of the shoe just before it hits the ground? The acceleration due to gravity is 9.81 m/s^2. When does the shoe reach the ground? Answer in units of s.
To find the speed at which the shoe hits the ground, we use the equation v = √(u²+2as). To determine the time it takes for the shoe to hit the ground, we use the formula t = √(2h/g). Substituting the given values into these equations, we can find the speed at which the shoe hits the ground and the time it takes to do so.
Explanation:The question is both a physics problem related to gravity and the free fall of an object. First, we need to find the final velocity of the shoe when it hits the ground. To calculate this, we can use the following equation from physics: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Since the shoe starts from the initial velocity of the parachutist (15.1 m/s), the acceleration a is the acceleration due to gravity (9.81 m/s²), and it falls a distance of 41.2m, we will need to use the equation v = √(u²+2as) to solve for v.
Next, to determine when the shoe hits the ground, we use the equation for time in free fall: t = √(2h/g), where h is the height (41.2 m), and g is acceleration due to gravity (9.81 m/s²). Substituting the given values, we can find the times when the shoe will hit the ground.
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An inflatable raft (unoccupied) floats down a river at an approximately constant speed of 5.6 m/s. A child on a bridge, 71 m above the river, sees the raft in the river below and attempts to drop a small stone onto the raft. The child releases the stone from rest. In order for the stone to hit the raft, what must be the horizontal distance between the raft and the bridge when the child releases the stone?
Answer:
21.28 m
Explanation:
height, h = 71 m
velocity of raft, v = 5.6 m/s
let the time taken by the stone to reach to raft is t.
use second equation of motion for stone
[tex]h = ut + \frac{1}{2}at^{2}[/tex]
u = 0 m/s, h = 71 m, g = 9.8 m/s^2
71 = 0 + 0.5 x 9.8 x t^2
t = 3.8 s
Horizontal distance traveled by the raft in time t
d = v x t = 5.6 x 3.8 = 21.28 m