Answer:
a) d= 1393 m
b) θ= 21º S of E.
Explanation:
a)
Since the second student goes due north, and the first student goes due east along 1300m till he turns directly northward, we conclude that when the second student turns, he is at 1300m west from the destination.Since he rode 1930 m due north, we can conclude also that the second student is 500 m past the destination in the north direction.So we can find the distance from the destination at the turning point, using the Pythagorean Theorem, taking the right triangle defined by the 1300 m segment due east, the 500 m segment due south, and which hypotenuse is the distance straight to the destination, as follows:[tex]d = \sqrt{(1300m)^{2} + (500m)^{2} } = 1393 m (1)[/tex]
b)
Taking the same right triangle than in (a), we can find the angle that makes the vector along the direction taken by the second student with the due east, applying the definition of tangent of an angle, as follows:[tex]tg \theta = \frac{500m}{1300m} = 0.385 (2)[/tex]
⇒ θ= tg⁻¹ (0.385) = 21º S of E.
What is the value of the angle of inclination of the slide?
Answer:
63°
that's my answer
but then I am sorry if I'm wrong
Explanation:
90-27 = 63°
A Typical operating voltage of an electron microscope is 50 kV. A Typical experimental operating voltage range of a Scanning electron microscope is 1kV to 30kV. Higher voltages can penetrate and causes deformation on the sample. Lets assume it operates at 10kV. (i)What is the smallest distance that it could possibly resolve
Answer:
y =[tex]\frac{1.22L}{D}[/tex] [tex]\sqrt{\frac{h^2 m}{2eV} }[/tex]
Explanation:
Let's solve this exercise in parts. Let's start by finding the wavelength of the electrons accelerated to v = 10 103 V, let's use the DeBroglie relation
λ= [tex]\frac{h}{p} = \frac{h}{mv}[/tex]
Let's use conservation of energy for speed
starting point
Em₀ = U = e V
final point
Em_f = K = ½ m v²
Em₀ = Em_f
eV = ½ m v²
v =[tex]\sqrt{\frac{2eV}{m} }[/tex]
we substitute
λ= [tex]\sqrt{ \frac{h^2 m}{2eV}}[/tex]
the diffraction phenomenon determines the minimum resolution, for this we find the first zero of the spectrum
a sin θ = m λ
first zero occurs at m = 1, also these experiments are performed at very small angles
sin θ = θ
θ = λ / a
This expression is valid for linear slits, in the microscope the slits are circular, when solving the polar coordinates we obtain
θ = 1.22 λ / D
where D is the diameter of the opening
we substitute
θ = [tex]\frac{1.22}{D}[/tex] \sqrt{ \frac{h^2 m}{2eV}}
this is the minimum angle that can be seen, if the distance is desired suppose that the distance of the microscope is L, as the angles are measured in radians
θ = y / L
when substituting
where y is the minimum distance that can be resolved for this acceleration voltage
y =[tex]\frac{1.22L}{D}[/tex] [tex]\sqrt{\frac{h^2 m}{2eV} }[/tex]
When the moving sidewalk at the airport is broken, as it often seems to be, it takes you 41 ss to walk from your gate to baggage claim. When it is working and you stand on the moving sidewalk the entire way, without walking, it takes 80 ss to travel the same distance. How long will it take you to travel from the gate to baggage claim if you walk while riding on the moving sidewalk?
Answer:
41
Explanation:
A student sits on a rotating stool holding two 1 kg objects. When his arms are extended horizontally, the objects are 0.9 m from the axis of rotation, and he rotates with angular speed of 0.76 rad/sec. The moment of inertia of the student plus the stool is 5 kg m2 and is assumed to be constant. The student then pulls the objects horizontally to a radius 0.33 m from the rotation axis.
Required:
a. Find the new angular speed of the student.
b. Find the kinetic energy of the student before and after the objects are pulled in.
Answer:
a) the new angular speed of the student is 0.9642 rad/s
b)
the kinetic energy of the student before the objects are pulled in is 1.9119 J
the kinetic energy of the student after the objects are pulled in is 2.4252 J
Explanation:
Given that;
mass of each object m = 1 kg
distance of objects from axis of rotation r = 0.9 m
Moment of inertia of each object initially [tex]I_{oi}[/tex]
[tex]I_{oi}[/tex] = mr² = 1kg ×(0.9m)² = 1 kg × 0.81 m² = 0.81 kg.m²
moment of inertia of each object finally [tex]I_{of}[/tex]
[tex]I_{of}[/tex] = mr² = 1kg × (0.33 m)² = 0.1089 kg.m²
Now
moment of inertia of student plus stool [tex]I_{}[/tex] = 5 kg.m²
initial angular speed ω₀ = 0.76 rad/sec
final angular speed ω = ?
Now using conservation of angular momentum;
([tex]I_{}[/tex] + 2 [tex]I_{oi}[/tex] )ω₀ = ([tex]I_{}[/tex] + 2 [tex]I_{of}[/tex] )ω
so we substitute
(5 + 2 (0.81) )0.76 = (5 + 2 (0.1089) )ω
5.0312 = 5.2178 ω
ω = 5.0312 / 5.2178
ω = 0.9642 rad/s
Therefore, the new angular speed of the student is 0.9642 rad/s
b)
K.E of student before = (0.5) ([tex]I_{}[/tex] + 2 [tex]I_{oi}[/tex] )ω₀²
= (0.5) (5 + 2 (0.81) )(0.76)²
= 0.5 × 6.62 × 0.5776
= 1.9119 J
Therefore, the kinetic energy of the student before the objects are pulled in is 1.9119 J
KE of student finally = (0.5) ([tex]I_{}[/tex] + 2 [tex]I_{of}[/tex] )ω²
= (0.5) (5 + 2 (0.1089) ) (0.9642)²
= 0.5 × 5.2178 × 0.9296
= 2.4252 J
Therefore, the kinetic energy of the student after the objects are pulled in is 2.4252 J
The x-component and y-component of two vectors A & B are Ax = 9, Ay = 12,Bx =
15 & By = 20. Find:
/A+B/
Answer:
40Explanation:
Given the following
Ax = 9,
Ay = 12,
Bx = 15
By = 20
Get A and B
A = √9²+12²
A= √81+144
A = √225
A = 15
Get B;
B = √15²+20²
B = √225+400
B = √625
B = 25
get /A+B/
A+B = 15+25
/A+B/ = /40/
Hence the value of /A+B/ is 40
Nikki was walking around a department store shopping one day, and did not realize that the shirt she was wearing looked just like the shirts worn by employees. When a stranger asked, "do you work here," she thought it was funny. The other customers' assumption that Nikki was a store employee demonstrates the Gestalt principle of _______.
Answer: Similarity
By.
When you jump, you push down on the earth and it pushes back up against you. The earth pushing up against you is what causes you to go into the air. Why doesn’t your push cause the earth to go down if your push on the earth is equal and opposite of the earth's push on you?
That's a great question !
The answer is: It does !
A push on an object causes the object to accelerate in the direction of the force.
The less mass the object has, the more the force accelerates it.
Now, when you jump, the forces on you and on the Earth are equal forces.
The up force on you causes you to accelerate up by some amount.
The down force on the Earth causes the Earth to accelerate down by some amount.
The Earth's mass is something like 5,972,000,000,000,000,000,000,000 kg, while your mass is something like 50 kg.
The Earth has something like 119,400,000,000,000,000,000,000 times as much mass as you have.
So your acceleration is something like 119,400,000,000,000,000,000,000 times as great as the Earth's acceleration.
==> The Earth's downward acceleration, caused by your jump, is there. It's just too small to notice.
BUT . . . That's the reason why seismometers (instruments to detect and measure the vibrations from distant earthquakes) have to be located as far as possible from cities and busy roads.
In places that are too close to cities and roads, the Earth's surface is always vibrating, wiggling, jiggling, heaving and weaving, in reaction to the forces of people walking around, cars and trucks driving around, even rain falling down. And kids jumping up and down !
In such places, these people-motions are louder and stronger than the vibrations coming from distant earthquakes. Seismometers wouldn't work there.
A woman on a bridge 84.5 m high sees a raft floating at a constant speed on the river below. She drops a stone from rest in an attempt to hit the raft. The stone is released when the raft has 6.00 m more to travel before passing under the bridge. The stone hits the water 4.00 m in front of the raft. Find the speed of the raft.
Answer:
V = 0.48 m/s
Explanation:
In this case, we need to analyze the given data by parts.
At first, we know that the woman is on a height of 84.5 m of a river. She drops a stone thinking that she may hit the raft that is traveling with a constant speed. When the raft is 6 m near the bridge, the woman drops the stone, and the stone hits the water when the raft is still 4 m far of the bridge.
With this given data, we can calculate the distance covered by the raft, because is traveling at a constant speed:
X = 6 - 4 = 2 m
And as it's traveling at constant speed then:
X = V.t
We have the distance of the raft, but not the time it took to cover that distance. This time will be the same time that the stone took to hit the water, therefore, if we can determine the time of the rock, well be determining the time of the raft to cover the distance, and then, we can determine it speed.
To determine the time of the rock, as the stone is going on a free fall, with an innitial speed of 0, the flight time of the rock will be:
y = gt²/2 ---> solving for t
2y/g = t²
t = √2y/g
If g = 9.8 m/s, and replacing the data we have that the flight time of the rock is:
t = √2*84.5 / 9.8
t = 4.15 s
So the rock took 4.15 s to hit the water, and it's also the time that the raft took to cover the distance of 2 m, then, it's speed:
V = X/t
V = 2 / 4.15
V = 0.48 m/sHope this helps
How old do you need to be in order to qualify to be a U.S. Senator
Answer: 30 Years Old
Explanation: The constitution has around three qualifications for service in the U.S. Senate, Your age must be at least 30 years.
It is common to see birds of prey rising upward on thermals. The paths they take may be spiral-like. You can model the spiral motion as uniform circular motion combined with a constant upward velocity. Assume a bird completes a circle of radius 6.00 m every 5.00 s and rises vertically at a rate of 3.00 m/s. Part APart complete Find the speed of the bird relative to the ground. Express your answer using three significant figures. v
Answer:
8.11 m/s
Explanation:
The bird has two velocity, one velocity is in the upward direction (+y direction) and the other velocity is in the x direction (that is along the radius of circle of the bird).
the vertical speed [tex]v_y=3\ m/s[/tex]
The horizontal speed ([tex]v_x[/tex]):
[tex]v_x=\frac{2\pi R}{T} =\frac{2\pi *6\ m}{5\ s} =7.54\ m/s[/tex]
The bird's speed relative to the ground (v) is given by:
v = [tex]\sqrt{v_x^2+v_y^2}[/tex]
[tex]v=\sqrt{7.54^2+3^2} \\\\v=8.11\ m/s[/tex]
what is a asteroid traveling rapidly called
Answer:
meteor
Explanation:
A asteroid stays still and a meteor goes fast
Answer:
meteor
Explanation:
or some people call it a shooting star
A pitching machine is programmed to pitch baseballs horizontally at a speed of 134 km/h. The machine is mounted on a truck and aimed forward. As the truck drives toward you at a speed of 85 km/h, the machine shoots a ball toward you. A pickup truck moves to the left at a constant velocity. A pitching machine sits in the bed of the pickup truck. The pitching machine launches a baseball to the right with a different constant velocity. A man with a baseball mitt stands at rest some distance to the right of the truck. For each of the object pairings listed, determine the correct relative speed. The speed of the pitching machine relative to the truck The speed of the pitched ball relative to the truck The speed of the pitching machine relative to you The speed of the pitched ball relative to you
Answer: 134 = 143 = 151 = 166 = 176
Hope this helps!!
Sorry if it's incorrect!!
:'(
A positively charged oil drop of mass is injected into a region of uniform electric field between two oppositely charged, horizontally oriented plates spaced 16 cm apart. If the mass of the drop is 8.0 10-15 kg and it remains stationary when the potential difference between the plates is 2.44 kV, what is the magnitude of the charge on the drop
Answer:
the magnitude of the charge on the drop is 5.163 × 10⁻¹⁸ C
Explanation:
Given that;
mass of the drop m = 8.0×10⁻¹⁵ kg
distance d = 16 cm = 0.16 m
potential difference between the plates V = 2.44 kV = 2440 v
acceleration of gravity g = 9.81 m/s²
the magnitude of the charge on the drop = ?
weight is balanced by the electrostatic force
weight = mg = 8.0×10⁻¹⁵ kg × 9.81 m/s² = 7.848 × 10⁻¹⁴
we know that; V = Ed
E = V/d = 2440 / 0.16 = 15200 v/m
Electrostatic force = qE
so weight = qE
q = weight / E
q = 7.848 × 10⁻¹⁴ / 15200
q = 5.163 × 10⁻¹⁸ C
Therefore, the magnitude of the charge on the drop is 5.163 × 10⁻¹⁸ C
Temperature and salinity difference in ocean water cause...
Answer:
high and low tied maybe? not sure
There are 5 planets visible to the naked eye in the sky.
True
False
First to answer gets brainliest
An airplane used to drop water on brushfires is flying horizontally in a straight line at 180 mi/h at an altitude of 450 ft. Determine the distance d at which the pilot should release the water so that it will hit the fire at B. The distance at which the pilot should release the water so that it will hit the fire at B is ft.
Answer:
1399.2 ft
Explanation:
The initial velocity = 180 mph = [(180 * 5280) / (1 * 3600)] ft/s = 264 ft/s
[tex]In\ the \ horizontal\ direction(x)\\\\Initial\ velocity = v_{ox}=264\ ft/s\\\\distance\ travelled\ in\ x \ direction(x) =v_{ox}t\\\\\\For\ the\ vertical\ direction:\\\\initial\ velocity(y_{oy})=0\\\\vertical\ distance(y)=y_{oy}t+0.5gt^2\\\\but\ g\ =-32\ ft/s^2. Hence:\\\\y=0t+0.5(-32)t^2\\\\y=-16t^2\\\\At\ point\ B, y=-450, therefore:\\\\-450=-16t^2\\\\t^2=28.125\\\\t=5.3\ s\\\\The\ distance\ at\ which\ the\ pilot\ should\ release\ the\ water=x=v_{ox}t=264*5.3\\\\x=1399.2\ ft[/tex]
Which would you choose to keep rods or cones? And why?
A dumped harmonic oscillator of a mass of 500 g has a period of 0.5 second. The amplitude of the oscillation is decreasing 2.0 % during each oscillation. a./ The initial amplitude is 10 cm. What will be the amplitude after 25 oscillation
Answer:
The answer is below
Explanation:
The amplitude decreases by 2% during each oscillation. Hence the decrease in amplitude can be represented by an exponential decay in the form:
y = abˣ; where x ad y are variables, a is the initial value and b is the factor.
Let y represent the amplitude after x oscillations. Since the initial amplitude is 10 cm, hence:
a = 10 cm, b = 2% = 0.02.
Therefore:
y = 10(0.02)ˣ
The amplitude after 25 oscillations is gotten by substituting x = 25 into the equation. Hence:
y = 10(0.02)²⁵
y= 3.355 * 10⁻⁴² cm
The amplitude after 25 oscillations is 3.355 * 10⁻⁴² cm
In a crash test, a car with a mass of 1600 kg is initially moving at a speed of 20 m/s just before it collides with a barrier. The final speed of the car after the collision is zero. The original length of the car is 4.50 m , but after the collision, the smashed car is only 3.60 m long.
Required:
a. What is the average speed Of the car during the period from first contact with the barrier to the moment the car comes to a stop? You may assume the force that the barrier exerts on the car is constant during this period.
b. How much time elapses between the moment the car makes first contact with the barrier and the moment it comes to a stop?
c. Making the very rough approximation that the large force that the barrier exerts on the car is approximately constant during contact, determine the approximate magnitude of this force?
Answer:
The answer to the given points can be defined as follows:
Explanation:
In point 1:
[tex]\bold{v_f^2= v_i^2+2as}\\\\\to v_f=0\\\\\to v_i=20 \frac{m}{s}\\\\\to s= 4.50\ m -3.60 \ m \\\\[/tex]
[tex]=0.9 \ m \\[/tex]
put the value in the above formula:
[tex]\to 0= 20^2+2 \times a \times 0.9\\\\\to -1.8\ a=400\\\\\to -a= \frac{400}{1.8} \\\\ \to a= -222.22\ \ \frac{m}{s^2}[/tex]
[tex]\bold{v_f=v_i+at}\\\\\to 0=20+ (-222.22)t\\\\\to 222.22t=20\\\\\to t=\frac{20}{222.22}\\\\\to t= 0.0900 \ s\\\\\to v_{avg}=\frac{s}{t}=\frac{0.9}{t}= 10\ \frac{m}{s}[/tex]
for point 2:
[tex]t= 0.0900 \ s -\text{found above}[/tex]
for point 3:
[tex]\to |a| = 222.22 \frac{m}{s^2} \text{found above}\\\\\to \bold{|F| = m \cdot |a|}\\\\[/tex]
[tex]=1600 \ kg \times 222.22 \ \frac{m}{s^2} \\\\= 3.55\times 10^{5} \ N[/tex]
where is a neutron located within an atom
Answer:
the neutron is located in the nucleus of an atom
Answer:
nucleus
Explanation:
Atoms are made up of protons and neutrons located within the nucleus, with electrons in orbitals surrounding the nucleus.
Two charged bees land simultaneously on flowers that are separated by a finite distance. For a few moments, the charged bees rest on the flowers. The charged bees both generate an electric field, and while the charged bees are resting on the flowers, the net electric field at some distance between them is zero.
(a) Do the bees have the same or opposite signs of charge?
Same � the electric fields point in opposite directions and therefore cancel at some midpoint.
Same � the electric fields multiply together to equal zero.
Opposite � the electric fields point in the same direction summing to zero.
Opposite � the net electric field due to the two bees points in a direction perpendicular to the direction from one bee to the other.
(b) Suppose the net electric field is zero at a distance that is closer to bee 1. Does bee 1 have a magnitude of charge greater than or less than that of bee 2?
greater than
less than
Answer:
a. Same � the electric fields point in opposite directions and therefore cancel at some midpoint.
b. bee 1 has a magnitude of charge less than bee 2
Explanation:
a. Do the bees have the same or opposite signs of charge?
They have the same charge. This is because since same charges would produce electric fields in opposite directions, that is the only way they can cancel out at some point. So, the charges are the same and the electric fields point in opposite directions and therefore cancel at some midpoint.
b. Suppose the net electric field is zero at a distance that is closer to bee 1. Does bee 1 have a magnitude of charge greater than or less than that of bee 2?
Let q be the charge on bee 1 and r its distance from the neutral electric field point. So, it electric field E = kq/r².
Also, let q' be the charge on bee 2 and d its distance from the neutral electric field point. So, it electric field E' = kq'/d².
Since E = E' at the neutral point.
kq/r² = kq'/d²
q/q' = r²/d² = (r/d)²
Given that r < d, so r/d < 1 and (r/d)² < 1
So, q/q' < 1
q < q'
So, the charge on bee 1 is less than that on bee 2
SERIOUSLY HELP NOWWW LIKE NOWW I REALLY NEED THIS ANSWERED
Which of the following is the best way to measure the age of the Earth?
Examine rocks collected in my backyard.
Analyze maps of all known fossil sites in the United States.
Test the rate of decay of specific elements in rock samples.
Explore leaves embedded in ash from recent volcanic eruptions.
Answer:
the second one Test the rate of decay of specific elements in rock samples.
This problem has been solved! See the answer A 6.0 kg object, initially at rest in free space, "explodes" into three segments of equal mass. Two of these segments are observed to be moving with equal speeds of 20 m/s with an angle of 60 degrees between their directions of motion. How much kinetic energy is released in this explosion?
Answer:
Explanation:
mass of each part = 6 / 3 = 2 kg .
momentum of each of given part = 2 x 20 = 40 kg m/s
Two momentum of 40 kg m/s , acting at angle 60 degree .
Resultant momentum = 2 x 40 cos 30 = 69.28 kg m/s
The third mass will have equal and opposite momentum to this momentum , following law of conservation of momentum .
If its velocity be v .
2 x v = 69.28
v = 34.64 m /s
Third mass will have velocity of 34.64 m /s
Total kinetic energy of all three mass
KE = 1/2 x 2 ( 20² + 20² + 34.64² )
= 400 + 400 + 1199.92
= 1999.93 J .
why would the bulb not light?
Im not sure this is correct but, i think thats because the wire connected at the negative side. The current go from positive to negative and thats why the bulb not light, also the bulb location is not at the center of the battery which explain why there's no current at the upper side.
Hope you understand. Sorry if my english sucks YEET
The light bulb isn't connected to the positive thing sticking out of the battery, both ends must connect back to the battery
When the skin produces hair, it is removing excess water and salts from the body.
True or false?
If you blow harder into a trumpet you will create a sound wave with a greater
A. pitch
B. amplitude
C wavelength
Answer is C
Explanation:
Answer:
C. wavelength
good luck, i hope this helps :)
A block of concrete has a mass of 48kg a crane lifts the block to a height of 12m above the ground calculate the gravitational potential energy stored by the block (gravitational field strength g=10N/kg)
Answer:
5760 J
Explanation:
From the question given above, the following data were obtained:
Mass of block = 48 kg
Height (h) = 12 m
Gravitational field strength (g) = 10 N/Kg
Gravitational potential energy (PE) =?
The gravitational potential energy stored by the block can simply be obtained as follow:
PE = mgh
PE = 48 × 10 × 12
PE = 5760 J
Therefore, the gravitational potential energy stored by the block is 5760 J
Energy cannot be created nor destroyed, only changed from one form to another. How does listening to music on a radio obey the law of conservation of energy?
Group of answer choices
Sound energy is changed into potential energy.
Electrical energy is converted into other forms of energy, such as sound.
Electrical energy remains unchanged.
Electrical energy is gradually destroyed as the radio plays.
Answer: Electrical energy is converted into other forms of energy, such as sound.
Explanation: Using the law of conservation of energy we know that energy can never be destryped it can be transferred or be transformed into from one form to another.
Electrical energy is converted into other forms of energy, such as sound.
The law of conservation of energy or matter, states that energy can neither be created nor destroyed but can be converted from one form to another.
When listening to music on radio, the electric energy supplied to the radio will be converted to mechanical energy of the moving parts of the radio which is then converted to sound energy.
Thus, we can conclude that electrical energy is converted into other forms of energy, such as sound.
Learn more here:https://brainly.com/question/2828402
) Explain why the foil is attracted at first by the charged rod. Consider any charge that exists in the uncharged foil. (You will consider later interaction between the rod and foil in the last question. Just think about the first interaction when answering this question.) [
Answer:
The attraction is due to the induced charge.
Explanation:
When we approach a charged rod to a sheet, an induced load is produced in the sheet that is of the same magnitude as the rod of opposite sign, this is because the charges of different sign attract each other, this explains the initial attraction.
This induced load occurs if importing the plate load
The attraction is due to the induced charge.