The magnitude of the charge on the particle is 0.0083 C.
The centripetal force acting on the charged particle is equal tp the magnetic force.
[tex]\frac{mv^{2} }{r}[/tex] = qB
q = [tex]\frac{mv}{rB}[/tex]
= [tex]\frac{(0.0020kg)(5.0m/s))}{(0.20 m)(60T)}[/tex]
= 8.33 × 10⁻³ C 0r 0.0083 C
A magnetic field is a vector discipline that describes the magnetic influence on transferring electric fees, electric currents, and magnetic materials. A shifting charge in a magnetic subject experiences a pressure perpendicular to its very own pace and to the magnetic area.
This force causes the magnets to attract or repel each other. Examples of magnetic force are a compass, a motor, the magnets that maintain stuff at the fridge, educate tracks, and new roller coasters. All transferring costs deliver upward thrust to a magnetic subject and the costs that circulate via its regions, revel in pressure.
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how far from the front of the car will the cm be when two people sit in the front seat 2.30 mm from the front of the car, and three people sit in the back seat 3.40 mm from the front? assume that each person has a mass of 60.0 kgkg . express your answer to three significant figures and include the appropriate units.
The center of mass exists a position defined relative to an object or system of objects.
The proper significant figures and proper rounding of digits exists 2.603 m.
What is meant by center of mass?The unique location where the weighted relative position of the scattered mass adds to zero is known as the center of mass in physics. Here is where a force can be applied to produce a linear acceleration without also producing an angular acceleration.
A position established in relation to an object or set of objects is the center of mass. It represents the system's average location as weighted by each component's mass. The center of mass for straightforward stiff objects with homogeneous density is found at the centroid.
Given:
Mass of car, m = 1150 kg
Distance, d = 2.35 m
The new position of center of mass will be:
[tex]$=\frac{1150 \times 2.35+160 \times 2.7+240 \times 3.75}{240+260+1150} \\[/tex]
simplifying the above equation, we get
= (2702.5 + 432 + 900) / 1650
= 4034.5 / 1650
= 2.603 m
Therefore, the proper significant figures and proper rounding of digits exists 2.603 m.
The complete question is:
The CM of an empty 1150 kg car is 2.35 m behind the front of the car. How far from the front of the car will the CM be when two people sit in the front seat 2.70 m from the front of the car, and three people sit in the back seat 3.75 m from the front? Assume that each person has a mass of 80.0 kg.
express answers in proper significant figures and proper rounding of digits.
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1. PE = ?
m = 0.6 kg
g= 9.8 m/s²
h = 35 m
WORKS
Answer:
Explanation:
Potential Energy = m * g * h = .6 * 9.8 * 35 = 205.8 J
Two people are playing soccer. The kinetic energy of each soccer player depends on the soccer players mass and speed. Compare the two soccer players
The two soccer players kinetic energy are 1 / 2 m1 v1² and 1/ 2 m2 v2² for soccer player 1 and 2 respectively
KE = 1 / 2 m v²
KE = Kinetic energy
m = Mass
v = Velocity
Let the mass and velocity be m1 and v1 respectively for soccer player 1. Let the mass and velocity be m2 and v2 respectively for soccer player 2.
For soccer player 1,
m = m1
v = v1
KE1 = 1 / 2 m1 v1²
For soccer player 2,
m = m2
v = v2
KE2 = 1 / 2 m2 v2²
KE ∝ m
KE ∝ v²
The kinetic energy is directly proportional to mass and velocity squared. So increase in any of these factors will increase the kinetic energy and similarly decrease in any of these factors will result in decrease of kinetic energy.
Therefore, the two soccer players kinetic energy are 1 / 2 m1 v1² and 1 / 2 m2 v2² for soccer player 1 and 2 respectively
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Answer:
Cristiano Ronaldo SUII
Explanation:
A concave mirror has a focal length of 0.50 m. If an object produces a virtual image 0.19 m from the mirror, where is it located?
In order to calculate the object's position, we can use the formula below:
[tex]\frac{1}{f}+\frac{1}{d_o}+\frac{1}{d_i}[/tex]Where f is the focal length, do is the object's position and di is the image's position.
Using f = 0.5 m and di = -0.19 m (we use a negative value because the image is virtual), we have:
[tex]\begin{gathered} \frac{1}{0.5}=\frac{1}{d_o}+\frac{1}{-0.19}\\ \\ 2=\frac{1}{d_o}-5.263\\ \\ \frac{1}{d_o}=2+5.263\\ \\ \frac{1}{d_o}=7.263\\ \\ d_o=\frac{1}{7.263}\\ \\ d_o=0.14\text{ m} \end{gathered}[/tex]Therefore the object is at 0.14 meters from the mirror.
A bar magnet is cut into two pieces from the middle as shown in the figure. Mark the poles on these magnets if it is further broken into smaller pieces?
Giving brainliest!
If we cut the magnet into smaller pieces as we can see in the image attached to the question then each of the ends of the smaller magnets would form a north and south pole respectively.
What is a magnet?The term magnet can be used to describe the material that is able to pick up any magnetizable material. We have to note that a magnet is able to create a magnetic field around it. The magnetic field is the vicinity where the impact of the magnet can be felt.
We know that a magnet has two poles, the north pole and the south pole. These are the points where the influence of the magnetic field is greatest In this discussion, we must keep it at the back of our minds that the influence of the magnetic force is stronger at the ends of the magnet than it is at the center of the magnet as it were.
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Two identical clocks are set to 12 noon time when on the ground at rest relative to the ground. One clock is placed in a spacecraft, sent into space and accelerated to a speed of v = 0.93 x 108 m/s. What will the moving clock read if the first clock reads 6 p.m.? Express your answer as the number of minutes from 6 pm.
First, for this problem, we will need to know the formula for time dilation
[tex]t=\frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]Where t is the time in the reference frame on earth
t0 is the time on the ship
v is the velocity relative to lightspeed
and c is light speed in a vacuum
For this problem, we will need to convert .93x10^8 to a fraction of light speed.
.93x10^8 = .31c
Now, we can plug in the numbers we get from the given
t0 = 6 hours since the ship has been in the air for 6 hours
[tex]t=\frac{6}{\sqrt{1-\frac{(.31c)^2}{c^2}}}[/tex]t = 6.31089
For the answer to be valid, we will convert .31089 to minutes by multiplying it by 60 (60 minutes in a hour)
18.654 minutes
14. suppose there is a 10 mbps microwave link between a geostationary satellite and its base station on earth. every minute the satellite takes a digital photo and sends it to the base station. assume a propagation speed of 2.4 x 108 m/s and the distance between satellite and station is 3.6 x 107 m. what is the propagation delay of the link?
The propagation delay of the link due to time dilatationis 1.152 x 10¹⁶ s.
We need to know about the time dilatation to solve this problem. Time dilatation represents that time is relative between the moving and static objects. From a static objects point of view, the time should be
t' = t / √(1 - v²/c²)
where t' is the interval time based on the static object, t is the interval time based on the moving object, v is velocity and c is the speed of light (3 x 10⁸ m/s),
From the question above, the given parameters are
v = 2.4 x 10⁸ m/s
D = 3.6 x 10⁷ m
Find the time taken based on moving objects
t = 2D . v
t = 2 . 3.6 x 10⁷ . 2.4 x 10⁸
t = 1.728 x 10¹⁶ s
Find the time interval based on static objects
t' = t / √(1 - v²/c²)
t' = 1.728 x 10¹⁶ / √(1 - (2.4 x 10⁸)²/(3 x 10⁸)²)
t' = 2.88 x 10¹⁶ s
Hence, the delay time is
Δt = t' - t
Δt = 2.88 x 10¹⁶ - 1.728 x 10¹⁶
Δt = 1.152 x 10¹⁶ s
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A tennis player can accelerate a tennis ball at 295m/s² using an average force of
58 N with his racquet. What is the mass of the tennis ball?
answer of the question is into the picture ️
While in a park, you walk west for 52 m, then you walk 31.1° north of west for 40 m, and finally you walk due north for 25 m. Find the components of your final displacement (in m), from your initial to final point, along the north and west directions.
The components of the final displacement (in m), from the initial to final point, along the north and west directions is given as:
88m west and 50m north
What is displacement?The displacement of an object is a vector whose length is the shortest distance from the initial to the final position of a point P undergoing motion.
we will find the components of each vector in x and y:
For Vector 1:
Fx = 52m
Fy = 0m
for Vector 2:
Fx = 40m*cos31.1
Fy = 40m * cos 31.1
For Vector 3:
Fx = 0m
Fy = 25m
we will then find the components of the final displacement acting in x and y:
For west direction:
Rx = 52m + 40m*cos31.1 + 0
Rx = 90m
For north direction:
Ry = 0 + 40m*sin31.1 + 25m
Ry = 12m.
In conclusion, displacement is a vector quantity that refers to how far out of place an object is. Displacement is known to be the object's overall change in position.
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In 2 - 3 sentences, explain the difference in Kinetic energy and Potential energy.
The main difference between Kinetic energy and Potential energy is that kinetic energy refers to movement while potential energy refers to storage.
What are Kinetic energy and Potential energy?Kinetic energy is a type of energy in motion or energy in movement such as a turbine (mechanical energy), while Potential energy refers to the energy that is stored to be used when required (e.g. chemical bonds of foods).
Therefore, with this data, we can see that Kinetic energy is used as movement, while Potential energy is stored to be used in the future.
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Write 2 uses of centripetal force.
An apple is resting on a table. Claire says that there are no forces acting on the apple because it is not moving. Is she right?
Answer:
no
Explanation:
there are forces , even when the apple isnt moving , for example gravity which is keeping the apple on the table
how to sketch the following:Sketch a ray diagram for a spherical convex lens with an object at Do = 2f. (Submit a file with a maximum size of 1 MB.)
Do = 2f means that object distance is 2 x focal length.
2 x focal length = radius of curvature
When an object is placed at the radius of curvature, a real, inverted image is formed at the other radius of curvature. The sizes of the image and the object are the same. The sketch is shown below
the floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.36 with the floor. if the train is initially moving at a speed of 45 km/h, in how short a distance can the train be stopped at constant acceleration without causing the crates to slide over the floor?
24.8 m is the shortest braking distance.
On the horizontal axis, we dread the force of friction (fr) that prevents the train from moving and accelerating.
Y axis
N- W = 0
N = W = mg
explanation?
In order to fix this issue, we must apply Newton's second law to the boxes, with the norm vertically up and the weight vertically down.
Write the equation for each axis. On the horizontal axis, we dread the force of friction (fr) that prevents the train from moving and accelerating.
Y axis
N- W = 0
N = W = mg
X axis
-Fr = m a
-μ N = m a
-μ mg = ma
a = μ g
a = - 0.36
a = - 3.14 m/s²
Utilizing the kinematics equations, we determine the distance.
(Vf2 - Vo2) / 2 a = Vf2 = Vo2 + 2 an
The speed of the train is zero (Vf = 0) when it stops.
Vo=45 km/h (1000 m/km) (1 h 3600 s)= 12.5m/s
x = ( 0 - 12.5²) / 2 (-3.14)
x= 24.88.8 m
The shortest braking distance is 24.8 m
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What average mechanical power (in W) must a 76.5 kg mountain climber generate to climb to the summit of a hill of height 305 m in 42.0 min? Note: Due to inefficiencies in converting chemical energy to mechanical energy, the amount calculated here is only a fraction of the power that must be produced by the climber's body.
Power is given by, P= W/t, where W is mechanical work done and t is time taken.
From work-energy relation, work done is equal to the change in potential or kinetic energy.
W = mg[tex]h_{2}[/tex] - mg[tex]h_{1}[/tex]
Given,
Mass of mountain climber, m =76.5kg
Height of mountain hill, [tex]h_{2}[/tex] = 305m
Time taken to climb, t = 42 min= 42×60s= 2520s.
Since climber starts from the down of hill, [tex]h_{1}[/tex] = 0m. So, [tex]h_{2} - h_{1}[/tex] = 305m.
Work done, W = 76.5kg × 9.8m/[tex]s^{2}[/tex] × 305m
⇒W = 22.866×[tex]10^{4}[/tex] J.
Average mechanical power, P = 22.866×[tex]10^{4}[/tex]J ÷ 2520s
⇒P = 90.7375 W (watt)
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if instead the same ball is used and the rope is shortened; that is, if rope 3 is half the length of rope 1, what is the tension t1 in the rope compared to the tension t3 in the shorter rope?
The tension in the shorter rope is double the tension in the longer rope T1<T3.
What is tension?
In physics, tension is defined as the pulling force that is transmitted axially by a string, rope, chain, or similar object, or through each end of a rod, truss member, or other comparable three-dimensional object. Tension can also be defined as the activity pair of forces acting at every end of the aforementioned elements. The polar opposite of compression might be tension. When atoms or molecules have been pulled apart from one another at the atomic level and gain potential energy while a restoring force is still present, the restoring force may produce what is also known as tension. When a string or rod is under such tension, each end may pull on the thing it is attached to in an attempt to release the tension and return the string or rod to its relaxed length.
The relation between the tension in the string and the length of the tetherball is,
mol=T The length of the rope 3 is half of the length of the rope 1.
Therefore mw^2l=T T∝l
The length of the rope 3 is half of the length of the rope 1.
Therefore l1=l3/2T1=T3/2T1<T3
Hence, the answer is T1<T3
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a projectile with a mass of 20 grams is launched with a horizontal linear momentum of 15000 g cm/s; after it falls 250 cm vertically what would be its horizontal displacement? assume horizontal velocity remains constant after launch.
The horizontal displacement of the projectile is determined as 5.3 m.
What is the horizontal displacement of the projectile?The horizontal displacement of the projectile is calculated using the following equation.
X = vt
where;'
v is the horizontal velocity if the projectilet is the time of motionThe time of motion of the projectile is calculated as follows;
t = √(2h/g)
where;
h is the height of fall of the projectile = 250 cm = 2.5 mg is acceleration due to gravityt = √(2 x 2.5 / 9.8)
t = 0.71 s
The horizontal velocity of the projectile is calculated as follows;
P = mv
where;
P is momentum of the projectilem is massv is horizontal velocityv = P/m
v = (15000 g cm/s) / (20 g)
v = 750 cm/s = 7.5 m/s
X = 7.5 x 0.71
X = 5.3 m
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__ is the rate at which an object changes position
Velocity is the rate at which an object changes position.
___________________________________________
"What is velocity?"Velocity is a vector quantity that refers to "the rate at which an object changes its position."
Hope this helps!
Want to learn more? Visit below for more information about velocity.
Look up "Speed versus Velocity" at The Physics Classroom.com
Mutations can occur randomly during DNA replication. Suppose DNA replication is occurring in a cell. If DNA replication occurs
correctly, the following strand will be produced.
If a point mutation occurs during DNA replication, which of these could represent the mutated strand produced instead?
OA.
ATG GAC CAT TGG C
ATG GAC CAA CTT TCG GC
ATG CGT CAT TTC GGC
ATG GAC CTT TTC GGC
B.
(C.
ATG GAC CAT TTC GGC
OD.
what is the principal difference between a radio wave and light? what is the principal difference between a radio wave and light? radio waves have significantly higher frequencies than light. a radio wave is a mechanical vibration of matter and light is a vibration of electric and magnetic fields. radio waves have significantly lower frequencies than light. light is a mechanical vibration of matter and radio wave is a vibration of electric and magnetic fields.
The principal difference between a radio wave and light is that radio waves have significantly lower frequencies than light.
Explanation:
Both radio waves and light are electromagnetic waves; their main difference is their frequency. Radio waves have a lower frequency and longer wavelength than visible light waves. Light waves have a lower frequency and longer wavelength than X-rays.
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A woman of mass 57.6 kg is standing in an elevator.If the elevator floor pushes up on her feet with a force of 395 N, what is the acceleration of the elevator? Assume the +y-direction to be up. Enter a negative answer if the acceleration is downward and a positive answer if the acceleration is upward.If the elevator maintains constant acceleration and is moving at 1.50 m/s as it passes the fourth floor on its way down, what is its speed 4.00 s later?
There are two forces acting on the woman: the normal force of the floor of the elevator and the weight of the woman.
First, find the weight of the woman by multiplying its mass times the acceleration of gravity:
[tex]\begin{gathered} W=mg \\ =(57.6\operatorname{kg})(9.8\frac{m}{s^2})=564.48N \end{gathered}[/tex]According to Newton's Second Law of Motion, the net force acting on an object relates to the acceleration of the object according to:
[tex]\Sigma F=ma[/tex]Since the normal force of the floor is exerted upwards and the weight is exerted downwards, then:
[tex]\begin{gathered} \Sigma F=N-W \\ \Rightarrow N-W=ma \end{gathered}[/tex]Isolate the acceleration from the equation:
[tex]\Rightarrow a=\frac{N-W}{m}[/tex]Replace the values of the normal force (395N), the weight and the mass of the woman to find the acceleration of the elevator:
[tex]\begin{gathered} \Rightarrow a=\frac{395N-564.48N}{57.6\operatorname{kg}}=-2.94236\ldots\frac{m}{s^2} \\ \therefore a\approx-2.9\frac{m}{s^2} \end{gathered}[/tex]Then, the acceleration of the elevator is 2.9 m/s^2 downwards.
Use the definition of acceleration to find the final speed of the elevator 4 seconds after the initial speed of 1.50 m/s downwards:
[tex]\begin{gathered} v_f=at+v_0 \\ =(-2.94\frac{m}{s^2})(4.00s)-1.5\frac{m}{s} \\ =-13.26\ldots\frac{m}{s} \\ \approx-13.3\frac{m}{s} \end{gathered}[/tex]Therefore, the answers are:
The acceleration of the elevator is 2.9 m/s^2 downwards.
The final speed of the elevator 4 seconds after it had a speed of 1.50 m/s downwards, is 13.3 m/s downwards.
You are visiting a friend from elementary school who now lives in a small town. One local amusement is the ice-cream parlor, where Stan, the short-order cook, slides his completed ice-cream sundaes down the counter at a constant speed of 2.0 m/s to the servers. (The counter is very well polished for this purpose.) If the servers catch the sundaes 7.0 cm from the edge of the counter, how far do they fall from the edge of the counter to the point at which the servers catch them?
The edge counter is 0.0060 meters or 0.60 cm far as they fall from the edge of the counter to the point at which the servers catch them.
What is Potential energy?Potential energy is a form of stored energy that is dependent on the arrangement of a system's components. When stretched or squeezed, a spring has higher potential energy.
When lifted above the ground, a steel ball has more potential energy than when it is brought to rest on the ground. It is capable of performing more work in the increased position.
From the given data of the question,
x = v/t
t = x/v (1)
y = -1/2gt²
Substitute value of t from (1)
y = -1/2 g(x/v)²
y = -1/2 (9.8)(0.70/2.0)²
y = 0.0060 meters, or,
y = 0.60 cm.
Therefore, it is 0.60 cm far from the edge of the counter.
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How much gravitational force is acting on a 5 kg bowling ball resting on the floor?
A. 0.2 N
B. 9.8 N
C. 1 N
D. 49 N
Answer:
Explanation:
F = m * a = 5 kg * 9.81 m/s^2 = ~ 49 N
A 12 kg box is pushed to the left by a 48 N force.
Find its acceleration.
Answer: 4m/2
Explanation: If you divide 48N and 12N, you will get 4m/s
a cats purple crinkle ball toy of mass 20g is throwwn straight up in the air with an initial speed of 4m/s. assume in this problem that air grad is negligible. what is the kinetic energy of the ball as it leaves the hand?
When the ball leaves the hand, it's kinetic energy is 160 Joules.
The kinetic energy of object is due to its virtue of motion. So, as the ball leaves the hand, it is in motion. The formula representing kinetic energy is: K.E. = 1/2×m×v², where K.E. is kinetic energy, m is mass and v is velocity.
Keep the values in formula to find the kinetic energy of the ball.
K.E. = 1/2×20×4²
Taking square on Right Hand Side of the equation
K.E. = 1/2×20×16
Performing multiplication and division on Right Hand Side of the equation
K.E. = 160 Joules
Thus, kinetic energy of ball is 160 Joules.
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An electric drill transfers some energy wastefully. Aside from heat, what else carries the energy that it transfers wastefully?
Answer:
Explanation:
Mechanical energy
Answer:
The right answer is "sound" or "sound waves"
Explanation:
Electric drills are noisy, but this is not useful, so sound is a wasted energy transfer.
Slim Jim, continually maintaining his svelte body, lifts a 70 kg barbell 1.4m above the grounda) How much energy did the barbell have when it was on the ground?b) How much energy does it have after being lifted 1.4m? What kind of energy does it have afterbeing lifted? Where did it come from?c) How much work did Jim do to the lift the object?dIf he lifted it in 1.5s how much power did he use?
Given,
The mass of the barbell, m=70 kg
The height to which Slim Jim lifts the barbell, h=1.4 m
a)
When the barbell was on the ground, it will have zero kinetic energy as it has no velocity. And if assume the height of ground as zero meters, then its potential energy is also zero.
Thus when the barbell was on the ground, its energy was zero joules.
b)
The energy of the barbell when it is at a height of h is given by,
[tex]E=\text{mgh}[/tex]Where g is the acceleration due to gravity.
On substituting the known values,
[tex]\begin{gathered} E=70\times9.8\times1.4 \\ =960.4\text{ J} \end{gathered}[/tex]Thus the energy that the barbell has after being lifted 1.4 m is 960.4 J
This energy is the energy stored in the barbell due to its position. Thus the energy stored in the barbell is the potential energy.
Slim Jim has to do some work to lift the barbell to the given height. This work done will be stored in the barbell in the form of potential energy. That is, the energy of the barbell is supplied to it from Slim Jim through the work.
c)
All the work done by Jim will be stored in the barbell in the form of potential energy. Thus, the work done by Jim is equal to the potential energy of the barbel.
Therefore, the work done by Jim is 960.4 J
d)
Given,
The time interval, t=1.5 s
The power is given by,
[tex]P=\frac{W}{t}[/tex]Where W is the work done by Jim.
On substituting the known values,
[tex]\begin{gathered} P=\frac{960.4}{1.5} \\ =640.27\text{ W} \end{gathered}[/tex]Thus the power used by Jim is 640.27 W
a flat coil of wire consisting of 17 turns, each with an area of 50 cm2, is positioned perpendicularly to a uniform magnetic field that increases its magnitude at a constant rate from 3 t to 6 t in 2.0 s. what is the magnitude of the emf (in volts) induced in the coil? your answer should be a number with two decimal places, do not include the unit.
The magnitude of electromotive force is 0.13 volt.
We need to know about the electromotive force of induction to solve this problem. The emf induction appears when there is any change in magnetic flux. The magnitude of emf can be determined by
ε = N dΦ / dt
where N is coil turns, ε is electromotive force, dΦ is change in magnetic flux and dt is time interval.
From the question above, the parameters given are
B1 = 3T
B2 = 6T
dt = 2 s
N = 17
A = 50 cm² = 5 x 10¯³ m²
Find the change in magnetic flux
dΦ = (Φ2 - Φ1)
dΦ = (B2 . A - B1 . A)
dΦ = A (B2 - B1)
dΦ = 5 x 10¯³(6 - 3)
dΦ = 15 x 10¯³ Tm²
By substituting the given parameters, we can calculate electromotive force
ε = N . dΦ / dt
ε = 17 . 15 x 10¯³ / 2
ε = 0.13 volt
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an ice hockey forward with mass 70.0 kg is skating due north with a speed of 5.5 m/s. as the forward approaches the net for a slap shot, a defensive player (mass 110 kg) skates toward him in order to apply a body-check. the defensive player is traveling south at 4.0 m/s just before they collide. if the two players become intertwined and move together after they collide, in what direction and at what speed do they move after the collision? friction between the two players and the ice can be neglected.
The speed of the two players after the collision is 0.31 m/s in a direction of due south.
What is the final speed of the two players?The final speed of the two players after the collision is calculated as follows.
Apply the principle of conservation of linear momentum;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
where;
m₁ is the mass of the first playerm₂ is the mass of the second playeru₁ is the initial velocity of the first playeru₂ is the initial velocity of the second playerv is the common speed of both players after the collision.(70 x 5.5) + (-110 x 4) = v(70 + 110)
385 - 440 = 180v
-55 = 180v
v = -55/180
v = -0.31 m/s
Thus, the direction of the speed of the two players after the collision is due south.
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the one component of the material of the milky way galaxy that prevents us from seeing and photographing the galactic center at optical wavelengths is the one component of the material of the milky way galaxy that prevents us from seeing and photographing the galactic center at optical wavelengths is very cold hydrogen gas. hot hydrogen gas. the glare of light from nearby stars. interstellar dust.
The one component of the material of the milky way galaxy that prevents us from seeing and photographing the galactic center at optical wavelengths is interstellar dust.
The Milky Way is a vast collection of stars dust and gas. It is called a spiral galaxy because it looks like a spinning windmill when viewed from above or below. The Sun is in one of the spiral arms about 25,000 light-years away from the center of the galaxy. Astronomers have released the first images of a supermassive black hole at the center of our Milky Way.
The results provide overwhelming evidence that this object is indeed a black hole and provide valuable clues about the workings of the giant galaxies thought to reside at the center of most galaxies. This extreme environment exposes you to intense UV and X-rays. But much of this activity is obscured from view by a huge plume of interstellar dust. The center of our Milky Way is hidden from the prying eyes of optical telescopes by clouds of dust and gas.
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