The net force's work is described as d=(3.82)(4.00)=15.3J for W=F net, where we made use of the fact that the canister began at rest and traveled horizontally while being affected by horizontal forces; the resulting effect is represented by the symbol F net.
Since all of these cases are in closed systems, the result is in a new preserved quantity.
Linear momentum and this new quantity, angular momentum, are comparable. In this chapter, we first define angular momentum and then examine it from a number of angles.
But first, we look into a single particle's angular momentum. This enables us to compute the force angular momentum of a rigid cylindrically symmetric body and a system of particles.
We used the fact that the canister started at rest and moved horizontally under the influence of horizontal forces; the ensuing effect is denoted by the symbol F (force) net.
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I’m not sure how to solve this equation but it’s about physics
The values of velocity at point A,C,E and G is 15m/s, the values of velocity at point B, D, F and H is 10m/s, 20m/s, 30m/s and 40m/s respectively.
The projectile is launched from the top of the cliff with a horizontal velocity of 15m/s.
Throughout the motion of the projectile the velocity will have two components. On along vertical and one along horizontal.
Here, the direction of the acceleration is only downwards because only gravity is there and that in only downwards direction.
There is no acceleration in the horizontal direction. So, the velocity in the horizontal direction will never change.
So, the velocity at A,C,E and G is 15m/s throughout the motion.
We know that,
t = √(2h/g)
h = 1/2gt²
Where,
H is height of the cliff at that point,
t is the time,
g is gravity which is equal to 10m/s².
So, the height at point,
B is 1/2(10)(1)² = 5m
D is 1/2(10)(2)² = 20m
F is 1/2(10)(3)² = 45m
H is 1/2(10)(4)² = 50m.
The vertical velocity can be found by using the formula,
V = √2gh
So, vertical velocity at point,
B is √2(10)(5) = 10m/s.
D is √2(10)(20) = 20m/s.
F is √2(10)(45) = 30m/s.
H is √2(10)(80) = 40m/s.
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1. How does the transfer of momentum change when the velocity of the marble change?
2. How does the transfer of momentum change when the mass (number of marbles) changes?
3. How could you calculate the momentum of a single marble?
4. If two marbles together are moving at the same speed as a single marble which will have greater momentum and why?
5. Can momentum be shown as a vector? Explain why or why not?
The transfer of momentum increases when the velocity of the marble increases and vice versa.
The transfer of momentum increases when the mass (number of marbles) increases and vice versa.
The momentum of a single marble can be calculated from the product of its mass and velocity.
If two marbles together are moving at the same speed as a single marble, the two marbles will have greater momentum because of their greater mass.
Momentum can be shown as a vector because the direction of the momentum of a body is specified.
What is the momentum of a body?The momentum of a body is the product of the mass and the velocity of the body.
Mathematically;
Momentum = mass * velocitySince the velocity of an object is a vector, the momentum of a body is also a vector.
The momentum of a body is directly proportional to its mass and velocity.
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3. Which kinematic equation should you use if you are trying to solve for time, but do not know displacement?
The kinematic equation should you use if you are trying to solve for time, but do not know displacement is v[f] = v[o] + at.
What is Kinematics?
Kinematics is the branch of mechanics concerned with the motion of objects without reference to the forces which cause the motion.
Given are the equations of motion.
If the displacement is not given, then we can solve for time using the first equation of motion given by -
v[f] = v[o] + at
In this equation, there is no term involving displacement. We can solve for time as follows -
v[f] = v[o] + at
v[f] - v[o] = at
t = (v[f] - v[o])/a
Therefore, the kinematic equation should you use if you are trying to solve for time, but do not know displacement is v[f] = v[o] + at.
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a car of mass 1850 kg is traveling at 22.5 m/s in a straight line. a truck of mass 3170 moving in the opposite direction collides with the car and the 2 vehicles stop immediately without sliding. the speed of the truck was
13.13 m/s is the velocity of the truck was when a truck of mass 3170 moving in the opposite direction collides with the car and the 2 vehicles stop immediately without sliding.
Momentum of car= 1850(22.5)=41625 kg m/s
Momentum of car = momentum of truck
Let the speed of truck = v m/s
Momentum of truck= v(3170)=3170 v kgm/s
3170v = 41625
v = 13.13 m/s
Speed of truck = 13.13 m/s
The pace at which a moving object changes position as seen from a certain point of view and as measured by a particular time standard is known as its velocity.
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If an astronaut weighs 148 N on the Moon and 893 N on Earth,
then what is his mass on Earth?
Answer:
So, at moon g’=1.62 m/s^2
Therefore, mg’=99N (as given , where m is the mass )
so, m=99/1.62 =61.11 kg
hence, weight on earth will be mg( where g is the acceleration due to gravity at earth, 9.8m/s^2)
61.11*9.8=598.878 N will be the weight on earth
Explanation:
hope it helps
Carlos does a chin-up in gym class and raises himself 0.8 m. If Carlos has a mass of 62 kg how much work does he accomplish?
Answer:486.1 J
Explanation:
2 and 23/75 as a percentage
2 and 23/75 as a percentage is 2.30666666666. Hope this helps! :)
A tennis player makes a return at a speed of 15m/s and at height of 3m to land in the court at a horizontal distanceof 12m from her. what are the possible angles of projection of the ball
With the use of range formula, also, in multiple of 180 degree, 59 and 239 degrees are the angles of projection.
What is a Projectile ?A projectile in kinematic under physics can be defined as any object that is projected or thrown to take a trajectory path.
Given that a tennis player makes a return at a speed of 15 m/s and at height of 3m to land in the court at a horizontal distance of 12 m from her.
Let us assume that the ball fall at height 3m. U = 0. Let us find the time
h = ut + 1/2gt²
3 = 0 + 1/2 × 9.8 × t²
3 = 4.9t²
t² = 3/4.9
t² = 0.612
t = √0.612
t = 0.782 s
Total time T = 2t
T = 2 × 0.782
T = 1.564 s
The possible angles of projection of the ball can be calculated by using the range formula
R = UcosФT
12 = 15cosФ × 1.564
12 = 23.46cosФ
cosФ = 12 / 23.46
cosФ = 0.512
Ф = [tex]Cos^{-1}[/tex] (0.512)
Ф = 59. 2 degree
Ф = 59°
Therefore, the possible angles of projection of the ball are 59° and 239°
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air is contained in a rigid tank with a volume of 2 m3, initially at 10 bar and 300 k. heat is transferred to the tank until its temperature reaches 400 k. during this process a pressure-relief valve allows air to escape and maintain the pressure constant at 10 bar. neglecting kinetic and potential energy variation e↵ects, and using ideal gas equations with constant specific heat at 300 k, determine, (a) the mass of air that escapes the tank (in kg), and (b) the amount of heat transferred to the tank.
Mass of air that escapes the tank is M =2.8075 kg and the amount of heat transferred to the tank is Q = 1008.942 kj
In physics, mass is a numerical representation of inertia, which is a basic characteristic of all matter. The resistance a body of matter offers to a change in its speed or location as a result of the application of a force is what it is in essence. The change a force applies produces a change that is smaller the more mass a body has. The kilogramme, which is defined in terms of Planck's constant as equal to 6.62607015 1034 joule second, is the unit of mass in the International System of Units (SI). One kilogramme times one square metre per second squared is equivalent to one joule. Accurate measurements of Planck's constant are used to determine the kilogramme since the second and the metre are already defined in terms of other physical constants.When a body breaks up into fragments, the mass also breaks up into bits, thus the total mass of all the pieces equals the original mass.
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at certain times of the year, five planets can be seen from earth, including mars and what other planet?
Mercury, Venus , Mars , Jupiter and Saturn are much seen planets.
What is planets ?
A planets is a body that is in orbit around the sun has sufficient Mars for it's self-gravity to overcome rigid body forces so that it assumes nearly round shape.
In simple words we can say any of the large bodies that revolve around the sun in the solar system called planet.
The current count orbiting our star there are eight planet but earth, mercury, Venus , Jupiter and Mars are much visible .
So,
At certain times of the year these 5 planets can be seen from earth .
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A gas occupies 1.56 L at 20 Pa. What will be the volume of this gas if the pressure becomes 40Pa?
Answer:
Hope the picture will help
The location of magnetic north changes over time in its relation to true north. This angular difference between magnetic north and true north is called.
The angular difference between true north and magnetic north is called declination.
What is declination?
Declination is the said to be the angle between magnetic north and geographic north in the horizontal plane that changes with time and geographic location. The true north or geographic north is a fixed point on the globe that points towards the north pole directly while the magnetic north unlike the true north changes over time, and it is the direction a compact needle points to as it aligns with the earth's magnetic field.
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1. what happens when electromagnetic radiation encounters a metal object? discuss two ways that can be used to reduce interference from electrical noise.
A full circuit is required for the flow of electrical current, and all electrical energy transmissions need the formation of a circuit with two wires—one going to the recipient and the other going back to the sender.
The construction of a circuit with two wires—one going to the recipient and the other back to the sender—is necessary for all electrical energy transmissions in order for current to flow. Due to the fact that all wireless signals are essentially delivered as electromagnetic signals rather than electrical signals, electromagnetic interference, often known as EMI, is a common problem in wireless communications.
This suggests that any interference with an electromagnetic signal (wave) may cause it to be amplified, destroyed, or lost (assuming constructive interference).
The interference brought on by electromagnetic disturbance is known as electromagnetic interference. Radio frequency interference is another name for it. We refer to interference that occurs in the radio frequency spectrum as radio frequency interference.
These days, wireless communication is expanding at an unprecedented rate. Over a large concentration, a small spectrum is available. Because of this, there is a greater likelihood of mutual interference. The performance of the device and the channel can both be impacted by a slight electromagnetic disturbance. It is crucial to keep interference at a minimal for better wireless performance.
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The most energetic region of the electromagnetic spectrum is the microwave region. True or false?.
False. The most energetic region of the electromagnetic spectrum is NOT the microwave region. It is the Gamma rays.
What are Gamma rays?
The electromagnetic radiation with the shortest wavelength and maximum energy is known as a gamma ray. The wavelengths of gamma radiation are often less than a few tenths of an angstrom (1010 meter), yet the energy of gamma photons exceed tens of thousands of electron volts.
Gamma rays are created when radioactive atomic nuclei break apart and when certain subatomic particles decay. Pair annihilation, in which two photons are formed along with the disappearance of an electron and its antiparticle, a positron, also results in the production of gamma rays. Additionally, they can be produced when some unstable subatomic particles, such the neutral pion, decay.
Therefore, The most energetic region of the electromagnetic spectrum are gamma rays.
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A car is coasting backward downhill at a speed of
3
.
0
m
/
s
when the driver gets the engine started. After
2
.
5
s
, the car is moving uphill at
4
.
5
m
/
s
. If uphill is chosen as the positive direction, what is the car’s average acceleration?
The car’s average acceleration is 3m/s².
What is the Car's average acceleration?
The average acceleration in the question measures the change in the velocity of the car by the time it used to cover the distance.
Average acceleration of the car is the change in the velocity of the car by the change in the time the car traveled.
ΔVelocity = 4.5m/s -( -3.0m/s)
= 7.5m/s
The initial velocity is negative 3.0m/s as the car started moving backwards.
The time is 2.5s
The
The average acceleration of the car = 7.5m/s ÷ 2.5s
= 3m/s²
In summary, the car coasting backward downhill at an initial speed of 3.0 m/s would have an average acceleration of 3m/s² After increasing the speed to 4.5 m/s at a time of 2.5s.
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a twist of 4 n m is applied to the handle of the screwdriver. resolve this couple moment into a pair of couple forces f exerted on the handle and p exerted on the blade. (10 points)
For nuclear meeting It should be used on a school bus driver's handle. Adapt to the outcome this couple force into a couple moment occurred to be that implies that we must act now to find a better situation.
Open to a nuclear meeting It is intended to be used on a school bus driver's handle. respect the outcome This couple of forces occurred to be in a couple at this time. That suggests that at this moment, we need to discover a better situation and start acting.
And at this time, 400 people are throwing up. Toe in toe is what is known as a couple of forty meters. As the radius of the handle is 30 millimeters in this instance, you can translate it to meters and use the force to draw up on a spot rather four upon it. Are you in favor of addressing this issue? 1 33 nuclear will be Therefore, one till people comprehend that is supposed f.
Now, we must determine a slow speed consent business. Once more, a boardroom mingled. I am. Carletto is a PVR as well. In contrast to the original five meters, Cal critic M is supplied four on a separated radius of five millimeters.
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25 Which wavefront is travelling at a speed closest to that of a sound wave through a solid? A one that moves 10m in 0.01 s B one that moves 50m in 0.5s C one that moves 1000 m in 100 s D one that moves 2000 m in 2000 s
The wavefront B is travelling at the speed(5000m/s) closest to that of a sound wave(6000m/s) travelling through solid.
The velocity of the sound waves in solids is nearly 6000 meters per second.
Let us check the velocity for all the three wave fronts,,
Wavefront A,
We know,
speed = distance/time taken
Distance travelled by the wavefront A is 10m,
Time taken by the wavefront A to travel the distance is 0.01 seconds,
Putting all the values,
Speed = 10/0.01
Speed = 1000m/s.
Speed of wavefront A is 1000m/s.
Wavefront B,
Again we know,
speed = distance/time taken
Distance travelled by the wavefront Bis 50m,
Time taken by the wavefront B to travel the distance is 0.01 seconds,
Putting all the values,
Speed = 50/0.01
Speed = 5000m/s.
Speed of wavefront B is 5000m/s.
Wavefront C,
We know,
speed = distance/time taken
Distance travelled by the wavefront C is 1000m,
Time taken by the wavefront C to travel the distance is 100 seconds,
Putting all the values,
Speed = 1000/100
Speed = 10m/s.
Speed of wavefront C is 10m/s.
Wavefront D,
Again using the formula,
speed = distance/time taken
Distance travelled by the wavefront is 2000m,
Time taken by the wavefront to travel the distance is 2000 seconds,
Putting all the values,
Speed = 2000/2000
Speed = 1m/s.
Speed of wavefront D is 1m/s.
As we can see, The speed of wavefront B is closest to the speed of sound waves in solids.
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On the kickoff, the football was in the air for 4 seconds before hitting the ground 53 meters away. How fast did the ball leave the foot of the kicker?
Answer:
Vx = 53 m / 4 sec = 13.25 m/s horizontal speed
Vy = 2 sec* 9.80 m/sec^2 = 19.6 m/s vertical speed
V = (13.25^2 + 19.6^2)^1/2 = 23.7 m/s
Check:
Range = V^2 sin 2θ / g
tan θ = 19.6 / 13.25 = 1.48 θ = 55.9 deg
R = 23.7^2 sin 112 deg / 9.8 = 53.1 m
a hockey puck with a mass of 0.11 kg is at rest on the horizontal frictionless surface of the rink. a player applies a horizontal force of 1 n to the puck. find the speed and the traveled distance 5 s later.
The solution to this ques is available in the image.
Given,
Force= 1N
Mass= 0.11kg
Time= 5sec
Force= mass X accelaration
Accelaration= velocity/ time
Speed=distance/ time
Hence, the speed is 45 m/s and the distance is 225 m.
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an airplane that is flying level needs to accelerate from a speed of 200 m/s to a speed of 240 m/s while it flies a distance of 1200 m. what must the constant acceleration of the plane be?
The constant acceleration of the plane is 7.33 m/[tex]s^{2}[/tex].
Use this formula of acceleration, [tex]v^{2}[/tex] = [tex]u^{2}[/tex] + 2as
[tex]v^{2}[/tex] = [tex]u^{2}[/tex] + 2as,
a = [tex]v^{2}[/tex] - [tex]u^{2}[/tex] / 2s
v = velocity
s = distance
a = acceleration
Put the values in this formula, a = [tex]v^{2}[/tex] = [tex]u^{2}[/tex] / 2s
a = (240× 240) - (200 × 200)/ 2 × 1200
a = 7.33 m/[tex]s^{2}[/tex]
Hence, the constant acceleration of the plane is 7.33 m/[tex]s^{2}[/tex].
Acceleration is the rate of change of velocity per unit time. It is vector quantity and its S.I. unit is m/[tex]s^{2}[/tex].
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if the satellite has a mass of 3500 kg , a radius of 4.4 m , and the rockets each add a mass of 220 kg , what is the required steady force of each rocket if the satellite is to reach 32 rpm in 5.0 min , starting from rest?
0.145 Newton is the required steady force of each rocket if the satellite is to reach 32 rpm in 5.0 min , starting from rest
mass of satellite ( M ) = 3500 kg
radius of satellite ( r ) = 4.4 m
mass added by each rocket ( m ) = 220 kg
Angular speed of satellite = 32 rpm
Time = 5 mins
moment of inertia
[tex]I=1/2Mr^2+4mr^2[/tex]
M = 3500 kg m = 220 kg, r =4.4 m
I=33880+17036
Insert values into equation ( 1 )
I =20416 kg-m^2
angular acceleration
∝ = ω / t
= ( 2π×32 ) / 3600×5
= 0.0112 rad/s²
steady force required by each rocket
Τ = I × ∝ ( 1 )
T = 4rF
Equation ( 1 ) becomes
4rF = I × ∝
Force = I×∝ / 4r
= ( 228×0.0112 ) / 4×4.4
=2.56/17.6
= 0.145 N .
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A swimmer needs to cross a river as quickly as possible. The swimmer’s speed in still water is 1.35 m/s.
(a) If the river’s current speed is 0.60 m/s and the river is 106.68 m wide, how long will it take the swimmer to cross the river if he swims so that his body is angled slightly upstream while crossing, and he ends up on the far bank directly across from where he started?
(b) If he points his body directly across the river and is therefore carried downstream, how long will it take to get across the river and how far downstream from his starting point will he end up?
A swimmer needs to cross a river as quickly as possible. The swimmer’s speed in still water is 1.35 m/s.
(a) If the river’s current speed is 0.60 m/s and the river is 106.68 m wide, it will take the time swimmer to cross the river is (t) = 72.57 S, if he swims so that his body is angled slightly upstream while crossing, and he ends up on the far bank directly across from where he started.
(b) If he points his body directly across the river and is therefore carried downstream, it will take time to get across the river is (t)=72.57 S, and the distance downstream from his starting point will he end up is (S) = 43.542m.
What is velocity?Velocity means that the covered distance in a given time. In other words mathematically distance divided by time called velocity. It can be measured in m/s and cm/s.
How can we calculate the time and distance?(a)To calculate the time for the first case we are using the formula, S=vt.
The swimmer’s speed in still water is 1.35 m/s.=v₁=1.35j m/s.
the river’s current speed is 0.60 m/s.=v₂=0.60i m/s.
Now, let us apply the rule of relative velocity:
v=v₁+v₂=1.35j+0.60i
Or, |v|=√(1.35)²+(0.60)²
Or, |v|=1.47 m/s
Now we know the velocity of the water, v=1.47 m/s.
The wide of the river, S=106.68 m
Now we have to calculate the time to travel of the swimmer= t s
Now we put the values in above equation,
S=vt
Or, t=S/v
Or, t= 106.68/1.47
Or, t=72.57 s.
From the above calculation we can say that, If the river’s current speed is 0.60 m/s and the river is 106.68 m wide, it will take the time swimmer to cross the river is (t)=72.57 S
(b) If he points his body directly across the river and is therefore carried downstream, The time taken by the swimmer is the same as the first case.
So the time would be (t)=72.57 S.
The distance downstream from his starting point will he end up,
S = vt
Or, S= 0.6*72.57
Or, S=43.542
So according to the solution, If he points his body directly across the river and is therefore carried downstream, it will take to get across the river is (t)=72.57 S, and the distance downstream from his starting point will he end up is (S)=43.542m.
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A steam turbine has an efficiency of 40.0%. A steam engine has an efficiency of
25.0%. Suppose both devices are provided with 1000 J of thermal energy. How much
more useful work will the steam turbine do? Show your work.
Answer:
150J (joules)
Explanation:
25% efficiency of 1000J = 250J
40% efficiency of 1000J= 400J
400-250=150
States that if no external forces act on a group of objects, their total momentum does not change
Answer: this is true
Explanation:
if the object is moving and there is nothing stooping it, it will keep moving
(Celestial body all natural bodies visible in the sky outside the earth atmosphere constitute the celestial bodies example stars planets their moon comet asteroid meteors)
Is artificial satellite a Celestial body
Celestial bodies are natural bodies so can we say artificial satellite is not a Celestial body??
An artificial satellite can be called a celestial body as it orbits around a planet.
Objects in space such as the sun, moon, planets, and stars are known as celestial bodies or heavenly bodies. Satellites are revolving bodies that orbit planets. They are the fundamental component of celestial bodies. These could be transmitted by humans or be of natural origin.
The moon orbits the earth as a natural satellite because it is held in place by the gravity of the planet. Additionally, man has positioned artificial or man-made satellites in orbit around the earth and other planets for communication and scientific research.
As a artificial satellite also orbits around a natural celestial body, it can be called as a celestial body.
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a wave has an amplitude of 2 cm (y-direction) and a frequency of 12 hz, and the distance (x-direction) from a crest to the nearest trough is measured to be 5 cm. determine the velocity of the wave.
120 centimeters per second, is the wave speed or wave velocity.
How fast do waves travel?Wave velocity is the amount of distance a wave travels in one unit of time. The rate at which a particle disturbance like a crest or trough or compression or rarefaction spreads in a medium is known as the particle's propagation speed.
Given,
Amplitude (A) = 2cm,
Frequency (f) = 12hz,
Time period [tex]T=\frac{1}{f} =\frac{1}{12} s[/tex]
The horizontal distance between the closest peak or trough, respectively, is referred to as the wavelength. A crest's distance from the closest tough was specified. The distance between a crest and the closest crest is therefore two times that distance.
This is the wave length :
λ=(2)(5)
=10cm
The wave velocity or wave speed is given by:
v = λf
v=(10)(12)
=120 cm/s
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A submarine begins a descent from the surface downward at an angle of 40.5 degrees with the water and continues along a straight-line path for a total distance of 397.0 meters before leveling off its path. How far (in meters) is the submarine below the surface of the water?
The submarine is 288.04 meters below the surface of the water.
As we can see,
The submarine is making an angle of 40.5° with the water surface and moving down till a displacement L of 397 meters.
The displacement L moved by the submarine will have two components because displacement is a vector quantity.
One along water surface Lcos(40.5°) and one along the line perpendicular to water surface Lsin(40.5°).
The depth D of the submarine below the water surface is along the line perpendicular to the water surface.
Here,
D = Lsin(40.5°)
Value of cos 40.5°is 0.76.
So,
D = 397×0.76
D = 288.04 meters.
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2. A feather is dropped on the moon from a height of 1.40 meters. The acceleration of gravity on the moon is 1.67 m/s2. Determine the time for the feather to fall to the surface of the moon.
The total time taken by the feather to fall to the surface of moon will be 1.29 seconds.
Given is a feather dropped on the moon from a height of 1.40 meters. The acceleration of gravity on the moon is 1.67 m/s².
State second equation of motion.The second equation of motion is -
S = ut + 1/2 at²
We can write -
[S] = 1.4 m
[a] = 1.67 m/s²
[u] = 0 m/s
Using the second equation of motion -
S = ut + 1/2 at²
S = 1/2 at²
1.4 = 1/2 x 1.67 x t²
1.4 = 0.835 x t²
t = 1.29 seconds
Therefore, the total time taken by the feather to fall to the surface of moon will be 1.29 seconds.
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What is the frequency of light when the energy of a single photon is 1. 48 × 10-15 j?.
The frequency of light is 3.88x10^23Hz
E = 1.55 x J/ mol
E = hv
v = (1.55 x /1mol x 1mol/6.023 x )/6.626 x 10^34
velocity = 3.88x10^23 s-1
since 1 Hz = 1/ s = 1
velocity =3.88x10^23 Hz
The portion of the electromagnetic spectrum contains the electromagnetic radiation that the human eye perceives as light or visible light. Visible light typically possesses wavelengths between 400 and 700 nanometers (nm) or frequencies between 750 and 420 terahertz, which are halfway between infrared and ultraviolet (with longer wavelengths) (with shorter wavelengths).
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Sample question: What is the frequency of light when the energy for a mole of photons is 1. 55 × 1013 j?.
a paramecium swimming through a fluid is moving at approximately a constant velocity as a result of wiggling its cilia. what can you say about the net force that is being exerted on it while it is doing this?
In the case of a paramecium moving at a constant speed, there is no net force acting on it.
The paramecium's cilia exerts a force on water that is equal to the force that the water exerts on the paramecium's cilia, according to Newton's third law.
The strength of the viscous force the water is exerting on the paramecium will be equal to the strength of the force the paramecium's cilia are exerting on the water.
How does Paramecium use cilia?
Cilia, or tiny strands like hair, cover the whole surface of parasites like cilia, which beat rhythmically to move them and guide bacteria and other food particles into their mouths.
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