Explanation:
Perspective- the art of drawing solid objects on a two-dimensional surface so as to give the right impression of their height, width, depth, and position in relation to each other when viewed from a particular point.
If the particles were moving with a speed much less than c, the magnitude of the momentum of the second particle would be twice that of the first. However, what is the ratio of the magnitudes of momentum for these relativistic particles?
Answer:
p₂ / p₁ = 2 (v₁ / v₂)
Explanation:
The moment is a very useful concept, since it is one of the quantities that is conserved during shocks and explosions, for which it had to be redefined to be consistent with special relativity,
p = m v / √[1+ (v/c)² ]
for the case of speeds much lower than the speed of light this expression is close to
p = m v
In this exercise they indicate that the moment of the second particle is twice the moment of the first, when their velocities are small
p₂ = 2 p₁
p₂/p₁ = 2
in consecuense
m v₂ = 2 m v₁
v₂ = 2 v₁
consider particles of equal mass.
By the time their speeds increase they enter the relativistic regime
p₂ = mv₂ /√(1 + v₂² /c²)
p₁ = m v₁ /√(1 + v₁² / c²)
let's look for the relationship between these two moments
p₂ / p₁ = mv₂ / mv₁ [√ (1+ v₁² / c²) /√ (1 + v₂² / c²)
from the initial statement
p₂ / p₁ = 2 √(c² + v₁²) / (c² + v₂²)
we take c from the root
p₂ / p₁ = 2 √ [(1+ v₁²) / (1 + v₂²)]
this is the exact result, to have an approximate shape suppose that the velocities are much greater than 1
p₂ / p₁ = 2 √ [v₁² / v₂²] = 2 √ [(v₁ / v₂)²]
p₂ / p₁ = 2 (v₁ / v₂)
we see the value of the moment depends on the speed of the particles
. A car going initially with a velocity 15 m/s accelerates at a rate of 2 m/s2 for 10 seconds. It then accelerates at a rate of -1.5 m/s until stop. Find the car’s maximum speed. Calculate the total distance traveled by the car.
Answer:
The maximum speed of the car is 35 m/s
The total distance traveled by the car is 658.33 m
Explanation:
Given;
initial velocity of the car, u = 15 m/s
acceleration of the car, a = 2 m/s²
time of car motion, t = 10 s
(i)
Initial distance traveled by the car is given by;
d₁ = ut + ¹/₂at²
d₁ = (15 x 10) + ¹/₂(2)(10)²
d₁ = 150 + 100
d₁ = 250 m
The maximum speed of the car during this is given by;
v² = u² + 2ad₁
v² = (15)² + (2 x 2 x 250)
v² = 1225
v = √1225
v = 35 m/s
(ii)
The final distance cover by the car during the deceleration of 1.5 m/s².
Note: the final or maximum speed of the car becomes the initial velocity during deceleration.
v² = u² + 2ad₂
where;
v is the final speed of the car when it stops = 0
0 = u² + 2ad₂
0 = (35²) + (2 x - 1.5 x d₂)
0 = 1225 - 3d₂
3d₂ = 1225
d₂ = 1225 / 3
d₂ = 408.33 m
The total distance traveled by the car is given by;
d = d₁ + d₂
d = 250 m + 408.33 m
d = 658.33 m
When a potential difference of 10 V is placed across a certain solid cylindrical resistor, the current through it is 2 A. If the diameter of this resistor is now tripled, the current will be:______.A) 18 A.
B) 2/3 A.
C) 3 A.
D) 2/9 A.
E) 2 A.
Answer:
sorry I wish I could it help you
A plane is heading due west and climbing at the rate of 80 km/hr. If its airspeed is 540 km/hr and there is a wind blowing 80 km/hr to the northwest, what is the ground speed of the plane?
Answer:
599.245km/hr
Explanation:
A plane is heading due west and climbing at the rate of 80 km/hr. If its airspeed is 540 km/hr and there is a wind blowing 80 km/hr to the northwest, what is the ground speed of the plane?
We solve the above question using vectors
In vector form Air speed is -540i + 0j Wind speed is (-80/√2)i + (80/√2)j
Vector notation wind speed is given as: -56.5685 i + 56.5685j
The vector for the ground speed of the plane =
-540i + 0j -56.5685i + 56.5685j
= -596.56854249i + 56.5685j
The the ground speed of the plane √[(596.56854249)² + (56.5685)²]
= √359094.021081 km/hr
= 599.24454197 km/hr
Approximately
= 599.245km/hr
A radio wave transmits 38.5 W/m2 of power per unit area. A flat surface of area A is perpendicular to the direction of propagation of the wave. Assuming the surface is a perfect absorber, calculate the radiation pressure on it.
Answer:
[tex]P=2.57\times 10^{-7}\ N/m^2[/tex]
Explanation:
Given that,
A radio wave transmits 38.5 W/m² of power per unit area.
A flat surface of area A is perpendicular to the direction of propagation of the wave.
We need to find the radiation pressure on it. It is given by the formula as follows :
[tex]P=\dfrac{2I}{c}[/tex]
Where
c is speed of light
Putting all the values, we get :
[tex]P=\dfrac{2\times 38.5}{3\times 10^8}\\\\=2.57\times 10^{-7}\ N/m^2[/tex]
So, the radiation pressure is [tex]2.57\times 10^{-7}\ N/m^2[/tex].
A plane mirror is placed to the right of an object. The image formed by the mirror will be a
real image that appears to be on the right of the mirror.
real image that appears to be on the left of the mirror.
virtual image that appears to be on the right of the mirror.
virtual image that appears to be on the left of the mirror.
Hamish is studying what happens when he sends a sound wave through different mediums, and he records his data in a table.
A 2-column table with 4 rows titled Hamish's Waves. The first column labeled Wave has entries 1, 2, 3, 4. The second column labeled Information has entries liquid, solid, gas, liquid.
Which statement could made about the data collected in Hamish’s table?
Wave 1 will move the fastest.
Wave 2 will move the slowest.
Wave 3 will move the slowest.
Wave 4 will move the fastest.
What is common between transverse waves and longitudinal waves?
Both include an amplitude, crest, and rarefactions
Both move faster at higher temperatures
Both move slower through densely packed molecules
Both include a wavelength from compression to compression
An angle of refraction is the angle between the refracted ray and the
incident ray.
normal.
medium.
boundary.
Answer:
A plane mirror is placed to the right of an object. The image formed by the mirror will be a virtual image that appears to be on the left of the mirror.
Explanation:
A woman standing before a cliff claps her hands, and 2.8s later she hears the echo. How far away is the cliff? The speed of sound in air a ordinary temperature is 343 m/s.
Answer:
480.2 m
Explanation:
The following data were obtained from the question:
Speed of sound (v) = 343 m/s.
Time (t) = 2.8 s
Distance (x) of the cliff =?
The distance of the cliff from the woman can be obtained as follow:
v = 2x /t
343 = 2x /2.8
Cross multiply
2x = 343 × 2.8
2x = 960.4
Divide both side by the coefficient of x i.e 2
x = 960.4/2
x = 480.2 m
Therefore, the cliff is 480.2 m away from the woman.
The distance should be 480.2 m
The calculation is as follows:Since A woman standing before a cliff claps her hands, and 2.8s later she hears the echo. And, there is the velocity of 343 m/s
[tex]v = 2x \div t\\\\343 = 2x \div 2.8\\\\2x = 343 \times 2.8[/tex]
2x = 960.4
x = 480.2 m
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In the winter sport of curling, players give a 20 kg stone a push across a sheet of ice. The Slone moves approximately 40 m before coming to rest. The final position of the stone, in principle, onlyndepends on the initial speed at which it is launched and the force of friction between the ice and the stone, but team members can use brooms to sweep the ice in front of the stone to adjust its speed and trajectory a bit; they must do this without touching the stone. Judicious sweeping can lengthen the travel of the stone by 3 m.1. A curler pushes a stone to a speed of 3.0 m/s over a time of 2.0 s. Ignoring the force of friction, how much force must the curler apply to the stone to bring it op to speed?A. 3.0 NB. 15 NC. 30 N
D. 150 N2The sweepers in a curling competition adjust the trajectory of the slope byA. Decreasing the coefficient of friction between the stone and the ice.
B. Increasing the coefficient of friction between the stone and the ice.C. Changing friction from kinetic to static.D. Changing friction from static to kinetic.3. Suppose the stone is launched with a speed of 3 m/s and travel s 40 m before coming to rest. What is the approximate magnitude of the friction force on the stone?A. 0 NB. 2 NC. 20 ND. 200 N4. Suppose the stone's mass is increased to 40 kg, but it is launched at the same 3 m/s. Which one of the following is true?A. The stone would now travel a longer distance before coming to rest.B. The stone would now travel a shorter distance before coming to rest.C. The coefficient of friction would now be greater.D. The force of friction would now be greater.
Answer:82. Since you have a distance and a force, then the easiest principle to use is energy, i.e. work.
The work done by friction is F * d. This work cancels out the kinetic energy of the stone (1/2)mv^2
Fd = (1/2)mv^2
F = (1/2)mv^2/d.
Plug in m = 20 kg, v = 3 m/sec, d = 40 m.
83. With more mass, the kinetic energy is higher now. The work needed is higher. W = F * d and F is the same.
Explanation:Hope I helped :)
How many significant figures are in 0.0067?
Answer:
2
Explanation:
there are 2 significant figures in there
In principle, when you fire a rifle, the recoil should push you backward. How big a push will it give? Let's find out by doing a calculation in a very artificial situation. Suppose a man standing on frictionless ice fires a rifle horizontally. The mass of the man together with the rifle is 70 kg, and the mass of the bullet is 10 g. If the bullet leaves the muzzle at a speed of 500 m/s, what is the final speed of the man?
Answer:
Explanation:
m1v1=m2v2
m1=70 kg
m2=10 g=0.01 kg
v2=500 m/s
m1v1=m2v2
v1=m2v2/m1
v1=0.01*500/70
v1=0.07
How should the magnetic field lines be drawn for the magnets shown below?
Answer:
Magnetic field lines can be drawn by moving a small compass from point to point around a magnet. At each point, draw a short line in the direction of the compass needle.When opposite poles of two magnets are brought together, the magnetic field lines join together and become denser between the poles.
Explanation:
A car moves forward up a hill at 12 m/s with a uniform backward acceleration of 1.6 m/s2. What is its displacement after 6 s?
Answer:
The displacement of the car after 6s is 43.2 m
Explanation:
Given;
velocity of the car, v = 12 m/s
acceleration of the car, a = -1.6 m/s² (backward acceleration)
time of motion, t = 6 s
The displacement of the car after 6s is given by the following kinematic equation;
d = ut + ¹/₂at²
d = (12 x 6) + ¹/₂(-1.6)(6)²
d = 72 - 28.8
d = 43.2 m
Therefore, the displacement of the car after 6s is 43.2 m
Power is the rate at which work is done true or false
Answer:
false
Explanation:
A 5.3 kg block rests on a level surface. The coefficient of static friction is μ_s=0.67, and the coefficient of kinetic friction is μ_k= 0.48 A horizontal force, x is applied to the block. As x is increased, the block begins moving. Describe how the force of friction changes as x increases from the moment the block is at rest to when it begins moving. Show how you determined the force of friction at each of these times ― before the block starts moving, at the point it starts moving, and after it is moving. Show your work.
As the pushing force x increases, it would be opposed by the static frictional force. As x passes a certain threshold and overcomes the maximum static friction, the block will start moving and will require a smaller magnitude x to maintain opposition to the kinetic friction and keep the block moving at a constant speed. If x stays at the magnitude required to overcome static friction, the net force applied to the block will cause it to accelerate in the same direction.
Let w denote the weight of the block, n the magnitude of the normal force, x the magnitude of the pushing force, and f the magnitude of the frictional force.
The block is initially at rest, so the net force on the box in the horizontal and vertical directions is 0:
n + (-w) = 0
n = w = m g = (5.3 kg) (9.80 m/s²) = 51.94 N
The frictional force is proportional to the normal force, so that f = µ n where µ is the coefficient of static or kinetic friction. Before the block starts moving, the maximum static frictional force will be
f = 0.67 (51.94 N) ≈ 35 N
so for 0 < x < 35 N, the block remains at rest and 0 < f < 35 N as well.
The block starts moving as soon as x = 35 N, at which point f = 35 N.
At any point after the block starts moving, we have
f = 0.48 (51.94 N) ≈ 25 N
so that x = 25 N is the required force to keep the block moving at a constant speed.
As x is increasing it will be opposed by a static frictional force and for the object to start moving and maintain its acceleration, the magnitude of x must exceed the magnitude of the static frictional force and kinetic frictional force
Magnitude of normal force ( object at rest ); n = 51.94 N Required magnitude of x before the movement of object ; x = 35 NMagnitude of x after object start moving x = 25 NGiven data :
mass of block at rest ( m ) = 5.3 kg
Coefficient of static friction ( μ_s ) =0.67
Coefficient of kinetic friction is ( μ_k ) = 0.48
Horizontal force applied to block = x
First step : magnitude of normal force ( n ) when object is at rest
n = w where w = m*g
n - w = 0
n - ( 5.3 * 9.81 ) = 0 ∴ n = 51.94 N
Second step : Required magnitude of x before the movement of object
F = μ_s * n
F = 0.67 * 51.94 = 34.79 N ≈ 35 N
∴ The object will start moving once F and x = 35 N
Final step : Magnitude of x after object start moving
F = μ_k * n
= 0.48 * 51.94 = 24.93 N ≈ 25 N
∴ object will continue to accelerate at a constant speed once F and x = 25N
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A rolling ball moves from x1 = 8.0 cm to x2 = -4.1 cm during the time from t1 = 2.9 s to t2 = 6.0 s .
Complete Question
A rolling ball moves from [tex]x_1 = 8.0 \ cm[/tex] to [tex]x_2 = - 4.1 \ cm[/tex] during the time from [tex]t_1 = 2.9 s[/tex] to [tex]t_2 = 6.0s[/tex]
What is its average velocity over this time interval?
Answer:
The velocity is [tex]v = 3.903 \ m/s[/tex]
Explanation:
From the question we are told that
The first position of the ball is [tex]x_1 = 8.0 \ cm[/tex]
The second position of the ball is [tex]x_2 = - 4.1 \ cm[/tex]
Generally the average velocity is mathematically represented as
[tex]v = \frac{ x_1 - x_2}{t_2 - t_1}[/tex]
=> [tex]v = \frac{ 8 - -4.1 }{ 6 - 2.9 }[/tex]
=> [tex]v = 3.903 \ m/s[/tex]
A car’s brakes decelerate it at a rate of -2.40 m/s2. If the car is originally travelling at 13 m/s and comes to a stop, then how far, in meters, will the car travel during that time?
Answer:
Approximately [tex]35.2\; \rm m[/tex].
Explanation:
Given:
Initial velocity: [tex]u = 13\; \rm m \cdot s^{-1}[/tex].
Acceleration: [tex]a = -2.40\; \rm m \cdot s^{-2}[/tex] (negative because the car is slowing down.)
Implied:
Final velocity: [tex]v = 0\; \rm m \cdot s^{-1}[/tex] (because the car would come to a stop.)
Required:
Displacement, [tex]x[/tex].
Not required:
Time taken, [tex]t[/tex].
Because the time taken for this car to come to a full stop is not required, apply the SUVAT equation that does not involve time:
[tex]\begin{aligned} x &= \frac{v^2 - u^2}{2\, a} \\ &= \frac{{\left(0\; \rm m \cdot s^{-1}\right)}^2 - {\left(13\; \rm m \cdot s^{-1}\right)}^2}{2\times \left(-2.40\; \rm m\cdot s^{-2}\right)} \approx 35.2\; \rm m \end{aligned}[/tex].
In other words, this car would travel approximately [tex]35.2\; \rm m[/tex] before coming to a stop.
A household refrigerator consumes electrical energy at the rate of 200 W. lf electricity costs 5 k per kWh, calculate the cost of operating the appliance for 30 days
Answer:
= 720000 [k]
Explanation:
The cost is equal to 5 [$/kW-h], kilowatt per hour, this value should be multiplied by the power, and then by the time.
[tex]5[\frac{k}{kw*h}]*200[w]*30[day]*24[\frac{h}{day} ][/tex]
= 720000 [k]
True or False when an object speeds up it gains momentum
Answer: True
Explanation:
An object with a mass of 3.0 kg has a
force of 9.0 newtons applied to it. What
is the resulting acceleration of the
object?
[tex] \LARGE{ \underline{ \tt{Required \: answer:}}}[/tex]
We have:
Mass of the object = 3 kgForce on the object = 9 NWe need to find:
Resulting accleration of the object?Solution:
According to Newton's 2nd law of motion, or quantitative measure of Force:
Force = Mass × AcclerationUsing this,
➝ F = ma
➝ 9N = 3 kg × a
➝ a = 9/3 m/s²
➝ a = 3 m/s²
Hence,
The resulting accleration of the object is 3 m/s². And we are done! :D⛱️ [tex] \large{ \blue{ \bf{FadedElla}}}[/tex]
A plane is flying due west at 34 m/s. It encounters a wind blowing at 19 m/s south. Find the resultant veloci
Answer:
The resultant velocity has a magnitude of 38.95 m/s
Explanation:
Vector Addition
Given two vectors defined as:
[tex]\vec v_1=(x_1,y_1)[/tex]
[tex]\vec v_2=(x_2,y_2)[/tex]
The sum of the vectors is:
[tex]\vec v=(x_1+x_2,y_1+y_2)[/tex]
The magnitude of a vector can be calculated by
[tex]d=\sqrt{x^2+y^2}[/tex]
Where x and y are the rectangular components of the vector.
We have a plane flying due west at 34 m/s. Its velocity vector is:
[tex]\vec v_1=(-34,0)[/tex]
The wind blows at 19 m/s south, thus:
[tex]\vec v_2=(0,-19)[/tex]
The sum of both velocities gives the resultant velocity:
[tex]\vec v =(-34,-19)[/tex]
The magnitude of this velocity is:
[tex]d=\sqrt{(-34)^2+(-19)^2}[/tex]
[tex]d=\sqrt{1156+361}=\sqrt{1517}[/tex]
d = 38.95 m/s
The resultant velocity has a magnitude of 38.95 m/s
What would happen if there is more male hyenas than female hyenas in a population?
Choices:
Male hyenas will compete to mate with the females.
Some male hyenas will die.
Male hyenas for wait for more females to join the population.
Answer:
Option 1
Explanation:
I always see animals do that
A jet airplane with a 75.0 m wingspan is flying at 260 m/s. What emf is induced between the wing tips in V if the vertical component of the Earth’s magnetic field is 3.00 × 10-5 T?
Answer:
0.585V
Explanation:
Given that:
B = 3.00 × 10-5 T
l = 75.0 m
v = 260 m/s
From Blv = emf between the wing tips
= 3.00 × 10-5 T × 75×260
= 117/200
= 0.585V
Hence, the emf between the wing tips is 0.585V
True or false. when objects collide , some momentum is lost
Answer:
It is neither false nor true. When they collide some of one of the objects goes to the other object.
Explanation:
Answer: True
Explanation:
A racecar accelerates from rest at 6.5 m/s2 for 4.1 s. How fast will it be going at the end of that time?
Answer:
The final velocity of the car is 26.65 m/s.
Explanation:
Given;
acceleration of the racecar, a = 6.5 m/s²
initial velocity of the car, u = 0
time of motion, t = 4.1 s
The final velocity of the car is given by;
v = u + at
where;
v is the final velocity of the car
suvstitute the givens
v = 0 + (6.5)(4.1)
v = 26.65 m/s.
Therefore, the final velocity of the car is 26.65 m/s.
When particles get close to the surface, they interact with atoms in
the
(Finish the sentence)
Need help ASAP..please help
Answer:
option 3
Explanation:
can i get brainliest
A 715 kg car stopped at an intersection is rear-ended by a 1490 kg truck moving with a speed of 12.5 m/s. If the car was in neutral and its brakes were off, so that the collision is approximately elastic, find the final speed of both vehicles after the collision.
Answer:
The final velocity of the car is 16.893 m/s
The final velocity of the truck is 4.393 m/s
Explanation:
Given;
mass of the car, m₁ = 715 kg
mass of the truck, m₂ = 1490 kg
initial velocity of the car, u₁ = 0
initial velocity of the truck, u₂ = 12.5 m/s
let the final velocity of the car, = v₁
let the final velocity of the truck, = v₂
Apply the principle of conservation of linear momentum for elastic collision;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(715 x 0) + (1490 x 12.5) = 715v₁ + 1490v₂
18625 = 715v₁ + 1490v₂ -----equation (1)
Apply one-directional velocity formula;
u₁ + v₁ = u₂ + v₂
0 + v₁ = 12.5 + v₂
v₁ = 12.5 + v₂
Substitute v₁ into equation (1)
18625 = 715(12.5 + v₂) + 1490v₂
18625 =8937.5 + 715v₂ + 1490v₂
18625 - 8937.5 = 715v₂ + 1490v₂
9687.5 = 2205v₂
v₂ = 9687.5 / 2205
v₂ = 4.393 m/s
solve for v₁
v₁ = 12.5 + v₂
v₁ = 12.5 + 4.393
v₁ = 16.893 m/s
Acceleration is sometimes expressed in multiples of g, where g = 9.8 m/s^2 is the magnitude of the acceleration due to the earth's gravity. In a test crash, a car's velocity goes from 26 m/s to 0 m/s in 0.15 s. How many g's would be experienced by a driver under the same conditions?
Answer:
Acceleration = 18g
Explanation:
Given the following data;
Initial velocity, u = 26m/s
Final velocity, v = 0
Time = 0.15 secs
To find the acceleration;
In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.
This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.
Hence, if we subtract the initial velocity from the final velocity and divide that by the time, we can calculate an object’s acceleration.
Mathematically, acceleration is given by the equation;
[tex]Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}[/tex]
Substituting into the equation, we have;
[tex]a = \frac{0 - 26}{0.15}[/tex]
[tex]a = \frac{26}{0.15}[/tex]
Acceleration = 173.33m/s2
To express it in magnitude of g;
Acceleration = 173.33/9.8
Acceleration = 17.7 ≈ 18g
Acceleration = 18g
How much work is done by the gravitational force on the block?
Answer:
Work = Mass * Gravity * Height and is measured in Joules. Imagine you find a 2 -Kg book on the floor and lift it 0.75 meters and put it on a table. Remember, that “force” is simply a push or a pull. If you lift 100 kg of mass 1-meter, you will have done 980 Joules of work.
Explanation:
What is magnet made of
Answer:
metals like iron or nickel
Explanation: