Answer:
35.6 m
Explanation:
The given ball possesses a projectile motion from its initial height. So, the required launch velocity of the ball is 6.55 m/s.
What is launch velocity?The horizontal component of velocity during the projection of an object is known as launch velocity. It is obtained when the horizontal range is known.
Given data -
The angle of projection is, [tex]\theta = 10.0 {^\circ}[/tex].
The initial height of the projection is, h = 1.50 m.
The horizontal displacement is, R = 3.00 m.
The mathematical expression for the horizontal displacement (Range) of the projectile is given as,
[tex]R = \dfrac{u^{2} \times sin2 \theta}{g}[/tex]
here,
u is the launch velocity.
g is the gravitational acceleration.
Solving as,
[tex]u =\sqrt{\dfrac{R \times g}{sin2 \theta}}\\\\\\u =\sqrt{\dfrac{1.50 \times 9.8}{sin(2 \times 10)}}\\\\\\u =\sqrt{\dfrac{1.50 \times 9.8}{sin20^\circ}}\\\\\\u=6.55 \;\rm m/s[/tex]
Thus, we can conclude that the required launch velocity of the ball is 6.55 m/s.
Learn more about the projectile motion here:
https://brainly.com/question/11049671
What is the change in internal energy of the system (∆U) in a process in which 10 kJ of heat energy is absorbed by the system and 70 kJ of work is done by the system?
Answer:
Explanation:
According to first law of thermodynamics:
∆U= q + w
= 10kj+(-70kJ)
-60kJ
, w = + 70 kJ
(work done on the system is positive)
q = -10kJ ( heat is given out, so negative)
∆U = -10 + (+70) = +60 kJ
Thus, the internal energy of the system decreases by 60 kJ.
A cylindrical coil of radius 6 cm is placed in magnetic field, which is changing in time with the rate 0.5 T/s. The magnetic field direction is parallel to the cylinder axis. What should be the number of turns in the coil to induce the emf
induced emf value is missing..
please correct the question
You are at the carnival with you your little brother and you decide to ride the bumper cars for fun. You each get in a different car and before you even get to drive your car, the little brat crashes into you at a speed of 3 m/s.
A. Knowing that the bumper cars each weigh 80 kg, while you and your brother weigh 60 and 30 kg,respectively, write down the equations you need to use to figure out how fast you and your brother are moving after the collision.
B. After the collision, your little brother reverses direction and moves at 0.36 m/s. How fast are you moving after the collision?
C. Assuming the collision lasted 0.05 seconds, what is the average force exerted on you during the collision?
D. Who undergoes the larger acceleration, you or your brother? Explain.
Answer:
a) The equation is [tex](m_{b}+m_{c} )u_{b} = (m_{y}+m_{c} )v_{y} + (m_{b}+m_{c} )v_{b}[/tex]
b) Your velocity after collision is 2.64 m/s
c) The force you felt is 7392 N
d) you and your brother undergo an equal amount of acceleration
Explanation:
Your mass [tex]m_{y}[/tex] = 60 kg
your brother's mass [tex]m_{b}[/tex] = 30 kg
mass of the car [tex]m_{c}[/tex] = 80 kg
your initial speed [tex]u_{y}[/tex] = 0 m/s (since you've not started moving yet)
your brother's initial velocity [tex]u_{b}[/tex] = 3 m/s
your final speed [tex]v_{y}[/tex] after collision = ?
your brother's final speed [tex]v_{b}[/tex] after collision = ?
a) equations you need to use to figure out how fast you and your brother are moving after the collision is
[tex](m_{y}+m_{c} )u_{y} + (m_{b}+m_{c} )u_{b} = (m_{y}+m_{c} )v_{y} + (m_{b}+m_{c} )v_{b}[/tex]
but [tex]u_{y}[/tex] = 0 m/s
the equation reduces to
[tex](m_{b}+m_{c} )u_{b} = (m_{y}+m_{c} )v_{y} + (m_{b}+m_{c} )v_{b}[/tex]
b) if your little brother reverses with velocity of 0.36 m/s it means
[tex]v_{b}[/tex] = -0.36 m/s (the reverse means it travels in the opposite direction)
then, imputing values into the equation, we'll have
[tex](m_{b}+m_{c} )u_{b} = (m_{y}+m_{c} )v_{y} + (m_{b}+m_{c} )v_{b}[/tex]
(30 + 80)3 = (60 + 80)[tex]v_{y}[/tex] + (30 + 80)(-0.36)
330 = 140[tex]v_{y}[/tex] - 39.6
369.6 = 140[tex]v_{y}[/tex]
[tex]v_{y}[/tex] = 369.6/140 = 2.64 m/s
This means you will also reverse with a velocity of 2.64 m/s
c) your initial momentum = 0 since you started from rest
your final momentum = (total mass) x (final velocity)
==> (60 + 80) x 2.64 = 369.6 kg-m/s
If the collision lasted for 0.05 s,
then force exerted on you = (change in momentum) ÷ (time collision lasted)
force on you = ( 369.6 - 0) ÷ 0.05 = 7392 N
d) you changed velocity from 0 m/s to 2.64 m/s in 0.05 s
your acceleration is (2.64 - 0)/0.05 = 52.8 m/s^2
your brother changed velocity from 3 m/s to 0.36 m/s in 0.05 s
his deceleration is (3 - 0.36)/0.05 = 52.8 m/s
you and your brother undergo an equal amount of acceleration. This is because you gained the momentum your brother lost
In your words, describe how momentum is related to energy.
Answer:
you need momentum in order to release energy. For example, if you need to push something heavy and you get a running head start, then it will be easier.
Explanation:
Consider a long rod of mass, m, and length, l, which is thin enough that its width can be ignored compared to its length. The rod is connected at its end to frictionless pivot.
a) Find the angular frequency of small oscillations, w, for this physical pendulum.
b) Suppose at t=0 it pointing down (0 = 0) and has an angular velocity of 120 (that is '(t = 0) = 20) Note that 20 and w both have dimensions of time-1. Find an expression for maximum angular displacement for the pendulum during its oscillation (i.e. the amplitude of the oscillation) in terms of 20 and w assuming that the angular displacement is small.
Answer:
Explanation:
The rod will act as pendulum for small oscillation .
Time period of oscillation
[tex]T=2\pi\sqrt{\frac{l}{g} }[/tex]
angular frequency ω = 2π / T
= [tex]\omega=\sqrt{\frac{g}{l} }[/tex]
b )
ω = 20( given )
velocity = ω r = ω l
Let the maximum angular displacement in terms of degree be θ .
1/2 m v ² = mgl ( 1 - cosθ ) ,
[ l-lcosθ is loss of height . we have applied law of conservation of mechanical energy .]
.5 ( ω l )² = gl( 1 - cos θ )
.5 ω² l = g ( 1 - cosθ )
1 - cosθ = .5 ω² l /g
cosθ = 1 - .5 ω² l /g
θ can be calculated , if value of l is given .
A skater of mass 45.0 kg standing on ice throws a stone of mass 7.65 kg with a speed of 20.9 m/s in a horizontal direction. Find:
a. The speed of the skater after throwing the stone.
b. The distance over which the skater will move in the opposite direction if the coefficient of kinetic friction between his skates and the ice is 0.03.
Answer:
Explanation:
know that there is no external force on skater and the stone so the total momentum of the system will remains constant
so we will have
here we have
so the skater will move back with above speed
now the deceleration of the skater is due to friction given as
Answer:
(a) 3.553 m/s
(b) 21.46 m
Explanation:
(a) Applying the law of of momentum,
Total momentum before collision = Total momentum after collision
mu+m'u' = mv+m'v'.................. Equation 1
Where m and m' are the mass of skater and stone respectively, u and u' are the initial velocity of skater and stone respectively, v and v' are the final velocity of the skater and the stone respectively.
Note, u = 0 m/s, u' = 0 m/s
Therefore,
0 = mv+m'v'
-mv = m'v'................ Equation 2
make v the subject of the equation
v = -m'v'/m............. Equation 3
Given: m = 45 kg, m' = 7.65 kg, v' = 20.9 m/s
Substitute into equation 3
v = 7.65(20.9)/45
v = -3.553 m/s
Hence the speed of the skater = 3.553 m/s
(b) F = mgμ..............Equation 4
But F = ma
Therefore,
ma = mgμ
a = gμ............... Equation 5
Where a = acceleration of the skater, g = acceleration due to gravity, μ = coefficient of kinetic friction
Given: μ = 0.03, g = 9.8 m/s²
Substitute into equation 5
a = 0.03(9.8)
a = 0.294 m/s²
Using the equation of motion,
v² = u²+2as............. Equation 6
Where s = distance moved by the skater.
note that u = 0 m/s.
therefore,
v² = 2as
s = v²/2a................ Equation 7
Given: v = 3.553 m/s, a = 0.294
Substitute into equation 7
s = 3.553²/(2×0.294)
s = 12.62/0.588
s = 21.46 m
A 25 kg box sliding to the left across a horizontal surface is brought to a halt in a distance of 15 cm by a horizontal rope pulling to the right with 15 N tension.
Required:
a. How much work is done by the tension?
b. How much work is done by gravity?
The work done by tensional force of the rope is 2.25 J and the work done by gravity is 36.75 J.
The given parameters;
mass of the box, m = 25 kgdistance traveled by the box, d = 15 cm = 0.15 mtension on the rope, T = 15 NThe work done by the tension is calculated as follows;
W = Fd
W = 15 x 0.15
W = 2.25 J
The work done by gravity is calculated as;
W = (25 x 9.8) x 0.15
W = 36.75 J
Thus, the work done by tensional force of the rope is 2.25 J and the work done by gravity is 36.75 J.
Learn more here: https://brainly.com/question/19498865
dandre expands 120w of power in moving a couch 15 meters in 5 seconds how much force does he exert ?
Answer:
The answer is 40 N for APX
Explanation:
A 34.8 kg runner has a kinetic energy of 1.09 x 10³ J. What is the speed of the runner?
Answer:
7.91 m/s
Explanation:
Speed: This can be defined as the ratio of this to time. Speed can also be the magnitude of a velocity or speed is velocity without direction.
The S.I unit of speed is m/s.
From the question,
K.E = 1/2(mv²)................ Equation 1
Where K.E = Kinetic Energy, m = mass of the runner, v = velocity of the runner.
make v the subject of the equation
v = √(2K.E/m).................Equation 2
Given: K.E = 1.09×10³ J, m = 34.8 kg.
Substitute into equation 2
v = √(2× 1.09×10³/34.8)
v = √(62.64)
v = 7.91 m/s.
Hence the speed of the runner = 7.91 m/s
Use I=∫r2 dm to calculate I of a slender uniform rod of length L, about an axis at one end perpendicular to the rod. note: a "slender rod" often refers to a rod of neglible cross sectional area, so that the volume is the Length, and the mass density X Length.
Answer:
The moment of inertia of a slender uniform rod of length L about an axis at one end perpendicular to the rod is [tex]I = \frac{1}{3}\cdot m \cdot L^{2}[/tex].
Explanation:
Let be an uniform rod of length L whose origin is located at one one end and axis is perpendicular to the rod, such that:
[tex]\lambda = \frac{dm}{dr}[/tex]
Where:
[tex]\lambda[/tex] - Linear density, measured in kilograms per meter.
[tex]m[/tex] - Mass of the rod, measured in kilograms.
[tex]r[/tex] - Distance of a point of the rod with respect to origin.
Mass differential can translated as:
[tex]dm = \lambda \cdot dr[/tex]
The equation of the moment of inertia is represented by the integral below:
[tex]I = \int\limits^{L}_{0} {r^{2}} \, dm[/tex]
[tex]I = \lambda \int\limits^{L}_{0} {r^{2}} \, dr[/tex]
[tex]I = \lambda \cdot \left(\frac{1}{3}\cdot L^{3} \right)[/tex]
[tex]I = \frac{1}{3}\cdot m \cdot L^{2}[/tex] (as [tex]m = \lambda \cdot L[/tex])
The moment of inertia of a slender uniform rod of length L about an axis at one end perpendicular to the rod is [tex]I = \frac{1}{3}\cdot m \cdot L^{2}[/tex].
A spherical balloon is inflated with gas at the rate of 500 cubic centimeters per minute. How fast is the radius of the balloon increasing at the instant the radius is 40 centimeters
Answer:
0.245cm/minExplanation:
The volume of the spherical balloon is expressed as V = 4/3πr³ where r is the radius of the spherical balloon. If the spherical balloon is inflated with gas at the rate of 500 cubic centimetres per minute then dV/dt = 500cm³.
Using chain rule to express dV/dt;
dV/dt = dV/dr*dr/dt
dr/dt is the rate at which the radius of the gallon is increasing.
From the formula, dV/dr = 3(4/3πr^3-1))
dV/dr = 4πr²
dV/dt = 4πr² *dr/dt
500 = 4πr² *dr/dt
If radius r = 40;
500 = 4π(40)² *dr/dt
500 = 6400π*dr/dt
dr/dt = 500/6400π
dr/dt = 5/64π
dr/dt = 0.245cm/min
Hence, the radius of the balloon is increasing at the rate of 0.245cm/min
The magnitude of the Poynting vector of a planar electromagnetic wave has an average value of 0.939 W/m2. The wave is incident upon a rectangular area, 1.5 m by 2.0 m, at right angles. How much total electromagnetic energy falls on the area during 1.0 minute
Answer:
The total energy is [tex]T = 169.02 \ J[/tex]
Explanation:
From the question we are told that
The Poynting vector (energy flux ) is [tex]k = 0.939 \ W/m^2[/tex]
The length of the rectangle is [tex]l = 1.5 \ m[/tex]
The width of the rectangle is [tex]w = 2.0 \ m[/tex]
The time taken is [tex]t = 1 \ minute = 60 \ s[/tex]
The total electromagnetic energy falls on the area is mathematically represented as
[tex]T = k * A * t[/tex]
Where A is the area of the rectangle which is mathematically represented as
[tex]A= l * w[/tex]
substituting values
[tex]A= 2 * 1.5[/tex]
[tex]A= 3 \ m^2[/tex]
substituting values
[tex]T = 0.939 * 3 * 60[/tex]
[tex]T = 169.02 \ J[/tex]
Huygens claimed that near the surface of the Earth the velocity downwards of an object released from rest, vy, was directly proportional to the square root of the distance it had fallen, . This is true if c is equal to
Answer:
the expression is False
Explanation:
From the kinematics equations we can find the speed of a body in a clean fall
v = v₀ - g t
v² = V₀² - 2 g y
If the body starts from rest, the initial speed is zero (vo = 0)
v= √ (2g y)
In the first equation it gives us the relationship between speed and time.
With the second equation we can find the speed in which the distance works, this is the expression, see that speed is promotional at the height of a delicate body.
Therefore the expression is False
What is the only force that acts on an object in free fall? centripetal friction gravity
Answer:
one with only gravity acting upon it
Explanation:
Edgenuity :)
A coil of area 0.320 m2 is rotating at 100 rev/s with the axis of rotation perpendicular to a 0.430 T magnetic field. If the coil has 700 turns, what is the maximum emf generated in it
Answer:
The maximum emf generated in the coil is 60527.49 V
Explanation:
Given;
area of coil, A = 0.320 m²
angular frequency, f = 100 rev/s
magnetic field, B = 0.43 T
number of turns, N = 700 turns
The maximum emf generated in the coil is calculated as,
E = NBAω
where;
ω is the angular speed = 2πf
E = NBA(2πf)
Substitute in the given values and solve for E
E = 700 x 0.43 x 0.32 x 2π x 100
E = 60527.49 V
Therefore, the maximum emf generated in the coil is 60527.49 V
wo parallel-plate capacitors C1 and C2 are connected in parallel to a 12.0 V battery. Both capacitors have the same plate area of 5.30 cm2 and plate separation of 2.65 mm. However, the first capacitor C1 is filled with air, while the second capacitor C2 is filled with a dielectric that has a dielectric constant of 2.10. (a) What is the charge stored on each capacitor
Complete Question
Two parallel-plate capacitors C1 and C2 are connected in parallel to a 12.0 V battery. Both capacitors have the same plate area of 5.30 cm2 and plate separation of 2.65 mm. However, the first capacitor C1 is filled with air, while the second capacitor C2 is filled with a dielectric that has a dielectric constant of 2.10.
(a) What is the charge stored on each capacitor
(b) What is the total charge stored in the parallel combination?
Answer:
a
i [tex]Q_1 = 2.124 *10^{-11} \ C[/tex]
ii [tex]Q_2 = 4.4604 *10^{-11} \ C[/tex]
b
[tex]Q_{eq} = 6.5844 *10^{-11} \ C[/tex]
Explanation:
From the question we are told that
The voltage of the battery is [tex]V = 12.0 \ V[/tex]
The plate area of each capacitor is [tex]A = 5.30 \ cm^2 = 5.30 *10^{-4} \ m^2[/tex]
The separation between the plates is [tex]d = 2.65 \ mm = 2.65 *10^{-3} \ m[/tex]
The permittivity of free space has a value [tex]\epsilon_o = 8.85 *10^{-12} \ F/m[/tex]
The dielectric constant of the other material is [tex]z = 2.10[/tex]
The capacitance of the first capacitor is mathematically represented as
[tex]C_1 = \frac{\epsilon * A }{d }[/tex]
substituting values
[tex]C_1 = \frac{8.85 *10^{-12 } * 5.30 *10^{-4} }{2.65 *10^{-3} }[/tex]
[tex]C_1 = 1.77 *10^{-12} \ F[/tex]
The charge stored in the first capacitor is
[tex]Q_1 = C_1 * V[/tex]
substituting values
[tex]Q_1 = 1.77 *10^{-12} * 12[/tex]
[tex]Q_1 = 2.124 *10^{-11} \ C[/tex]
The capacitance of the second capacitor is mathematically represented as
[tex]C_2 = \frac{ z * \epsilon * A }{d }[/tex]
substituting values
[tex]C_1 = \frac{ 2.10 *8.85 *10^{-12 } * 5.30 *10^{-4} }{2.65 *10^{-3} }[/tex]
[tex]C_1 = 3.717 *10^{-12} \ F[/tex]
The charge stored in the second capacitor is
[tex]Q_2 = C_2 * V[/tex]
substituting values
[tex]Q_2 = 3.717*10^{-12} * 12[/tex]
[tex]Q_2 = 4.4604 *10^{-11} \ C[/tex]
Now the total charge stored in the parallel combination is mathematically represented as
[tex]Q_{eq} = Q_1 + Q_2[/tex]
substituting values
[tex]Q_{eq} = 4.4604 *10^{-11} + 2.124*10^{-11}[/tex]
[tex]Q_{eq} = 6.5844 *10^{-11} \ C[/tex]
In the far future, astronauts travel to the planet Saturn and land on Mimas, one of its 62 moons. Mimas is small compared with the Earth's moon, with mass Mm = 3.75 ✕ 1019 kg and radius Rm = 1.98 ✕ 105 m, giving it a free-fall acceleration of g = 0.0636 m/s2. One astronaut, being a baseball fan and having a strong arm, decides to see how high she can throw a ball in this reduced gravity. She throws the ball straight up from the surface of Mimas at a speed of 41 m/s (about 91.7 mph, the speed of a good major league fastball).
(a) Predict the maximum height of the ball assuming g is constant and using energy conservation. Mimas has no atmosphere, so there is no air resistance.
(b) Now calculate the maximum height using universal gravitation.
(c) How far off is your estimate of part (a)? Express your answer as a percent difference and indicate if the estimate is too high or too low.
Answer:
a) h = 13,205.4 m
b) r_f = 2.12 106 m
c) e% = 0.68%
Explanation:
a) This is an exercise we are asked to use energy conservation,
Starting point. On the surface of Mimas
Em₀ = K = ½ m v²
Final point. Where the ball stops
[tex]Em_{f}[/tex] = U = m g h
Em₀ = Em_{f}
½ m v² = m g h
h = ½ v² / g
let's calculate
h = ½ 41² / 0.0636
h = 13,205.4 m
b) For this part we are asked to use the law of universal gravitation, write the energy
starting point. Satellite surface
Em₀ = K + U = ½ m v² - GmM / r_o
final point. Where the ball stops
[tex]Em_{f}[/tex]= U = - G mM / r_f
Em₀ = Em_{f}
½ m v² - G m M / r_o = - G mM / r_f
In this case all distances are measured from the center of the satellite
1 / rf = 1 / GM (-½ v² + G M / r_o)
let's calculate
1 / rf = 1 / (6.67 10⁻¹¹ 3.75 10¹⁹) (- ½ 41 2 + 6.67 10⁻¹¹ 3.75 10¹⁹ / 1.98 105)
1 / r_f = 3,998 10⁻¹¹(-840.5 + 12.63 10³)
1 / r_f = 4,714 10⁻⁷
r_f = 1 / 4,715 10⁻⁷
r_f = 2.12 106 m
to measure this distance from the satellite surface
r_f ’= r_f - r_o
r_f ’= 2.12 106 - 1.98 105
r_f ’= 1,922 106 m
c) the percentage difference is
e% = 13 205.4 / 1,922 106 100
e% = 0.68%
The estimate of part a is a little low
If you were to calculate the pull of the Sun on the Earth and the pull of the Moon on the Earth, you would undoubtedly find that the Sun's pull is much stronger than that of the Moon, yet the Moon's pull is the primary cause of tides on the Earth. Tides exist because of the difference in the gravitational pull of a body (Sun or Moon) on opposite sides of the Earth. Even though the Sun's pull is stronger, the difference between the pull on the near and far sides is greater for the Moon.
Required:
a. "Let F(r) be the gravitational force exerted on one mass by a second mass a distance r away. Calculate dF(r)/dr to show how F changes as r is changed.
b. Evaluate this expression for dF(r) jdr for the force of the Sun at the Earth's center and for the Moon at the Earth's center.
c. Suppose the Earth-Moon distance remains the same, but the Earth is moved closer to the Sun. Is there any point where dF(r)/dr for the two forces has the same value?
Answer:effective
Explanation:
g Power is _________________the force required to push something the work done by a system the speed of an object the rate that the energy of a system is transformed the energy of a system
Answer:
the rate that the energy of a system is transformed
Explanation:
We can define energy as the capacity or ability to do work. Power is defined as the rate of doing work or the rate at which energy is transformed. It can also be regarded as the time rate of energy transfer. In older physics literature, power is sometimes referred to as activity.
Power is given by energy/time. Its unit is watt which is defined as joule per second. Another popular unit of power is horsepower. 1 horsepower = 746 watts.
Very large magnitude of power is measured in killowats and megawatts.
If a 950 kg merry-go-round platform of radius 4.5 meters is driven by a mechanism located 2.0 meters from its center of rotation, how much force must the mechanism provide to get the platform moving at 5.5 revolutions per minute after 12 seconds if it were initially at rest
Answer:
F = 213.75 N
Explanation:
First we need to calculate the angular acceleration of merry-go-round. For that purpose we use 1st equation of motion in angular form.
ωf = ωi + αt
where,
ωf=final angular velocity=(5.5 rev/min)(2π rad/1 rev)(1 min/60 s)=0.58 rad/s
ωi =initial angular velocity = 0 rad/s
t = time = 12 s
α = angular acceleration = ?
Therefore,
0.58 rad/s = 0 rad/s + α(12 s)
α = (0.58 rad/s)/(12 s)
α = 0.05 rad/s²
Now, we shall find the linear acceleration of the merry-go-round:
a = rα
where,
a = linear acceleration = ?
r = radius = 4.5 m
Therefore,
a = (4.5 m)(0.05 rad/s²)
a = 0.225 m/s²
Now, the force is given by Newton;s 2nd Law:
F = ma
where,
F = Force = ?
m = mass pf merry-go-round = 950 kg
Therefore,
F = (950 kg)(0.225 m/s²)
F = 213.75 N
A very fine thread is placed between two glass plates on one side and the other side is touching to form a wedge. A beam of monochromatic light of wavelength 600 nm illuminates the wedge and 178 bright fringes are observed. What is the thickness of the thread?
Answer:
53.3micro meters
Explanation:
See attached file
Rank the electromagnetic radiation from lowest to highest in the simulation in terms of energy, wavelength, and frequency.
a. Energy
b. Wavelength
c. Frequency
Answer:
A.ENERGY: Radio<microwaves<infrared<visible light<ultraviolet<xrays<gammarays
B. WAVELENGTH: Radio>microwaves> infrared>visible light>ultraviolet>xray> gammarays
C. FREQUENCY: Radio<microwaves<infrared<visible light<ultraviolet<xray< gammarays
Explanation:
THIS IS BECAUSE OF THE FOLLOWING EQUATIONS
1.ENERGY (E)= hX freqency
So as energy of radiation increases frequency also increases
2. Velocity (v) = wavelength x frequency
So as wavelength increases frequency decreases and vice versa
A 0.400-kg ice puck, moving east with a speed of 5.86 m/s , has a head-on collision with a 0.900-kg puck initially at rest.
A. Assume that the collision is perfectly elastic, what will be the speed of the 0.300 kg object after the collision?
B. What will be the direction of the 0.300 kg object after the collision?
C. What will be the speed of the 0.900 kg object after the collision?
Answer:
a) The final speed of the 0.400-kg puck after the collision is 2.254 meters per second, b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards, c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.
Explanation:
a) Since collision is perfectly elastic and there are no external forces exerted on pucks system, the phenomenon must be modelled after the Principles of Momentum and Energy Conservation. Changes in gravitational potential energy can be neglected. That is:
Momentum
[tex]m_{1}\cdot v_{1,o} + m_{2}\cdot v_{2,o} = m_{1}\cdot v_{1,f} + m_{2}\cdot v_{2,f}[/tex]
Energy
[tex]\frac{1}{2}\cdot (m_{1}\cdot v_{1,o}^{2}+ m_{2}\cdot v_{2,o}^{2})=\frac{1}{2}\cdot (m_{1}\cdot v_{1,f}^{2}+ m_{2}\cdot v_{2,f}^{2})[/tex]
[tex]m_{1}\cdot v_{1,o}^{2} + m_{2}\cdot v_{2,o}^{2} = m_{1}\cdot v_{1,f}^{2} + m_{2}\cdot v_{2,f}^{2}[/tex]
Where:
[tex]m_{1}[/tex], [tex]m_{2}[/tex] - Masses of the 0.400-kg and 0.900-kg pucks, measured in kilograms.
[tex]v_{1,o}[/tex], [tex]v_{2,o}[/tex] - Initial speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Final speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.
If [tex]m_{1} = 0.400\,kg[/tex], [tex]m_{2} = 0.900\,kg[/tex], [tex]v_{1,o} = +5.86\,\frac{m}{s}[/tex], [tex]v_{2,o} = 0\,\frac{m}{s}[/tex], the system of equation is simplified as follows:
[tex]2.344\,\frac{kg\cdot m}{s} = 0.4\cdot v_{1,f} + 0.9\cdot v_{2,f}[/tex]
[tex]13.736\,J = 0.4\cdot v_{1,f}^{2}+0.9\cdot v_{2,f}^{2}[/tex]
Let is clear [tex]v_{1,f}[/tex] in first equation:
[tex]0.4\cdot v_{1,f} = 2.344 - 0.9\cdot v_{2,f}[/tex]
[tex]v_{1,f} = 5.86-2.25\cdot v_{2,f}[/tex]
Now, the same variable is substituted in second equation and resulting expression is simplified and solved afterwards:
[tex]13.736 = 0.4\cdot (5.86-2.25\cdot v_{2,f})^{2}+0.9\cdot v_{2,f}^{2}[/tex]
[tex]13.736 = 0.4\cdot (34.340-26.37\cdot v_{2,f}+5.063\cdot v_{2,f}^{2})+0.9\cdot v_{2,f}^{2}[/tex]
[tex]13.736 = 13.736-10.548\cdot v_{2,f} +2.925\cdot v_{2,f}^{2}[/tex]
[tex]2.925\cdot v_{2,f}^{2}-10.548\cdot v_{2,f} = 0[/tex]
[tex]2.925\cdot v_{2,f}\cdot (v_{2,f}-3.606) = 0[/tex]
There are two solutions:
[tex]v_{2,f} = 0\,\frac{m}{s}[/tex] or [tex]v_{2,f} = 3.606\,\frac{m}{s}[/tex]
The first root coincides with the conditions before collision and the second one represents a physically reasonable solution.
Now, the final speed of the 0.400-kg puck is: ([tex]v_{2,f} = 3.606\,\frac{m}{s}[/tex])
[tex]v_{1,f} = 5.86-2.25\cdot (3.606)[/tex]
[tex]v_{1,f} = -2.254\,\frac{m}{s}[/tex]
The final speed of the 0.400-kg puck after the collision is 2.254 meters per second.
b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards.
c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.
A mass M = 4 kg attached to a string of length L = 2.0 m swings in a horizontal circle with a speed V. The string maintains a constant angle \theta\:=\:θ = 35.4 degrees with the vertical line through the pivot point as it swings. What is the speed V required to make this motion possible?
Answer:
The velocity is [tex]v = 2.84 1 \ m/s[/tex]
Explanation:
The diagram showing this set up is shown on the first uploaded image (reference Physics website )
From the question we are told that
The mass is m = 4 kg
The length of the string is [tex]L = 2.0 \ m[/tex]
The constant angle is [tex]\theta = 35.4 ^o[/tex]
Generally the vertical forces acting on the mass to keep it at equilibrium vertically is mathematically represented as
[tex]Tcos (\theta ) - mg = 0[/tex]
=> [tex]mg = Tcos (\theta )[/tex]
Now let the force acting on mass horizontally be k so from SOHCAHTOA rule
[tex]sin (\theta ) = \frac{k }{T}[/tex]
=> [tex]k = T sin \theta[/tex]
Now this k is also equivalent to the centripetal force acting on the mass which is mathematically represented as
[tex]F_v = \frac{m v^2}{r}[/tex]
So
[tex]k = F_v[/tex]
Which
=> [tex]T sin \theta= \frac{ m v^2}{ r }[/tex]
So
[tex]\frac{Tsin (\theta )}{Tcos (\theta )} = \frac{mg}{ \frac{mv^2}{r} }[/tex]
=> [tex]Tan (\theta ) = \frac{v^2}{ r * g }[/tex]
=> [tex]v = \sqrt{r * g * tan (\theta )}[/tex]
Now the radius is evaluated using SOHCAHTOA rule as
[tex]sin (\theta) = \frac{ r}{L}[/tex]
=> [tex]r = L sin (\theta)[/tex]
substituting values
[tex]r = 2 sin ( 35.4 )[/tex]
[tex]r = 1.1586 \ m[/tex]
So
[tex]v = \sqrt{1.1586* 9.8 * tan (35.4 )}[/tex]
[tex]v = 2.84 1 \ m/s[/tex]
Estimate the radiation pressure due to a bulb that emits 25 W of EM radiation at a distance of 6.5 cm from the center of the bulb. Assume that light is completely absorbed.
..........................................................
The radiation pressure due to a bulb that emits 25 W of EM radiation at a distance of 6.5 cm from the center of the bulb and the light is completely absorbed is 1.5707x10⁻⁶ N/m².
What is the Radiation pressure?Radiation pressure was defined as the mechanical pressure exerted upon any surface due to the exchange of momentum between the object and the electromagnetic field.
Radiation pressure always includes the Momentum of light or electromagnetic radiation of any wavelength that can be absorbed, reflected, or otherwise emitted by matter on any scale.
E.g: Black-body radiation
Given the values are,
Wattage of bulb = W = 25 W
distance = d = 6.5 cm = 0.065 m
To know the Radiation Pressure,
It can be given by
P = I/c
Where, c = 299792458 m/s is the speed of light,
I is the intensity of radiation and given by
I = W/4πd²
Where W is the Wattage of bulb and d is the distance
I = 25/4π*0.065²
I = 470.872 w/m²
so, the radiation pressure becomes
P = 470.872/299792458
P = 1.5707x10⁻⁶ N/m²
Therefore, the radiation pressure due to a 25 W bulb at a distance of 6.5 cm is 1.5707x10⁻⁶ N/m²
To know more about the radiation pressure,
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7. How many 1.00 µF capacitors must be connected in parallel to store a charge of 1.00 C with a potential of 110 V across the capacitors?
Answer:
q = C V charge on 1 capacitor
q = 1 * 10E-6 * 110 = 1.1 * 10E-4 C per capacitor
N = Q / q = 1 / 1.1 * 10E-4 = 9091 capacitors
9.09 × 10³ capacitors must be connected in parallel.
How to calculate the number of capacitors connected in parallel?
Given C = 1.00μF = 1 × 10⁻⁶ F
q = 1.00 C
V = 110 V
The equivalent capacitance is given by
Ceq = q/V
where q = total charge on all the capacitors
V = potential difference
For N number of identical capacitors in parallel,
Ceq = NC
Therefore,
NC = q/V
N = q/VC
Putting on the values in the above formula,
N = 1/ (110)(1 × 10⁻⁶)
= 1 / 110 × 10⁻⁶
= 9.09 × 10³
Learn more about the capacitors here:
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if F, V, and we're chosen as fundamental unit of force, velocity and time respectively , the dimensions of mass would be represented as
Answer:
The dimension of mass can be represented as: [tex][F^{1} T^{1} V^{-1} ][/tex]
Explanation:
We have Force = Mass X Acceleration
= Mass X [tex]\frac{Change in Velocity}{Time Taken}[/tex]
or, Mass = Force x [tex]\frac {Time Taken } { Change in Velocity }[/tex]
So, dimensions of mass = [tex]\frac{[F][T]}{[V]}[/tex]
= [tex][F^{1} T^{1} V^{-1} ][/tex]
The angle of incidence of a light beam in air onto a reflecting surface is continuously variable. The reflected ray is found to be completely polarized when the angle of incidence is 64.2° .What is the index of refraction of the reflecting material?
Answer:
n = 2.0686
Explanation:
When an unpolarized ray of light is reflected on a surface, the reflected ray is partially polarized, complete polarization occurs when it is true that between the transmitted and reflected ray one has 90, the relationship is
n = so tea
let's calculate
n = tan 64.2
n = 2.0686
A 10-m-long glider with a mass of 680 kg (including the passengers) is gliding horizontally through the air at 26 m/s when a 60 kg skydiver drops out by releasing his grip on the glider.
What is the glider's speed just after the skydiver lets go?
Answer:
The glider’s speed after the skydiver lets go is 26 m/s
Explanation:
To calculate the glider’s speed just after the skydiver lets go, we will need to use the conservation of momentum
Mathematically;
mv = mv + mv
so 680 * 26 = (680-60)v + 60 * 26
17680 = 620v + 1560
17680-1560 = 620v
16120 = 620v
v = 16120/620
v = 26 m/s
Do women like when men shave their pubic hair?
Answer:
Maybe
Explanation:
Tbh it’s different for every women. Most women would say yes because having a bush is a bit disturbing for some and/or could be in return for a women shaving down there (if they do)