Answers
Explanation:
1. Impulse = change in momentum
J = Δp
J = mΔv
In the x direction:
Jₓ = mΔvₓ
Jₓ = (0.40 kg) (30 m/s cos 45° − (-20 m/s))
Jₓ = 16.5 kg m/s
In the y direction:
Jᵧ = mΔvᵧ
Jᵧ = (0.40 kg) (30 m/s sin 45° − 0 m/s)
Jᵧ = 8.49 kg m/s
The magnitude of the impulse is:
J = √(Jₓ² + Jᵧ²)
J = 18.5 kg m/s
The average force is:
FΔt = J
F = J/Δt
F = 1850 N
2. Momentum is conserved.
m₁u₁ + m₂u₂ = (m₁ + m₂) v
In the x direction:
(1000 kg) (0 m/s) + (1500 kg) (-12 m/s) = (1000 kg + 1500 kg) vₓ
vₓ = -7.2 m/s
In the y direction:
(1000 kg) (20 m/s) + (1500 kg) (0 m/s) = (1000 kg + 1500 kg) vᵧ
vᵧ = 8 m/s
The magnitude of the final speed is:
v = √(vₓ² + vᵧ²)
v = 10.8 m/s
3. Momentum is conserved.
m₁u₁ + m₂u₂ = (m₁ + m₂) v
In the x direction:
(0.8 kg) (18 m/s cos 45°) + (0.36 kg) (9.0 m/s) = (0.8 kg + 0.36 kg) vₓ
vₓ = 11.6 m/s
In the y direction:
(0.8 kg) (-18 m/s sin 45°) + (0.36 kg) (0 m/s) = (0.8 kg + 0.36 kg) vᵧ
vᵧ = -8.78 m/s
The magnitude of the final speed is:
v = √(vₓ² + vᵧ²)
v = 14.5 m/s
Related Questions
A coil of 160 turns and area 0.20 m2 is placed with its axis parallel to a magnetic field of initial magnitude 0.40 T. The magnetic field changes uniformly from 0.40 T in the +x direction to 0.40 T in the -x direction in 2.0 s. If the resistance of the coil is 16 Ω, at what rate is power generated in the coil?
Answers
Answer:
The rate at which power is generated in the coil is 10.24 Watts
Explanation:
Given;
number of turns of the coil, N = 160
area of the coil, A = 0.2 m²
magnitude of the magnetic field, B = 0.4 T
time for field change = 2 s
resistance of the coil, R = 16 Ω
The induced emf in the coil is calculated as;
emf = dΦ/dt
where;
Φ is magnetic flux = BA
emf = N (BA/dt)
emf = 160 (0.4T x 0.2 m²)/dt
emf = 12.8 V/s
The rate power is generated in the coil is calculated as;
P = V²/ R
P = (12.8²) / 16
P = 10.24 Watts
Therefore, the rate at which power is generated in the coil is 10.24 Watts
Which of the following is true about a planet orbiting a star in the uniform circular motion? B. The speed of the plant is always changing. C. The velocity of the plant is constant. D. The acceleration vector of the plant points towards the center of the circle.
Answers
Explanation:
If an object moves in a circular path, this type of motion is called uniform circular motion. The speed of the object is constant in this type of motion.As velocity is a vector quantity, it keeps on changing. It can be given by drawing tangent to the circle.Acceleration acts towards the center of circle. Hence, out of the given four options, the correct statement about a planet orbiting a star in uniform circular motion is (d) i.e. The acceleration vector of the plant points towards the center of the circle.Answer:
D. The acceleration vector of the plant points towards the center of the circle. (Works for Apex)
Explanation:
Velocity is always changing because to have velocity you require direction and the direction is always changing when orbiting therefore A and C are not true. And B is not true because the speed of the planet is not accelerating when orbiting around a star.
9. You are given a number of 10 Ω resistors, each capable of dissipating only 1.0 W without being destroyed. What is the minimum number of such resistors that you need to combine in series or in parallel to make a 10 Ω resistance that is capable of dissipating at least 5.0 W?
Answers
Answer:
here are 9 resistors, forming a group of 3 resistors in parallel and each group in series with the other.
Explanation:
Let's work carefully this exercise, they indicate that the total resistance 10 ohm and dissipates 5W, so we can use the power equation to find the circuit current
P = Vi = i² R
i = √ P / R
i = √ (5/10)
i = 0.707 A
This is the current that must circulate in the circuit.
Let's build a circuit with three resistors in series and each resistor in series has three resistors in parallel
The equivalent resistance is
1 /[tex]R_{equi}[/tex] = 1/10 + 1/10 + 1/10 = 3/10
Requi = 10/3
Requi = 3.3 Ω
The current in the three series resistors is I = 0.707 A, and this is divided into three equal parts for the parallel resistors
current in each residence in parallel
i_P = 0.707 / 3
I_p = 0.2357 A
now let's look at the power dissipated in each resistor
P = R i²
P = 10 0.2357²
P = 0.56 W
the power dissipated by each resistance is within the range of 1 A, let's see the total power that the 9 resistors dissipate
P_total = 9 P
P = total = 9 0.56
P_total = 5 W
we see that this combination meets the specifications of the problem.
Therefore there are 9 resistors, forming a group of 3 resistors in parallel and each group in series with the other.
A soap bubble is 115 nm thick and illuminated by white light incident perpendicular to its surface. What wavelength (in nm) and color of visible light is most constructively reflected, assuming the same index of refraction as water (nw = 1.33)?
Answers
Answer:
The wavelength is [tex]\lambda = 612nm[/tex] and the color is Orange
Explanation:
from the question we are told that
The thickness is [tex]D = 115 nm = 115 *10^{-9} \ m[/tex]
The refractive index of water is [tex]n_w = 1.33[/tex]
Generally the condition for constrictive interference is
[tex]2 * D = \frac{\lambda _n}{2}[/tex]
Where [tex]\lambda _n[/tex] is the wavelength of light in a particular medium
Now considering the medium of water(soap bubble )
The wavelength of light in this medium is mathematically represented as
[tex]\lambda = \frac{\lambda }{n }[/tex]
So
[tex]2 * D = \frac{\frac{\lambda }{n} }{2}[/tex]
[tex]2 * D = \frac{\lambda }{2 * n }[/tex]
=> [tex]\lambda = 4 *n * D[/tex]
substituting values
[tex]\lambda = 4 *1.33 * 115*10^{-9}[/tex]
[tex]\lambda = 6.118 *10^{-7} \ m[/tex]
[tex]\lambda = 612nm[/tex]
The color is orange because the wavelength range of yellow is
590–625 nm
The distance measured between five successive crests of a wave motion executed by a photon of an electromagnetic radiation is 2.4cm.
1. Determine the wavelength in micrometers
2. what is the frequency of the photon
3. what is the energy possessed by the photon.
Answers
Answer:
1. λ = 0.48 cm = 4800 μm
2. υ = 6.25 x 10¹⁰ Hz
3. E = 4.14 x 10⁻²³ J
Explanation:
1.
Since, the wavelength is defined as the distance between two consecutive or successive crests or troughs. Therefore, in this case the wavelength will be equal to:
Wavelength = λ = Distance between 5 successive crests/5
λ = 2.4 cm/5
λ = 0.48 cm = 4800 μm
2.
The frequency of photon can be given as:
υ = c/λ
where,
υ = frequency of photon = ?
c = speed of light = 3 x 10⁸ m/s
λ = wavelength = 0.48 cm = 0.0048 m
Therefore,
υ = (3 x 10⁸ m/s)/(0.0048 m)
υ = 6.25 x 10¹⁰ Hz
3.
Now, the energy of photon is given as:
E = hυ
where,
E = Energy = ?
h = Plank's Constant = 6.626 x 10⁻³⁴ J.s
Therefore,
E = (6.626 x 10⁻³⁴ J.s)(6.25 x 10¹⁰ Hz)
E = 4.14 x 10⁻²³ J
1. Notice that the voltmeter moves in response to the coil entering or leaving the magnetic gap.
2. Let's apply Faraday's Law to this situation. Faraday's Law says that the induced voltage (or emf )in a loop of wire caused by a changing magnetic field is:
€ = 1
Where is the magnetic flux which is
Q = BA
In this case, the flux density B is not changing. Instead, the changing flux is due to the motion of the coil as it enters or leaves the magnetic gap:
do = BdA
Given that the area immersed in the gap is changing as the coil enters the gap, what is the correct expression of Faraday's Law for this situation?
Answers
Answer:
Explanation:
let the coil of length l and breathe b entering the magnetic field B with speed v.
So, the magnetic flux through the coil is
Ф = B(l×b)
length × breathe = area
Ф = BA
dФ = BdA
therefore induced emf is given as
ε = [tex]Bl(\frac{db}{dt})[/tex]
note: [tex]\frac{db}{dt} = v[/tex]
ε[tex]= Blv[/tex]
attached is the diagram for the solution
A car is moving along a road at 28.0 m/s with an engine that exerts a force of
2,300.0 N on the car to balance the drag and friction so that the car maintains a
constant speed. What is the power output of the engine?
Answers
Answer:
Power = Force × Distance/time
Power = Force × Velocity
Power = 2,300.0 N × 28.0 m/s²
Power = 64400 Nm/s
Explanation:
First show the formula of Power
Re-arrange formula and used to work out Power
Pretty simple stuff!
Hope this Helps!!
A Buchner funnel uses _______ when separating a(n) _______ by filtration.
Answers
Explanation:
A Buchner funnel uses perforatet glass plate when separating a(n) solide from liquid by filtration.
[tex]hope \: this \: helps[/tex]
A parallel-plate capacitor in air has circular plates of radius 2.8 cm separated by 1.1 mm. Charge is flowing onto the upper plate and off the lower plate at a rate of 5 A. Find the time rate of change of the electric field between the plates.
Answers
Answer:
The time rate of change of the electric field between the plates is [tex]\frac{E }{t} = 2.29 *10^{14} \ N \cdot C \cdot s^{-1}[/tex]
Explanation:
From the question we are told that
The radius is [tex]r = 2.8 \ cm = 0.028 \ m[/tex]
The distance of separation is [tex]d = 1.1 \ mm = 0.0011 \ m[/tex]
The current is [tex]I = 5 \ A[/tex]
Generally the electric field generated is mathematically represented as
[tex]E = \frac{q }{ \pi * r^2 \epsilon_o }[/tex]
Where [tex]\epsilon_o[/tex] is the permitivity of free space with a value
[tex]\epsilon_o = 8.85*10^{-12 }\ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]
So the time rate of change of the electric field between the plates is mathematically represented as
[tex]\frac{E }{t} = \frac{q}{t} * \frac{1 }{ \pi * r^2 \epsilon_o }[/tex]
But [tex]\frac{q}{t } = I[/tex]
So
[tex]\frac{E }{t} = * \frac{I }{ \pi * r^2 \epsilon_o }[/tex]
substituting values
[tex]\frac{E }{t} = * \frac{5 }{3.142 * (0.028)^2 * 8.85 *10^{-12} }[/tex]
[tex]\frac{E }{t} = 2.29 *10^{14} \ N \cdot C \cdot s^{-1}[/tex]
The 2-Mg truck is traveling at 15 m/s when the brakes on all its wheels are applied, causing it to skid for 10 m before coming to rest. The total mass of the boat and trailer is 1 Mg. Determine the constant horizontal force developed in the coupling C, and the friction force developed between the tires of the truck and the road during this time.
Answers
Answer:
constant horizontal force developed in the coupling C = 11.25KN
the friction force developed between the tires of the truck and the road during this time is 33.75KN
Explanation:
See attached file
The friction force between the tires of the truck and the road is 22500 N.
Calculating the friction force:It is given that a 2 Mg truck ( m = 2000 Kg) is initially moving with a speed of u = 15 m/s.
Distance traveled before coming to rest, s = 10m
The final velocity of the truck will be zero, v = 0
When the breaks are applied, only the frictional force is acting on the truck and it is opposite to the motion of the truck.
The frictional force is given by:
f = -ma
the acceleration of the truck = -a
The negative sign indicates that the acceleration is opposite to the motion.
Applying the third equation of motion we get:
v² = u² -2as
0 = 15² - 2×a×10
225 = 20a
a = 11.25 m/s²
So the magnitude of frictional force is:
f = ma = 2000 × 11.25 N
f = 22500 N
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What is the only force that acts on an object in free fall? centripetal friction gravity
Answers
Answer:
one with only gravity acting upon it
Explanation:
Edgenuity :)
Two objects are in all respects identical except for the fact that one was coated with a substance that is an excellent reflector of light while the other was coated with a substance that is a perfect absorber of light. You place both objects at the same distance from a powerful light source so they both receive the same amount of energy U from the light. The linear momentum these objects will receive is such that:
Answers
Answer:
absorbent p = S / c
reflective p = 2S/c
Explanation:
The moment of radiation on a surface is
p = U / c
where U is the energy and c is the speed of light.
In the case of a fully absorbent object, the energy is completely absorbed. The energy carried by the light is given by the Poynting vector.
p = S / c
in the case of a completely reflective surface the energy must be absorbed and remitted, therefore there is a 2-fold change in the process
p = 2S/c
An unknown additional charge q3q3q_3 is now placed at point B, located at coordinates (0 mm, 15.0 mm ). Find the magnitude and sign of q3q3q_3 needed to make the total electric field at point A equal to zero.
Answers
Answer:
0.3nanocouloumb
Explanation:
Pls see attached file
"Pipe A has length L and is open at one end and closed at the other. Pipe B is open at both ends and has length 2L. Which harmonic of pipe B matches in frequency the fundamental of pipe A?"
Answers
Answer:
Explanation:
length of pipe A is L
frequency of fundamental note of pipe A which is closed organ pipe
= velocity of sound / 4 x length of pipe
= V / 4 x L where V is velocity of sound .
Similarly frequency of fundamental note of pipe B which is open organ pipe
= velocity of sound / 2 x length of pipe
= V / (2 x 2L) = V / 4L , This is also called first harmonic
so fundamental frequency of pipe A will be equal to first harmonic of pipe B .
Determine the slit spacing d. Explain which measurement you made, show your calculation and your result for the slit spacing. There are several measurements you can make.
Answers
Answer:
The quantities to measure are:
* the distance to the screen
* The distance from the central maximum to each interference
* in order of interference
* wavelength
Explanation:
To determine the gap spacing we must use the constructive interference equation
d sin θ = m λ
as the angles are small
tan θ = sin θ / cos θ
tan θ = sin θ
and the definition of tangent is
tan θ = y / L
Thus
sin θ = y / L
when replacing
d y / L = m λ
d = m λ L / y
with this equation we can know what parameter should be measured.
The quantities to measure are:
* the distance to the screen
* The distance from the central maximum to each interference
* in order of interference
* wavelength
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?
Answers
Answer:
53.3micro meters
Explanation:
See attached file
A solid disk of radius 1.4 cm and mass 430 g is attached by a wire to one of its circular faces. It is twisted through an angle of 10 o and released. If the spring has a torsion constant of 370 N-m/rad, what is the frequency of the motion
Answers
Answer:
f= 4,186 10² Hz
Explanation:
El sistema descrito es un pendulo de torsión que oscila con con velocidad angular, que esta dada por
w = √ k/I
donde ka es constante de torsion de hilo e I es el momento de inercia del disco
El momento de inercia de indican que giran un eje que pasa por enronqueces
I= ½ M R2
reduzcamos las cantidades al sistema SI
R= 1,4 cm = 0,014 m
M= 430 g = 0,430 kg
substituimos
w= √ (2 k/M R2)
calculemos
w = RA ( 2 370 / (0,430 0,014 2)
w = 2,963 103 rad/s
la velocidad angular esta relacionada con la frecuencia por
w =2pi f
f= w/2π
f= 2,963 10³/ (2π)
f= 4,186 10² Hz
Red light is bent the least of all colors as it passes through a prism. What does this tell you about red light? It has a short wavelength. It has a long wavelength. It has a high intensity. It has a low intensity.
Answers
Answer:
Longest wavelength, lowest intensity
Explanation:
Answer:
It has a long wavelength
Explanation:
GRADPOINT
A 10 kg mass car initially at rest on a horizontal track is pushed by a horizontal force of 10 N magnitude. If we neglect the friction force between the car and the track, calculate how much the car travels in 10 s
Answers
Answer:
50 m
Explanation:
F = ma
10 N = (10 kg) a
a = 1 m/s²
Given:
v₀ = 0 m/s
a = 1 m/s²
t = 10 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (0 m/s) (10 s) + ½ (1 m/s²) (10 s)²
Δx = 50 m
Two large, flat, horizontally oriented plates are parallel to each other, a distance d apart. Half-way between the two plates the electric field field has a magnitude E. If the separation of the plates is reduced to d/2 what is the magnitude of the electric field half-way between the plates
Answers
Answer:
Explanation:
Let the potential difference between the plate is V . Then in the first case
Electric field E between plate
E₁ = V / d
where d is separation between plate
When the plate separation becomes d / 2
Electric field E between plate
E₂ = V / d /2
= 2 V / d =2E₁
Or twice the earlier field
Isaac drop ball from height og 2.0 m, and it bounces to a height of 1.5 m what is the speed before and after the ball bounce?
Answers
Explanation:
It is given that, Isaac drop ball from height of 2.0 m, and it bounces to a height of 1.5 m.
We need to find the speed before and after the ball bounce.
Let u is the initial speed of the ball when he dropped from height of 2 m. The conservation of energy holds here. So,
[tex]\dfrac{1}{2}mu^2=mgh\\\\u=\sqrt{2gh} \\\\u=\sqrt{2\times 9.8\times 2} \\\\u=6.26\ m/s[/tex]
Let v is the final speed when it bounces to a height of 1.5 m. So,
[tex]\dfrac{1}{2}mv^2=mgh\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 1.5} \\\\v=5.42\ m/s[/tex]
So, the speed before and after the ball bounce is 6.26 m/s and 5.42 m/s respectively.
An electron moves in a circular path perpendicular to a magnetic field of magnitude 0.285 T. If the kinetic energy of the electron is 2.10 10-19 J, find the speed of the electron and the radius of the circular path. (a) the speed of the electron
Answers
Answer:
The speed of the electron is 6.79 x 10⁵ m/s
The radius of the circular path is 1.357 x 10⁻⁵ m
Explanation:
Given;
magnetic field, B = 0.285 T
energy of electron, E = 2.10 x 10⁻¹⁹ J
The kinetic energy of the electron is calculated as;
[tex]K.E = \frac{1}{2} m_eV^2[/tex]
Where;
[tex]m_e[/tex] is the mass of electron = 9.11 x 10⁻³¹ kg
V is the speed of the electron
[tex]K.E = \frac{1}{2} m_eV^2\\\\V^2 = \frac{2.K.E}{m_e} \\\\V = \sqrt{\frac{2K.E}{m_e} } \\\\V = \sqrt{\frac{2*(2.1*10^{-19})}{9.11*10^{-31}} }\\\\V = 6.79 *10^{5} \ m/s[/tex]
The radius of the circular path is given by;
[tex]R = \frac{M_eV}{qB}[/tex]
where;
q is the charge of the electron = 1.6 x 10⁻¹⁹ C
[tex]R = \frac{M_eV}{qB} \\\\R = \frac{9.11 *10^{-31}*6.79 *10^{5}}{1.6*10^{-19}*0.285} \\\\R = 1.357 *10^{-5} \ m[/tex]
which of the following statements is not true Negatively charged objects attract other negatively charged objects. Positively charged objects attract negatively charged objects. Positively charged objects attract neutral objects. Negatively chargers objects attract neutral objects.
Answers
Answer:
negatively charged object attract other negatively objects
Explanation:
opposites attract
Answer:
negativelycharged objects attract other negatively charged objects
Explanation:
unlike charges attract like charges repel
A tennis player swings her 1000 g racket with a speed of 12 m/s. She hits a 60 g tennis ball that was approaching her at a speed of 15 m/s. The ball rebounds at 40 m/s.
A) How fast is her racket moving immediately after the impact? You can ignore the interaction of the racket with her hand for the brief duration of the collision.
_________m/s
B) If the tennis ball and racket are in contact for 7.00, what is the average force that the racket exerts on the ball?
_________N
Answers
The velocity and force are required.
The speed of the racket is 8.7 m/s
The required force is 471.43 N.
[tex]m_1[/tex] = Mass of racket = 1000 g
[tex]m_2[/tex] = Mass of ball = 60 g
[tex]u_1[/tex] = Initial velocity of racket = 12 m/s
[tex]u_2[/tex] = Initial velocity of ball = -15 m/s
[tex]v_1[/tex] = Final velocity of racket
[tex]v_2[/tex] = Final velocity of ball = 40 m/s
[tex]\Delta t[/tex] = Time = 7 ms
The equation of the momentum will be
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\Rightarrow v_1=\dfrac{m_1u_1+m_2u_2-m_2v_2}{m_1}\\\Rightarrow v_1=\dfrac{1\times 12+0.06\times (-15)-0.06\times 40}{1}\\\Rightarrow v_1=8.7\ \text{m/s}[/tex]
Force is given by
[tex]F=m_2\dfrac{v_2-u_2}{\Delta t}\\\Rightarrow F=0.06\times \dfrac{40-(-15)}{7\times 10^{-3}}\\\Rightarrow F=471.43\ \text{N}[/tex]
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A Young'sdouble-slit interference experiment is performed with monochromatic light. The separation between the slits is 0.44 mm. The interference pattern on the screen 4.2 m away shows the first maximum 5.5 mm from the center of the pattern. What is the wavelength of the light in nm
Answers
Answer:
Explanation:
The double slit interference phonemene is described for the case of constructive interference
d sin θ= m λ (1)
let's use trigonometry to find the sinus
tan θ = y / L
in general in interference phenomena the angles are small
tan θ = sin θ / cos θ = sin θ
The double slit interference phonemene is described for the case of constructive interference
d sin θ = m lam (1)
let's use trigonometry to find the sinus
tan θ = y / L
in general in interference phenomena the angles are small
tan θ = sin θ / cos θ = sin θ
we substitute
sin θ = y / L
we substitute in equation 1
d y / L = m λ
λ = dy / L m
let's reduce the magnitudes to the SI system
d = 0.44 mm = 0.44 10⁻³ m
y = 5.5 mm = 5.5 10⁻³ m
L = 4.2m
m = 1
let's calculate
λ = 0.44 10⁻³ 5.5 10⁻³ / (4.2 1)
λ = 5.76190 10-7 m
let's reduce to num
lam = 5.56190 10-7 m (109 nm / 1m)
lam = 556,190 nmtea
we substitute
without tea = y / L
we substitute in equation 1
d y / L = m lam
lam = dy / L m
let's reduce the magnitudes to the SI system
d = 0.44 me = 0.44 10-3 m
y = 5.5 mm = 5.5 10-3
L = 4.2m
m = 1
let's calculate
lam = 0.44 10⁻³ 5.5 10⁻³ / (4.2 1)
lam = 5.76190 10⁻⁷ m
let's reduce to num
lam = 5.56190 10⁻⁷ m (109 nm / 1m)
lam = 556,190 nm
The wire of the potentiometer has resistance 4 ohms and length 1 m. It is connected to a cell of e.m.f. 2 volts and internal resistance 1 ohm. If a cell of e.m.f. 1.2 volts is balanced by it, the balancing length will be:
A) 90 cm
B) 60 cm
C) 50 cm
D) 75 cm
Answers
Answer:
Potential gradient = 1.6 v/m
E.M.F. = potential gradient × balancing length 1.2 = 1.6 × l
l=1.2\1.6
=3\4
3\4=0.75m
convert m into cm 0.75x10
0.75m=75 cm
ans=75 cm
Explanation:
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?
Answers
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]
A system loses 510 J of potential energy. In the process, it does 430 J of work on the environment and the thermal energy increases by 100 J.
Required:
Find the change in kinetic energy.
Answers
Answer:
20J
Explanation:
The computation of the change in kinetic energy is shown below:
As we know that
Work was done on system = change in potential energy + change in kinetic energy + change in thermal energy
-430J = -510J + change in kinetic energy + 100J
-430J = -410J + change in kinetic energy
So, the change in kinetic energy is 20J
We simply applied the above formula to find out the change in kinetic energy
A sinusoidal wave travels along a string. The time for a particular point to move from maximum displacement to zero is 0.17 s. What are the (a) period and (b) frequency? (c) The wavelength is 1.5 m; what is the wave speed?
Answers
Answer:
31
Explanation:
which of the following is a physical change?
A. a newspaper burns when placed in a fire.
B.an iron chair rusts when left outside
C.a sample of water boils and releases gas.
D.a plant changes carbon dioxide and water into sugar
