Answer:
None of the above.
The correct answer would be momentum
Answer:
Momentum (None of the above)Explanation:
The two objects free-fall at the same rate of acceleration, thus giving them the same speed when they hit the ground. The heavier object however has more momentum since momentum takes into account both the speed and the mass of the object (p=m*v).
A circular coil 17.0 cm in diameter and containing nine loops lies flat on the ground. The Earth's magnetic field at this location has magnitude 5.50×10−5T and points into the Earth at an angle of 56.0∘ below a line pointing due north. A 7.80-A clockwise current passes through the coil.
Determine the torque on the coil.
Answer:
The torque in the coil is 4.9 × 10⁻⁵ N.m
Explanation:
T = NIABsinθ
Where;
T is the torque on the coil
N is the number of loops = 9
I is the current = 7.8 A
A is the area of the circular coil = ?
B is the Earth's magnetic field = 5.5 × 10⁻⁵ T
θ is the angle of inclination = 90 - 56 = 34°
Area of the circular coil is calculated as follows;
[tex]A = \frac{\pi d^2}{4} \\\\A = \frac{\pi 0.17^2}{4} =0.0227 m^2[/tex]
T = 9 × 7.8 × 0.0227 × 5.5×10⁻⁵ × sin34°
T = 4.9 × 10⁻⁵ N.m
Therefore, the torque in the coil is 4.9 × 10⁻⁵ N.m
A 70-kg circus performer is fired from a cannon that is elevated at an angle of 37° above the horizontal. The cannon uses strong elastic bands to propel the performer, much in the same way that a slingshot fires a stone. Setting up for this stunt involves stretching the bands by 3.8 m from their unstrained length. He takes 4.1 s to travel between the launch point (where he is free from the bands) and the net into which he is shot. Assume the launch and landing points are at the same height and do not neglect the change in height during stretching.
The above is not complete. Couldn't find the it online but i have found a similar question which would help you solve the above
Answer:
Explanation:
From the question we are given that
Mass = 70kg
Angle of elevation = 40°
Length of stretch = 3.00 m
Height of performer above floor = Height of the net into the into which he is shot
Time traveled = 2.14 s
Distance traveled = 26.8 m
First obtain the horizontal velocity = [tex]\frac{Distance}{time} =\frac{26.8}{2.14} = 12.523 \ m/s[/tex]
To obtain the initial we would divide the final velocity by the cosine of the angle of elevation
Initial velocity = [tex]\frac{12.523}{cos\ 40} = 16.348m/s[/tex]
Next is to obtain the initial kinetic energy
This is equal to = [tex]\frac{1}{2} mv^2= \frac{1}{2} *70*16.348^2 = 9354 \ Joules[/tex]
Looking at the diagram in the second uploaded the cannon raised the performer before releasing him is [tex]30sin(40) = 19284 \ m[/tex]
So the potential energy given by the cannon is = mgh [tex]=70 *9.80*1.9284 = 1323 \ Joules[/tex]
Hence the total energy the band gives the performer i.e the total energy stored in the band is = 9354 +1323 = 10677 Joules
To obtain the Spring constant we would use the stored energy formula
i.e Stored Energy [tex]=\frac{1}{2}kx^2[/tex]
And we have calculated the stored energy as 10677
Substituting
[tex]10677 = \frac{1}{2} *k * 3.00^2[/tex]
[tex]k = 2373 \ N/m[/tex]
The circus performer's stunt includes aspects of elastic potential energy and projectile motion. The performer is launched due to the elastic potential energy in the cannon bands and then follows the laws of projectile motion. The performer's mass does not influence the time of flight.
Explanation:This problem demonstrates the principles of projectile motion and elastic potential energy applied to a real-world situation. The 70-kg circus performer is propelled out of the cannon because of the elastic potential energy stored in the stretched bands. Once launched, the performer's motion is an example of projectile motion.
In projectile motion, the horizontal and vertical motions are independent of each other. Therefore, it's possible to find the horizontal and vertical components of the performer's velocity separately and use them to construct the overall motion. The angle of 37° is used to calculate these components.
While the performer is in flight, the vertical motion is under constant acceleration due to gravity, while the horizontal motion has a constant velocity because air resistance is negligible. This allows us to calculate various aspects of the motion, such as the time it takes for the performer to land in the net.
Lastly, the performer's mass does not influence the time of flight in a vacuum. This illustrates an important principle of physics known as the independence of motion.
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A wire with current I = 15 A flows at an angle of 30 degrees to the direction of a uniform magnetic field of 0.9 T. What is the magnitude of the force exerted on a 10 m length of wire? (in units of N)
Answer:
67.5
Explanation:
[tex]F= BILsin\theta[/tex] where F is the magnitude of the force, B is the magnetic field, I is current, [tex]\theta[/tex] is the angle of inclination, L is the length of the wire.
Substituting 0.9 T for magnetic field B, 10 m for length, 15 A for current I and 30 degrees for [tex]\theta[/tex] then
[tex]F=0.9\times 15\times 10\times sin 30^{\circ}=67.5[/tex]
suppose that throughout the united states, 350.0 x 10^6 suck braking processes occur in the course of a given day. calculate the average rate (megawatts) at which energy us being
Full Question:
Consider an automobile with a mass of 5510 lbm braking to a stop from a speed of 60.0 mph.
How much energy (Btu) is dissipated as heat by the friction of the braking process?
______Entry field with incorrect answer Btu
Suppose that throughout the United States, 350.0 x 10^6 such braking processes occur in the course of a given day. Calculate the average rate (megawatts) at which energy is being dissipated by the resulting friction.
______Entry field with incorrect answer MW
Answer:
Energy dissipated as heat = 852.10 Btu
Average rate of energy dissipation = 3.64 MW
Explanation:
1 lb = 0.453592 Kg
Mass = 5510 lb = 0.453592 * 5510 Kg = 2499.29 Kg
1 mph = 0.44704 m/s
Velocity = 60 mph = 60 * 0.44704 m/s = 26.82 m/s
Let E be equal to energy acquired
[tex]E = 1/2 MV^{2}[/tex]
[tex]E = 0.5 * 2499.29 * 26.88^{2} \\E = 899.01 * 10^{3} Joules[/tex]
Energy dissipated as heat:
[tex]E= 8.99*10^{3} = 0.0009478 * 8.99*10^{3}\\E = 852.10 tu[/tex]
E = 852.10 Btu
Energy dissipated throughout the United States
[tex]E = 350 * 10^{6} * 899.01 *10^{3} \\E=3.1466 * 10^1^4 Joules\\[/tex]
Average power rate in MW, P
[tex]P = (3.1466*10^{14} )/(24*60*60)\\P=3.64*10^{9} \\P=3.64MW[/tex]
A woman walks due west on the deck of a ship at 4 mi/h. The ship is moving north at a speed of 20 mi/h. Find the speed and direction of the woman relative to the surface of the water. (Round your answers to one decimal place.)
Answer:
The speed of woman is 20.39 mi/h at and angle of 11.3 degrees wrt the surface of the water.
Explanation:
Given that,
Speed of women due west, [tex]v_w=4\ mi/h[/tex]
Speed of women due north, [tex]v_n=20\ mi/h[/tex]
We need to find the speed and direction of the woman relative to the surface of the water. The resultant speed is given by :
[tex]v=\sqrt{v_w^2+v_n^2}[/tex]
[tex]v=\sqrt{4^2+20^2}[/tex]
[tex]v=20.39\ mi/h[/tex]
Let [tex]\theta[/tex] is the direction of speed. It is given by :
[tex]\tan\theta=\dfrac{4}{20}[/tex]
[tex]\theta=11.3^{\circ}[/tex]
So, the speed of woman is 20.39 mi/h at and angle of 11.3 degrees wrt the surface of the water.
The speed of the woman relative to the surface of the water is approximately 20.4 mi/h, and she is moving approximately 11.3° south of west.
Explanation:To find the speed and direction of the woman relative to the surface of the water, we can use vector addition. Since the woman is walking due west, her velocity vector is 4 mi/h west. The ship's velocity vector is 20 mi/h north. By adding these two vectors using vector addition, we can find the resultant velocity vector. The magnitude of the resultant velocity represents the speed of the woman relative to the surface of the water, while the direction represents the direction she is moving relative to the surface of the water.
Using the Pythagorean theorem, we can find the magnitude of the resultant velocity:
Resultant Velocity: √(4^2 + 20^2) = √(16 + 400) = √416 ≈ 20.4 mi/h
Using trigonometry, we can find the direction of the resultant velocity:
Direction: tan^(-1)(4/20) ≈ 11.3° south of west
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The tip of one prong of a tuning fork undergoes SHM of frequency L000 Hz and amplitude 0.40 mm. For this tip, what is the magnitude of the (a) maximum acceleration, (b) maxi- mum velocity, (c) acceleration at tip displacement 0.20 ffiffi, and (d) velocity at tip displacement 0.20 mm
The tip of one prong of a tuning fork undergoes SHM of frequency 1000 Hz and amplitude 0.40 mm. For this tip, what is the magnitude of the
(a) maximum acceleration,
(b) maximum velocity,
(c) acceleration at tip displacement 0.20 mm, and
(d) velocity at tip displacement 0.20 mm?
Answer:(a) 15795.5m/s²
(b) 2.5m/s
(c) 7897.7 m/s²
(d) 2.2m/s
Explanation:The displacement, y, of a body undergoing simple harmonic motion (SHM) is given by
y = A sin (ωt + φ) ------------------(i)
Where;
A = maximum displacement or amplitude of the body
ω = angular frequency of the body
t = time taken for the displacement
φ = phase constant
The velocity, v, of the body can be found by differentiating equation (i) as follows;
v = Aω cos (ωt + φ) ------------------(ii)
Where;
Aω = maximum velocity or amplitude of the velocity of the body
Also, the acceleration of the body can be found by differentiating equation (ii) as follows;
a = -Aω² sin(ωt + φ) --------------------(iii)
Where;
-Aω² = maximum acceleration or amplitude of the acceleration of the body
(a) From equation (iii), the magnitude of the maximum acceleration [tex]a_{max}[/tex] is given by;
[tex]a_{max}[/tex] = Aω² ----------------(iv)
Where;
A = amplitude = 0.40mm = 0.00040m
ω = 2 π f [Take π = 3.142. Also, f = frequency of the motion = 1000Hz]
=> ω = 2 x 3.142 x 1000 = 6284 rad/s
Substitute these values into equation (iv) as follows;
[tex]a_{max}[/tex] = 0.00040 x 6284² = 15795.5m/s²
Therefore, the magnitude of the maximum acceleration is 15795.5m/s²
========================================================
(b) From equation (ii), the magnitude of the maximum velocity [tex]v_{max}[/tex], is given by;
[tex]v_{max}[/tex] = Aω ----------------(v)
Where;
A = amplitude = 0.40mm = 0.00040m
ω = 2 π f [Take π = 3.142. Also, f = frequency of the motion = 1000Hz]
=> ω = 2 x 3.142 x 1000 = 6284 rad/s
Substitute these values into equation (v) as follows;
[tex]v_{max}[/tex] = 0.00040 x 6284 = 15795.5m/s²
Therefore, the magnitude of the maximum velocity is 2.5m/s
========================================================
(c) Comparing equations (i) and (iii), equation (iii) can be written as;
a = -ω² y -------------------(vi)
Therefore, to get the acceleration at tip displacement of 0.20mm, substitute y = 0.20mm = 0.00020m and ω = 6284rad/s into equation (vi) as follows;
a = - 6284² x 0.00020
a = - 7897.7 m/s²
Therefore, the magnitude of the acceleration at the tip displacement is 7897.7 m/s²
========================================================
(d) Recall that;
sin²θ + cos²θ = 1
=> cos²θ = 1 - sin²θ
=> cosθ = √(1 - sin²θ )
=> cos (ωt + φ) = √(1 - sin² (ωt + φ))
Substitute this value into equation (ii) as follows;
v = Aω √(1 - sin² (ωt + φ))
v = ω√(A² - A²sin² (ωt + φ))
Now, comparing the equation above and equation (i), the equation above can be written as;
v = ω√(A² - y²) -------------(vii)
Therefore, to get the velocity at tip displacement of 0.20mm, substitute y = 0.20mm = 0.00020m, ω = 6284rad/s and A = 0.00040m into equation (vii) as follows;
v = 6284√(0.00040² - 0.00020²)
v = 6284√(0.00000060)
v = 2.2m/s
Therefore, the magnitude of the velocity at the tip displacement is 2.2 m/s
In a location in outer space far from all other objects, a nucleus whose mass is 3.969554 × 10−25 kg and that is initially at rest undergoes spontaneous alpha decay. The original nucleus disappears, and two new particles appear: a He-4 nucleus of mass 6.640678 × 10−27 kg (an alpha particle consisting of two protons and two neutrons) and a new nucleus of mass 3.902996 × 10−25 kg. These new particles move far away from each other, because they repel each other electrically (both are positively charged). Because the calculations involve the small difference of (comparatively) large numbers, you need to keep seven significant figures in your calculations, and you need to use the more accurate value for the speed of light, 2.9979246e8 m/s. Choose all particles as the system. Initial state: Original nucleus, at rest. Final state: Alpha particle + new nucleus, far from each other.
Answer:
The sum of the kinetic energies of the alpha particle and the new nucleus = (1.359098 × 10⁻¹²) J
Explanation:
We will use the conservation of energy theorem for extremely small particles,
Total energy before split = total energy after split
That is,
Total energy of the original nucleus = (total energy of the new nucleus) + (total energy of the alpha particle)
Total energy of these subatomic particles is given as equal to (rest energy) + (kinetic energy)
Rest energy = mc² (Einstein)
Let Kinetic energy be k
Kinetic energy of original nucleus = k₀ = 0 J
Kinetic energy of new nucleus = kₙ
Kinetic energy of alpha particle = kₐ
Mass of original nucleus = m₀ = (3.969554 × 10⁻²⁵) kg
Mass of new nucleus = mₙ = (3.902996 × 10⁻²⁵) kg
Mass of alpha particle = mₐ = (6.640678 × 10⁻²⁷) kg
Speed of light = (2.9979246 × 10⁸) m/s
Total energy of the original nucleus = m₀c² (kinetic energy = 0, since it was originally at rest)
Total energy of new nucleus = (mₙc²) + kₙ
Total energy of the alpha particle = (mₐc²) + kₐ
(m₀c²) = (mₙc²) + kₙ + (mₐc²) + kₐ
kₙ + kₐ = (m₀c²) - [(mₙc²) + (mₐc²)
(kₙ + kₐ) = c² (m₀ - mₙ - mₐ)
(kₙ + kₐ) = (2.9979246 × 10⁸)² [(3.969554 × 10⁻²⁵) - (3.902996 × 10⁻²⁵) - (6.640678 × 10⁻²⁷)]
(kₙ + kₐ) = (8.98755191 × 10¹⁶)(1.5122 × 10⁻²⁹) = (1.35909760 × 10⁻¹²) J
Two satellites, one in geosynchronous orbit (T = 24 hrs) and one with a period of 12 hrs, are orbiting Earth. How many times larger than the radius of Earth is the distance between the orbits of the two satellites.
The distance between the orbits of two satellites is 7.97 m.
Explanation:
Johannes Kepler was the first to propose three laws for the planetary motion. According to him, the orbits in which planets are rotating are elliptical in nature and Sun is at the focus of the ellipse. Also the area of sweeping is same.
So based on these three assumptions, Kepler postulated three laws. One among them is Kepler's third law of planetary motion. According to the third law, the square of the time taken by a planet to cover a specified region is directly proportional to the cube of the major elliptical axis or the radius of the ellipse.
So, [tex]T^{2} = r^{3}[/tex]
Thus, for the geosynchornous satellite, as the time taken is 24 hours, then the radius or the major axis of this satellite is
[tex](24)^{2}= r^{3} \\(2*2*2*3)^{2} = r^{3}\\r = \sqrt[\frac{2}{3} ]{2*2*2*3} =(2)^{2} * (6)^{\frac{2}{3} } =13.21 m[/tex]
Similarly, for the another satellite orbiting in time period of 12 hours, the major axis of this satellite is
[tex](12)^{2}= r^{3} \\(2*2*3)^{2} = r^{3}\\r = \sqrt[\frac{2}{3} ]{12} =5.24 m[/tex]
So, the difference between the two radius will give the distance between the two orbits, 13.21-5.24 = 7.97 m.
So the distance between the orbits of two satellites is 7.97 m.
Two large parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm. If the surface charge density for each plate has magnitude 47.0 nC/m2, what is the magnitude of E⃗ in the region between the plates?
Answer:
[tex]5.3\times 10^3 N/C[/tex]
Explanation:
We are given that
Distance between plates=d=2.2 cm=[tex]2.2\times 10^{-2} m[/tex]
[tex]1 cm=10^{-2} m[/tex]
[tex]\sigma=47nC/m^2=47\times 10^{-9}C/m^2[/tex]
Using [tex]1 nC=10^{-9} C[/tex]
We have to find the magnitude of E in the region between the plates.
We know that the electric field for parallel plates
[tex]E=\frac{\sigma}{2\epsilon_0}[/tex]
[tex]E_1=\frac{\sigma}{2\epsilon_0}[/tex]
[tex]E_2=\frac{\sigma}{2\epsilon_0}[/tex]
[tex]E=E_1+E_2[/tex]
[tex]E=\frac{\sigma}{2\epsilon_0}+\frac{\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0}[/tex]
Where [tex]\epsilon_0=8.85\times 10^{-12}C^2/Nm^2[/tex]
Substitute the values
[tex]E=\frac{47\times 10^{-9}}{8.85\times 10^{-12}}[/tex]
[tex]E=5.3\times 10^3 N/C[/tex]
Hence, the magnitude of E in the region between the plates=[tex]5.3\times 10^3 N/C[/tex]
The magnitude of the electric field E in the region between two parallel conducting plates carrying opposite charges of equal magnitude can be calculated using the formula E = σ / ε0. Given the surface charge density σ of 47.0 nC/m2 and the permittivity of free space ε0 as 8.85 × 10-12 F/m, the electric field E results as 5.31 × 103 N/C.
Explanation:
The system mentioned in the question is known as a parallel-plate capacitor. It is important to understand that the electric field between two large parallel conducting plates carrying opposite charges of equal magnitude is perpendicular to the plates and is constant in both magnitude and direction. The electric field E between the plates of a capacitor can be calculated by the formula E = σ / ε0, where σ is the surface charge density on one plate, and ε0 is the permittivity of free space, which is approximately 8.85 × 10-12 F/m.
Given in the question, the surface charge density σ is 47.0 nC/m2. To convert nanocoulombs to coulombs, multiply by 10-9, so σ = 47.0 × 10-9 C/m2. Applying the values to the formula, we get E = (47.0 × 10-9 C/m2) / (8.85 × 10-12 F/m) = 5.31 × 103 N/C. Therefore, the electric field E in the region between the plates is 5.31 × 103 N/C.
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They are originally neutral. A positively charged rod is brought near (but not touching) the far end of A. While the charged rod is still close, A and B are separated. The charged rod is then withdrawn. Is the sphere then positively charged, negatively charged, or neutral?
Answer:
The sphere returns back to the neutral.
Explanation:
When a positively charged rod is brought near the far end of the rod A a redistribution of the charges occurs in the rod A. The free electrons in the rod A are attracted towards the positively charged rod. This causes positive charges to be shifted towards the other end of the rod A. The sphere B is in contact with the positively charged end of rod A. So it also acquires a positive charge.
When the positively charged rod is removed from the faf end of rod A, the electrons return to their original state and the charges redistributes themselves and a neutral state is re-established .
A 1 225.0 kg car traveling initially with a speed of 25.000 m/s in an easterly direction crashes into the back of a 9 700.0 kg truck moving in the same direction at 20.000 m/s. The velocity of the car right after the collision is 18.000 m/s to the east.(a) What is the velocity of the truck right after the collision? (Give your answer to five significant figures.) m/s east (b) What is the change in mechanical energy of the cartruck system in the collision? J (c) Account for this change in mechanical energy.
Answer:
The answers to the questions are;
(a) The velocity of the truck right after the collision is 20.884 m/s
(b) The change in mechanical energy of the car truck system in the collision is -9076.4384 J
(c) The change in mechanical energy is due to energy consumed by the collision process.
Explanation:
(a) From the principle of conservation of linear momentum, we have
m₁·v₁+m₂·v₂ = m₁·v₃ + m₂·v₄
Where:
m₁ = Mass of the car = 1225.0 kg
m₂ = Mass of the truck = 9700.0 kg
v₁ = Initial velocity of the car = 25.000 m/s
v₂ = Initial velocity of the truck = 20.000 m/s
v₃ = Final velocity of the car right after collision = 18.000 m/s
v₄ = Final velocity of the truck right after collision
Therefore
1225.0 kg × 25.000 m/s + 9700.0 kg × 20.000 m/s = 1225.0 kg × 18.000 m/s + 9700.0 kg × v₄
That is 30625 kg·m/s + 194000 kg·m/s = 22050 kg·m/s + 9700.0 kg × v₄
Making v₄ the subject of the formula yields
v₄ = (202575 kg·m/s)÷9700.0 kg = 20.884 m/s
The velocity of the truck right after the collision to five significant figures = 20.884 m/s
(b) The change in mechanical energy of the car truck system in the collision can be found by
The change in kinetic energy of the car truck system
Change in kinetic energy, ΔK.E. = Sum of final kinetic energy - Sum of initial kinetic energy
That is ΔK.E. = ∑ Final K.E -∑ Initial K.E.
ΔK.E. = [tex](\frac{1}{2} m_1v_3^{2}+\frac{1}{2} m_2v_4^{2}) - (\frac{1}{2} m_1v_1^{2} +\frac{1}{2} m_2v_2^{2} )[/tex]
= ([tex]\frac{1}{2}[/tex]·1225·18²+ [tex]\frac{1}{2}[/tex]·9700·20.884²) - ([tex]\frac{1}{2}[/tex]·1225·25²+[tex]\frac{1}{2}[/tex]·9700·20²)
= 2313736.0616 kg·m²/s² - 2322812.5 kg·m²/s² = -9076.4384 kg·m²/s²
1 kg·m²/s² = 1 J ∴ -9076.4384 kg·m²/s² = -9076.4384 J
(c) The energy given off by way of the 9076.4384 J is energy transformed into other forms including
1) Frictional resistance between the tires and the road for the truck and car
2) Frictional resistance in the transmission system of the truck to increase its velocity
3) Sound energy, loud sound heard during the collision
4) Energy absorbed when the car and the truck outer frames are crushed
5) Heat energy in the form of raised temperatures at the collision points of the car and the truck.
6) Energy required to change the velocity of the car over a short distance.
(a) The velocity of the truck right after the collision is 20.95258 m/s east. (b)The change in mechanical energy of the car-truck system in the collision is -2729.87 J. (c) This reduction in mechanical energy is due to energy conversion into heat, sound, and deformation.
Let's analyze the collision using the principles of conservation of momentum and mechanical energy.
The law of conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision.
Before the Collision:
Momentum of car: (1225.0 kg)(25.000 m/s) = 30625.0 kg·m/sMomentum of truck: (9700.0 kg)(20.000 m/s) = 194000.0 kg·m/sTotal momentum: 30625.0 kg·m/s + 194000.0 kg·m/s = 224625.0 kg·m/sAfter the Collision:
Momentum of car: (1225.0 kg)(18.000 m/s) = 22050.0 kg·m/sLet the final velocity of the truck be vt.Total momentum after the collision: 22050.0 kg·m/s + (9700.0 kg)(vt)
Setting initial and final momenta equal:
224625.0 kg·m/s = 22050.0 kg·m/s + 9700.0 kg * vt
Solving for vt:
vt = (224625.0 kg·m/s - 22050.0 kg·m/s) / 9700.0 kg = 20.952577 m/s
Therefore, the velocity of the truck right after the collision is 20.95258 m/s (east).
Kinetic energy is calculated using the formula KE = (1/2)mv².
Before the Collision:
KE of car: (1/2)(1225.0 kg)(25.000 m/s)² = 382812.5 JKE of truck: (1/2)(9700.0 kg)(20.000 m/s)² = 1940000.0 JTotal KE: 382812.5 J + 1940000.0 J = 2322812.5 JAfter the Collision:
KE of car: (1/2)(1225.0 kg)(18.000 m/s)² = 198450.0 JKE of truck: (1/2)(9700.0 kg)(20.95258 m/s)² = 2121632.5 JTotal KE: 198450.0 J + 2121632.5 J = 2320082.5 JChange in mechanical energy: 2320082.5 J - 2322812.5 J = -2729.87 J
The mechanical energy of the system decreases by 2729.87 J.
The decrease in mechanical energy is due to some energy being converted into other forms such as heat, sound, and deformation of the vehicles during the collision.
suppose we have two masses m1=2000 g and m2=4000g, where m1 is moving with initial velocity v1,i=24m/s and m2 is at rest at t=0s. they collide with each other at some time t=t1 after the collision they stick together and move as one. what is the final velocity, v1f, (magnitude and direction)
Answer:
The final velocity is 8 m/s and its direction is along the positive x-axis
Explanation:
Given :
Mass, m₁ = 2000 g = 2 kg
Mass, m₂ = 4000 g = 4 kg
Initial velocity of mass m₁, v₁ = 24i m/s
Initial velocity of mass m₂, v₂ = 0
According to the problem, after collision the two masses are stick together and moving with same velocity, that is, [tex]v_{1f}[/tex].
Applying conservation of momentum,
Momentum before collision = Momentum after collision
[tex]m_{1} v_{1} +m_{2} v_{2} =(m_{1}+m_{2}) v_{1f}[/tex]
Substitute the suitable values in the above equation.
[tex]2\times24 +4\times0 =(2+4}) v_{1f}[/tex]
[tex]v_{1f}=8i\ m/s[/tex]
A very long thin wire carries a uniformly distributed charge, which creates an electric field. The electric field is (2300 N/C , toward the wire) 4.10 cm from the wire Consider a 1.00-cm-long segment of the wire. How much charge (in nC) is on this segment ?
Answer:
λ= 5.24 × 10 ⁻² nC/cm
Explanation:
Given:
distance r = 4.10 cm = 0.041 m
Electric field intensity E = 2300 N/C
K = 9 x 10 ⁹ Nm²/C
To find λ = linear charge density = ?
Sol:
we know that E= 2Kλ / r
⇒ λ = -E r/2K (-ve sign show the direction toward the wire)
λ = (- 2300 N/C × 0.041 m) / 2 × 9 x 10 ⁹ Nm²/C
λ = 5.24 × 10 ⁻⁹ C/m
λ = 5.24 nC/m = 5.24 nC/100 cm
λ= 5.24 × 10 ⁻² nC/cm
For a particular pipe in a pipe-organ, it has been determined that the frequencies 576 Hz and 648 Hz are two adjacent natural frequencies. Using 343 m/s as the speed of sound in air, determine the following.(a) fundamental frequency for this pipe (b) Is the pipe is open at both ends or closed at one end?(c) length of the pipe
Answer:
(A) Fo = 72 Hz
(B) The pipe is open at both ends
(C) The length of the pipe is 2.38m
This problem involves the application of the knowledge of standing waves in pipes.
Explanation:
The full solution can be found in the attachment below.
For pipes open at both ends the frequency of the pipe is given by
F = nFo = nv/2L where n = 1, 2, 3, 4.....
For pipes closed at one end the frequency of the pipe is given by
F = nFo = nv/4L where n = 1, 3, 5, 7...
The full solution can be found in the attachment below.
One ring of radius a is uniformly charged with charge +Q and is placed so its axis is the x-axis. A second ring with charge –Q is placed concentric with the first and in the same plane. The radius of this ring is a/2. If a = 1m and Q = 3µC, what force is exerted on an electron 5m to the right of these along their common axis?
Answer:
The force exerted on an electron is [tex]7.2\times10^{-18}\ N[/tex]
Explanation:
Given that,
Charge = 3 μC
Radius a=1 m
Distance = 5 m
We need to calculate the electric field at any point on the axis of a charged ring
Using formula of electric field
[tex]E=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}[/tex]
[tex]E_{1}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}[/tex]
Put the value into the formula
[tex]E_{1}=\dfrac{9\times10^{9}\times3\times10^{-6}\times5}{(1^2+5^2)^{\frac{3}{2}}}[/tex]
[tex]E_{1}=1.0183\times10^{3}\ N/C[/tex]
Using formula of electric field again
[tex]E_{2}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}[/tex]
Put the value into the formula
[tex]E_{2}=\dfrac{9\times10^{9}\times(-3\times10^{-6})\times5}{((0.5)^2+5^2)^{\frac{3}{2}}}[/tex]
[tex]E_{2}=-1.064\times10^{3}\ N/C[/tex]
We need to calculate the resultant electric field
Using formula of electric field
[tex]E=E_{1}+E_{2}[/tex]
Put the value into the formula
[tex]E=1.0183\times10^{3}-1.064\times10^{3}[/tex]
[tex]E=-0.045\times10^{3}\ N/C[/tex]
We need to calculate the force exerted on an electron
Using formula of electric field
[tex]E = \dfrac{F}{q}[/tex]
[tex]F=E\times q[/tex]
Put the value into the formula
[tex]F=-0.045\times10^{3}\times(-1.6\times10^{-19})[/tex]
[tex]F=7.2\times10^{-18}\ N[/tex]
Hence, The force exerted on an electron is [tex]7.2\times10^{-18}\ N[/tex]
The electric potential difference between two locations can exist when there is only a source charge. Even when there is no TEST charge q, the potential difference can exist between two points in the space.a. trueb. false
Answer:
False
Explanation:
The potential difference between two points exists solely because of a source charge and depends on the source charge distribution. For a potential energy to exist, there must be a system of two or more charges. The potential energy belongs to the system and changes only if a charge is moved relatively to the rest of the system.
Suppose you are on a cart, initially at rest, which rides on a frictionless horizontal track. You throw a ball at a vertical surface that is firmly attached to the cart. If the ball bounces straight back as shown in the picture, will the cart be put into motion after the ball bounces back from the surface?
A.Yes, and it moves to the right.
B.Yes, and it moves to the left.
C.No, it remains in place
Yes, the cart will move to the left after the ball bounces back from the surface. Thus, option (B) is correct.
When you throw the ball at the vertical surface attached to the cart, the ball's momentum changes due to the collision. Initially, both you and the cart are at rest, so the total momentum of the system (you + cart + ball) is zero.
As the ball collides with the vertical surface and bounces back, it changes its direction and gains momentum in the opposite direction. This change in momentum is due to the impulse imparted to the ball during the collision.
According to the law of conservation of momentum, when the ball bounces back from the vertical surface, it exerts a backward force on the cart due to the change in momentum.
Since momentum is conserved, the cart will experience an equal and opposite forward force, causing it to move to the left.
Thus, option (B) is correct.
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The cart will move due to conservation of momentum. When the ball bounces back the cart moves to the left following Newton's Third Law of Motion.
Explanation:This question relates to the law of conservation of momentum in the realm of physics. When you throw a ball towards a vertical surface on the cart, both the ball and the cart will experience a change in momentum. As the ball bounces back (with momentum in the opposite direction the law of conservation of momentum applies the total momentum before the event (throwing the ball) must equal the total momentum after the event (ball bouncing back).
The answer to the scenario is B. Yes and it moves to the left. When the ball hits the surface and bounces back it applies a force to the cart and due to Newton's third law the cart will move in the opposite direction hence to the left assuming the ball was thrown to the right.
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Two objects, one having twice the mass of the other, are initially at rest. Two forces, one twice as big as the other, act on the objects in opposite directions as shown above.1)Which of the following statements about the acceleration of the center of mass of the system is true?a. a = F/M to the rightb. a = F/(3M) to the rightc. a = 0d. a = F/(3M) to the lefte. a = F/M to the left
Note: There is no image to take as a reference, so I'm assuming F2 directed to the right and F1 to the left, and F2=2F1
Answer:
[tex]\displaystyle a=\frac{F}{3M}[/tex]
to the right
Explanation:
Net Force
When several forces are applied to a particle or a system of particles, the net force is the sum of them all, considering each force as a vector. As for the second Newton's law, the total force equals the product of the mass by the acceleration of the system:
[tex]\vec F_n=m\cdot \vec a[/tex]
If the net force is zero, then the system of particles keeps at rest or at a constant velocity.
The system of particles described in the question consists of two objects of masses m1=M and m2, where
[tex]m_2=2m_1=2M[/tex]
Two forces F1=F and F2 act individually on each object in opposite directions and
[tex]F_2=2F_1=2F[/tex]
We don't get to see any image to know where the forces are applied to, so we'll assume F2 to the right and F1 to the left.
The net force of the system of particles is
[tex]F_n=2F-F=F[/tex]
The mass of the system is
[tex]m_t=m_1+m_2=3M[/tex]
Thus, the acceleration of the center of mass of the system is
[tex]\displaystyle a=\frac{F}{3M}[/tex]
Since F2 is greater than F1, the direction of the acceleration is to the right.
Note: If the forces were opposite than assumed, the acceleration would be to the left
A paper clip will be suspended from a string inside a Styrofoam cup surrounded by aluminum foil. Predict what will happen to the paperclip when a charged rod is brought near the cup.
Answer:
The clip will experience no force from the charged rod.
Explanation:
If a charged rod is brought near the cup, it wil cause that opposite charges travel to the outer surface of the aluminum foil, and like charges move to the inner surface, as charges can move freely in a conductor, and the aluminum foil must remain neutral.The charges on the inner surface, will attract opposite charges to the surface of the cup, and will repel like charges in the opposite direction, towards the paper clip.As the Styrofoam can be considered as a good insulator, charges can move very little distance within it, so the part of the cup facing the paper clip, will remain neutral, for normal cup thickness.As consequence, the paper clip will experience no force due to the presence of the charged rod.Final answer:
When a charged rod is brought near a suspended paper clip inside a cup with aluminum foil, polarization of the paper clip occurs, resulting in its attraction to the charged rod due to induced charges.
Explanation:
When a charged rod is brought near a paper clip suspended from a string inside a Styrofoam cup surrounded by aluminum foil, the concept of polarization is observed. As has been demonstrated with aluminum foil, the presence of the charged rod will induce a separation of charges within the materials of the paper clip and aluminum foil. This is due to electrons being free to move within the metals. If the rod is negatively charged, it will repel the electrons in the metal towards the far end, leaving the near end with a net positive charge, which is attracted towards the rod. Conversely, if the rod is positively charged, it will attract electrons towards itself, leaving the far end positively charged. The force of attraction to the near end will be greater than the force of repulsion from the far end because the near end is closer to the rod. Consequently, the net force on the paper clip will be towards the rod, resulting in the attraction of the paper clip towards the charged rod.
This concept can be applied generally to conductive objects. A very small amount of charge, on the order of a pC (picocoulomb, 1×10⁻¹² coulombs), is sufficient to significantly affect the electrical potential of such objects. Hence, the paper clip, which is a conductive object, is likely to become polarized and attracted to the rod based on these principles.
A velocity selector in a mass spectrometer uses an electric field of 4.4 x 105 V/m. What magnetic field strength (in Tesla) is needed to select a speed of 4.8 x 106 m/s
Answer:
B = 9.16 10⁻² T
Explanation:
The speed selector is a configuration where the electric and magnetic force has the opposite direction, which for a specific speed cancel
q v B = q E
v = E / B
B = E / v
Let's calculate
B = 4.4 10⁵ / 4.8 10⁶
B = 9.16 10⁻² T
To select a speed of 4.8 x 10^6 m/s in a velocity selector, a magnetic field strength of 9.17 x 10^-2 T is needed.
Explanation:A velocity selector in a mass spectrometer uses a combination of electric field and magnetic field to select charged particles based on their speed. The electric field and magnetic field exert forces on the particles in opposite directions, allowing only particles with a specific speed to pass through unaffected. To determine the magnetic field strength needed to select a speed of 4.8 x 106 m/s, we can use the equation qvB = qE, where q is the charge of the particle, v is its speed, B is the magnetic field strength, and E is the electric field strength.
Given the electric field strength of 4.4 x 105 V/m, we can rearrange the equation to solve for B:
B = E/v = (4.4 x 105 V/m) / (4.8 x 106 m/s) = 9.17 x 10-2 T
Therefore, the magnetic field strength needed to select a speed of 4.8 x 106 m/s is 9.17 x 10-2 Tesla.
The 1.0-kg collar slides freely on the fixed circular rod. Calculate the velocity v of the collar as it hits the stop at B if it is elevated from rest at A by the action of the constant 58-N force in the cord. The cord is guided by the small fixed pulleys.
Answer:
6.21 m/s
Explanation:
Using work energy equation then
[tex]U_{1-2}=T_B- T_A\\58d-mgh=0.5m(v_b^{2}-v_a^{2})[/tex]
where d is displacement from initial to final position, v is velocity and subscripts a and b are position A and B respectively, m is mass of collar, g is acceleration due to gravity
Substituting 1 Kg for m, 0.4m for h, [tex]v_a[/tex] as 0, 9.81 for g then
[tex]58(\sqrt{0.4^{2}+0.3^{2}}-0.1)-(1\times 9.81\times 0.4)=0.5\times 1\times (v_b^{2}-v_a^{2})\\19.276=0.5\times 1v_b^{2}\\v_b=6.209025688 m/s\approx 6.21 m/s[/tex]
A certain parallel-plate capacitor is filled with a dielectric for which k ???? 5.5. The area of each plate is 0.034 m2, and the plates are separated by 2.0 mm. The capacitor will fail (short out and burn up) if the electric field between the plates exceeds 200 kN/C. What is the maximum energy that can be stored in the capacitor?
Answer:
The maximum energy that can be stored in the capacitor is 6.62 x 10⁻⁵ J
Explanation:
Given;
dielectric constant k = 5.5
the area of each plate, A = 0.034 m²
separating distance, d = 2.0 mm = 2 x 10⁻³ m
magnitude of the electric field = 200 kN/C
Capacitance of the capacitor is calculated as follows;
[tex]C = \frac{k \epsilon A}{d} = \frac{5.5*8.85*10^{-12}*0.034}{2*10^{-3}} = 8.275 *10^{-10} \ F[/tex]
Maximum potential difference:
V = E x d
V = 200000 x 2 x 10⁻³ = 400 V
Maximum energy that can be stored in the capacitor:
E = ¹/₂CV²
E = ¹/₂ x 8.275 x 10⁻¹⁰ x (400)²
E = 6.62 x 10⁻⁵ J
Therefore, the maximum energy that can be stored in the capacitor is 6.62 x 10⁻⁵ J
The mass luminosity relation L M 3.5 describes the mathematical relationship between luminosity and mass for main sequence stars. It describes how a star with a mass of 2 M⊙ would have a luminosity of _____________ L⊙ while a star with luminosity of 3,160 L⊙ would have an approximate mass of ________________ M⊙.
Answer:
(a) 11.3 L
(b) 10 M
Explanation:
The mass-luminosity relationship states that:
Luminosity ∝ Mass^3.5
Luminosity = (Constant)(Mass)^3.5
So, in order to find the values of luminosity or mass of different stars, we take the luminosity or mass of sun as reference. Therefore, write the equation for a star and Sun, and divide them to get:
Luminosity of a star/L = (Mass of Star/M)^3.5 ______ eqn(1)
where,
L = Luminosity of Sun
M = mass of Sun
(a)
It is given that:
Mass of Star = 2M
Therefore, eqn (1) implies that:
Luminosity of star/L = (2M/M)^3.5
Luminosity of Star = (2)^3.5 L
Luminosity of Star = 11.3 L
(b)
It is given that:
Luminosity of star = 3160 L
Therefore, eqn (1) implies that:
3160L/L = (Mass of Star/M)^3.5
taking ln on both sides:
ln (3160) = 3.5 ln(Mass of Star/M)
8.0583/3.5 = ln(Mass of Star/M)
Mass of Star/M = e^2.302
Mass of Star = 10 M
Answer:
1. 11
2. 10
Explanation:
1. Using the formula L=M^3.5, and the fact that we were given the Mass,
L=2^3.5
L=11.3137085
L=11
Thus, the luminosity of the star would be 11 solar lumen
2. If we invert the formula, M=L^(2/7)
M=3160^(2/7)
M=9.99794159
M=10
Thus, the mass of the starwould be 10 solar masses.
A generator has a square coil consisting of 269 turns. The coil rotates at 113 rad/s in a 0.222-T magnetic field. The peak output of the generator is 79.2 V. What is the length of one side of the coil?
Answer:
length of one side of the coil is 0.1083 m
Explanation:
given data
no of turn = 269
coil rotates = 113 rad/s
magnetic field = 0.222-T
peak output = 79.2 V
solution
we will apply here formula for maximum emf induced in a coil that is
εo = N × A × B × ω ........................1
and here Area A = l²
so
l = [tex]\sqrt{A}[/tex]
put here value of A from equation 1
I = [tex]\sqrt{\frac{\epsilon _o}{N\times B\times \omega} }[/tex]
put here value and we will get
l = [tex]\sqrt{\frac{79.2}{269\times 0.222\times 113}}[/tex]
l = 0.1083 m
Answer:
Explanation:
number of turns, n = 269
angular velocity, ω = 113 rad/s
magnetic field, B = 0.222 T
Voltage, V = 79.2 V
Let the area of the coil is A and the side of the square is s.
The maximum emf generated by the coil is
V = n x B x A x ω
79.2 = 269 x 0.222 x A x 113
A = 0.01174 m²
so, s x s = 0.01174
s = 0.108 m
Thus, the side of the square coil is 0.108 m.
A wire carrying a 2-A current is placed at an angle of 60° with the respect to a magnetic field of strength 0.2 T. If the length of the wire is 0.6 m what is the magnitude of the magnetic force acting on the wire?
Answer:
0.208 N
Explanation:
Parameters given:
Current, I = 2A
Angle, A = 60°
Magnetic field strength, B = 0.2 T
Length, L = 0.6 m
Magnetic force is given as:
F = I * L * B * sinA
F = 2 * 0.6 * 0.2 * sin60
F = 0.24 * sin60.
F = 0.208 N
Answer:
0.20784 N
Explanation:
The force on conductor carrying current in a magnetic Field is given as,
F = BILsinθ..................... Equation 1
Where F = force on wire carrying current, B = magnetic Field, I = current, L = Length of the wire, θ = angle between the wire and the magnetic Field.
Given: B = 0.2 T, I = 2 A, L = 0.6 m, θ = 60°
Substitute into equation 1
F = 0.2(2)(0.6)sin60°
F = 0.24×0.8660
F = 0.20784 N
A certain elastic conducting material is stretched into a circular loop of 15.0 cm radius. It is placed with its plane perpendicular to a uniform 0.600 T magnetic field. When released, the radius of the loop starts to shrink at an instantaneous rate of 68.0 cm/s. What emf is induced in the loop at that instant?
The emf induced in the loop is 0.38 V.
Explanation:
The emf induced in any coil is directly proportional to the rate of change in flux linked with the coil as stated by Faraday's law.
Flux can be obtained by the product of magnetic lines of force with cross sectional area. As flux is the measure of magnetic lines of force crossing a certain cross sectional area.
As here the region is considered as circular loop, so the area will be area of the circle.
Then, [tex]emf=\frac{d(phi)}{dt}[/tex]
phi = Magnetic field * Area
[tex]emf = \frac{d(BA)}{dt} =B*2\pi*r\frac{dr }{dt}[/tex]
[tex]emf = 0.6*2*3.14*0.15*0.68=0.38 V[/tex]
So, the emf induced in the loop is 0.38 V.
Calculate the centripetal force on the end of a 100 m (radius) massless cable that is rotating at 0.5 rev/s. Assume the mass on the end of the cable is 4 kg.
Answer:
[tex]F_{cp}=3947.84N[/tex]
Explanation:
From the formula for centripetal force and acceleration we can deduce that:
[tex]F_{cp}=ma_{cp}=m\frac{v^2}{r}=m\frac{(r\omega)^2}{r}=mr\omega^2=mr\omega^2[/tex]
Since one revolution is [tex]2\pi\ rad[/tex], 0.5 revolutions are [tex]\pi\ rad[/tex], so we have:
[tex]F_{cp}=(4kg)(100m)(\pi\ rad/s)^2=3947.84N[/tex]
In preparing to shoot an arrow, an archer pulls a bow string back 0.424 m by exerting a force that increases uniformly from 0 to 223 N. What is the equivalent spring constant of the bow?
Answer:
525.94N/m
Explanation:
According to Hooke's law, the extension or compression of an elastic material is proportional to an applied force provided that the elastic limit of the material is not exceeded.
[tex]F=ke..................(1)[/tex]
where F is the applied force or load, k is the elastic constant or stiffness of the material and e is the extension.
In this problem, the bow string is assumed to behave like an elastic material that is stretched not beyond the elastic limit.
Given;
F = 223N;
e = 0.424m
k = ?
We make substitutions into equation (1) and then solve for k.
[tex]223=k*0.424\\k=\frac{223N}{0.424m}\\k = 525.94N/m[/tex]
The turd is launched at a speed of 72 m/s at an angle of 12 derees above the horizontal. at exactly what time after launch should the ground be covered in a portal to intercept the turd before it hits the ground?
The time the projectile (or 'turd') takes to hit the ground can be calculated using the formula (T = 2u*sin(θ)/g). This formula uses the initial velocity, the projection angle, and gravitational acceleration.
Explanation:This question relates to the physics concept of projectile motion. Given the launch speed, in this case, 72 m/s, and the angle of projection, 12 degrees, you can calculate the time of flight, i.e., the time taken by the projectile (in this humorous context, a 'turd') to hit the ground.
The time of flight (T) for a projectile can be calculated using the formula: T = 2u*sin(θ)/g, where u is the initial velocity, θ is the projection angle, and g is the gravitational acceleration (approximately 9.81 m/s² on Earth). Substituting the given values, we get T = 2*72*sin(12)/9.81, which gives the time after which the ground should be covered by a hypothetical 'portal' to intercept the turd just before it hits the ground.
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A 1050 kg sports car is moving westbound at 13.0 m/s on a level road when it collides with a 6320 kg truck driving east on the same road at 14.0 m/s . The two vehicles remain locked together after the collision.
a. What is the velocity (magnitude and direction) of the two vehicles just after the collision?
b. At what speed should the truck have been moving so that both it and the car are stopped in the collision?
c. Find the change in kinetic energy of the system of two vehicles for the situations of part (a) and part (b). For which situation is the change in kinetic energy greater in magnitude?
Answer:
(A) V = 9.89m/s
(B) U = -2.50m/s
(C) ΔK.E = –377047J
(D) ΔK.E = –257750J
Explanation:
The full solution can be found in the attachment below. The east has been chosen as the direction for positivity.
This problem involves the principle of momentum conservation. This principle states that the total momentum before collision is equal to the total momentum after collision. This problem is an inelastic kind of collision for which the momentum is conserved but the kinetic energy is not. The kinetic energy after collision is always lesser than that before collision. The balance is converted into heat by friction, and also sound energy.
See attachment below for full solution.