A track consists of a frictionless arc XY, which is a quarter-circle of radius R, and a rough horizontal section YZ. Block A of mass M is released from rest at point X, slides down the curved section of the track, and collides instantaneously and inelastically with identical block B at point Y. The two blocks move together to the right, sliding past point P, which is a distance l from point Y. The coefficient of kinetic friction between the blocks and the horizontal part of the track is ì. Express your answers in terms of M, R, ì, R, and g.
Answers
The question is not complete, so i have attached an image with the complete question.
Answer:
A) speed of block a before it hits block B;v = √2gR
B) speed of combined blocks after collision;V_ab = ½√2gR
C) kinetic energy lost = ½MgR
D) temperature change; ∆t = ½gR/c
E) Additional thermal energy;W_f = 2μMgL
Explanation:
A) From conservation of energy, potential energy = kinetic energy.
Thus; P.E = K.E
So, mgr = ½mv²
m will cancel out to give;
gr = v²
So, v = √2gR
B) from conservation od momentum, momentum before collision = momentum after collision.
Thus;
M_a•V_a = M_ab•V_ab
Where;
M_a is the mass of block A
V_a is speed of block A
M_ab is the mass after collision
V_ab is speed after collision
So, making V_ab the subject, we have;
V_ab = M_a•V_a/M_ab
Now from answer in a above,
V_a = V = √2gR
Also, M_a = M and M_ab = 2M because it's the sum of 2 masses after collision.
Thus;
V_ab = (M√2gR)/(2M)
M will cancel out to give;
V_ab = ½√2gR
C) From the work-kinetic energy theorem, the net work done on the object is equal to the change in the kinetic energy of the object.
Thus;
W_net = K_f - K_i = ½m_ab•v_ab²- ½m_a•v_a²) = ∆K.
We have seen that:
M_a = M and M_ab = 2M
V_a = √2gR and V_ab = ½√2gR
Thus;
∆K = ½(2M•(½√2gR)²) - ½(M•(√2gR)²)
∆K = ½MgR - MgE
∆K = -½MgR
Negative sign means loss of energy.
Thus kinetic energy lost = ½MgR
D) The formula for heat energy is given by;
Q = m•c•∆t
Thus, change in temperature is;
∆t = Q/Mc
Q is the heat energy, thus Q = ½MgR
Thus;∆t = ½MgR/(Mc)
M will cancel out to give;
∆t = ½gR/c
E) Additional thermal energy is gotten from;
Work done by friction;W_f = F_f x d
Where;
F_f is frictional force given by μF_n. F_n is normal force
d is distance moved
Thus;
W_f = μF_n*d
Mass after collision was 2M,thus, F_n = 2Mg
We are told to express distance in terms of L
Thus;
W_f = μ2Mg*L
W_f = 2μMgL
The problem involves principles of energy conservation and kinetic friction. As the block slides down, it converts potential energy to kinetic, collides with another block, and moves together along a rough surface. The work done against friction equals energy at the collision point yielding the expression for velocity after collision V' = sqrt(2µgl).
Explanation:The important concept here is conservation of energy. As block A slides down, it is converting potential energy into kinetic energy. At point X, the potential energy is maximum because it's at the highest point. The potential energy is given by M*g*R. By the time it reaches point Y, the potential energy has turned into kinetic energy which results in the block's speed (½MV²) at point Y.
After the collision, the two blocks move together with a new mass of 2M and new velocity V': MV = 2MV'. In the horizontal section YZ, due to kinetic friction, the blocks lose energy as work done against friction which causes them to stop.
The work done against friction is: F *d = µ*2M*g*l. Equating the energy at point Y after collision to work done against friction, we get: 2*½MV'² = µ*2M*g*l. Resolving gives the expression V' = sqrt(2µgl).
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Related Questions
As time progresses a capacitor hooked up to a battery begins to act like a. a resistor draining power from the battery b. Another battery but working against the first battery c. Another battery aiding the first battery d. An inverse resistor giving power to the system
Answers
Answer:
D
Explanation:
An inverse resistor giving power to the system
A capacitor hooked up to a battery acts as another battery aiding the first battery as time progresses.
As time progresses, a capacitor hooked up to a battery begins to act as another battery aiding the first battery. The capacitor acts as a temporary storehouse of energy and can drive its collected charge through a second circuit.
A spring (70 N/m ) has an equilibrium length of 1.00 m. The spring is compressed to a length of 0.50 m and a mass of 2.2 kg is placed at its free end on a frictionless slope which makes an angle of 41° with respect to the horizontal. The spring is then released.(A) If the mass is attached to the spring, how far up the slope from the compressed point will the mass move before coming to rest?Express your answer using two significant figures(B) Now the incline has a coefficient of kinetic friction μk. If the block, attached to the spring, is observed to stop just as it reaches the spring's equilibrium position, what is the coefficient of friction μk?
Answers
Answer:
A) = 0.63 m
B)This is approximately 0.21
Explanation:
Part A)
Now if spring is connected to the block then again we can use energy conservation
so we will have
[tex]\frac{1}{2}kx^2 = mg(x + x')sin\theta + \frac{1}{2}kx'^2[/tex]
so we will have
[tex]\frac{1}{2}(70)(0.5^2) = 2(9.81) (0.50 + x') sin41 + \frac{1}{2}(70)x'^2[/tex]
[tex]8.75 = 6.43 + 12.87 x' + 35 x'^2[/tex]
[tex]x' = 0.13 m[/tex]
so total distance moved upwards is
[tex]L = 0.5 + 0.13 = 0.63 m[/tex]
Part B
Now the incline has a coefficient of kinetic friction μk. If the block, attached to the spring, is observed to stop just as it reaches the spring's equilibrium position, what is the coefficient of friction μk
KE = 8.75 – 10.78 * sin 41
This is approximately 1.678 J. This is the kinetic energy at the equilibrium position. For the block to stop moving at this position, this must be equal to the work that is done by the friction force.
Ff = μ * 10.78 * cos 41
Work = 0.5 * μ * 10.71 * cos 41
μ * 10.78 * cos 41 = 8.75 – 10.78 * sin 41
μ = (10 – 8.82 * sin 41) ÷ 8.82 * cos 41
μ = 1.678/8.136
μ = 0.206
This is approximately 0.21
Answer:
A) d = 0.596m
B) μ = 0.206
Explanation:
A) The potential energy stored in a spring when it is compressed is given as;
U = (1/2)kx² - - - - - - (eq1)
Where
U is potential energy stored in spring
k is spring constant
x is compressed length of the spring
Let the height the mass moved before coming to rest be h.
Thus, the potential energy at this height is;
U = mgh - - - - - - (eq2)
Since the mass is connected to the spring, according to the principle of conservation of energy, initial potential energy of the spring is equal to the sum of the final potential energy in the spring and the potential energy of the mass. Thus, we have;
(1/2)K(x1)² = (1/2)K(x2)² + mgh
Where x1 is the initial compressed length and x2 is the final compressed length.
Now, h will be dsin41 while x2 will be d - x1
Where d is the distance the mass moves up before coming to rest
Thus, we now have;
(1/2)K(x1)² = (1/2)K(d - x1)² + mg(dsin41)
(1/2)K(x1)² = (1/2)K(d² - 2dx1 + (x1)²) + mg(dsin41)
(1/2)K(x1)² = (1/2)Kd² - Kdx1 + (1/2)K(x1)² + mg(dsin41)
(1/2)Kd² - Kdx1 + mg(dsin41) = 0
(1/2)Kd² + d[mg(sin41) - Kx1] = 0
From the question,
k = 70 N/m
m = 2.2 kg
x1 = 0.5m
g = 9.8 m/s²
Thus, plugging in these values, we now have;
(1/2)(70)d² + d[(2.2•9.8•0.6561) - (70•0.5)] = 0
35d² - 20.8545d = 0
35d² = 20.8545d
Divide both sides by d to get;
35d = 20.8545
d = 20.8545/35
d = 0.596m
B) Here we are looking for the coefficient of friction.
First of all, let's find the kinetic energy at the equilibrium position;
The potential energy of the spring;
P.E = (1/2)K(x1)² = (1/2)(70)(0.5)² = 8.75J
The energy that will cause the block to decelerate = mg(x1)sin41 = 2.2 x 9.8 x 0.5 x 0.6561 = 7.073
So,
Net KE = 8.75 – 7.073 = 1.677J
Now, for the block to stop moving at this equilibrium position, the work done by the frictional force must be equal to KE of 1.677J
Thus,
F_f(x1) = 1.677J
F_f = μmgcos 41
Where, μ is the coefficient of friction;
So, F_f = μ(2.2 x 9.8 x 0.7547)
Ff = 16.271μ
Thus,
16.271μ x 0.5 = 1.677J
8.136μ = 1.677
μ = 1.677/8.136
μ = 0.206
Suppose you are finding the weight of a beaker of water, and you have it on the balance and are ready to record the weight. Your partner dips the tip of his or her pen in the water, but does not touch the beaker or the balance. Does the presence of the pen affect the reading on the balance? If so, how? If not, why not?
Answers
Answer:
Yes it would
Explanation:
The submerged portion of the pen would be affected by water buoyancy force which equals to the water weight that it displaces. By Newton's 3rd law, the pen would then make a reaction force on the water itself, which in turn affect the whole reading on the balance.
Yes, it would: The submerged part of the pen would be influenced by water buoyancy force which equals the water weight that it replaces.
What is Newton's Third Law?
The submerged part of the pen would be influenced by water buoyancy force which equals the water weight that it replaces. According to Newton's 3rd law, the pen would then make a reaction force on the water itself, which in turn impacts the whole reading on the balance.
Newton's Third Law refers to There are two forces resulting from this interaction. A force on the chair and also a force on your body. These two forces are called action and reaction forces and also are the subject of Newton's third law of motion.
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Three small objects are arranged along a uniform rod of mass m and length L. one of mass m at the left end, one of mass m at the
center, and one of mass 2m at the right end. How far to the left or right of the rod's center should you place a support so that the rod
with the attached objects will balance there?
Answers
Final answer:
To balance the rod with the attached objects, you need to place a support at a certain distance from the rod's center. The support should be placed at (2/3) times the length of the rod to the right of the rod's center.
Explanation:
To balance the rod with the attached objects, you need to place a support at a certain distance from the rod's center. Let's call this distance 'x'. The torque on the rod is balanced when the clockwise torque is equal to the counterclockwise torque.
The torque on the left end of the rod is given by the force on the left object times the distance from the left end of the rod to the support, which is (m * x). The torque on the right end of the rod is given by the force on the right object times the distance from the right end of the rod to the support, which is ((2m) * (L - x)). Since the torque must be balanced, we set the two torques equal to each other:
(m * x) = ((2m) * (L - x))
Simplifying this equation gives us:
x = (2/3) * L
Therefore, you should place the support at a distance of (2/3) times the length of the rod to the right of the rod's center.
A machine part has the shape of a solid uniform sphere of mass 240 g and diameter 2.50 cm . It is spinning about a frictionless axle through its center, but at one point on its equator it is scraping against metal, resulting in a friction force of 0.0200 N at that point. Part A Find its angular acceleration. Let the direction the sphere is spinning be the positive sense of rotation.
Answers
Answer:
[tex]-16.6 rad/s^2[/tex]
Explanation:
The torque exerted on a rigid body is related to the angular acceleration by the equation
[tex]\tau = I \alpha[/tex] (1)
where
[tex]\tau[/tex] is the torque
I is the moment of inertia of the body
[tex]\alpha[/tex] is the angular acceleration
Here we have a solid sphere: the moment of inertia of a sphere rotating about is centre is
[tex]I=\frac{2}{5}MR^2[/tex]
where
M = 240 g = 0.240 kg is the mass of the sphere
[tex]R=\frac{2.50}{2}=1.25 cm = 0.0125 m[/tex] is the radius of the sphere
Substituting,
[tex]I=\frac{2}{5}(0.240)(0.0125)^2=1.5\cdot 10^{-5} kg m^2[/tex]
The torque exerted on the sphere is
[tex]\tau = Fr[/tex]
where
F = -0.0200 N is the force of friction
r = 0.0125 m is the radius of the sphere
So
[tex]\tau=(-0.0200)(0.0125)=-2.5\cdot 10^{-4} Nm[/tex]
Substituting into (1), we find the angular acceleration:
[tex]\alpha = \frac{\tau}{I}=\frac{-2.5\cdot 10^{-4}}{1.5\cdot 10^{-5}}=-16.6 rad/s^2[/tex]
Find the shear stress and the thickness of the boundary layer (a) at the center and (b) at the trailing edge of a smooth flat plate 10 ft wide and 2 ft long parallel to the flow, immersed in 60F water (=1.938 slug/ft3, =1.217×10-5ft2/s =2.359×10-5 lbf.s/ft2) flowing at an undisturbed velocity of 3 ft/s. Assume laminar boundary layer over the whole plate. Also (f) find the total friction drag on one side of the plate.
Answers
Answer:
a) The shear stress is 0.012
b) The shear stress is 0.0082
c) The total friction drag is 0.329 lbf
Explanation:
Given by the problem:
Length y plate = 2 ft
Width y plate = 10 ft
p = density = 1.938 slug/ft³
v = kinematic viscosity = 1.217x10⁻⁵ft²/s
Absolute viscosity = 2.359x10⁻⁵lbfs/ft²
a) The Reynold number is equal to:
[tex]Re=\frac{1*3}{1.217x10^{-5} } =246507, laminar[/tex]
The boundary layer thickness is equal to:
[tex]\delta=\frac{4.91*1}{Re^{0.5} } =\frac{4.91*1}{246507^{0.5} } =0.0098[/tex] ft
The shear stress is equal to:
[tex]\tau=0.332(\frac{2.359x10^{-5}*3 }{1} )(246507)^{0.5} =0.012[/tex]
b) If the railing edge is 2 ft, the Reynold number is:
[tex]Re=\frac{2*3}{1.215x10^{-5} } =493015.6,laminar[/tex]
The boundary layer is equal to:
[tex]\delta=\frac{4.91*2}{493015.6^{0.5} } =0.000019ft[/tex]
The sear stress is equal to:
[tex]\tau=0.332(\frac{2.359x10^{-5}*3 }{2} )(493015.6^{0.5} )=0.0082[/tex]
c) The drag coefficient is equal to:
[tex]C=\frac{1.328}{\sqrt{Re} } =\frac{1.328}{\sqrt{493015.6} } ==0.0019[/tex]
The friction drag is equal to:
[tex]F=Cp\frac{v^{2} }{2} wL=0.0019*1.938*(\frac{3^{2} }{2} )(10*2)=0.329lbf[/tex]
Two concentric circular coils of wire lie in a plane. The larger coil has N1 = 7200 turns and radius of a = 75.50 cm. The smaller coil has N2 = 920 turns and radius of b = 1.00 cm. Ultimately we will find the Mutual Inductance, M, and then the induced emf, ϵmf. So let's take it step by step.
First, what is the magnitude of the B-field at the smaller coil due to the larger coil?
Answers
Note: The values of the current and the radii are not given, substitute whatever the value of the current and radius are to the given solution to obtain the magnitude.
Answer:
magnitude of the B-field at the smaller coil due to the larger coil is [tex]B = \frac{7200\mu_{0}I }{2a}[/tex]
Explanation:
N₁ = 7200 turns
N₂ = 920 turns
a = 75.50 cm = 0.755 m
b = 1.00 cm = 0.01 m
From the given data, b<<a, and it is at the center of the larger coil,
so we are safe to assume that the magnetic field at the smaller coil is constant.
The formula for magnetic field due to circular loop at the center of a coil is given by:
[tex]B = \frac{N\mu_{0}I }{2R}[/tex]
Therefore, magnetic field in the smaller coil due to the larger coil of radius a will be given by:
[tex]B = \frac{7200\mu_{0}I }{2a}[/tex]
Where [tex]\mu_{0} = 4\pi * 10^{-7} Wb/A-m[/tex]
I = current in the larger coil
The magnitude of the B-field at the smaller coil due to the larger coil will be "4π × 10⁻⁷ Wb/A-m".
Magnetic fieldAccording to the question,
Number of turns, N₁ = 7200 turns
N₂ = 920 turns
Radius, a = 75.50 cm or,
= 0.755 m
b = 1.00 cm or,
= 0.01 m
We know the formula,
Magnetic field, B = [tex]\frac{N \mu_0 I}{2R}[/tex]
here, μ₀ = 4π × 10⁻⁷ Wb/A-m
By substituting the values,
= [tex]\frac{7200 \mu_0 I}{2a}[/tex]
Thus the above answer is correct.
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In a double-slit interference experiment, the slit separation is 2.41 μm, the light wavelength is 512 nm, and the separation between the slits and the screen is 4.45 m. (a) What is the angle between the center and the third side bright fringe? If we decrease the light frequency to 94.5% of its initial value, (b) does the third side bright fringe move along the screen toward or away from the pattern's center and (c) how far does it move?
Answers
Answer:
Using equation 2dsinФ=n*λ
given d=2.41*10^-6m
λ=512*10^-12m
θ=52.64 degrees
Answer:
a
The angle between the center and the third side bright fringe is
[tex]\theta = 39.60^o[/tex]
b
The third side bright fringe move away from the pattern's center
c
The distance by which it moves away is [tex]\Delta z=0.3906 m[/tex]
Explanation:
From the question the
The wavelength is [tex]\lambda = 512nm[/tex]
In the first question we a asked to obtain the angle between the center and the third side bright fringe
since we are considering the third side of the bright fringe the wavelength of light on the three sides would be evaluated as
[tex]\lambda_{3} = 3 * 512nm[/tex]
The slit separation is given as [tex]d = 2.41 \mu m[/tex]
The angle between the center and the third side bright fringe is
[tex]\theta = sin^{-1} (\frac{\lambda_3}{d} )[/tex]
[tex]\theta = sin^{-1} (\frac{3 *512*10^{-9}}{2.24*10^{-6}} )[/tex]
[tex]= sin^{-1} (0.6374)[/tex]
[tex]\theta = 39.60^o[/tex]
When the frequency of the light is reduced the wavelength is increased
i.e [tex]f = \frac{c}{\lambda}[/tex]
and this increase would cause the third side bright to move away from the pattern's center
Now from the question frequency is reduce to 94.5% this mean that the wavelength would also increase by the same as mathematically represented below
[tex]\lambda_{new} = \frac{512 *10^{-9}}{0.945}[/tex]
[tex]= 0.542 \mu m[/tex]
The angle between the center and the third side bright fringe is for new wavelength
[tex]\theta = sin^{-1} (\frac{3 *512*10^{-9}}{2.41*10^{-6}} )[/tex]
[tex]= 42.46^o[/tex]
The distance traveled away from the pattern's center is mathematically represented as
[tex]z = A tan \theta[/tex]
Where A is the separation between the slits and the screen
[tex]\Delta z = 4.45(tan 42.46 - tan39.60 )[/tex]
[tex]\Delta z=0.3906 m[/tex]
HELP ASAP PLEASE!!!
Which of the following is the least important property of a mineral?
A. color
B. luster
C. streak
D. hardness
Answers
Answer:
Color
Explanation:
The most obvious property of a mineral, its color, and is unfortunately also the least diagnostic.
Answer:
color
Explanation:
The pupil of a cat's eye narrows to a vertical slit of width 0.540 mm in daylight. Assume the average wavelength of the light is 471 nm. What is the angular resolution for horizontally separated mice
Answers
Answer:
Resolution of the light will be equal to [tex]872.22\times 10^{-6}radian[/tex]
Explanation:
We have given a pupil of cat width of the slit d = 0.540 mm = [tex]0.540\times 10^{-3}m[/tex]
Average wavelength of the light [tex]\lambda =471nm=471\times 10^{-9}m[/tex]
We have to find the angular resolution for horizontally separated mice
Angular resolution is given by [tex]\Theta =\frac{\lambda }{d}=\frac{471\times 10^{-9}}{0.540\times 10^{-3}}=872.22\times 10^{-6}radian[/tex]
So resolution of the light will be equal to [tex]872.22\times 10^{-6}radian[/tex]
An explosion in a rigid pipe shoots three balls out of its ends. A 6 gam ball comes out the right end. A 4 gram ball comes out the left end with twice the speed of the 6 gram ball. From which end does the third ball emerge?
Answers
Answer:
The third ball emerges from the right side.
Explanation:
This is a conservation of Momentum problem
In an explosion or collision, the momentum is always conserved.
Momentum before explosion = Momentum after explosion
Since the rigid pipe was initially at rest,
Momentum before explosion = 0 kgm/s
- Taking the right end as the positive direction for the velocity of the balls
- And calling the speed of the 6 g ball after explosion v
- This means the velocity of the 4 g ball has to be -2v
- Mass of the third ball = m
- Let the velocuty of the third ball be V
Momentum after collision = (6)(v) + (4)(-2v) + (m)(V)
Momentum before explosion = Momentum after explosion
0 = (6)(v) + (4)(-2v) + (m)(V)
6v - 8v + mV = 0
mV - 2v = 0
mV = 2v
V = (2/m) v
Note that since we have established that the sign on m and v at both positive, the sign of the velocity of the third ball is also positive.
Hence, the velocity of the third ball according to our convention is to the right.
Hope this Helps!!!
Answer:
the third ball emerge from right end .
Explanation:
initially the three balls are at rest so u = 0
Given m1 = 6g
v1 = v(i) (Right end indicates positive X axis so it is represented as (i))
m2 = 4g
v2 = 2(v)(-i) (left end indicates negative X axis so it is represented as (-i))
m3 = m
v3 = v'
Applying conservation of momentum
m1(u1) + m2(u2) + m3(u3) = m1(v1) + m2(v2) + m3(v3)
u1 = u2 = u3 = u = 0 ( at rest)
0 = (6×10⁻³)(v)(i) + (4×10⁻³)(2v)(-i) + m(v')
m(v') = (2×10⁻³)(v)(i)
v' = (2×10⁻³)(v)(i)/(m)
So positive (i) indicates right end
So the third ball emerge from right end .
QUESTION 2
Objectives:
I . identify potential and kinetic energy in a situation and draw corresponding energy bar charts,
II . calculate gravitational and elastic potential energy,
III . draw & analyze potential energy functions, and (d) use conservation of energy to relate the total energy at one time to total energy at another time.
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Bungee Jump: you will step off with zero initial vertical velocity from a platform a height h above the ground. The bungee cord will act like a giant extensional spring that will, you hope, provide an upward force on becoming taut. After weighing you (you have a mass M), the operator has selected a bungee cordwith an un-stretched length of d and a spring constant of k.
Consider yourself to be a single point – i.e., use the particle model.
Choosing the ground as your origin (and the z-axis directed upwards), answer the following questions about your bungee jumping adventure in terms of M, h, d, k, z, and the gravitational field strength, g. Answer with variables.
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a)
Write an expression for the stretching ?L of the cord in terms of d and z and for the total potential energy U of the jumper-bungee-Earth system for each situation (consider the latter two situations together).
b)
Which type of potential energy ( UG or US ) is largest for large z (early in the fall)? For small z (late in the fall)?
c)
Sketch a graph of your gravitational potential energy, UG(z) vs your height, z, from z = 0 to z = h, on the left plot. Then sketch a graph of your elastic potential energy, US(z) on the center plot. Finally on the rightmost plot, sketch a graph of your total potential energy1 U(z) = UG(z) + US(z). Do these plots on your own without the help of a computer or calculator.
Answers
Answer:
See explaination and attachment please.
Explanation:
Potential energy is defined as the energy possessed by a body by virtue of its position relative to others, stresses within itself, electric charge, and other factors.
Kinetic energy on the other hand is defined as the energy that it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity.
These understand will foster our knowledge to the best way of solving the question.
Please kindly check attachment for a step by step approach on the graph.
Final answer:
In a bungee jumping scenario, the stretching of the cord can be expressed in terms of length and height. Gravitational potential energy dominates early in the fall, while elastic potential energy is more significant later. Graphs can visually represent how potential energy changes with height during the jump.
Explanation:
a) Stretching of the cord: The stretching ?L of the cord can be expressed as ?L = d - z. The total potential energy U can be represented as U = UG + US, where UG is the gravitational potential energy and US is the elastic potential energy.
b) Largest potential energy: For large z (early in the fall), UG is largest. For small z (late in the fall), US is largest.
c) Graphs: Sketch a graph of UG(z) on the left plot, US(z) on the center plot, and U(z) = UG(z) + US(z) on the rightmost plot.
A baseball of mass m1 = 0.26 kg is thrown at another ball hanging from the ceiling by a length of string L = 1.45 m. The second ball m2 = 0.64 kg is initially at rest while the baseball has an initial horizontal velocity of V1 = 2.5 m/s. After the collision the first baseball falls straight down (no horizontal velocity).
Answers
The angle is approximately 50°
Explanation:
We know,
h = L – L * cos θ
= 1.55 – 1.55 (cos θ )
To determine the maximum height, we use conservation potential and kinetic energy. As 2nd ball rises to its maximum height, the increase of its potential energy is equal to the decrease of its kinetic energy. Below is the equation for determining the value of h from the 2nd ball’s initial velocity.
Potential energy, PE = mgh = m X 9.8 X h
Kinetic energy, KE = [tex]\frac{1}{2} mv^2[/tex]
Set PE equal to KE and solve for h.
h = v²/19.6
To determine the initial kinetic energy of the second ball, we need to the velocity of the 2nd ball immediately after the collision. To do this we need to use conservation of momentum. Below is the equation for determining the value of h from the 2nd ball’s initial velocity.
For the 1st ball, horizontal momentum = 0.26 X 2.5 = 0.65
Since this ball falls straight down after the collision, its final horizontal momentum is 0.
For the 2nd ball, horizontal momentum = 0.61 X vf
0.64 * vf = 0.65
vf = 0.65/.0.64
vf = 0.02m/s
h = (0.65/0.64)²/19.6
h = 0.00002m
0.00002 = 1.45 – 1.45 * cos θ
Subtract 1.55 from both sides.
0.00002 – 1.45 = -1.45 * cos θ
θ = 50°
You know very well that the classical trajectory of a charged particle (charge q, mass m) in the uniform magnetic field B (directed, say, along the z-axis) is helical: the particle moves with constant velocity along the field lines and executes a circular motion in the perpendicular (xy) plane with the cyclotron frequency
Answers
Answer:
Check the explanation
Explanation:
Stationary states might as well be illustrated in a simpler form of the Schrödinger equation, It utilizes the theory of energy conservation (Kinetic Energy + Potential Energy = Total Energy) to acquire information about an electron’s behavior that is been bound to a nucleus.
Kindly check the attached image below to get the step by step explanation to the above question.
Please help me. Calculate the average travel time for each distance, and then use the results to calculate.
A 6-column table with 3 rows in the second, third, and fifth columns and 1 row in the other columns. The first column labeled Number of Washers has entry 1 washer mass = 4.9 grams. The second column labeled Trial has entries trial number 1, trial number 2, trial number 3. The third and fourth columns labeled Time to travel 0.25 meters t subscript 1 (seconds) have entries in the third column 2.24, 2.21, 2.23 and average in the fourth column. The fifth and sixth columns labeled time to travel 0.5 meters t subscript 2 (seconds) have entries in the fifth column 3.16, 3.08, 3.15 and in the sixth column average.
The average time that it takes for the car to travel the first 0.25m is
s.
The average time to travel just between 0.25 m and 0.50 m is
s.
Given the time taken to travel the second 0.25 m section, the velocity would be
m/s.
Answers
Answer:
What is the average velocity of the car over the first 0.25m?
⇒ 0.11 m/s
What is the average velocity of the car over the second 0.25m?
⇒ 0.28 m/s
Explanation:
Just did this problem! :)
Answer:
⇒ 2.23
⇒ 0.90
⇒ 0.28
Explanation:
Just did the problem (⌐■_■)
6) The electron volt (eV) is a convenient unit of energy for expressing atomic-scale energies. It is the amount of energy that an electron gains when subjected to a potential of 1 volt; 1 eV = 1.602 × 10–19 J. Using the Bohr model, determine the energy, in electron volts, of the photon produced when an electron in a hydrogen atom moves from the orbit with n = 5 to the orbit with n = 2. Show your calculations.
Answers
Answer:
The energy of the photon is [tex]x = 2.86 eV[/tex]
Explanation:
From the question we are told that
The first orbit is [tex]n_1 = 5[/tex]
The second orbit is [tex]n_2 = 2[/tex]
According to Bohr model
The energy of difference of the electron as it moves from on orbital to another is mathematically represented as
[tex]\Delta E = k [\frac{1}{n^2 _1} + \frac{1}{n^2 _2} ][/tex]
Where k is a constant which has a value of [tex]k = -2.179 *10^{-18} J[/tex]
So
[tex]\Delta E = - 2.179 * 10^{-18} [\frac{1}{5^2 _1} + \frac{1}{2^2 _2} ][/tex]
[tex]= 4.576 *10^{-19}J[/tex]
Now we are told from the question that
[tex]1 eV = 1.602 * 10^{-19} J[/tex]
so x eV = [tex]= 4.576 *10^{-19}J[/tex]
Therefore
[tex]x = \frac{4.576*10^{-19}}{1.602 *10^{-19}}[/tex]
[tex]x = 2.86 eV[/tex]
The energy of the photon produced by an electron in a Hydrogen atom moving from the 5th orbit to the 2nd orbit is 2.85 electron Volts as calculated via the Rydberg formula.
Explanation:The energy of the photon produced by the transition of an electron in a hydrogen atom from the 5th orbit to the 2nd orbit can be calculated using Rydberg's formula.
Rydberg's formula for energy is given as E = 13.6 * (1/n1^2 - 1/n2^2), where n1 and n2 are the initial and final energy levels respectively, and E is the energy difference in eV (electron volts). Here n1 = 2 and n2 = 5.
So, substituting these values in the formula E = 13.6 * (1/2^2 - 1/5^2) = -13.6 * (-0.21) = 2.85 eV. Notice that the energy is negative which signifies a transition down to a lower energy level (which is exothermic), however, we are interested in the magnitude of the energy which is 2.85 eV.
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A wire loop with a current flowing through it is also within a uniform magnetic field. While the loop may have a net force equal to zero, in most cases, it will have an induced __________ on it.
Answers
Answer:
Current
Explanation:
A wire loop with a current flowing through it is also within a uniform magnetic field. While the loop may have a net force equal to zero, in most cases, it will have an induced current on it.
A student’s backpack, full of textbooks, is hung from a spring scale attached to the ceiling of an elevator. When the elevator is accelerating downward at 3.8m/s2, the scale reads 60N. a) What is the mass of the backpack? b) What does the scale read if the elevator moves upward while slowing down at a rate 3.8 m/s2? c) What does the scale read if the elevator moves upward at constant velocity? d) If the elevator had no brakes and the cable supporting it were to break loose so that the elevator could fall freely, what would the spring scale read? 6
Answers
a) 10 kg
b) 60 N
c) 98 N
d) 0 N
Explanation:
a)
In this problem, there are two forces acting on the backpack:
- The restoring force from the spring, upward, of magnitude [tex]F=60 N[/tex]
- The weight of the backpack, downward, of magnitude [tex]mg[/tex], where
m = mass of the backpack
[tex]g=9.8 m/s^2[/tex] acceleration due to gravity
So the net force on the backpack is (taking upward as positive direction)
[tex]F_{net}=F-mg[/tex]
According to Newton's second law of motion, the net force must be equal to the product between the mass of the backpack and its acceleration, so
[tex]F-mg=ma[/tex]
where
[tex]a=-3.8 m/s^2[/tex] is the acceleration of the backpack and the elevator, downward (so, negative)
If we solve the formula for m, we can find the mass of the backpack:
[tex]F=m(a+g)\\m=\frac{F}{a+g}=\frac{60}{-3.8+9.8}=10 kg[/tex]
b)
In this case, the elevator is moving upward, and it is slowing down at a rate of [tex]3.8 m/s^2[/tex].
Since the elevator is slowing down, it means that the direction of the acceleration is opposite to the direction of motion: and since the elevator is moving upward, this means that the direction of the acceleration is downward: so the acceleration is negative,
[tex]a=-3.8 m/s^2[/tex]
The net force acting on the backpack is still:
[tex]F_{net}=F-mg[/tex]
where
F is the restoring force in the spring, which this time is unknown (it corresponds to the reading on the scale)
Using again Newton's second law of motion,
[tex]F-mg=ma[/tex]
Therefore in this case, the reading on the scale will be:
[tex]F=m(g+a)=(10)(9.8-3.8)=60 N[/tex]
So the reading is the same as in part a).
c)
In this case, the elevator is moving at a constant velocity.
The net force on the backpack is still:
[tex]F_{net}=F-mg[/tex]
However, since here the elevator is moving at constant velocity, and acceleration is the rate of change of velocity, this means that the acceleration of the elevator is zero:
[tex]a=0[/tex]
So Newton's second law of motion can be written as:
[tex]F_{net}=0[/tex]
So
[tex]F-mg=0[/tex]
Which means that the reading on the scale is equal to the weight of the backpack:
[tex]F=mg=(10)(9.8)=98 N[/tex]
d)
In this case, the elevator had no brakes and the cable supporting it breaks loose.
This means that the elevator is now in free fall.
So its acceleration is simply the acceleration due to gravity (which is the acceleration of an object in free fall):
[tex]a=-9.8 m/s^2[/tex]
And the direction is downward, so it has a negative sign.
The net force on the backpack is still
[tex]F_{net}=F-mg[/tex]
So Newton's second law can be rewritten as
[tex]F-mg=ma[/tex]
Therefore, we can re-arrange the equation to find F, the reading on the scale, and we find:
[tex]F=m(g+a)=(10)(9.8-9.8)=0 N[/tex]
So, the reading on the scale is 0 N.
Water at 45 C and 10 kPa enters an adiabatic pump and is discharged at a pressure of 8600 kPa.The pump efficiency is 75%. Calculate the work for the pump, the temperature change of the water and the entropy change. For water at 45C: V = 1010 cm3/kg, β= 485*10(-6) K-1and Cp=4.178kJ/(kgK)
Answers
Answer:
Explanation:
find the solution below
The graph shows a heating curve for water. Between which points on the graph would condensation occur?
A) Between R and T
B) Between S and T
C) Between R and S
D) Between R and Q (not shown)
Answers
Answer:
Between R and S
Explanation:
Answer:
C) Between R and S
Explanation:
wo cars start moving from the same point. One travels south at 60 mi/h and the other travels west at 25 mi/h. At what rate is the distance between the cars increasing three hours later?
Answers
Answer:
[tex]\frac{dz}{dt} = 65 mi/h[/tex]
Explanation:
let distance between two cars is = z mi
we have to find =[tex]\frac{dz}{dt}[/tex]
One travels south at = 60 mi/h = [tex]\frac{dx}{dt}[/tex] (given)
the other travels west at =25 mi/h.= [tex]\frac{dy}{dt}\\[/tex] (given)
since both car have constant speed
at t = 3 hrs
x = 3× 60 = 180 mi/h
y = 3 × 25 = 75 mi/h
from the figure (i) we get
[tex]z = \sqrt{( x^2+ y^2)}[/tex] ...............(i)
put x and y values
we get
[tex]z = \sqrt{(180)^2 + 75^2}[/tex]
[tex]z = \sqrt{32400 + 5625} \\z = \sqrt{38025} \\z = 195 mi/h[/tex]
differentiate the equation (i) w r to t
[tex]z^2 = x^2 +y^2[/tex]
[tex]2z\frac{dz}{dt} = 2x\frac{dx}{dt}+ 2y\frac{dy}{dt}\\[/tex]
put each values
[tex]2 \times195\frac{dz}{dt} = 2 \times 180\frac{dx}{dt}+2 \times75\frac{dy}{dt}\\[/tex]
[tex]2 \times195\frac{dz}{dt} = 2 \times 180\times 60}+2 \times75\times25\\\frac{dz}{dt} = \frac{{2 \times 180\times 60+2 \times75\times25}}{ 2 \times195}\\\frac{dz}{dt} = 65 mi/h[/tex]
Answer: Both cars have equal kinetic energy
Explanation:
A solenoid with 500 turns has a radius of 0.040 m and is 40 cm long. If this solenoid carries a current of 12 A, what is the magnitude of the magnetic field near the center of the solenoid? (μ 0 = 4π × 10-7 T · m/A)
Answers
Answer:
[tex]18.8\times 10^{-3} T[/tex]
Explanation:
We are given that
Number of turns,N=500
Radius,r=0.04 m
Length of solenoid,L=40 cm=[tex]\frac{40}{100}=0.4 m[/tex]
1 m=100 cm
Current,I=12 A
We have to find the magnitude of magnetic field near the center of the solenoid.
Number of turns per unit length,n=[tex]\frac{500}{0.4}=1250[/tex]
Magnetic field near the center of the solenoid,B=[tex]\mu_0 nI[/tex]
Where [tex]\mu_=0=4\pi\times 10^{-7}Tm/A[/tex]
[tex]B=4\pi\times 10^{-7}\times 1250\times 12=18.8\times 10^{-3} T[/tex]
[tex]B=18.8\times 10^{-3} T[/tex]
In Example 10.1 (p. 234), if the pebble was instead launched at a 45 degree angle above the horizontal, how would the total mechanical energy of the system change compared to when the pebble was launched directly upwards?
Answers
Answer:
The total mechanical energy of the system would stay the same
Explanation:
The law of conservation energy states:
In a closed system, (a system that isolated from its surroundings) the total energy of the system is conserved.
For instance, the air resistance is negligible. So non conservative force is acting on the system and so the energy is conserved for the system.
The total initial energy must be equal to the total final energy , if the energy is conserved.
Therefore, when the pebbles was launched at 45 degrees to the horizontal the total mechanical energy stay the same.
Two taut strings of identical mass and length are stretched with their ends fixed, but the tension in one string is 1.10 times greater than in the other. Waves on the string with the lower tension propagate at 35.2 m/s. The fundamental frequency of that string is 258 Hz. What is the beat frequency when each string is vibrating at its fundamental frequency?
Answers
Answer:
The beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz
Explanation:
Given;
velocity of wave on the string with lower tension, v₁ = 35.2 m/s
the fundamental frequency of the string, F₁ = 258 Hz
velocity of wave on the string with greater tension;
[tex]v_1 = \sqrt{\frac{T_1}{\mu }[/tex]
where;
v₁ is the velocity of wave on the string with lower tension
T₁ is tension on the string
μ is mass per unit length
[tex]v_1 = \sqrt{\frac{T_1}{\mu} } \\\\v_1^2 = \frac{T_1}{\mu} \\\\\mu = \frac{T_1}{v_1^2} \\\\ \frac{T_1}{v_1^2} = \frac{T_2}{v_2^2}\\\\v_2^2 = \frac{T_2v_1^2}{T_1}[/tex]
Where;
T₁ lower tension
T₂ greater tension
v₁ velocity of wave in string with lower tension
v₂ velocity of wave in string with greater tension
From the given question;
T₂ = 1.1 T₁
[tex]v_2^2 = \frac{T_2v_1^2}{T_1} \\\\v_2 = \sqrt{\frac{T_2v_1^2}{T_1}} \\\\v_2 = \sqrt{\frac{1.1T_1*(35.2)^2}{T_1}}\\\\v_2 = \sqrt{1.1(35.2)^2} = 36.92 \ m/s[/tex]
Fundamental frequency of wave on the string with greater tension;
[tex]f = \frac{v}{2l} \\\\2l = \frac{v}{f} \\\\thus, \frac{v_1}{f_1} =\frac{v_2}{f_2} \\\\f_2 = \frac{f_1v_2}{v_1} \\\\f_2 =\frac{258*36.92}{35.2} \\\\f_2 = 270.6 \ Hz[/tex]
Beat frequency = F₂ - F₁
= 270.6 - 258
= 12.6 Hz
Therefore, the beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz
The fundamental frequency for the second string is found to be 272 Hz, resulting in a beat frequency of 14 Hz.
To find the beat frequency when two strings are vibrating at their fundamental frequencies, follow these steps:
Use the formula for wave speed in a string: v = √(T/μ), where T is the tension and μ is the mass per unit length.Given that the speed of waves in the string with lower tension is 35.2 m/s, and the fundamental frequency is 258 Hz, use the relationship: v = f × λ to calculate the wavelength: λ = v / f = 35.2 m/s / 258 Hz ≈ 0.136 m.The tension in the second string is 1.10 times that in the first string. So, if T1 is the tension in the first string, the tension in the second string T₂ is 1.10 × T₁The speed of waves in the second string is: v₂ = √(T₂/μ) = √(1.10 × T₁/μ). Since v₁ = √(T₁/μ) = 35.2 m/s, v₂ = 35.2 × √1.10 ≈ 37.0 m/s.Calculate the fundamental frequency of the second string: f₂ = v₂ / λ = 37.0 m/s / 0.136 m ≈ 272 Hz.The beat frequency is the absolute difference between the two fundamental frequencies: beat frequency = |f₂ - f₁| = |272 Hz - 258 Hz| = 14 Hz.Therefore, the beat frequency when both strings vibrate at their fundamental frequencies is 14 Hz.
A wheel rotating with a constant angular acceleration turns through 23 revolutions during a 3 s time interval. Its angular velocity at the end of this interval is 16 rad/s. What is the angular acceleration of the wheel
Answers
Answer
Given,
Revolution of wheel = 23 rev.
= 23 x 2π = 46 π
Time = 3 s
final angular velocity = 16 rad/s
angular acceleration of wheel = ?
Now, Calculating the initial angular speed of the wheel
Angular displacement = [tex]\dfrac{1}{2}[/tex](initial velocity + final velocity) x time.
[tex]46 π = \dfrac{\omega_o+16}{2}\times 3[/tex]
[tex]\omega_0 = 80.29\ rad/s[/tex]
now, angular acceleration
[tex]\alpha = \dfrac{\omega-\omega_0}{t}[/tex]
[tex]\alpha = \dfrac{16-80.29}{3}[/tex]
[tex]\alpha = -21.43\ rad/s^2[/tex]
Hence, the angular acceleration of wheel is negative means wheel is decelerating.
The angular acceleration of the wheel is negative (-).
Angular acceleration:The angular acceleration would be the temporal ratio during which the angular speed changes and therefore is commonly denoted by alpha (α) as well as written throughout radians/sec.
According to the question,
Revolutions, 23 rev or,
23 × 2π = 46π
Time, 3 seconds
Final angular velocity, 16 rad/s
We know the formula,
→ Angular displacement = [tex]\frac{1}{2}[/tex] (Initial velocity + Final velocity) × Time
By substituting the values,
46 = [tex]\frac{\omega_o +16}{2}[/tex] × 3
[tex]\omega_o[/tex] = 80.29 rad/s
hence,
The angular acceleration will be:
→ α = [tex]\frac{\omega - \omega_o}{T}[/tex]
= [tex]\frac{16-80.29}{3}[/tex]
= -21.43 rad/s²
Thus the above response is correct.
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A mass of 148 g stretches a spring 13 cm. The mass is set in motion from its equlibrium position with a downward velocity of 10 cm/s and no damping is applied.
(a) Determine the position u of the mass at any time t. Use 9.8 m/s as the acceleration due to gravity. Pay close attention to the units.
(b) When does the mass first return to its equilibrium position?
Answers
Answer:
[tex]u(t)=1.15 \sin (8.68t)cm[/tex]
0.3619sec
Explanation:
Given that
Mass,m=148 g
Length,L=13 cm
Velocity,u'(0)=10 cm/s
We have to find the position u of the mass at any time t
We know that
[tex]\omega_0=\sqrt{\frac{g}{L}}\\\\=\sqrt{\frac{980}{13}}\\\\=8.68 rad/s[/tex]
Where [tex]g=980 cm/s^2[/tex]
[tex]u(t)=Acos8.68 t+Bsin 8.68t[/tex]
u(0)=0
Substitute the value
[tex]A=0\\u'(t)=-8.68Asin8.68t+8.68 Bcos8.86 t[/tex]
Substitute u'(0)=10
[tex]8.68B=10[/tex]
[tex]B=\frac{10}{8.68}=1.15[/tex]
Substitute the values
[tex]u(t)=1.15 \sin (8.68t)cm[/tex]
Period =T = 2π/8.68
After half period
π/8.68 it returns to equilibruim
π/8.68 = 0.3619sec
Objects 1 and 2 attract each other with a electrostatic force of 18.0 units. If the charge of Object 1 is one-third the original value AND the charge of object 2 is doubled AND the distance then the new electrostatic force will be _____ unit
Answers
Answer:
F'=(8/3)F
Explanation:
to find the change in the force you take into account that the electric force is given by:
[tex]F=k\frac{q_1q_2}{(18.0u)^2}=k\frac{q_1q_2}{324.0}u^2[/tex]
However, if q1'=1/3*q, q2'=2*q2 and the distance is halved, that is 18/=9.o unit:
[tex]F'=k\frac{q_1'q_2'}{(9u)^2}=k\frac{(1/3)q_1(2)q_2}{81.0u^2}=\frac{2}{3}k\frac{q_1q_2}{81.0u^2}[/tex]
if you multiply this result by 4 and divide by 4 you get:
[tex]F'=\frac{8}{3}k\frac{q_1q_2}{324.0u^2}=\frac{8}{3}F[/tex]
hence, the new force is 8/3 of the previous force F.
A vertical bar magnet is dropped through the center of a horizontal loop of wire, with its north pole leading. At the instant when the midpoint of the magnet is in the plane of the loop, the induced current in the loop, viewed from above, is:
Answers
Answer:
When the midpoint of the magnet is in the plane of the loop, the induced current in the loop viewed from above is essentially zero.
Explanation:
At some point far away from the plane of the loop of the wire, there's no flux and hence, no induced emf, and no current in the loop. As the magnet descends and gets close to the entrance of the coil, some of the magnetic field from the magnet threads the top few imaginary disks of the coil. The time changing flux induces some EMF and subsequently some induced current.
As the magnet continues to fall, and enters the coil, more of its magnetic field is threading imaginary disks in the coil, so as it moves, the time rate of change of total flux increases, so the EMF goes up. Note that the field lines above and below the magnet's midpoint point in the same direction from the the North pole to the south.
As the bar moves through the plane of the coil, the North end first, flux is added by the motion of the magnet and flux is removed by the motion of South end.
At some point, the bar reaches the middle of the coil. At this point, the amount of flux added to the top half of the coil by a small motion of the magnet is equal to the amount of flux removed from the bottom half. Therefore, at this point the EMF is zero. And hence, there is no induced current observed in the wire at this point.
Hope this Helps!!!
When viewed from the above, there is no induced current observed in the wire at this point due to zero EMF.
The given problem is based on the concept of electric flux and induced emf. The electric flux is dependent on induced emf, which is produced to current. Since, point far away from the plane of the loop of the wire, there's no flux and hence, no induced emf, and no current in the loop.
As the magnet descends and gets close to the entrance of the coil, some of the magnetic field from the magnet threads the top few imaginary disks of the coil. The time changing flux induces some EMF and subsequently some induced current.When magnet continues to fall, and enters the coil, more of its magnetic field is threading imaginary disks in the coil, so as it moves, the time rate of change of total flux increases, so the EMF goes up.
Note: - The field lines above and below the magnet's midpoint point in the same direction from the the North pole to the south.
As the bar moves through the plane of the coil, the North end first, flux is added by the motion of the magnet and flux is removed by the motion of South end.
When the bar reaches the middle of the coil, then at this point the amount of flux added to the top half of the coil by a small motion of the magnet is equal to the amount of flux removed from the bottom half. Therefore, at this point the EMF is zero. And hence, there is no induced current observed in the wire at this point.
Thus, we conclude that when viewed from the above, there is no induced current observed in the wire at this point due to zero EMF.
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Why do plant cells need chloroplasts?
Answers
Answer:
Chloroplasts convert light energy into sugars that can be used by the cells. This conversion creates 'food' for the plant.
Explanation:
A fairground ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follow has a 6.75 m radius, at how many revolutions per minute are the riders subjected to a centripetal acceleration equal to that of gravity
Answers
Answer: The riders are subjected to 11.5 revolutions per minute
Explanation: Please see the attachments below