Final answer:
The velocity of the defensive tackle before the collision is -6 m/s (toward the left).
Explanation:
According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. Since the two players stick together after the collision, their velocities will be equal but opposite in direction.
Let's assume that the velocity of the defensive tackle before the collision is v. The total momentum before the collision is given by:
m1 * v1 + m2 * v2 = (m1 + m2) * v
Where:
m1 = mass of the running back = 86 kg
v1 = velocity of the running back = 9 m/s (toward the right)
m2 = mass of the defensive tackle = 129 kg
v2 = velocity of the defensive tackle (to be determined)
Using the above equation, we can solve for v:
(86 kg * 9 m/s) + (129 kg * v2) = (86 kg + 129 kg) * 0
774 kg*m/s + 129 kg * v2 = 0 kg*m/s
v2 = -6 m/s
Therefore, the velocity of the defensive tackle before the collision is -6 m/s (toward the left).
A knife thrower throws a knife toward a 300 g target that is sliding in her direction at a speed of 2.45 m/s on a horizontal frictionless surface. She throws a 22.5 g knife at the target with a speed of 36.0 m/s. The target is stopped by the impact and the knife passes through the target. Determine the speed of the knife (in m/s) after passing through the target.
Answer:
Explanation:
We shall apply conservation of momentum here because it is a case of inelastic collision
u₁ , u₂ initial velocity of knife and target . v₁ , v₂ be their final velocity.
m₁ u₁ + m₂u₂ = m₁v₁ + m₂v₂
u₂ will be negative as it is coming from opposite direction.
36 x 22.5 - 300 x 2.45 = 22.5 x v₁ + 0
810 - 735 = 22.5 x v₁
v₁ = 3.33 m /s
The surface of the dock is 6 feet above the water. If you pull the rope in at a rate of 2 ft/sec, how quickly is the boat approaching the dock at the moment when there is 10 feet of rope still left to pull in
Answer:
The boat is approaching the dock at a rate of 2.5 ft/s.
Explanation:
Let the rope length be 'l' at any time 't', the distance of boat from dock be 'b' at any time 't'.
Given:
The height of dock above water (h) = 6 feet
Rate of pull of rope or rate of change of rope is, [tex]\frac{dl}{dt}=2\ ft/s[/tex]
As clear from the question, the height is fixed and only the length 'l' and distance 'b' varies with time 't'.
Now, the above situation represents a right angled triangle as shown below.
Using Pythagoras Theorem, we have:
[tex]l^2=h^2+b^2\\\\l^2=6^2+b^2\\\\l^2=36+b^2----------(1)[/tex]
Now, differentiating the above equation with time 't', we get:
[tex]2l\frac{dl}{dt}=0+2b\frac{db}{dt}\\\\l\frac{dl}{dt}=b\frac{db}{dt}\\\\\frac{db}{dt}=\frac{l}{b}\frac{dl}{dt}------(2)[/tex]
Now, the distance 'b' can be calculated using 'l=10 ft' in equation (1). This gives,
[tex]b^2=10^2-36\\\\b=\sqrt{64}=8\ ft[/tex]
Now, substituting all the given values in equation (2) and solve for [tex]\frac{db}{dt}[/tex]. This gives,
[tex]\frac{db}{dt}=\frac{10}{8}\times 2\\\\\frac{db}{dt}=2.5\ ft/s[/tex]
Therefore, the boat is approaching the dock at a rate of 2.5 ft/s.
In the picture below, a 7.00 kg piece of aluminum hanging from a spring scale is immersed in water. The total mass of the water is 3.00 kg, and it is contained in a beaker with a mass of 2.00 kg. The beaker sits on top of another scale. Find the readings on both scales. Take the density of aluminum to be 2700 kg/m3 and the density of water to be 1000 kg/m3.
Answer:
Upper scale reads 4.4kg while lower scale reads 7.6 kg
Explanation:
The buoyant force of the water on the aluminum piece would equal to the weight of the water displaced by the aluminum.
The mass of the water displaced by aluminum is
[tex]m_w = V_w \rho_w = \frac{m_a}{\rho_a}\rho_w = \frac{7}{2700}1000 = 2.6 kg[/tex]
As 7kg aluminum piece is supported by 2.6 kg buoyant force, the scale that the aluminum is hung on would read
Mass of aluminum - mass of water displaced
7 - 2.6 = 4.4 kg
This buoyant force would also created a reaction on the lower scale of 2.6 kg (according to Newton's 3rd law). So the lower scale would read:
Mass of water + mass of bleak + mass of buoyant reaction force
3 + 2 + 2.6 = 7.6 kg
An electron and a proton, moving side by side at the same speed, enter a 0.020-T magnetic field. The electron moves in a circular path of radius9.0 mm. What is the radius of the proton?
Answer:
16.5 m
Explanation:
Given,
Magnetic field = 0.02 T
radius of electron = 9 mm
speed of electron = speed of proton
radius of proton = ?
We know,
[tex]F_e = q_e vB[/tex].........(1)
[tex]F_p = q_p vB[/tex]
using newton second law
[tex]F = m a = m\dfrac{v^2}{r}[/tex]
equating Force due to electron and proton
[tex]F_e = F_p[/tex]
[tex]\dfrac{m_ev^2}{r_e}=\dfrac{m_pv^2}{r_p}[/tex]
[tex]\dfrac{m_e}{r_e} = \dfrac{m_p}{r_p}[/tex]
m_ e = 9.1 x 10⁻³¹ Kg and m_p = 1.67 x 10⁻²⁷ Kg
[tex]r_p = \dfrac{m_p}{m_e}\times r_e[/tex]
[tex]r_p = \dfrac{1.67\times 10^{-27}}{9.1 \times 10^{-31}}\times 9 \times 10^{-3}[/tex]
[tex]r_p = 16.5 m[/tex]
Hence, the radius of proton is equal to 16.5 m.
in a mass spectrometer ionized molecules are accelerated from rest through a potential difference V into a uniform magnetic field. only those molecules with the correct mass will have the travel radius that allows them to go through the hole and into the detector. what potential differeence is needed so that singly ionized carbon dioxide molecules
Answer:
Potential difference = 245 V
Explanation:
First, we have to find velocity at which ionized molecules are accelerated into a magnetic field. Here, magnetic force acts as a centripetal force due to circular path of molecules towards detector slot.
B= 0.25 T,d= 12 cmmass = (12+16+16)×1.67×10-27kgNow, the magnetic force to the centripetal force equates as:
qvB = [tex]\frac{mv2}{r}[/tex] = [tex]\frac{mv2}{d/2}[/tex]
For Velocity (v) :
v = [tex]\frac{qBd}{2m}[/tex]
According to law of conservation of energy, we know that the initial potential energy of the molecules is converted into kinetic energy as they enter into the magnetic field. So,
[tex]qV=\frac{1}{2}mv2[/tex]
Substitute the value of v in above equation:
qV = [tex]\frac{1}{2}[/tex] m [tex]\frac{q2B2d2}{4m2}[/tex]
V = [tex]\frac{qB2d2}{8m}[/tex]
V= 245 V
A vice pushes on a system of three boards, each oriented vertically and held up by horizontal forces. The outer boards weigh 90 N and the inner board weighs 118 N. The coefficient of friction between the inner and the outer boards is 0.67. What is the magnitude of one of the compression forces acting on either side of the inner board?
Answer:
88.1 N
Explanation:
As shown in the free body diagram attached to the question, the only forces acting on the inner block include in the required vertical direction.
- Its Weight.
- Frictional forces as a result of the Two compression forces.
If the block is not to slip off, the weights must match the two frictional forces
Let the compressive forces be F and frictional force be Fr
Fr = μ F
Sum of force acting on the inner block in the y-direction
Fr + Fr - W = 0
μ F + μ F = mg
2 μF = 118
2(0.67) F = 118
F = 118/1.34
F = 88.1 N
Each of the compression forces is 88.1 N
Hope this Helps!!!
The magnitude of one of the compression forces acting on either side of the inner board in a vice setup is 88.1 N
Given that the coefficient of friction () between the inner and the outer boards is 0.67 and the normal force (weight of the inner board) is 118 N, we can use the relationship for static friction to find the frictional force. Since the inner board is at equilibrium and not moving, the magnitude of one of the compression forces that acts on either side of the inner board must be equal to the static frictional force.
Thus,
Let the compressive forces be F and the frictional force be Fr
Fr = μ F
The sum of force acting on the inner block in the y-direction
Fr + Fr - W = 0
μ F + μ F = mg
2 μF = 118
2(0.67) F = 118
F = 118/1.34
F = 88.1 N
What is the electric flux through one side of a cube that has a single point charge of -3.80 µC placed at its center? Hint: You do not need to integrate any equations to get the answer.
? N·m^2/C
Answer:
φ = -7.16 × 10⁴ Nm²/C
Explanation:
q = -3.80 × 10⁻⁶ C
ε₀ = 8.85 10⁻¹² C²/Nm²
By Gauss's law
φ = q/ε₀
as the cube has 6 faces so for one side the flux will
φ =1/6 ( q/ε₀)
φ = 1/6 ( -3.80 × 10⁻⁶ C / 8.85 10⁻¹² C²/Nm²)
φ = -71,563.088 Nm²/C
φ = -7.16 × 10⁴ Nm²/C
-7.00 x 10⁴Nm²/C
Explanation:Let the single point charge of -3.80µC be Q
i.e
Q = -3.80µC = -3.80 x 10⁻⁶C
From Gauss's law, the total electric flux,φ, through an enclosed surface is equal to the quotient of the enclosed charge Q, and the permittivity of free space, ε₀. i.e;
φ = Q / ε₀ ----------------------(i)
Where;
ε₀ = known constant = 8.85 x 10⁻¹²F/m
Now, If the total flux through the enclosed cube is given by equation (i), then the flux, φ₁, through one of the six sides of the cube is found by dividing the equation by 6 and is given as follows;
φ₁ = Q / 6ε₀ --------------------(ii)
Substitute the values of Q and ε₀ into equation (ii) as follows;
φ₁ = -3.80 x 10⁻⁶ / (6 x 8.85 x 10⁻¹²)
φ₁ = -0.07 x 10⁶
φ₁ = -7.00 x 10⁴
Therefore, the electric flux through one side of the cube is -7.00 x 10⁴ Nm²/C
Note:
The result (flux) above is negative to show that the electric field lines from the point charge points radially inwards in all directions.
A parallel-plate capacitor in air has a plate separation of 1.51 cm and a plate area of 25.0 cm2. The plates are charged to a potential difference of 260 V and disconnected from the source. The capacitor is then immersed in distilled water. Assume the liquid is an insulator.
(a) What is the charge on the plates before immersion?
(b) What is the charge on the plates after immersion?
Answer:
(a) 380.96 pC
(b) 3.25V
Explanation:
(a) Before immersion,
[tex]C_{air}[/tex] = [tex]\frac{E_{0}A }{d}[/tex]
⇒(8.85E-12× 25E-4× 260) ÷(0.0151)
= 380.96 pC
(b) Charge on the plates after immersion can be calculated by,
Q = ΔV×C
= Δ[tex]V_{air}[/tex] ÷ K
where K is the constant for distilled water
= 260 ÷ 80
= 3.25V
Answer:
Explanation:
(a) Charge on the plates before immersion
as we know that,
C = εA/d
C = 8.85×[tex]10^{-12}[/tex]× 25×[tex]10^{-4}[/tex]/ 1.51×[tex]10^{-2}[/tex]
= 1.46×[tex]10^{-12}[/tex]
Q = CV
= (1.46×[tex]10^{-12}[/tex]) (260)
= 3.796×[tex]10^{-10}[/tex] C
(b) Charge on the plates after immersion
Q = 3.796×[tex]10^{-10}[/tex] C
The charge will remain the same, as the capacitor was disconnected before it was immersed.
A 37 cm long solenoid, 1.8 cm in diameter, is to produce a 0.50 T magnetic field at its center. If the maximum current is 4.4 A, how many turns must the solenoid have? Express your answer using two significant figures.
Answer:
33,458.71 turns
Explanation:
Given: L = 37 cm = 0.37 m, B= 0.50 T, I = 4.4 A, n= number of turn per meter
μ₀ = Permeability of free space = 4 π × 10 ⁻⁷
Solution:
We have B = μ₀ × n × I
⇒ n = B/ (μ₀ × I)
n = 0.50 T / ( 4 π × 10 ⁻⁷ × 4.4 A)
n = 90,428.94 turn/m
No. of turn through 0.37 m long solenoid = 90,428.94 turn/m × 0.37
= 33,458.71 turns
Given Information:
Diameter of solenoid = d = 1.8 cm = 0.018 m
Length of solenoid = L = 37 cm = 0.37 m
Current = I = 4.4 A
Magnetic field = B = 0.50 T
Required Information:
Number of turns = N = ?
Answer:
Number of turns ≈ 33,498 or 33,458
Step-by-step explanation:
The magnetic field at the center of the solenoid is given by
B = μ₀NI/√ (L²+4r²)
N = B√ (L²+4r²)/μ₀I
Where L is the length and r is the radius of the solenoid, N is the number of turns and B is the magnetic field.
r = d/2 = 0.018/2 = 0.009 m
N = 0.50√ (0.37)²+(4*0.009²)/4πx10⁻⁷*4.4
N ≈ 33,498 Turns
Please note that we can also use a more simplified approximate model for this problem since the length of the solenoid is much greater than the radius of the solenoid
L = 0.37 >> r = 0.009
The approximate model is given by
B = μ₀NI/L
N = BL/μ₀I
N = 0.50*0.37/4πx10⁻⁷*4.4
N ≈ 33,458 Turns
As you can notice the results with the approximate model are very close to the exact model.
Consider the expression below. Assume m is an integer. 6m(2m + 18) Enter an expression in the box that uses the variable m and makes the equation true. (Simplify your answer completely. If no expression exists, enter DNE.)
The expression 6m(2m + 18) simplifies to 12m² + 108m. Therefore, the equivalent expression using the variable 'm' is 12m² + 108m.
Explanation:The given expression is 6m(2m + 18). First, we distribute 6m across the terms in the parentheses, resulting in 12m² + 108m. So, an expression that uses the variable 'm' and makes the equation true when the given expression is simplified would be 12m² + 108m. This answer is fully simplified, and the variable 'm' is assumed to be an integer.
The expression 6m(2m + 18) simplifies to 12m² + 108m. Therefore, the equivalent expression using the variable 'm' is 12m² + 108m.
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Europe and North America are moving apart by about 5 m per century. As the continents separate, new ocean floor is created along the mid-Atlantic Rift. If the rift is 5000 km long, what is the total area of new ocean floor created in the Atlantic each century?
Answer:
Area = 25km² in a century.
Explanation:
Given
Spread Rate = 5m/century
Length of rift = 5000km
Convert to metres
Length of rift = 5000 * 1000m
Length of rift = 5,000,000m
In one century, the additional area to Atlantis = (Rift Length) * (Spread rate) * 1 century
Atea = 5,000,000 m * 5m/century * 1 century
Area = 25,000,000m² in a century
----- Convert to km²
Area = 25,000,000km² * 1m²/1,000,000km²
Area = 25km² in a century.
Hence, the total area of new ocean floor created in the Atlantic each century is 25km²
Final answer:
The total area of new ocean floor created along the mid-Atlantic Rift each century is 5,000,000 m², or 25 km² of new oceanic crust formed every century.
Explanation:
The question asks how much new ocean floor is created in the Atlantic Ocean each century along the mid-Atlantic Rift, which is known to be 5000 km long, as Europe and North America drift apart by about 5 m per century.
To calculate the total area of the new ocean floor created, we consider the length of the rift (5000 km) and the rate of separation (5 m per century).
First, we need to convert kilometers to meters for uniform units:
5000 km * 1000 m/km = 5,000,000 m.
Now we can calculate the area:
Area = Length * Width
Area = 5,000,000 m * 5 m
Area = 25,000,000
So, every century, an area of 25,000,000 m2 of new ocean floor is created along the mid-Atlantic Rift. This is equivalent to 25 km2 of new oceanic crust, since 1 km2 = 1,000,000 m2.
High-altitude mountain climbers do not eat snow, but always melt it first with a stove. To see why, calculate the energy absorbed from a climber's body under the following conditions. The specific heat of ice is 2100 J/kg⋅C∘, the latent heat of fusion is 333 kJ/kg, the specific heat of water is 4186 J/kg⋅C∘
Answer:
The conditions are not given in the question. Here is the complete question.
High-altitude mountain climbers do not eat snow, but always melt it first with a stove. To see why, calculate the energy absorbed from a climber's body under the following conditions. The specific heat of ice is 2100 J/kg⋅C∘, the latent heat of fusion is 333 kJ/kg, the specific heat of water is 4186 J/kg⋅C∘
a) eat 1.0kg of -15 degrees Celsius snow which your body warms to body temperature of 37 degrees Celsius ?
b) melt 1.0kg of -15 degrees Celsius snow using a stove and drink the resulting 1.0kg of water at 2 degrees Celsius, which your body has to warm to 37 degrees Celsius ?
Explanation:
Let's calculate the heat required to convert ice from -15°C to 0°C and then the heat required to convert it into the water.
a)
Heat required to convert -15°C ice to 0°C.
[tex]ms_{ice}[/tex]ΔT = (1.0)(2.100×10³)(15) = 3.150×[tex]10^{4}[/tex]J
Heat required to convert 1.0 kg ice to water.
[tex]mL_{ice} = 1[/tex]×[tex]3.33[/tex]×[tex]10^{5}[/tex] = 3.33×[tex]10^{5}[/tex]J
Heat required to convert 1.0 kg water at 0°C to 37°C.
[tex]ms_{water}[/tex]ΔT = 1×4.186×[tex]10^{3}[/tex]×37 = 1.548×[tex]10^{5}[/tex] J
Total heat required = 3.150×[tex]10^{4}[/tex] + 3.33×[tex]10^{5}[/tex] + 1.548×[tex]10^{5}[/tex]
= 5.19×[tex]10^{5}[/tex] J
b)
Heat required to warm 1.0 kg water at 2°C to water at 37°C.
[tex]ms_{water}[/tex]ΔT = 1×4.186×[tex]10^{3}[/tex]×35
= 1.465×[tex]10^{5}[/tex]J
At the start of meiosis II, the cell exhibits characteristics of a haploid cell preparing for mitosis. It has one set of homologous chromosomes, each with two chromatids, equivalent to a haploid cell in the G₂ phase of interphase.
Explanation:At the beginning of meiosis II, a cell appears much like a haploid cell preparing to undergo mitosis. After completion of meiosis I, the cell does not duplicate its chromosomes, so at the onset of meiosis II, each dividing cell has only one set of homologous chromosomes, each with two chromatids. This results in half the number of sister chromatids to separate out as a diploid cell undergoing mitosis. In terms of chromosomal content, cells at the start of meiosis II are similar to those of haploid cells in the G₂ phase of interphase, where they are preparing for mitosis.
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A rod of length L has a charge Q uniformly distributed along its length. The rod lies along the y axis with one end at the origin. (a) Find an expression for the electric potential as a function of position along the x axis.
Answer:
V = k Q/l ln [(l +√(l² + x²)) / x]
Explanation:
The electrical potential for a continuous distribution of charges is
V = k ∫ dq / r
Let's apply this expression to our case, define a linear charge density for the bar
λ = dq / dy
dq = λ dy
The distance from a point on the bar to the x-axis is
r = √ (x² + y²)
Let's replace
V = K ∫ λ dy /√ (x² + y²)
We integrate
V = k λ ln (y + √ (x² + y²))
Let's evaluate between y = 0 and y = l
V = k λ [ ln (l +√(x² + l²) - ln x]
We substitute the linear density
V = k Q/l ln [(l +√(l² + x²)) / x]
Final answer:
To calculate the electric potential at a point on the x-axis due to a uniformly charged rod, we use the concept of charge density and integrate the contributions from each infinitesimal charge element along the rod with respect to the position.
Explanation:
The student is asking about the electric potential due to a uniformly charged rod on the x-axis. To find an expression for the electric potential at a point on the x-axis, we can imagine dividing the rod into infinitesimally small segments of charge dq. Given that the charge Q is evenly distributed along the rod of length L, the linear charge density λ is Q/L. The potential dV due to a small element dq at a distance x from the rod is given by Coulomb's law as dV = k * dq / r, where k is Coulomb's constant and r is the distance from the charge element to the point on the x-axis.
To find the total potential V, we integrate this expression from one end of the rod to the other. The distance r varies along the rod, so we integrate with respect to y, the position along the rod. Setting up the integral, we have V = k * ∫_{-L/2}^{L/2} (dq / √((L/2 - y)^2 + x^2)) where the limits of integration account for the rod's position along the y-axis with one end at the origin and λ = Q/L. After substituting dq with λdy, performing the integration, and simplifying, we obtain the electric potential V(x) as a function of position along the x-axis.
A cube of mass m = 0.49 kg is set against a spring with a spring constant of k1 = 606 N/m which has been compressed by a distance of 0.1 m. Some distance in front of it, along a frictionless surface, is another spring with a spring constant of k2 = 233 N/m. How far d2, in meters, will the second spring compress when the block runs into it?
Answer:0.161 m
Explanation:
Given
mass of cube [tex]m=0.49\ kg[/tex]
Spring constant [tex]k_1=606\ N/m[/tex]
compression in the spring [tex]x_1=0.1\ m[/tex]
When this cube is released then it will compress another spring of spring constant [tex]k_2=233\ N/m[/tex]
Conserving energy
[tex]\frac{1}{2}k_1x^2=\frac{1}{2}mv^2=\frac{1}{2}k_2x'^2[/tex]
[tex]\frac{x'}{x}=\sqrt{\frac{k_1}{k_2}}[/tex]
[tex]x'=0.1\times \sqrt{\frac{606}{233}}[/tex]
[tex]x'=0.1\times 1.61[/tex]
[tex]x'=0.161\ m[/tex]
When its 75 kW (100 hp) engine is generating full power, a small single-engine airplane with mass 700 kg gains altitude at a rate of 2.5 m>s 1150 m>min, or 500 ft>min2. What fraction of the engine power is being used to make the airplane climb
Answer:
0.2289
Explanation:
Power required to climb= Fv where F is force and v is soeed. We know that F= mg hence Power, P= mgv and substituting 700 kg for m, 9.81 for g and 2.5 m/s for v then
P= 700*9.81*2.5=17167.5 W= 17.1675 kW
To express it as a fraction of 75 kw then 17.1675/75=0.2289 or 22.89%
A tugboat tows a ship with a constant force of magnitude F1. The increase in the ship's speed during a 10 s interval is 3.0 km/h. When a second tugboat applies an additional constant force of magnitude F2 in the same direction, the speed increases by 11 km/h during a 10 s interval. How do the magnitudes of F1 and F2 compare? (Neglect the effects of water resistance and air resistance.)
Answer:
The magnitude of F₁ is 3.7 times of F₂
Explanation:
Given that,
Time = 10 sec
Speed = 3.0 km/h
Speed of second tugboat = 11 km/h
We need to calculate the speed
[tex]v_{1}=\dfrac{3.0\times10^{3}}{3600}[/tex]
[tex]v_{1}=0.833\ m/s[/tex]
The force F₁is constant acceleration is also a constant.
[tex]F_{1}=ma_{1}[/tex]
We need to calculate the acceleration
Using formula of acceleration
[tex]a_{1}=\dfrac{v}{t}[/tex]
[tex]a_{1}=\dfrac{0.833}{10}[/tex]
[tex]a_{1}=0.083\ m/s^2[/tex]
Similarly,
[tex]F_{2}=ma_{2}[/tex]
For total force,
[tex]F_{3}=F_{2}+F_{1}[/tex]
[tex]ma_{3}=ma_{2}+ma_{1}[/tex]
The speed of second tugboat is
[tex]v=\dfrac{11\times10^{3}}{3600}[/tex]
[tex]v=3.05\ m/s[/tex]
We need to calculate total acceleration
[tex]a_{3}=\dfrac{v}{t}[/tex]
[tex]a_{3}=\dfrac{3.05}{10}[/tex]
[tex]a_{3}=0.305\ m/s^2[/tex]
We need to calculate the acceleration a₂
[tex]0.305=a_{2}+0.083[/tex]
[tex]a_{2}=0.305-0.083[/tex]
[tex]a_{2}=0.222\ m/s^2[/tex]
We need to calculate the factor of F₁ and F₂
Dividing force F₁ by F₂
[tex]\dfrac{F_{1}}{F_{2}}=\dfrac{m\times0.83}{m\times0.22}[/tex]
[tex]\dfrac{F_{1}}{F_{2}}=3.7[/tex]
[tex]F_{1}=3.7F_{2}[/tex]
Hence, The magnitude of F₁ is 3.7 times of F₂
Consider a spring that does not obey Hooke’s law very faithfully. One end of the spring is fixed. To keep the spring stretched or compressed an amount x, a force along the x-axis with x-component Fx=kx−bx2+cx3 must be applied to the free end. Here k = 100 N/m, b=700N/m2, and c=12,000N/m3. Note that x < 0 when the spring is stretched and x > 0 when it is compressed. How much work must be done
(a) to stretch this spring by 0.050 m from its unstretched length?
(b) To compress this spring by 0.050 m from its unstretched length?
(c) Is it easier to stretch or compress this spring? Explain why in terms of the dependence of Fx on x. (Many real springs behave qualitatively in the same way.)
Final answer:
To find the work done when stretching or compressing a non-Hookean spring by 0.050 m, we calculate the definite integral of the force function over the respective intervals. The work done for stretching is the integral from 0 to -0.050 m, and for compression, it is the integral from 0 to 0.050 m. The ease of stretching versus compressing depends on the non-linear force constants.
Explanation:
Work Done to Stretch or Compress a Non-Hookean Spring
To compute the work (W) required to stretch or compress the spring an amount x, we need to integrate the force function Fx = kx - bx2 + cx3 from the unstretched length (0) to the final position (x).
For stretching (x < 0), substitute the given constants (k = 100 N/m, b = 700 N/m2, c = 12,000 N/m3) and integrate from 0 to -0.050 m.
For compression (x > 0), use the same constants and integrate from 0 to 0.050 m.
Analyze the dependence of Fx on x to determine which process (stretching or compressing) requires less work.
The work done for both stretching and compressing can be calculated by evaluating the definite integral of Fx over the interval [0, x].
(a) Work done to stretch:
W = ∫0-0.050 (100x - 700x² + 12,000x3) dx
(b) Work done to compress:
W = ∫00.050 (100x - 700x² + 12,000x3) dx
(c) Comparison of ease:
The comparison is based on the shape of the force-displacement curve. With the given non-linear characteristics, it might be easier or harder to stretch or compress the spring depending on the values of b and c which affect the quadratic and cubic terms, respectively.
A mass is connected to a spring on a horizontal frictionless surface. The potential energy of the system is zero when the mass is centered on x = 0, its equilibrium position. If the potential energy is 3.0 J when x = 0.050 m, what is the potential energy when the mass is at x = 0.10 m?
Answer:
U = 12 J.
Explanation:
The potential energy in a spring is given by the following formula
[tex]U = \frac{1}{2}kx^2[/tex]
where k is the spring force constant and x is the displacement from the equilibrium.
If U = 3 J when x = 0.05 m, then k is
[tex]3 = \frac{1}{2}k(0.05)^2\\k = 2400~N/m[/tex]
Using this constant, we can calculate the potential energy at x = 0.10 m:
[tex]U = \frac{1}{2}(2400)(0.1)^2 = 12 ~J[/tex]
The potential energy when the mass is at [tex]\( x = 0.10 \)[/tex] m is 12 J.
The potential energy (U) of a mass-spring system is given by the equation:
[tex]\[ U = \frac{1}{2} k x^2 \][/tex]
Given that the potential energy is 3.0 J when \( x = 0.050 \) m, we can use this information to find the spring constant \( k \). Plugging the values into the equation, we get:
[tex]\[ 3.0 \, \text{J} = \frac{1}{2} k (0.050 \, \text{m})^2 \] \[ k = \frac{2 \times 3.0 \, \text{J}}{(0.050 \, \text{m})^2} \] \[ k = \frac{6.0 \, \text{J}}{0.0025 \, \text{m}^2} \] \[ k = 2400 \, \text{N/m} \][/tex]
Now that we have the spring constant, we can find the potential energy when [tex]\( x = 0.10 \)[/tex]m using the same formula:
[tex]\[ U = \frac{1}{2} k x^2 \] \[ U = \frac{1}{2} (2400 \, \text{N/m}) (0.10 \, \text{m})^2 \] \[ U = \frac{1}{2} (2400 \, \text{N/m}) (0.010 \, \text{m}^2) \] \[ U = 1200 \, \text{N/m} \times 0.010 \, \text{m}^2 \] \[ U = 12 \, \text{J} \][/tex]
Therefore, the potential energy when the mass is at[tex]\( x = 0.10 \)[/tex]m is 12 J."
In a series RCL circuit the generator is set to a frequency that is not the resonant frequency. This nonresonant frequency is such that the ratio of the inductive reactance to the capacitive reactance of the circuit is observed to be 5.68. The resonant frequency is 240 Hz. What is the frequency of the generator
Answer:
Explanation:
Resonant frequency is 240
4π² x 240² = 1 / LC
230400π² = 1 / LC
Let the required frequency = n
inductive reactance = 2 πn L
capacitative reactance = 1 / 2 π n C
inductive reactance / capacitative reactance
= 4π² x n ² x LC = 5.68
4π² x n ² = 1 / LC x 5.68
= 230400π² x 5.68
4n ²= 230400 x 5.68
n ²= 57600 x 5.68
n ² = 327168
n = 572 approx
The generator's frequency in this RLC circuit, where the ratio of inductive to capacitive reactance is 5.68 and the resonant frequency is 240 Hz, is approximately 571 Hz.
To solve this problem, we need to use the relationship between inductive and capacitive reactance in an RLC circuit. Given that the ratio of the inductive reactance (XL) to the capacitive reactance (XC) is 5.68,
X_L / X_C = 5.68Reactance is frequency-dependent, with the following formulas for inductive and capacitive reactance:
X_L = 2πfLX_C = 1 / (2πfC)Given the resonant frequency as 240 Hz, we start by calculating the resonance reactances:
At resonance: X_L = X_C, so 2π(240)L = 1 / (2π(240)C)Using the non-resonant frequency, we use the reactance ratio:
2πfL / (1 / (2πfC)) = 5.68Rearranging gives us:
f² = 5.68 / (2π)²(LC)Since 240 Hz is the resonant frequency, substituting f:
f² = 5.68 * (240)²Simplifying yields:
f = 240 * √(5.68)f ≈ 571 HzTherefore, the generator frequency is approximately 571 Hz.
5. From Gauss’s law, the electric field set up by a uniform line of charge is E S 5 a l 2pP0r b r^ where r^ is a unit vector pointing radially away from the line and l is the linear charge density along the line. Derive an expression for the potential difference between r 5 r1 and r 5 r2.
Answer:
ΔV = 2k λ ln (r₁ / r₂)
Explanation:
The electric potential is
ΔV = - ∫ E. ds
The expression for the electric field of a charge line is given
E = 2k λ / r
Let's replace and integrate
ΔV = - 2k λ ∫ dr / r
ΔV = -2 k λ ln r
Let's evaluate
ΔV = - 2k λ ln r₂- ln r₁
ΔV = -2k λ ln (r₂ / r₁)
ΔV = 2k λ ln (r₁ / r₂)
After your school's team wins the regional championship, students go to the dorm roof and start setting off fireworks rockets. The rockets explode high in the air and the sound travels out uniformly in all directions. If the sound intensity is 1.97 10-6 W/m2 at a distance of 113 m from the explosion, at what distance from the explosion is the sound intensity half this value?
Answer:
160 m
Explanation:
The intensity, I, of the sound is inversely proportional to the square of the distance, r, from the source.
[tex]I\propto \dfrac{1}{r^2}[/tex]
Hence,
[tex]I_1r_1^2 = I_2r_2^2[/tex]
[tex]r_2 = r_1\sqrt{\dfrac{I_1}{I_2}}[/tex]
From the question, [tex]I_2[/tex] is half of [tex]I_1[/tex]
[tex]r_2 = r_1\sqrt{\dfrac{I_1}{0.5I_1}}[/tex]
[tex]r_2 = r_1\sqrt{2}[/tex]
[tex]r_2 = 113\text{ m}\sqrt{2} = 160 \text{ m}[/tex]
Answer:
Distance ,d= 159.81m
Explanation:
The intensity, I, of the sound is inversely proportional to the square of the distance, d,from the source.
Using the equation d/do=sqrt2
d= dosqrt2
Where d=113m
d= 113sqrt2
d= 159.81m
Suppose a person whose mass is m is being held up against the wall with a constant tangential velocity v greater than the minimum necessary. Find the magnitude of the frictional force between the person and the wall
The magnitude of the frictional force between the person and the wall is
[tex]\[ f = \frac{mv^2}{r} \][/tex]
To find the magnitude of the frictional force between the person and the wall, we can use the concept of centripetal force.
First, let's define the variables:
m = mass of the person
v = tangential velocity of the person
R = radius of the circular path (distance from the person to the axis of rotation, in this case, the distance from the person to the wall)
The centripetal force [tex](\( F_c \))[/tex] required to keep the person moving in a circular path is given by:
[tex]\[ F_c = \frac{{mv^2}}{R} \][/tex]
Since the person is held against the wall, the frictional force [tex](\( F_f \))[/tex] provides the centripetal force. Therefore,
[tex]\[ F_f = F_c \][/tex]
[tex]\[ F_f = \frac{{mv^2}}{R} \][/tex]
Therefore, the magnitude of the frictional force between the person and the wall is [tex]\( \frac{{mv^2}}{R} \)[/tex].
The slit-to-screen distance is D = 200 cm , and the laser wavelength is 633 nm, use the formula for single-slit diffraction minima to find the slit width a.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The slit width [tex]a = \frac{2L \lambda}{W}[/tex]
Explanation:
Assuming the unit on the graph is cm
Given that the slit to screen distance is D = 200 cm = 20 000 m
The wavelength [tex]\lambda[/tex] = 633 nm = [tex]633*10^{-9}m[/tex]
slit width a = ?
The width of the spot that is the width of the peak from the graph is
W = 1.6 × 2 = 3.2 cm
Where the 1.6 is the distance from 0 to the right end point of the peak
The change in y i.e [tex]\Delta y[/tex] has a formula
[tex]\Delta y[/tex] = Ltanθ
An the width of the spot is 2 × [tex]\Delta y[/tex]
W = 2Ltanθ
Applying this formula qsinθ = m[tex]\lambda[/tex]
where m = 1 because we a focused on the first zeros ,using small angle approximation we have y
[tex]a\theta = (1) \lambda[/tex]
[tex]\theta = \frac{\lambda}{a}[/tex]
Substituting this into W = 2ltanθ
Using small angle approximation
W = 2ltanθ = 2Lθ
[tex]W = 2L\frac{\lambda}{a}[/tex]
[tex]a = \frac{2L \lambda}{W}[/tex] and this is the slit width
A string of mass m is under tension, and the speed of a wave in the string is v. What will be the speed of a wave in the string if the mass of the string is increased to 2m but with no change in the length or tension?
A) v/ sq. rt. of 2
B) v/2
C) 2v
D) v * sq. rt. of 2
E) 4v
Answer:
A) v/ sq. rt. of 2
Explanation:
The speed of the wave in the string is defined as:
[tex]v=\sqrt{\frac{T}{\mu}}[/tex]
Where T is the tension on the string and [tex]\mu[/tex] is the linear density, that is, the mass per unit length:
[tex]\mu=\frac{m}{L}[/tex]
Where m is the mass of the string and L its length. We have [tex]m'=2m[/tex], [tex]T'=T[/tex] and [tex]L'=L[/tex]:
[tex]v'=\sqrt\frac{T'}{m'/L'}\\v'=\sqrt\frac{T}{2m/L}\\v'=\frac{1}{\sqrt2}\sqrt\frac{T}{m/L}\\v'=\frac{v}{\sqrt2}[/tex]
Explanation:
Below is an attachment containing the solution.
You throw a football straight up. Air resistance can be neglected. (a) When the football is 4.00 m above where it left your hand, it is moving upward at 0.500 m/s. What was the speed of the football when it left your hand
To find the initial speed of the football, you can use the kinematic equation. Given that the football is moving upward at 0.500 m/s and the displacement is 4.00 m, the equation can be used to find the initial velocity.
Explanation:To find the initial speed of the football, we can use the kinematic equation that relates the final velocity, initial velocity, acceleration, and displacement: vf = vi + at. Given that the football is moving upward at 0.500 m/s and the displacement is 4.00 m, we can plug in the values to solve for the initial velocity:
0.500 m/s = vi + (-9.8 m/s2) x t
Since the football is thrown straight up, the acceleration due to gravity is negative. Let's assume the time taken for the football to reach a height of 4.00 m is t. Since the football goes up and then comes back down, the time taken to reach the height of 4.00 m in the upward direction will be half of the total time. Let's call this time tu. Since the motion is symmetric, the time taken to reach the height of 4.00 m on the way down will also be tu. Therefore, the total time taken for the football's motion is 2tu.
Given that the total time is 2tu, we can substitute it into the equation:
0.500 m/s = vi + (-9.8 m/s2) x 2tu
Now, we can solve for vi by isolating it:
vi = 0.500 m/s - (-9.8 m/s2) x 2tu
Since the time taken for the motion is unknown, we cannot determine the precise initial velocity without more information.
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After the NEAR spacecraft passed Mathilde, on several occasions rocket propellant was expelled to adjust the spacecraft's momentum in order to follow a path that would approach the asteroid Eros, the final destination for the mission. After getting close to Eros, further small adjustments made the momentum just right to give a circular orbit of radius 45 km (45 × 103 m) around the asteroid. So much propellant had been used that the final mass of the spacecraft while in circular orbit around Eros was only 550 kg. The spacecraft took 1.04 days to make one complete circular orbit around Eros. Calculate what the mass of Eros must be.
Answer:
[tex]6.68\times 10^{15}\ kg[/tex]
Explanation:
G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²
R = Radius of orbit = 45 km
T = Time period = 1.04 days
Mass of Eros would be given by the following equation
[tex]M=\dfrac{4\pi^2R^3}{GT^2}\\\Rightarrow M=\dfrac{4\pi^2\times (45\times 10^3)^3}{6.67\times 10^{-11}\times (1.04\times 24\times 3600)^2}\\\Rightarrow M=6.68\times 10^{15}\ kg[/tex]
The mass of Eros is [tex]6.68\times 10^{15}\ kg[/tex]
Two froghoppers sitting on the ground aim at the same leaf, located 35 cm above the ground. Froghopper A jumps straight up while froghopper B jumps at a takeoff angle of 58° above the horizontal.
Which froghopper experiences the greatest change in kinetic energy from the start of the jump to when it reaches the leaf?
Answer: A
Explanation: We can use the concept of conservation energy which implies that the kinetic energy of the froghoppers equals it potential energy from the ground level.
Where potential energy = mgh
Where m = mass of the object , g = acceleration due gravity and h = height from ground level.
The value of potential energy will reduce when the height is inclined at an angle.
Let us assume equal mass for both froghoppers, say m , g = 10 m/s^2 and a value of h.
For the first froghopper, potential energy = m×9.8×h = 9.8 mh
For the second froghopper, potential energy = m×9.8×hsin58 ( hsin58 is the vertical componet of height h inclined at angle 58),
potential energy = 8.3109 mh
As we can see , froghopper A has more potential energy than froghoppers B which implies that A has more kinetic energy than B
Final answer:
Both froghoppers experience the same change in kinetic energy, as the change in potential energy is the same for both due to reaching the same height.
Explanation:
To determine which froghopper experiences the greatest change in kinetic energy we can consider their jumps from an energy perspective. As both froghoppers are aiming for the same leaf at a height of 35 cm, the change in gravitational potential energy (UG) from the ground to the leaf will be the same for both, since potential energy depends only on the height and mass, which are constant for both froghoppers. Assuming that both froghoppers start at rest (kinetic energy Ki = 0), the change in kinetic energy during the jump (ΔK) will equal the change in potential energy (ΔUG) at the top of their trajectories where their velocity is zero.
If we follow the work-energy principle, the work done by the froghoppers' muscles will convert to potential energy at the peak of their jumps. Thus, the greatest change in kinetic energy will be equivalent to the potential energy gained at the leaf's height, which is same for both, meaning that neither froghopper experiences a greater change in kinetic energy than the other when reaching the leaf. The takeoff angle does not affect the change in kinetic energy in this case since both reached the same height.
A quarter-wave monopole radio antenna (also called a Marconi antenna) consists of a long conductor of one quarter the length of the transmitted wavelength. The lower end of the antenna is connected to a conducting ground plane (often simply the Earth) which reflects the transmitted signal, creating an effective antenna length of a half wavelength
(a) What must be the height of the antenna (in m) for a radio station broadcasting at 604 kHz?
(b) What must be the height of the antenna (in m) for radio stations broadcasting at 1,710 kHz?
Answer:
a) Height of the antenna (in m) for a radio station broadcasting at 604 kHz = 124.17 m
b)Height of the antenna (in m) for radio stations broadcasting at 1,710 kHz =43.86 m
Explanation:
(a) Radiowave wavelength= λ = c/f
As we know, Radiowave speed in the air = c = 3 x 10^8 m/s
f = frequency = 604 kHz = 604 x 10^3 Hz
Hence, wavelength = (3x10^8/604x10^3) m
λ = 496.69 m
So the height of the antenna BROADCASTING AT 604 kHz = λ /4 = (496.69/4) m
= 124.17 m
(b) As we know , f = 1710 kHz = 1710 x 10^3 Hz (1kHZ = 1000 Hz)
Hence, wavelength = λ = (3 x 10^8/1710 x 10^3) m
λ= 175.44 m
So, height of the antenna = λ /4 = (175.44/4) m
= 43.86 m
If each frame of a motion picture film is 35 mm high, and 24 frames go by in a second, estimate how many frames are needed to show a two hour long movie.
1 hour = 3600 seconds
2 hrs = 7200 sec
(24 frame/sec) x (72 sec) =
172,800 frames.
If each frame of a motion picture film is 35 mm high, and 24 frames go by in a second, then the number of frames needed to show a two-hour-long movie would be 172800 frames
What is multiplication?Finding the product of two or more numbers in mathematics is done by multiplying the numbers. It is one of the fundamental operations in mathematics that we perform on a daily basis.
As given in the problem If each frame of a motion picture film is 35 mm high, and 24 frames go by in a second,
The number of seconds in 2 hours = 2 ×3600
=7200 seconds
The number of frames goes by in one second = 24 frames
the number of frames in a 2-hour movie = 24×7200
=172800 frames
Thus, the number of frames needed to show a two-hour-long movie would be 172800 frames
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On the moon, all free-fall distance functions are of the forms(t)=0.81t^2 where t is in seconds and s is in meters. An object is dropped from a height of 200
meters above the moon. After 9 sec, consider parts (a) through(d) below.
a)
How far has the object fallen?
b)
How fast is it traveling?
c)
What is its acceleration?
d)
Explain the meaning of the second derivative of this free-fall function.
After 9 seconds, an object dropped on the moon will have fallen 65.61 meters, be traveling at 14.58 meters per second, with an acceleration of 1.62 meters per second squared. The second derivative of the distance function represents the moon's gravitational acceleration.
Explanation:An object is dropped on the moon from a height of 200 meters, and the distance function of free-fall due to lunar gravity is given by s(t)=0.81t^2. We are asked to determine various properties of the object's motion after 9 seconds.
a) Distance Fallen
To find how far the object has fallen, we plug t=9 into the distance function to get s(9)=0.81(9)^2=65.61 meters. This is the distance the object has fallen, not the distance from its starting height.
b) Velocity
The object's velocity can be determined by finding the derivative of the distance function, which gives us v(t) = 2(0.81)t = 1.62t. Substituting t=9, we find v(9) = 1.62(9) = 14.58 meters per second.
c) Acceleration
The acceleration of the object is constant and equal to twice the coefficient of t^2 in the distance function, which is 1.62 meters per second squared on the moon.
d) Meaning of the Second Derivative
The second derivative of the free-fall distance function represents the acceleration due to gravity on the moon. It's constant at 1.62 meters per second squared, indicating uniform acceleration during free-fall.