Answer:
Mass in kg will be 60 kg
Explanation:
We have given initial speed of the man u = 100 m/sec
And final speed v = 0 m/sec
Time is given, t = 10 sec
So acceleration [tex]a=\frac{v-u}{t}=\frac{0-100}{10}=-10m/sec^2[/tex]
Force is given , F= 600 N
From Newton's law we know that F = ma
So [tex]600=m\times 10[/tex]
mass = 60 kg
So mass in kg will be 60 kg
Earth turns on its axis about once every 24 hours. The Earth's equatorial radius is 6.38 x 106 m. If some catastrophe caused Earth to suddenly come to a screeching halt, with what speed would Earth's inhabitants who live at the equator go flying off Earth's surface?
To solve this problem it is necessary to apply the concepts related to the Period of a body and the relationship between angular velocity and linear velocity.
The angular velocity as a function of the period is described as
[tex]\omega = \frac{2\pi}{T}[/tex]
Where,
[tex]\omega =[/tex]Angular velocity
T = Period
At the same time the relationship between Angular velocity and linear velocity is described by the equation.
[tex]v = \omega r[/tex]
Where,
r = Radius
Our values are given as,
[tex]T = 24 hours[/tex]
[tex]T = 24hours (\frac{3600s}{1 hour})[/tex]
[tex]T = 86400s[/tex]
We also know that the radius of the earth (r) is approximately
[tex]6.38*10^6m[/tex]
Usando la ecuación de la velocidad angular entonces tenemos que
[tex]\omega = \frac{2\pi}{T}[/tex]
[tex]\omega = \frac{2\pi}{86400}[/tex]
[tex]\omega = 7.272*10^{-5}rad/s[/tex]
Then the linear velocity would be,
[tex]v = \omega *r[/tex]
x[tex]v = \omega *r[/tex]
[tex]v= 463.96m/s[/tex]
The speed would Earth's inhabitants who live at the equator go flying off Earth's surface is 463.96
A runner of mass 56.0kg runs around the edge of a horizontal turntable mounted on a vertical, frictionless axis through its center. The runner's velocity relative to the earth has magnitude 3.10m/s . The turntable is rotating in the opposite direction with an angular velocity of magnitude 0.210rad/s relative to the earth. The radius of the turntable is 3.10m , and its moment of inertia about the axis of rotation is 81.0kg?m2
Question: Find the final angular velocity of the system if the runner comes to rest relative to the turntable. (You can treat the runner as a particle.)
This problem can be resolved using the principle of conservation of angular momentum. Initially, the positive momentum of runner and negative of turntable cancel each other, making the total momentum zero. When the runner stops, the angular momentum of the turntable is expected to compensate for the total momentum of the system.
Explanation:In Physics, this is a classic problem involving the conservation of angular momentum. Angular momentum is given as L = Iω for an object rotating about an axis, where I is the moment of inertia and ω is the angular velocity. The total initial angular momentum of the system is zero because the positive runner's angular momentum cancels out the negative turntable's angular momentum.
When the runner comes to rest, he no longer contributes an angular momentum. The turntable has to account for all of the angular momentum. We therefore use the conservation of angular momentum to solve for the new angular momentum of the turntable after the runner stops:
Equating moment of inertia of the runner (considering the runner as a point mass, moment of inertia, I = m*r^2) and moment of inertia of the turntable and using the equation L_initial = L_final, we can find the final angular velocity of the system:
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The final angular velocity of the system is 6.853 rad/s.
The initial angular momentum [tex]\(L_{i}\)[/tex] of the turntable is:
[tex]\[ L_{i} = I\omega_{i} \][/tex]
where I is the moment of inertia of the turntable.
The initial angular momentum [tex]\(L_{runner,i}\)[/tex] of the runner is:
[tex]\[ L_{runner,i} = mvr \][/tex]
since the angular momentum of a particle moving in a circle is the product of its mass, velocity, and the radius of the circle.
The final angular momentum [tex]\(L_{f}\)[/tex] of the system is:
[tex]\[ L_{f} = I\omega_{f} + mvr_{f} \][/tex]
where [tex]\(r_{f}\)[/tex] is the final radius at which the runner is at rest relative to the turntable, which is the same as the initial radius r since the runner is still on the turntable.
By conservation of angular momentum:
[tex]\[ L_{i} + L_{runner,i} = L_{f} \] \[ I\omega_{i} + mvr = I\omega_{f} + mvr \][/tex]
Since the runner is at rest relative to the turntable in the final state, (vr = 0), and the equation simplifies to:
[tex]\[ I\omega_{i} + mvr = I\omega_{f} \][/tex]
Now we can solve for [tex]\(\omega_{f}\)[/tex]:
[tex]\[ \omega_{f} = \frac{I\omega_{i} + mvr}{I} \] \[ \omega_{f} = \omega_{i} + \frac{mvr}{I} \][/tex]
Given:
[tex]\(m = 56.0\) kg\\ \(v = 3.10\) m/s\\ \(r = 3.10\) m\\ \(I = 81.0\) kg\(\cdot\)m\(^2\)\\ \(\omega_{i} = 0.210\) rad/s[/tex]
Plugging in the values:
[tex]\[ \omega_{f} = 0.210 + \frac{56.0 \times 3.10 \times 3.10}{81.0} \] \[ \omega_{f} = 0.210 + \frac{56.0 \times 9.61}{81.0} \] \[ \omega_{f} = 0.210 + \frac{538.16}{81.0} \] \[ \omega_{f} = 0.210 + 6.643 \] \[ \omega_{f} = 6.853 \][/tex]
A felt-covered beanbag is fired into a empty wooden crate that sits on a concrete floor and is open on one side. The beanbag is moving exactly horizontally when it strikes and sticks to the inside wall of the crate, which is lined with velcro. As a result of this impact, the crate slides backwards 0.935 meters along the concrete, which has a coefficient of kinetic friction of \mu_k = 0.480μ k =0.480. The beanbag has a mass of 0.354 kg and the empty crate has a mass of 3.77 kg. Calculate the speed of the beanbag vv at the moment it strikes the crate. You may neglect the effect of air resistance.
Answer:
31.42383 m/s
Explanation:
g = Acceleration due to gravity = 9.81 m/s²
[tex]\mu[/tex] = Coefficient of kinetic friction = 0.48
s = Displacement = 0.935 m
[tex]m_1[/tex] = Mass of bean bag = 0.354 kg
[tex]m_2[/tex] = Mass of empty crate = 3.77 kg
[tex]v_1[/tex] = Speed of the bean bag
[tex]v_2[/tex] = Speed of the crate
Acceleration
[tex]a=-\frac{f}{m}\\\Rightarrow a=-\frac{\mu mg}{m}\\\Rightarrow a=-\mu g[/tex]
[tex]a=--9.81\times 0.48=4.7088\ m/s^2[/tex]
From equation of motion
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 4.7088\times 0.935+0^2}\\\Rightarrow v=2.96739\ m/s[/tex]
In this system the momentum is conserved
[tex]m_1v_1=(m_1+m_2)v_2\\\Rightarrow v_1=\frac{(m_1+m_2)v_2}{m_1}\\\Rightarrow u=\frac{(0.354+3.77)\times 2.69739}{0.354}\\\Rightarrow u=31.42383\ m/s[/tex]
The speed of the bean bag is 31.42383 m/s
The speed of the beanbag when it strikes the crate is 34.46 m/s.
Inelastic collision:
In an inelastic collision, two bodies collide, stick together and move with the same velocity. The momentum is conserved while the kinetic energy is not.
As the beanbag sticks to the crate they move together with the same velocity until they stop after 0.935 m due to frictional force.
Let M be the combined mass of the crate and beanbag. the frictional force acting on the system is:
[tex]F=-\mu_k Mg\\\\Ma=-\mu_k Mg\\\\a=-\mu g\\\\a=-0.480\times9.8\;m/s^2\\\\a=4.7\;m/s^2[/tex]will be the acceleration of the system.
Let the initial velocity of the system be u and the final velocity is 0, then from the third equation of motion:
[tex]0=u^2-2as\\\\u=\sqrt[]{2as}\\\\u=\sqrt{2\times4.7\times0.935}\;m/s\\\\u=2.96\;m/s[/tex]
So the momentum after collision is
Mu = (0.354 + 3.77)×2.96 = 12.2 kgm/s
Which must be equal to the initial momentum that is contributed only by the beanbag, since the crate is on rest. Let the speed of the bean bag be v then initial momentum should be:
12.2 = 0.354 × v
v = 34.46 m/s
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A new roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radius of the loop is 11.8 m, with what minimum speed must the car traverse the loop so that the rider does not fall out while upside down at the top? Assume the rider is not strapped to the car.
To solve this problem it is necessary to apply the concepts based on Newton's second law and the Centripetal Force.
That is to say,
[tex]F_c = F_w[/tex]
Where,
[tex]F_c =[/tex]Centripetal Force
[tex]F_w =[/tex]Weight Force
Expanding the terms we have to,
[tex]mg = \frac{mv^2}{r}[/tex]
[tex]gr = v^2[/tex]
[tex]v = \sqrt{gr}[/tex]
Where,
r = Radius
g = Gravity
v = Velocity
Replacing with our values we have
[tex]v = \sqrt{(9.8)(11.8)}[/tex]
[tex]v = 10.75m/s[/tex]
Therefore the minimum speed must the car traverse the loop so that the rider does not fall out while upside down at the top is 10.75m/s
How does the magnitude of the electrical force compare between a pair of charged particles when they are brought to half their original distance of separation? To one-quarter their original distance? To four times their original distance? (What law guides your answers?)
Answer:
a) 4 times larger. b) 16 times larger. c) 16 times smaller. d) Coulomb´s Law
Explanation:
Between any pair of charged particle, there exists a force, acting on the line that join the charges (assuming they can assimilated to point charges) directed from one to the other, which is directly proportional to the product of the charges, and inversely proportional to the square of the distance between them.
F= k q1q2 / (r12)2
a) If the distance is reduced to the half of the original distance of separation, and we introduce this value in the force equation, we get:
F(r/2) = k q1q2 / (r12/2)2 = k q1q2 /((r12)2/4) = 4 F(r)
b) By the same token, if r= r/4, we will have F(r/4) = 16 F(r)
c) If the distance increases 4 times, as the force is inversely proportional to the square of the distance, the force will be the original divided by 16, i.e., 16 times smaller.
The empirical law that allows to find out easily these values, is the Coulomb´s Law.
Explanation:
Tarzan is running with a horizontal velocity, along level ground. While running, he encounters a 2.21 m vine of negligible mass, the vine was hanging vertically from a tall tree limb and the end of the vine is directly level with his center of mass. Tarzan decides to grab the vine and swing which causes him to experience uniform circular motion. At the maximum height, Tarzan notices that the angle the vine makes with the vertical is 19 degrees. Tarzan mass is 65kg. Ignoring any elastic properties of the vine may have.a) Determine the magnitude of Tarzan's centripetal accelerationb) Determine the magnitude of the tension at the beginning of the swing
Answer:
a) [tex]a_c = 1.09m/s^2[/tex]
b) T = 720.85N
Explanation:
With a balance of energy from the lowest point to its maximum height:
[tex]m*g*L(1-cos\theta)-1/2*m*V_o^2=0[/tex]
Solving for [tex]V_o^2[/tex]:
[tex]V_o^2=2*g*L*(1-cos\theta)[/tex]
[tex]V_o^2=2.408[/tex]
Centripetal acceleration is:
[tex]a_c = V_o^2/L[/tex]
[tex]a_c = 2.408/2.21[/tex]
[tex]a_c = 1.09m/s^2[/tex]
To calculate the tension of the rope, we make a sum of forces:
[tex]T - m*g = m*a_c[/tex]
Solving for T:
[tex]T =m*(g+a_c)[/tex]
T = 720.85N
In a double-star system, two stars of mass 4.6 x 1030 kg each rotate about the system's center of mass at radius 1.9 x 1011 m.
(a) What is their common angular speed?
(b) If a meteoroid passes through the system's center of mass perpendicular to their orbital plane, what minimum speed must it have at the center of mass if it is to escape to "infinity" from the two-star system?
To carry out this exercise, it is necessary to use the equations made to Centripetal Force and Gravitational Energy Conservation.
By definition we know that the Centripetal Force is estimated as
[tex]F_c = M\omega^2R[/tex]
Where,
M = mass
[tex]\omega =[/tex] Angular velocity
R = Radius
From the 'linear' point of view the centripetal force can also be defined as
[tex]F_c = \frac{GM^2}{R^2}[/tex]
PART A ) Equating both equations we have,
[tex]\frac{M}{\omega^2R}=\frac{GM^2}{R^2}[/tex]
Re-arrange to find \omega
[tex]\omega = \sqrt{\frac{Gm}{r^3}}[/tex]
Replacing with our values
[tex]\omega = \sqrt{\frac{(6.67*10^{-11})(4.6*10^{30})}{(1.9*10^{11})^3}}[/tex]
[tex]\omega = 2.115*!0^{-7}rad/s[/tex]
Therefore the angular speed is [tex]\omega = 2.115*!0^{-7}rad/s[/tex]
PART B) For energy conservation we have to
[tex]KE_{min} = PE_{cm}[/tex]
Where,
[tex]KE_{min} =[/tex] Minimus Kinetic Energy
[tex]PE_{cm} =[/tex] Gravitational potential energy at the center of mass
Then,
[tex]\frac{1}{2} mv^2_{min} = \frac{2GMm}{R}[/tex]
Re-arrange to find v,
[tex]v_min = \sqrt{\frac{4GM}{R}}[/tex]
[tex]v_min = \sqrt{\frac{4(6.67*10^{-11})(2.2*10^{30})}{(1.9*10^{11})}}[/tex]
[tex]v_min = 5.55*10^4m/s[/tex]
Therefore the minimum speed must it have at the center of mass if it is to escape to "infinity" from the two-star system is [tex]v_min = 5.55*10^4m/s[/tex]
A 0.750-kg object hanging from a vertical spring is observed to oscillate with a period of 1.50 s. When the 0.750-kg object is removed and replaced by a 1.50-kg object, what will be the period of oscillation
Answer:
New time period, [tex]T_2=2.12\ s[/tex]
Explanation:
Given that,
Mass of the object 1, [tex]m_1=0.75\ kg[/tex]
Time period, [tex]T_1=1.5\ s[/tex]
If object 1 is replaced by object 2, [tex]m_2=1.5\ kg[/tex]
Let [tex]T_2[/tex] is the new period of oscillation.
The time period of oscillation of mass 1 is given by :
[tex]T_1=2\pi \sqrt{\dfrac{m_1}{k}}[/tex]
[tex]1.5=2\pi \sqrt{\dfrac{0.75}{k}}[/tex]............(1)
The time period of oscillation of mass 2 is given by :
[tex]T_2=2\pi \sqrt{\dfrac{m_2}{k}}[/tex]
[tex]T_2=2\pi \sqrt{\dfrac{1.5}{k}}[/tex]............(2)
From equation (1) and (2) we get :
[tex](\dfrac{T_1}{T_2})^2=\dfrac{m_1}{m_2}[/tex]
[tex](\dfrac{1.5}{T_2})^2=\dfrac{0.75}{1.5}[/tex]
[tex]\dfrac{1.5}{T_2}=0.707[/tex]
[tex]T_2=2.12\ s[/tex]
So, the new period of oscillation is 2.12 seconds. Hence, this is the required solution.
Final answer:
To find the new period of oscillation when the mass attached to a spring is increased, one can use the formula for the period T = 2π√(m/k). When the mass is doubled from 0.750 kg to 1.50 kg, the new period of oscillation is approximately 2.12 seconds.
Explanation:
The period of oscillation for a mass m suspended from a spring is given by the formula T = 2π√(m/k), where T is the period, m is the mass, and k is the spring constant. Using this relationship, we can compare the periods for different masses to find the new period when the mass changes.
When the mass was 0.750 kg, the period was 1.50 s. Now, if the mass is increased to 1.50 kg, we can expect the period to change. Noting that the spring constant k remains the same, we use the ratio of the periods T1/T2 = √(m1/m2). Solving for the new period T2 gives us T2 = T1 * √(m2/m1).
If we plug in the numbers, we get:
T1 = 1.50 s (given for 0.750 kg)
m1 = 0.750 kg
m2 = 1.50 kg
T2 = 1.50 s * √(1.50 kg / 0.750 kg) = 1.50 s * √2 ≈ 2.12 s
The new period of oscillation for the 1.50-kg object is therefore approximately 2.12 seconds.
Suppose that there are N=108 two-state systems, each with energy difference E=6 × 10-21 J between the two states. The environment temperature is 300 K. The number of systems in their high energy states is M. Therefore, the free energy is a function of M: F(M).What is dU/dM?
Answer:
The value of [tex]\dfrac{dU}{dM}[/tex] is [tex]6\times10^{-29}\ J/unit[/tex].
Explanation:
Given that,
Number [tex]N=10^{8}[/tex]
Energy difference = 6\times10^{-21}\ J[/tex]
Temperature T =300 K
We need to calculate the value of [tex]\dfrac{dU}{dM}[/tex]
We know that,
[tex]\dfrac{dU}{dM}=\dfrac{energy\ difference}{Change\ in\ number\ of\ system}[/tex]
[tex]\dfrac{dU}{dM}=\dfrac{6\times10^{-21}}{10^{8}}[/tex]
[tex]\dfrac{dU}{dM}=6\times10^{-29}\ J/unit[/tex]
Hence, The value of [tex]\dfrac{dU}{dM}[/tex] is [tex]6\times10^{-29}\ J/unit[/tex].
The average density of the body of a fish is 1080kg/m^3 . To keep from sinking, the fish increases its volume by inflating an internal air bladder, known as a swim bladder, with air. By what percent must the fish increase its volume to be neutrally buoyant in fresh water? Use 1.28kg/m^3 for the density of air at 20 degrees Celsius.
To solve this problem it is necessary to apply the concept related to density and its definition with respect to mass and volume.
Density can be expressed as
[tex]\rho = \frac{m}{V}[/tex]
Where,
m = mass
V = Volume
The average density of fish air system is equal to the neutral bouyancy when the water displaced by the fish air system can be expressed as
[tex]\rho_{b} = \frac{m_a+m_f}{V_a+V_f}[/tex]
Where,
[tex]m_a[/tex]= Mass of air
[tex]m_f =[/tex] Mass of fish
[tex]V_a[/tex]= Volume of air
[tex]V_f[/tex]= Volume of fish
As we know that m = \rho V we can replacing the equivalent value for the mass fro fish and air, then
[tex]\rho_{b} = \frac{m_a+m_f}{V_a+V_f}[/tex]
[tex]\rho_{b} = \frac{\rho_a V_a+\rho_f V_f}{V_a+V_f}[/tex]
[tex]\rho_{b}(V_a+V_f)=\rho_a V_a+\rho_f V_f[/tex]
[tex]\rho_{b} V_a+\rho_{b}V_f = \rho_a V_a+\rho_f V_f[/tex]
[tex]\rho_{b} V_a- \rho_a V_a=\rho_f V_f-\rho_{b}V_f[/tex]
[tex]V_a(\rho_{b}-\rho_a)=V_f(\rho_f-\rho_{b})[/tex]
[tex]\frac{V_a}{\rho_f}=\frac{(\rho_f-\rho_{b})}{(\rho_{b}-\rho_a)}[/tex]
Replacing with our values we have that
[tex]\frac{V_a}{\rho_f}=\frac{(1000-1080)}{(1.2-1000)}[/tex]
[tex]\frac{V_a}{\rho_f}=0.08[/tex]
[tex]\frac{V_a}{\rho_f}=8\%[/tex]
Therefore the percentage of Volume is increased by 8%
To achieve neutral buoyancy in fresh water, a fish with an initial density of 1080 kg/m^3 needs to increase its volume by 8%, utilizing its swim bladder.
Explanation:The solution to this problem involves understanding Archimedes' Principle and the concept of density. In order to achieve neutral buoyancy, the fish needs to match its density to the density of freshwater, which is approximately 1000 kg/m³. It can do this by increasing its volume through inflating its swim bladder, thus reducing its overall density.
Since the average density of an object is mass divided by its volume, we can set up an equation involving the initial and final densities of the fish, with the initial density being 1080 kg/m³ and the final density being 1000 kg/m³. Let the initial volume of the fish be V1, the final volume be V2, and the mass of the fish (which doesn't change) be m. Therefore, m/V1 = 1080 and m/V2 = 1000. We can then solve these equations to find that V2 = 1.08V1.
In terms of percent increase, (V2 - V1) / V1 = 0.08 or 8%. Hence, the fish needs to increase its volume by 8% to achieve neutral buoyancy in freshwater.
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a kg object and a kg object are separated by m. find the magnitude of the net gravitational force exerted by these objects on a kg object placed midway between them. at what position other than an infinitely remote one can the kg object be placed so as to experience a net force of zero from the other two objects?
Answer:
F₂ = 0 N
x = 0.5*d = 0.5 m
Explanation:
Given
m₁ = m₂ = m₃ = m = 1 Kg
d = distance between m₁ and m₃ = d₁₃ = 1 m
d₁₂ = 0.5*d = 0.5*(1 m) = 0.5 m = d₃₂
G = 6.673*10⁻¹¹ N*m²/ Kg²
a) Find the magnitude of the net gravitational force exerted by these objects on a kg object (F₃) placed midway between them.
we can apply
F₂ = F₁₂ + F₃₂
then we use the formula
F₁₂ = G*m₁*m₂ / d₁₂² = G*m² / (0.5*d)² = 4*G*m² / d²
⇒ F₁₂ = 4*G*m² / d² = 4*(6.673*10⁻¹¹ N*m²/ Kg²)*(1 Kg)² / (1 m)²
⇒ F₁₂ = 2.6692*10⁻¹⁰ N (←)
and
F₃₂ = G*m₃*m₂ / d₃₂² = G*m² / (0.5*d)² = 4*G*m² / d²
⇒ F₃₂ = 4*G*m² / d² = 4*(6.673*10⁻¹¹ N*m²/ Kg²)*(1 Kg)² / (1 m)²
⇒ F₃₂ = 2.6692*10⁻¹⁰ N (→)
we get
F₂ = F₁₂ + F₃₂ = (- 2.6692*10⁻¹⁰ N) + (2.6692*10⁻¹⁰ N) = 0 N
b) At what position other than an infinitely remote one can the kg object be placed so as to experience a net force of zero from the other two objects?
From a) It is known that x = 0.5*d = 0.5*(1 m) = 0.5 m
Nevertheless, we can find the distance as follows
If
F₂ = 0 ⇒ F₁₂ + F₃₂ = 0 ⇒ F₁₂ = - F₃₂
If
x = distance between m₁ and m₂
d - x = distance between m₂ and m₃
we get
F₁₂ = G*m² / x²
F₃₂ = G*m² / (d - x)²
If F₁₂ = - F₃₂
⇒ G*m² / x² = - G*m² / (d - x)²
⇒ 1 / x² = - 1 / (d - x)²
⇒ (d - x)² = - x²
⇒ d²- 2*d*x + x² = - x²
⇒ 2*x² - 2*d*x + d² = 0
⇒ 2*x² - 2*1*x + 1² = 0
⇒ 2*x² - 2*x + 1 = 0
Solving the equation we obtain
x = 0.5 m = 0.5*d
A cannon fires a cannonball of mass 16.0 kg by applying a force of 2750 N along the 1.25 m length of the barrel. (a) How much work does the cannon do on the cannonball? (b) The cannon is aimed at a 25.0° angle above the horizontal. Assume gravity is the only other force acting on the cannonball as it moves through the cannon barrel. (That is, ignore all frictional forces.) What is the net work done by these 2 forces on the cannonball while it is in the cannon barrel?
To develop this problem it is necessary to apply the concepts related to Work and energy conservation.
By definition we know that the work done by a particle is subject to the force and distance traveled. That is to say,
[tex]W = F*d[/tex]
Where,
F= Force
d = Distance
On the other hand we know that the potential energy of a body is given based on height and weight, that is
[tex]PE = -mgh[/tex]
The total work done would be given by the conservation and sum of these energies, that is to say
[tex]W_{net} = W+PE[/tex]
PART A) Applying the work formula,
[tex]W = F*d\\W = 2750*1.25\\W = 3437.5J[/tex]
PART B) Applying the height equation and considering that there is an angle in the distance of 25 degrees and the component we are interested in is the vertical, then
[tex]PE = -mgh*sin25[/tex]
[tex]PE = -16*9.81*1.15sin25[/tex]
[tex]PE = -82.9J[/tex]
The net work would then be given by
[tex]W_{net} = W+PE[/tex]
[tex]W_{net} = 3437.5J-82.9J[/tex]
[tex]W_{net} = 3354.6J[/tex]
Therefore the net work done by these 2 forces on the cannonball while it is in the cannon barrel is 3355J
A continuous succession of sinusoidal wave pulses are produced at one end of a very long string and travel along the length of the string. The wave has frequency 62.0 Hz, amplitude 5.20 mm and wavelength 0.560 m.
(a) How long does it take the wave to travel a distance of 8.40 m along the length of the string?
(b) How long does it take a point on the string to travel a distance of 8.40 m, once the wave train has reached the point and set it into motion?
(c) In parts (a) and (b), how does the time change if the amplitude is doubled?
a) The time taken for the wave to travel a distance of 8.40 m is 0.24 s
b) The time taken for a point on the string to travel a distance of 8.40 m is 6.46 s
c) The time in part a) does not change; the time in part b) is halved (3.23 s)
Explanation:
a)
In order to solve this part, we have to find the speed of the wave, which is given by the wave equation:
[tex]v=f \lambda[/tex]
where
v is the speed
f is the frequency
[tex]\lambda[/tex] is the wavelength
For the wave in this problem,
[tex]f = 62.0 Hz\\\lambda =0.560 m[/tex]
So its speed is
[tex]v=(62.0)(0.560)=34.7 m/s[/tex]
Now we want to know how long does it take for the wave to travel a distance of
d = 8.40 m
Since the wave has a constant velocity, we can use the equation for uniform motion:
[tex]t=\frac{d}{v}=\frac{8.40}{34.7}=0.24 s[/tex]
b)
In this case, we want to know how long does it take for a point on the string to travel a distance of
d = 8.40 m
A point on the string does not travel along the string, but instead oscillates up and down. The amplitude of the wave corresponds to the maximum displacement of a point on the string from the equilibrium position, and for this wave it is
A = 5.20 mm = 0.0052 m
The period of the wave is the time taken for a point on the string to complete one oscillation. It can be calculated as the reciprocal of the frequency:
[tex]T=\frac{1}{f}=\frac{1}{62.0}=0.016 s[/tex]
During one oscillation (so, in a time corresponding to one period), the distance covered by a point on the string is 4 times the amplitude:
[tex]d=4A=4(0.0052)=0.0208 m[/tex]
Since d is the distance covered by a point on the string in a time T, then we can find the time t needed to cover a distance of
d' = 8.40 m
By setting up the following proportion:
[tex]\frac{d}{T}=\frac{d'}{t}\\t=\frac{d'T}{d}=\frac{(8.40)(0.016)}{0.0208}=6.46 s[/tex]
c)
As we can see from the equation used in part a), the speed of the wave depends only on its frequency and wavelength, not on the amplitude: therefore, if the amplitude is changed, the speed of the wave is not affected, therefore the time calculated in part a) remains the same.
On the contrary, the answer to part b) is affected by the amplitude. In fact, if the amplitude is doubled:
A' = 2A = 2(0.0052) = 0.0104 m
Then the distance covered during one period is
[tex]d=4A' = 4(0.0104)=0.0416 m[/tex]
The time period does not change, so it is still
T = 0.016 s
Therefore, the time needed to cover a distance of
d' = 8.40 m
this time is
[tex]\frac{d}{T}=\frac{d'}{t}\\t=\frac{d'T}{d}=\frac{(8.40)(0.016)}{0.0416}=3.23 s[/tex]
So, the time taken has halved.
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The time taken for a wave to travel a distance of 8.40 m is approximately 0.120 seconds. The time taken for a point on the string to travel 8.40 m once the wave train has reached it is also approximately 0.120 seconds. Doubling the amplitude of the wave does not affect the time taken.
Explanation:To determine how long it takes for the wave to travel a distance of 8.40 m, we can use the formula:
Time = Distance / Speed
The speed of a wave is equal to the product of its frequency and wavelength:
Speed = Frequency x Wavelength
Substituting the given values into the formulas, we find that it takes approximately 0.120 seconds for the wave to travel 8.40 m along the string.
To calculate the time it takes for a point on the string to travel a distance of 8.40 m once the wave train has reached the point and set it into motion, we need to divide the distance by the speed of the wave pulse:
Time = Distance / Speed
Since the speed of the wave pulse is the same as the speed of the wave, the time it takes for a point on the string to travel 8.40 m is also approximately 0.120 seconds.
If the amplitude of the wave is doubled, it does not affect the time it takes for the wave to travel a distance of 8.40 m or for a point on the string to travel 8.40 m. The only thing that changes is the maximum displacement of the particles in the wave, but it does not impact the time taken.
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A space shuttle in orbit around the Earth carries its payload with its mechanical arm. Suddenly, the arm malfunctions and releases the payload. What will happen to the payload?
a. It will fall straight down and hit the Earth.
b. It will follow a curved path and eventually hit the Earth.
c. It will drift out into deep space.
d. It will remain in the same orbit with the shuttle.
We must remember that a body in Earth orbit is under perpetual free fall, that is, that the body moves under a constant speed in constant around the earth.
The moment one of the charges is released, its displacement speed will be equal to that of the ship, so it will tend to continue rotating in the orbit in which the ship was traveling.
The correct answer is D. The payload will remain in the same orbit with the space shuttle.
An object is at rest on top of a smooth sphere with a radius of ???? = 15.3 m that is buried exactly halfway under the ground. It then begins to slide down. At what height from the ground is the object no longer in contact with the sphere?
Answer:
10.2 m .
Explanation:
Let object falls by angle θ .
At any moment after the fall , there are two forces acting on the sphere
1 ) mg cosθ which is a component of weight towards the centre 2 ) normal reaction of the surface R .
mgcosθ - R is net force acting, which provides centripetal force
mgcosθ - R = mv² / r
But v² = 2g r( 1-cosθ ) [ object falls by height ( r - r cosθ ).
mgcosθ - R = m / r x 2g r( 1-cosθ )
When the object is no longer in touch with sphere,
R = 0
mgcosθ = m / r x 2g r( 1-cosθ )
3 gr cosθ = 2gr
cosθ = 2/3
height of fall
= r ( 1-cosθ )
r ( 1 - 2/3 )
1/3 r
1/3 x 15.3
5.1 m
Height from the ground
15.3 - 5.1
10.2 m .
The object sliding down the sphere loses contact at a point where its gravitational and centripetal forces become equal, corresponding to an angle θ where cosθ = 2/3. Given the sphere's radius of 15.3m, this corresponds to a height from the ground of 10.2 m.
Explanation:To solve this problem, we need to apply the concept of conservative forces and energy conservation. Initially, the object is at rest at the top of the sphere, thus it possesses gravitational potential energy and no kinetic energy. As the object slides down, its potential energy converts into kinetic energy until a point where the object no longer remains in contact with the sphere.
This occurs when the gravitational force acting towards the center of the sphere becomes equal to the centripetal force required for the object in circular motion. Mathematically, this correlates to the equation: mgcosθ = mv²/r.
By simplifying this equation, considering all the potential energy converts into kinetic energy at the point of leaving contact (mgh = 1/2mv²), and knowing the radius of the sphere, we can deduce that the object loses contact with the sphere at a point where the angle θ (from the vertical axis) is cosθ = 2/3. The height (h) it loses contact can then be determined by the relationship h = r(1 - cosθ). Given the radius r = 15.3m, we find the height from the ground to be 10.2 m.
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If a farsighted person has a near point that is 0.600 m from the eye, what is the focal length f2 of the contact lenses that the person would need to be able to read a book held at 0.350 m from the person's eyes? Express your answer in meters.
Answer:
f = 0.84 m
Explanation:
given,
near point of the distance from eye(di) = 0.60 m
focal length calculation = ?
distance he can read (d_o)= 0.350 m
now,
Using lens formula
[tex]\dfrac{1}{f} = \dfrac{1}{d_0} + \dfrac{1}{d_i}[/tex]
[tex]\dfrac{1}{f} = \dfrac{1}{0.35} + \dfrac{1}{-0.6}[/tex]
[tex]\dfrac{1}{f} = 1.19[/tex]
[tex] f = \dfrac{1}{1.19}[/tex]
f = 0.84 m
The focal length of the lens f = 0.84 m
A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from 0 to 5.10 m/s in 0.820 s. (a) What is the magnitude of the linear impulse experienced by a 66.0-kg passenger in the car during the time the car accelerates?
Answer:
[tex]I=336.6kgm/s[/tex]
Explanation:
The equation for the linear impulse is as follows:
[tex]I=F\Delta t[/tex]
where [tex]I[/tex] is impulse, [tex]F[/tex] is the force, and [tex]\Delta t[/tex] is the change in time.
The force, according to Newton's second law:
[tex]F=ma[/tex]
and since [tex]a=\frac{v_{f}-v_{i}}{\Delta t}[/tex]
the force will be:
[tex]F=m(\frac{v_{f}-v_{i}}{\Delta t})[/tex]
replacing in the equation for impulse:
[tex]I=m(\frac{v_{f}-v_{i}}{\Delta t})(\Delta t)[/tex]
we see that [tex]\Delta t[/tex] is canceled, so
[tex]I=m(v_{f}-v_{i})[/tex]
And according to the problem [tex]v_{i}=0m/s[/tex], [tex]v_{f}=5.10m/s[/tex] and the mass of the passenger is [tex]m=66kg[/tex]. Thus:
[tex]I=(66kg)(5.10m/s-0m/s)[/tex]
[tex]I=(66kg)(5.10m/s)[/tex]
[tex]I=336.6kgm/s[/tex]
the magnitude of the linear impulse experienced the passenger is [tex]336.6kgm/s[/tex]
You’re in a mall and you need some money to buy a nose ring, but you’re broke. You decide to scoop some quarters out of the fountain. The water in the fountain is one foot deep. How far below the water do the quarters appear to be? The index of refraction of water is 4/3.
Answer: 0.75 ft
Explanation:
This situation is due to Refraction, a phenomenon in which a wave (the light in this case) bends or changes its direction when passing through a medium with an index of refraction different from the other medium. In this context, the index of refraction is a number that describes how fast light propagates through a medium or material.
In addition, we have the following equation that states a relationship between the apparent depth [tex]{d}^{*}[/tex] and the actual depth [tex]d[/tex]:
[tex]{d}^{*}=d\frac{{n}_{1}}{{n}_{2}}[/tex] (1)
Where:
[tex]n_{1}=1[/tex] is the air's index of refraction
[tex]n_{2}=\frac{4}{3}=1.33[/tex] water's index of refraction.
[tex]d=1 ft[/tex] is the actual depth of the quarters
Now. when [tex]n_{1}[/tex] is smaller than [tex]n_{2}[/tex] the apparent depth is smaller than the actual depth. And, when [tex]n_{1}[/tex] is greater than [tex]n_{2}[/tex] the apparent depth is greater than the actual depth.
Let's prove it:
[tex]{d}^{*}=1 ft\frac{1}{1.33}[/tex] (2)
Finally we find the aparent depth of the quarters, which is smaller than the actual depth:
[tex]{d}^{*}=0.75 ft[/tex]
A fisherman’s scale stretches 3.3 cm when a 2.6-kg fish hangs from it. The fish is pulled down 2.5 cm more and released so that it oscillates up and down with a simple harmonic motion (SHM). ? (a) What is the spring stiffness constant? (b) What is be the amplitude? (c) What is the period and frequency of oscillations?
a) The spring constant is 772.1 N/m
b) The amplitude is 0.025 m
c) The period is 0.365 s, the frequency is 2.74 Hz
Explanation:
a)
At equilibrium, the weight of the fish hanging on the spring is equal to the restoring force of the spring. Therefore, we can write:
[tex]mg=kx[/tex]
where
(mg) is the weight of the fish
kx is the restoring force
m = 2.6 kg is the mass of the fish
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
k is the spring constant
x = 3.3 cm = 0.033 m is the stretching of the spring
Solving for k, we find:
[tex]k=\frac{mg}{x}=\frac{(2.6)(9.8)}{0.033}=772.1 N/m[/tex]
b)
Later, the fish is pulled down by 2.5 cm (0.025 m), and the system starts to oscillate.
The amplitude of a simple harmonic motion is equal to the maximum displacement of the system. In this case, when the fish is pulled down by 2.5 cm and then released, immediately after the releasing the fish moves upward, until it reaches the same displacement on the upper side (+2.5 cm).
This means that the amplitude of the motion corresponds also to the initial displacement of the fish when it is pulled down: therefore, the amplitude is
A = 2.5 cm = 0.025 m
c)
The period of oscillation of a mass-spring system is given by
[tex]T=2\pi \sqrt{\frac{m}{k}}[/tex]
where
m is the mass
k is the spring constant
Here we have
m = 2.6 kg
k = 772.1 N/m
So, the period is
[tex]T=2\pi \sqrt{\frac{2.6}{772.1}}=0.365 s[/tex]
And the frequency is given by the reciprocal of the period, therefore:
[tex]f=\frac{1}{T}=\frac{1}{0.365}=2.74 Hz[/tex]
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A bicycle is traveling north at 5.0 m/s. The mass of the wheel, 2.0 kg, is uniformly distributed along the rim, which has a radius of 20 cm. What are the magnitude and direction of the angular momentum of the wheel about its axle?
Answer:2
Explanation:
Given
Velocity of bicycle is 5 m/s towards north
radius of rim [tex]r=20 cm[/tex]
mass of rim [tex]m=2 kg[/tex]
Angular momentum [tex]\vec{L}=I\cdot \vec{\omega }[/tex]
[tex]I=mr^2=2\times 0.2^2=0.08 kg-m^2[/tex]
[tex]\omega =\frac{v}{r}=\frac{5}{0.2}=25 rad/s[/tex]
[tex]L=0.08\times 25=2kg-m^2/s[/tex]
direction of Angular momentum will be towards west
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 6.72. The resonant frequency is 240 Hz.
What is the frequency of the generator?
Answer: 624 Hz
Explanation:
If the ratio of the inductive reactance to the capacitive reactance, is 6.72, this means that it must be satified the following expression:
ωL / 1/ωC = 6.72
ω2 LC = 6.72 (1)
Now, at resonance, the inductive reactance and the capacitive reactance are equal each other in magnitude, as follows:
ωo L = 1/ωoC → ωo2 = 1/LC
So, as we know the resonance frequency, we can replace LC in (1) as follows:
ω2 / ωo2 = 6.72
Converting the angular frequencies to frequencies, we have:
4π2 f2 / 4π2 fo2 = 6.72
Simplifying and solving for f, we have:
f = 240 Hz . √6.72 = 624 Hz
As the circuit is inductive, f must be larger than the resonance frequency.
A 910-kg object is released from rest at an altitude of 1200 km above the north pole of the Earth. If we ignore atmospheric friction, with what speed does the object strike the surface of the earth? (G = 6.67 × 10-11 N · m2/kg2, MEarth = 5.97 × 1024 kg, the polar radius of the earth is 6357 km)
In order to develop this problem it is necessary to use the concepts related to the conservation of both potential cinematic as gravitational energy,
[tex]KE = \fract{1}{2}mv^2[/tex]
[tex]PE = GMm(\frac{1}{r_1}-(\frac{1}{r_2}))[/tex]
Where,
M = Mass of Earth
m = Mass of Object
v = Velocity
r = Radius
G = Gravitational universal constant
Our values are given as,
[tex]m = 910 Kg[/tex]
[tex]r_1 = 1200 + 6371 km = 7571km[/tex]
[tex]r_2 = 6371 km,[/tex]
Replacing we have,
[tex]\frac{1}{2} mv^2 = -GMm(\frac{1}{r_1}-\frac{1}{r_2})[/tex]
[tex]v^2 = -2GM(\frac{1}{r_1}-\frac{1}{r_2})[/tex]
[tex]v^2 = -2*(6.673 *10^-11)(5.98 *10^24) (\frac{1}{(7.571 *10^6)} -\frac{1}{(6.371 *10^6)})[/tex]
[tex]v = 4456 m/s[/tex]
Therefore the speed of the object when striking the surface of earth is 4456 m/s
The speed of a 910-kg object striking the Earth after falling from 1200 km above the North Pole can be found using conservation of mechanical energy and the formula for gravitational potential energy.
Explanation:To determine the speed at which a 910-kg object strikes the surface of the Earth when released from rest at an altitude of 1200 km above the North Pole, we can use the conservation of mechanical energy. The total mechanical energy (kinetic plus potential) at the initial and final points must be equal since no external work is done on the system (we ignore atmospheric friction).
Initially, the object has potential energy due to its altitude above Earth and no kinetic energy since it's at rest. When it strikes the Earth, it has kinetic energy and minimal potential energy. Setting the initial and final total energies equal gives:
Ui + Ki = Uf + Kf,
where U is potential energy and K is kinetic energy. The potential energy can be calculated using the formula U = -GMEarthm/r, where m is the object's mass, r is the distance from the object to the center of Earth, and G is the gravitational constant. The kinetic energy is given by K = (1/2)mv2, where v is the velocity of the object.
The initial distance ri to the center of Earth is 6357 km + 1200 km (release altitude), and when the object strikes Earth, rf = 6357 km (polar radius of Earth). We know Ki = 0 and Uf is small enough to be negligible at the Earth's surface.
By plugging in the values, solving the equation for v, and taking the square root, we find the speed of the object when it strikes the ground.
The average density of the body of a fish is 1080 kg/m3. To keep from sinking, the fish increases its volumeby inflating an internal air bladder. By what percentage must the fish increase its volume in order to beneutrally buoyant in fresh water? The density of air at room temperature is 1.2 kg/m3.
Answer:
f = 8 %
Explanation:
given,
density of body of fish = 1080 kg/m³
density of air = 1.2 Kg/m³
density of water = 1000 kg/m²
to protect the fish from sinking volume should increased by the factor f
density of fish + density of water x increase factor = volume changes in water
1080 +f x 1.2 =(1 + f ) x 1000
1080 + f x 1.2 = 1000 + 1000 f
998.8 f = 80
f = 0.0800
f = 8 %
the volume increase factor of fish will be equal to f = 8 %
Tyrone, an 85-kg teenager, runs off the end of a horizontal pier and lands on a free-floating 130-kg raft that was initially at rest. After he lands on the raft, the raft, with him on it, moves away from the pier at 1.6 m/s. What was Tyrone's speed as he ran off the end of the pier?
Answer:u=4.04 m/s
Explanation:
Given
Mass m=85 kg
mass of Raft M=130 kg
velocity of raft and man v=1.6 m/s
Let initial speed of Tyrone is u
Conserving Momentum as there is no external Force
[tex]mu=(M+m)v[/tex]
[tex]85\times u=(85+130)\cdot 1.6[/tex]
[tex]u=\frac{215}{85}\cdot 1.6[/tex]
[tex]u=2.529\cdot 1.6=4.04 m/s[/tex]
Final answer:
Tyrone's speed as he ran off the end of the pier was approximately 4.04 m/s.
Explanation:
To calculate Tyrone's speed as he ran off the end of the pier, we can use the principle of conservation of momentum. According to this principle, the initial momentum of the system (Tyrone and the raft) is equal to the final momentum. Tyrone's initial momentum is given by his mass (85 kg) multiplied by his speed (which we need to find). The mass of the raft is 130 kg and its initial velocity is 0 m/s since it was at rest. The final momentum of the system is given by the mass of Tyrone and the raft (215 kg) multiplied by their final velocity (1.6 m/s).
Using the equation for conservation of momentum, we have:
(85 kg)(v) = (215 kg)(1.6 m/s)
Solving for the velocity v:
v = (215 kg)(1.6 m/s) / 85 kg
v ≈ 4.04 m/s
Therefore, Tyrone's speed as he ran off the end of the pier was approximately 4.04 m/s.
The matter that makes up a planet is distributed uniformly so that the planet has a fixed, uniform density. How does the magnitude of the acceleration due to gravity g at the planet surface depend on the planet radius R ? (Hint: how does the total mass scale with radius?) g ∝ 1 / R g ∝ R g ∝ √ R g ∝ R 2
Answer:
the acceleration due to gravity g at the surface is proportional to the planet radius R (g ∝ R)
Explanation:
according to newton's law of universal gravitation ( we will neglect relativistic effects)
F= G*m*M/d² , G= constant , M= planet mass , m= mass of an object , d=distance between the object and the centre of mass of the planet
if we assume that the planet has a spherical shape, the object mass at the surface is at a distance d=R (radius) from the centre of mass and the planet volume is V=4/3πR³ ,
since M= ρ* V = ρ* 4/3πR³ , ρ= density
F = G*m*M/R² = G*m*ρ* 4/3πR³/R²= G*ρ* 4/3πR
from Newton's second law
F= m*g = G*ρ*m* 4/3πR
thus
g = G*ρ* 4/3π*R = (4/3π*G*ρ)*R
g ∝ R
The driver of a car, which is initially at rest at the top A of the grade, releases the brakes and coasts down the grade with an acceleration in feet per second squared given by a = 3.22-0.004v2 , where v is the velocity in feet per second. Determine the velocity vB at the bottom B of the grade.
Answer:
28.3 ft/s
Explanation:
We are given that
Initial velocity of a car=[tex]u=0[/tex]
Acceleration of the car=[tex]a=3.22-0.004v^2[/tex]
We have to find the velocity [tex]v_B[/tex] at the bottom of the grade.
Distance covered by car=600 ft
We know that
Acceleration=a=[tex]\frac{dv}{dt}=\frac{dv}{ds}\times \frac{ds}{dt}=v\frac{dv}{ds}[/tex]
[tex]v=\frac{ds}{dt}[/tex]
[tex]ds=\frac{vdv}{a}[/tex]
Taking integration on both side and taking limit of s from 0 to 600 and limit of v from 0 to [tex]v_B[/tex]
[tex]\int_{0}^{600}ds=\int_{0}^{v_B}\frac{vdv}{3.22-0.004v^2}[/tex]
[tex][S]^{600}_{0}=-\frac{1}{0.008}[ln(3.22-0.004v^2)]^{v_B}_{0}[/tex]
Using substitution method and [tex]\int f(x)dx=F(b)-F(a)[/tex]
[tex]600-0=-\frac{1}{0.008}[ln(3.22-0.004v^2_B)-ln(3.22)][/tex]
[tex]600=-\frac{1}{0.008}(ln(\frac{3.22-0.04v_B}{3.22}))[/tex]
[tex]ln(m)-ln(n)=ln(\frac{m}{n})[/tex]
[tex]600\times 0.008=-ln(\frac{3.22-0.04v_B}{3.22})[/tex]
[tex]-4.8=ln(\frac{3.22-0.04v_B}{3.22})[/tex]
[tex]\frac{3.22-0.04v_B}{3.22}=e^{-4.8}[/tex]
[tex]ln x=y\implies x=e^y[/tex]
[tex]3.22-0.004v^2_B=3.22e^{-4.8}[/tex]
[tex]3.22-0.004v^2_B=0.0265[/tex]
[tex]3.22-0.0265=0.004v^2_B[/tex]
[tex]0.004v^2_B=3.22-0.0265=3.1935[/tex]
[tex]v_B=\sqrt{\frac{3.1935}{0.004}}=28.3 ft/s[/tex]
Hence, the velocity [tex]v_B[/tex] at the bottom of the grade=28.3 ft/s
The velocity of the car at the bottom of the grade is 28.37 ft/s.
What is acceleration?Acceleration is the rate of change of velocity with time.
Motion of the car from top to bottom of the gradeThe velocity of the car increases as the car moves downwards and it will be maximum when the car reaches equilibrium position. At equilibrium position (bottom of the grade) the acceleration will be zero but the velocity will be maximum.
a = 3.22 - 0.004v²
0 = 3.22 - 0.004v²
0.004v² = 3.22
v² = 3.22/0.004
v² = 805
v = √805
v = 28.37 ft/s
Thus, the velocity of the car at the bottom of the grade is 28.37 ft/s.
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A daredevil is attempting to jump his motorcycle over a line of buses parked end to end by driving up a 32° ramp, measured from the horizontal, at a speed of 40.0 m/s (144 km/h). The top of the ramp is at the same height as the roofs of the buses and each bus is 20.0 m long.
Answer:
Dare devil can cross 7 buses.
Explanation:
given,
angle of inclination of ramp = 32°
speed of the motorcycle = 40 m/s
length of bus = 20 m
how many buses daredevil can clear =?
to solve this we need to calculate the range of the daredevil
considering it as projectile
the range of motorcyclist
[tex]R = \dfrac{V^2 sin (2\theta)}{g}[/tex]
[tex]R = \dfrac{40^2 sin (2\times 32^0)}{9.8}[/tex]
[tex]R =163.26 \times sin(2\times 32^0)[/tex]
[tex]R =146.74\ m[/tex]
length of bus is given as 20 m
Number of bus daredevil can cross
[tex]N = \dfrac{147.74}{20}[/tex]
[tex]N =7.34[/tex]
Dare devil can cross 7 buses.
Answer:
The number of buses are 7.
Explanation:
Given that,
Angle =32°
Speed = 40.0 m/s
Length of bus = 20.0
We need to calculate the range of bus
Using formula of range
[tex]R=\dfrac{v^2\sin2\theta}{g}[/tex]
Where, g = acceleration due to gravity
v = initial velocity
Put the value into the formula
[tex]R=\dfrac{(40.0)^2\times\sin(2\times32)}{9.8}[/tex]
[tex]R=150.20\ m[/tex]
We need to calculate the number of buses
Using formula of number of buses
[tex]N=\dfrac{R}{L}[/tex]
Where, R = range
L = length of bus
[tex]N=\dfrac{150.20}{20.0}[/tex]
[tex]N=7[/tex]
Hence, The number of buses are 7.
A (B + 25.0) g mass is hung on a spring. As a result, the spring stretches (8.50 A) cm. If the object is then pulled an additional 3.0 cm downward and released, what is the period of the resulting oscillation? Give your answer in seconds with 3 significant figures. A = 13 B = 427
Answer:
Time period of the osculation will be 2.1371 sec
Explanation:
We have given mass m = (B+25)
And the spring is stretched by (8.5 A )
Here A = 13 and B = 427
So mass m = 427+25 = 452 gram = 0.452 kg
Spring stretched x= 8.5×13 = 110.5 cm
As there is additional streching of spring by 3 cm
So new x = 110.5+3 = 113.5 = 1.135 m
Now we know that force is given by F = mg
And we also know that F = Kx
So [tex]mg=Kx[/tex]
[tex]K=\frac{mg}{x}=\frac{0.452\times 9.8}{1.135}=3.90N/m[/tex]
Now we know that [tex]\omega =\sqrt{\frac{K}{m}}[/tex]
So [tex]\frac{2\pi }{T} =\sqrt{\frac{K}{m}}[/tex]
[tex]\frac{2\times 3.14 }{T} =\sqrt{\frac{3.90}{0.452}}[/tex]
[tex]T=2.1371sec[/tex]
The period of the resulting oscillation, named T, given a mass of (B + 25.0) g (with B = 427 g) and a spring displacement of (8.50 A) cm (with A = 13), when the object is pulled an additional 3.0 cm downward and released, is 2.345 seconds, when rounded to three significant figures.
Explanation:The problem given can be solved using the concept of simple harmonic motion, which is encountered in Physics. The object attached to the spring can be visualized as a simple harmonic oscillator. The period of the resulting oscillation can be calculated using the formula T = 2π√(m/k), where T is the period, m is the mass and k is the spring constant.
First, we compute the mass in grams, which is given by (B + 25.0) g, where B = 427 g. Therefore, the mass is (427 g + 25.0 g) = 452 g or 0.452 kg (since 1 kg = 1000 g).
Next, the spring constant k is calculated using Hooke’s law (F=kx), where F is the force exerted by the mass on the spring (F=mg), m is the mass and g is the acceleration due to gravity, and x is the displacement of the spring. From the problem, m = 0.452 kg, g = 9.8 m/s², and x = (8.50 A) cm = (8.50 x 13) cm = 110.5 cm or 1.105 m (since 1 m = 100 cm). Substituting these values into Hooke’s law gives: k = mg/x = (0.452 kg x 9.8 m/s²) / 1.105 m = 4.00 N/m.
Finally, the period T of the resulting oscillation is found using the formula T = 2π√(m/k) = 2π√(0.452 kg/4.00 N/m) = 2.345 s, rounded to three significant figures.
Learn more about Simple Harmonic Motion here:https://brainly.com/question/28208332
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A Texas cockroach of mass 0.117 kg runs counterclockwise around the rim of a lazy Susan (a circular disk mounted on a vertical axle) that has a radius 17.3 cm, rotational inertia 5.20 x 10-3 kg·m2, and frictionless bearings. The cockroach's speed (relative to the ground) is 1.91 m/s, and the lazy Susan turns clockwise with angular velocity ω0 = 2.87 rad/s. The cockroach finds a bread crumb on the rim and, of course, stops. (a) What is the angular speed of the lazy Susan after the cockroach stops? (b) Is mechanical energy conserved as it stops?
Answer
given,
mass of cockroach = 0.117 Kg
radius = 17.3 cm
rotational inertia = 5.20 x 10⁻³ Kg.m²
speed of cockroach = 1.91 m/s
angular velocity of Susan (ω₀)= 2.87 rad/s
final speed of cockroach = 0 m/s
Initial angular velocity of Susan
L_s = I ω₀
L_s = 5.20 x 10⁻³ x 2.87
L_s=0.015 kg.m²/s
initial angular momentum of the cockroach
L_c = - mvr
L_c = - 0.117 x 1.91 x 0.173
L_c = - 0.0387 kg.m²/s
total angular momentum of Both
L = 0.015 - 0.0387
L = - 0.0237 kg.m²/s
after cockroach stop inertia becomes
I_f = I + mr^2
I_f = 5.20 x 10⁻³+ 0.117 x 0.173^2
I_f = 8.7 x 10⁻³ kg.m²/s
final angular momentum of the disk
L_f = I_f ω_f
L_f = 8.7 x 10⁻³ x ω_f
using conservation of momentum
L_i = L_f
-0.0237 =8.7 x 10⁻³ x ω_f
[tex]\omega_f = \dfrac{-0.0237}{8.7 \times 10^{-3}}[/tex]
[tex]\omega_f = -2.72\ rad/s[/tex]
angular speed of Susan is [tex]\omega_f = -2.72\ rad/s[/tex]
The value is negative because it is in the opposite direction of cockroach.
[tex]|\omega_f |= 2.72\ rad/s[/tex]
b) the mechanical energy is not conserved because cockroach stopped in between.
If the radius of a blood vessel drops to 80.0% of its original radius because of the buildup of plaque, and the body responds by increasing the pressure difference across the blood vessel by 10.0%, what will have happened to the flow rate? The flow rate will have changed to _______ % of its original value.
Answer:
40.96%
Solution:
As per the question:
If the original radius be 'r' and the initial pressure difference be 'P':
After the drop, radius:
r' = 0.8r
After rise in pressure:
P' = 0.1 P
Now,
Rate of flow is given by:
R ∝ [tex]r^{4}P[/tex]
Thus
[tex]\frac{R'}{R} = (\frac{0.8r}{r})^{4}\frac{0.1P}{P}[/tex]
R' = 40.96%