A wheel with a weight of 395 N comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at an angular velocity of 23.1 rad/s . The radius of the wheel is 0.652 m and its moment of inertia about its rotation axis is 0.800 MR2. Friction does work on the wheel as it rolls up the hill to a stop, at a height of h above the bottom of the hill; this work has a magnitude of 3470 J .
Answers
Answer:
h=12.04m
Explanation:
1) Notation and some important concepts
[tex]\omega_i[/tex] represent the initial angular velocity
[tex]v_i[/tex] represent the initial velocity
[tex]h[/tex] represent the final height reached by the mass
[tex]M[/tex] represent the mass of the object
[tex]W_f[/tex] represent the work due the friction force (variable of interest)
[tex]KE_{rot}[/tex] represent the rotational energy
[tex]KE_{tra}[/tex] represent the transitional kinetic energy
[tex]PE=mgh[/tex] represent the potential energy
[tex]I=0.8MR^2[/tex] represent the rotational inertia
W= 395 N is the weight of the object
For this problem we can use the principle of energy conservation, this principle states that "the total energy of an isolated system remains constant; it is said to be conserved over time".
At the begin the wheel had rotational energy defined as "The kinetic energy due to rotational motion. Is a scalar quantity and have units of energy usually Joules". And this energy is represented by the following formula: [tex]KE_{rot}=\frac{1}{2}I\omega^2_i[/tex]
At the starting point the wheel also had kinetic energy defined as "The energy of mass in motion" and is given by the formula : [tex]KE_{tran}=\frac{1}{2}mv_i^2_i[/tex]
At the end of the movement we have potential energy since the mass is at height h the potential energy is defined as "The energy stored within an object, due to the object's position, arrangement or state" and is given by the formula [tex]PE=mgh[/tex].
Since we have friction acting we have a work related to the force of friction and we need to subtract this from the formula of conservation of energy
2) Formulas to use
The figure attached is an schematic draw for the problem
Using the principle of energy conservation we have:
[tex]KE_{rot}+KE_{tran}-W_f =PE[/tex]
Replacing the formulas for each energy w ehave:
[tex]\frac{1}{2}I\omega^2_i+\frac{1}{2}mv_i^2_i-W_f =Mgh[/tex] (1)
We also know that [tex]v_i =\omega_i R[/tex] and [tex]I=0.8MR^2[/tex] so if we replace this into equation (1) we got:
[tex]0.8(\frac{1}{2})M(R\omega_i)^2 +\frac{1}{2}M(\omega_i R)^2-W_f=Mgh[/tex] (2)
We also know that the weight is defined as [tex]W=mg[/tex] so then [tex]M=\frac{W}{g}[/tex], so if we replace this into equation (2) we have:
[tex]0.8(\frac{1}{2})\frac{W}{g}(R\omega_i)^2 +\frac{1}{2}\frac{W}{g}(\omega_i R)^2-W_f=Wh[/tex] (3)
So then if we solve for h we got:
[tex]h=\frac{0.8(\frac{1}{2})\frac{W}{g}(R\omega_i)^2 +\frac{1}{2}\frac{W}{g}(\omega_i R)^2-W_f}{W}[/tex] (4)
3) Solution for the problem
Now we can replace the values given into equation (4):
[tex]h=\frac{0.8(\frac{1}{2})\frac{395N}{9.8\frac{m}{s^2}}(0.652m(23.1\frac{rad}{s}))^2 +\frac{1}{2}\frac{395}{9.8\frac{m}{s^2}}(23.1\frac{rad}{s}(0.652m))^2-3470J}{395N}=12.04m[/tex] (4)
So then our final answer would be h=12.04m
Related Questions
Young's double slit experiment is one of the classic tests for the wave nature of light. In an experiment using red light (λ = 641 nm) the second dark fringe on either side of the central maximum is θ = 4.4 degrees relative to the central bright fringe.
(a) Write an expression for the separation distance between the slits.
Answers
Answer:
The separation distance between the slits is 16710.32 nm.
Explanation:
Given that,
Wavelength = 641 nm
Angle =4.4°
(a). We need to calculate the separation distance between the slits
Using formula of young's double slit
[tex]d\sin\theta=m\lambda[/tex]
[tex]d=\dfrac{m\lambda}{\sin\theta}[/tex]
Where, d = the separation distance between the slits
m = number of order
[tex]\lambda[/tex] =wavelength
Put the value into the formula
[tex]d=\dfrac{2\times641\times10^{-9}}{\sin4.4}[/tex]
[tex]d=0.00001671032\ m[/tex]
[tex]d=16710.32\ nm[/tex]
Hence, The separation distance between the slits is 16710.32 nm.
The separation distance between the slits in Young's double slit experiment can be calculated using the formula d = λx / sin(θ), where d is the separation distance, λ is the wavelength of light, θ is the angle of the fringe, and x is the distance between the fringe and the screen.
Explanation:In Young's double slit experiment, the separation distance between the slits can be determined using the formula:
d = λx / sin(θ)
Where d is the separation distance between the slits, λ is the wavelength of light, θ is the angle of the fringe, and x is the distance between the fringe and the screen. In this example, the second dark fringe on either side of the central maximum is given as θ = 4.4 degrees. To calculate the separation distance between the slits, we also need to know the wavelength of the red light, which is given as λ = 641 nm.
Using the formula, we have:
d = (641 nm) * x / sin(4.4 degrees)
To find the value of d, we need to know the value of x, which is the distance between the fringe and the screen.
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In which of the following cases does a car have a negative velocity and a positive acceleration? A car that is traveling in the:
a. +x direction decreasing in speed.
b. -x direction at a constant 20 m/s.
c. -x direction decreasing in speed.
d. +x direction increasing in speed.
e. -x direction increasing in speed.
Answers
Answer:
option C
Explanation:
The correct answer is option C.
For negative velocity
Negative velocity is when the velocity of the object in a negative direction.
so, for the velocity of the car to be negative, it should be in -x-direction.
For positive acceleration
Positive acceleration is the change of the velocity of the object with respect to time in a positive direction
but when the velocity of the object is in negative direction the slowing down of the vehicle will give positive acceleration.
hence, the correct answer is an option -x-direction decreasing in speed.
A 10-g bullet moving horizontally with a speed of 2.0 km/s strikes and passes through a 4.0-kg block moving with a speed of 4.2 m/s in the opposite direction on a horizontal frictionless surface. If the block is brought to rest by the collision, what is the kinetic energy of the bullet as it emerges from the block
Answers
Answer:
[tex]K=512J[/tex]
Explanation:
Since the surface is frictionless, momentum will be conserved. If the bullet of mass [tex]m_1[/tex] has an initial velocity [tex]v_{1i}[/tex] and a final velocity [tex]v_{1f}[/tex] and the block of mass [tex]m_2[/tex] has an initial velocity [tex]v_{2i}[/tex] and a final velocity [tex]v_{2f}[/tex] then the initial and final momentum of the system will be:
[tex]p_i=m_1v_{1i}+m_2v_{2i}[/tex]
[tex]p_f=m_1v_{1f}+m_2v_{2f}[/tex]
Since momentum is conserved, [tex]p_i=p_f[/tex], which means:
[tex]m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}[/tex]
We know that the block is brought to rest by the collision, which means [tex]v_{2f}=0m/s[/tex] and leaves us with:
[tex]m_1v_{1i}+m_2v_{2i}=m_1v_{1f}[/tex]
which is the same as:
[tex]v_{1f}=\frac{m_1v_{1i}+m_2v_{2i}}{m_1}[/tex]
Considering the direction the bullet moves initially as the positive one, and writing in S.I., this gives us:
[tex]v_{1f}=\frac{(0.01kg)(2000m/s)+(4kg)(-4.2m/s)}{0.01kg}=320m/s[/tex]
So kinetic energy of the bullet as it emerges from the block will be:
[tex]K=\frac{mv^2}{2}=\frac{(0.01kg)(320m/s)^2}{2}=512J[/tex]
In what phase of matter do molecules vibrate but stay in position?
A. gas
B.liquid
c.solid
d.plasma
Answers
Answer:
solid
Explanation:
they molecules are closely together
A 0.01-kg object is initially sliding at 9.0 m/s. It goes up a ramp (increasing its elevation by 1.5 m), and then moves horizontally before striking a spring of force constant k = 100 N/m. The spring is compressed by 5.0 cm as it completely stops the object.
How much heat energy was created during this motion?
Answers
Answer:
During this motion, 0.133 J of heat energy was created
Explanation:
Hi there!
Let´s calculate the energy of the object in each phase of the motion.
At first, the object has only kinetic energy (KE):
KE = 1/2 · m · v²
Where:
m = mass of the object.
v = velocity.
KE = 1/2 · 0.01 kg · (9 m/s)²
KE = 0.405 J
When the object goes up the ramp, it gains some gravitational potential energy (PE). Due to the conservation of energy, the object must convert some of its kinetic energy to obtain potential energy. By calculating the potential energy that the object acquires, we can know the loss of kinetic energy:
PE = m · g · h
Where:
m = mass of the object.
g = acceleration due to gravity (9.81 m/s²)
h = height.
PE = 0.01 kg · 9.81 m/s² · 1.5 m
PE = 0.147 J
The object "gives up" 0.147 J of kinetic energy to be converted into potential energy.
Then, after going up the ramp, the kinetic energy of the object will be:
0.405 J - 0.147 J = 0.258 J
When the object reaches the spring, kinetic energy is used to compress the spring and the object obtains elastic potential energy (EPE). Let´s calculate the EPE obtained by the object:
EPE = 1/2 · k · x²
Where:
k = spring constant.
x = compression of the spring
EPE = 1/2 · 100 N/m · (0.05 m)² = 0.125 J
Then, only 0.125 J of kinetic energy was converted into elastic potential energy. The object is at rest at the end of the motion, i.e., the object does not have kinetic energy when it compresses the spring by 5.0 cm. Since energy can´t be lost, the rest of the kinetic energy, that was not used to compress the spring, had to be converted into heat energy:
Heat energy = initial kinetic energy - obtained elastic potential energy
Heat energy = 0.258 J - 0.125 J = 0.133 J
During this motion, 0.133 J of heat energy was created.
Final answer:
The question asks to calculate the heat energy produced when a 0.01-kg object sliding at 9.0 m/s is stopped by a spring after going up a ramp. Using conservation of energy, the heat energy can be found by subtracting the spring's elastic potential energy from the total initial kinetic and gravitational potential energies.
Explanation:
The student's question involves determining the amount of heat energy created during the motion of a 0.01-kg object as it slides, goes up a ramp, and is stopped by a compressed spring. To solve this problem, we need to apply the principles of conservation of energy and mechanics.
Initially, the object has kinetic energy since it is sliding at 9.0 m/s. As it goes up a ramp with a height increase of 1.5 m, it gains gravitational potential energy and loses some kinetic energy. Finally, when it strikes and compresses the spring (of constant k = 100 N/m) by 5.0 cm, it loses all its kinetic energy, which gets converted into the energy stored in the spring (elastic potential energy) and heat energy due to non-conservative forces.
The formula to find the elastic potential energy (Ep) stored in a compressed spring is:
Ep = 1/2 * k * x2
Where k is the spring constant and x is the compression length. Since the initial kinetic energy and the gravitational potential energy are converted into the elastic potential energy and heat, we can find the heat energy by subtracting the elastic potential energy from the sum of the initial kinetic and gravitational potential energies.
Heat energy = Initial kinetic energy + Gravitational potential energy - Elastic potential energy
A golf ball and an equal-mass bean bag are dropped from the same height and hit the ground. The bean bag stays on the ground while the golf ball rebounds. Which experiences the greater impulse from the ground?
a. The golf ball.
b. Not enough information.
c. The bean bag.
d. Both the same.
Answers
Answer:
option A
Explanation:
The correct answer is option A
given,
mass of ball and bean bag is same
golf ball rebound with certain velocity where as the bean bag stops.
Impulse = Change in momentum
I = m v_f - m v_i
When the golf ball rebound there will be negative velocity on the ball.
Which will add up and increase the impulse of the golf ball.
But in the case of beam bag velocity after the collision is zero the impulse will be less.
The golf ball experiences the greater impulse from the ground.
What is impulse?Impulse refers to the force acting over time to change the momentum of an object. It is represented by J and usually expressed in Newton-seconds or kg m/s so we can conclude that the golf ball experiences the greater impulse from the ground because of the structure and elasticity.
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In 1851, Jean Bernard Léon Foucault suspended a pendulum (later named the Foucault pendulum) from the dome of the Panthéon in Paris. The mass of the pendulum was 28.00 kg and the length of the rope was 67.00 m. The acceleration due to gravity in Paris is 9.809 m/s2. Calculate the period of the pendulum.
Answers
To solve this problem, it is necessary to apply the concepts of the Simple Pendulum Period. Under this definition it is understood as the time it takes for the pendulum to pass through a point in the same direction. It is also defined as the time it takes to get a complete swing. Its value is determined by:
[tex]T = 2\pi \sqrt{\frac{l}{g}}[/tex]
Where,
T= Period
l = Length
g = Gravitaitonal Acceleration
With our values we have tat
[tex]T = 2\pi \sqrt{\frac{l}{g}}[/tex]
[tex]T = 2\pi \sqrt{\frac{67}{9.809}}[/tex]
[tex]T = 16.413s[/tex]
Therefore the period of the pendulum is 16.4s
The period of a pendulum is calculated using the formula T = 2π √(L/g), where T is the period, L is the length and g is the acceleration due to gravity. In the given scenario, values are provided for L and g, which can be substituted into the formula to find the period. Importantly, the mass of the pendulum does not affect the period.
Explanation:The period of a pendulum can be calculated using the formula T = 2π √(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity. In the given problem, the length, L, of the pendulum is 67.00 m and the acceleration due to gravity, g, in Paris is 9.809 m/s². Using these values in the formula, the period T would be T = 2π √(67.00 m / 9.809 m/s²).
It's important to note, as shown in the formula, that the period of a pendulum depends solely on its length and the acceleration due to gravity. The mass of the pendulum does not affect the period. This is one of the unique properties of a pendulum and is the principle behind its use in clocks and other timing devices.
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A 100 kg object hangs from two steel cables, both with radius 1.2 mm. The first cable is 2.5 m long and 2 mm shorter than the second, but the object is horizontal (so they are now the same length). What is the force on the first cable? Young's Modulus for steel is 2.0 x 1011 N/m2.A :470 NB :500 NC :850 ND :1000 N
Answers
Answer:
850N
Explanation:
The step by step is in the attachment.
A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an angular acceleration of 0.745 rad/s2. The child is standing 4.65 m from the center of the merry-go-round. What is the magnitude of the linear acceleration of the child?
Answers
Answer:
So the acceleration of the child will be [tex]8.05m/sec^2[/tex]
Explanation:
We have given angular speed of the child [tex]\omega =1.25rad/sec[/tex]
Radius r = 4.65 m
Angular acceleration [tex]\alpha =0.745rad/sec^2[/tex]
We know that linear velocity is given by [tex]v=\omega r=1.25\times 4.65=5.815m/sec[/tex]
We know that radial acceleration is given by [tex]a=\frac{v^2}{r}=\frac{5.815^2}{4.65}=7.2718m/sec^2[/tex]
Tangential acceleration is given by
[tex]a_t=\alpha r=0.745\times 4.65=3.464m/sec^[/tex]
So total acceleration will be [tex]a=\sqrt{7.2718^2+3.464^2}=8.05m/sec^2[/tex]
The magnitude of the linear acceleration of the child is mathematically given as
a=8.05m/sec^2
The magnitude of the linear acceleration
Question Parameters:
A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an angular acceleration of 0.745 rad/s2.
Generally the equation for the linear velocity is mathematically given as
v=wr
Therefore
v=1.25*4.65
v=5.815
radial acceleration is given by
a=v^2/r
Hence
a=5.815/4.65
a=7.2718
Tangential acceleration is
a_t=\alpha r
a_t=0.745*4.65
a_t=3.464m/sec
Hence, total acceleration will be
[tex]a=\sqrt{7.2718^2+3.464^2}[/tex]
a=8.05m/sec^2
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7. A block of copper of unknown mass has an initial temperature of 65.4oC. The copper is immersed in a beaker containing 95.7g of water at 22.7oC. When the two substances reach thermal equilibrium, the final temperature is 24.2oC. What is the mass of the copper block?
Answers
Answer:
37.34372 kg
Explanation:
m = Mass
[tex]\Delta T[/tex] = Change in temperature
1 denotes water
2 denotes copper
c = Heat capacity
Heat is given by
[tex]Q=mc\Delta T[/tex]
In this case the heat transfer will be equal
[tex]m_1c_1\Delta T_1=m_2c_2\Delta T_2\\\Rightarrow m_2=\frac{m_1c_1\Delta T_1}{c_2\Delta T_2}\\\Rightarrow m_2=\frac{95.7\times 4.18(24.2-22.7)}{0.39(65.4-24.2)}\\\Rightarrow m_2=37.34372\ kg[/tex]
Mass of copper block is 37.34372 kg
A boy sits on a boat in a lake. In his boat he has a pile of rocks, each with mass 0.538 kg. He throws the rocks off the back of the boat at a rate of one rock every 0.964 sec, with a horizontal velocity of 11.6 m/s. What is the thrust on the boat from this "rock-throwing propulsion"? That is, what is the magnitude of the average force experienced by the boat during this process?
Answers
Answer:
6.47385 N
Explanation:
m = Mass of stone = 0.538 kg
t = Time taken to hit the rock = 0.964 seconds
v = Velocity of rock = 11.6 m/s
Force is given by
[tex]F=\frac{mv}{t}\\\Rightarrow F=\frac{0.538\times 11.6}{0.964}\\\Rightarrow F=\frac{6.2408}{0.964}\\\Rightarrow F=6.47385\ N[/tex]
The magnitude of the average force experienced by the boat during this process is 6.47385 N
The momentum of each thrown rock is 6.241 kg·m/s. The force or thrust on the boat is the rate of change of this momentum, which equates to an average force of 6.47 N.
Explanation:The topic of this question falls under the domain of Physics, specifically momentum. The phenomenon at hand is the principle of conservation of momentum which states that the momentum of an isolated system remains constant if no external forces are acting on it.
The momentum of each rock thrown is given by the product of its mass and its velocity, which is (0.538 kg)(11.6 m/s)= 6.241 kg·m/s. Given that a rock is thrown every 0.964 second, the rate of change of momentum, which is equal to the force, can be calculated by dividing the momentum of each rock by the time between each throw. So, the average force experienced by the boat is 6.241 kg·m/s / 0.964 s = 6.47 N (rounded to 2 decimal places).
Therefore, each time a rock is thrown out of the boat, there is an equal and opposite thrust on the boat which averages to a force of 6.47 N.
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An electromagnet produces a magnetic field of 0.520 T in a cylindrical region of radius 2.40 cm between its poles. A straight wire carrying a current of 10.5 A passes through the center of this region and is perpendicular to both the axis of the cylindrical region and the magnetic field.
What magnitude of force is exerted on the wire?
Answers
Answer:
Magnetic force, F = 0.262 N
Explanation:
It is given that,
Magnetic field of an electromagnet, B = 0.52 T
Length of the wire, [tex]l = 2r =2\times 2.4=4.8\ cm=0.048\ m[/tex]
Current in the straight wire, i = 10.5 A
Let F is the magnitude of force is exerted on the wire. The magnetic force acting on an object of length l is given by :
[tex]F=ilB[/tex]
[tex]F=10.5\ A\times 0.048\ m\times 0.52\ T[/tex]
F = 0.262 N
So, the magnitude of force is exerted on the wire is 0.262 N.
A bar on a hinge starts from rest and rotates with an angular acceleration α=(10+6t) rad/s^2, where t is in seconds. Determine the angle in radians through which the bar turns in the first 4.00s.
Answers
Answer:
[tex]\theta=144\ rad[/tex]
Explanation:
given,
α=( 10+6 t ) rad/s²
[tex]\alpha =\dfrac{d\omega}{dt}[/tex]
[tex]d\omega= \alpha dt[/tex]
integrating both side
[tex]\omega= \int (10+6 t )dt[/tex]
[tex]\omega=10 t+6\dfrac{t^2}{2}[/tex]
[tex]\omega=10 t+3t^2[/tex]
we know
[tex]\omega =\dfrac{d\theta}{dt}[/tex]
[tex]d\theta= \alpha dt[/tex]
integrating both side
[tex]\theta= \int (10 t+3t^2 )dt[/tex]
[tex]\theta=10\dfrac{t^2}{2}+3\dfrac{t^3}{3}[/tex]
[tex]\theta=5 t^2+t^3[/tex]
now, at t = 4 s θ will be equal to
[tex]\theta=5\times 4^2+4^3[/tex]
[tex]\theta=144\ rad[/tex]
Case 1: A Styrofoam cup holds an unknown amount of lemonade (which is essentially water) at 20.5 °C. A 0.0550-kg ice cube at -10.2°C is placed in the lemonade. When thermal equilibrium is reached, all the ice has melted and the final temperature of the mixture is measured to be 11.8 °C. Assume that the mass of the cup is so small that it absorbs a negligible amount of heat, and ignore any heat lost to the surroundings. The latent heat of fusion for water is 3.4 X10 J/kg. The specific heat capacity for water and lemonade is the same, 4186 J/(kg °C). The specific heat capacity for ice is 2000 J/(kg °C). The cold ICE absorbs heat in 3 steps: Step 1: COLD ICE warms up from -10.2 °C. to 0.0 °C. (Remember: cold ice DOESN't Melt before reaching 0.0 °C.) Apply Q = mcAT, to the cold ice. Calculate the heat absorbed by the ICE in this step. Keep 2 decimal places. Report all heat as POSITIVE. Enter a number Submit (5 attempts remaining) Step 2: ICE at 0.0 °C melts into water at 0.0 °C. Calculate the heat absorbed by the ICE in this step. Keep 2 decimal places. Report all heat as POSITIVE. Enter a number Submit (5 attempts remaining) Step 3: The water at 0.0 °C from melted ice warms up to 11.8 °C. Apply Q = mcAT, to the water. Calculate the heat absorbed by water in this step. Keep 2 decimal places. Report all heat as POSITIVE. Enter a number Submit (5 attempts remaining) Calculate the total heat absorbed by ICE in the above 3 steps. Enter a number J Submit (5 attempts remaining) The warm lemonade releases heat in cooling down from 20.5 °C to 11.8 °C. Apply Q = mcAT, to the lemonade. Keep mass m as "space hoder. Keep all heat as POSITIVE. Let Q released by lemonade = Total heat absorbed by ICE Find the mass of the lemonade. Keep 2 decimal places. Enter a number kg Submit (5 attempts remaining)
Answers
Answer:
[tex]m_l=0.619\ kg[/tex]
Explanation:
Given:
initial temperature of water(lemonade), [tex]T_{il}=20.5^{\circ}C[/tex]mass of ice, [tex]m=0.055\ kg[/tex]initial temperature of ice, [tex]T_{ii}=-10.2^{\circ}C[/tex]final temperature of the mixture, [tex]T_f=11.8^{\circ}C[/tex]specific heat capacity of ice, [tex]c_i=2000\ J.kg^{-1}. ^{\circ}C^{-1}[/tex]specific heat capacity of water, [tex]c_w=4186\ J.kg^{-1}. ^{\circ}C^{-1}[/tex]Latent heat of fusion of ice, [tex]L=340000\ J.kg^{-1}[/tex]For the whole ice to melt in lemonade and result a temperature of 11.8°C the total heat lost by the lemonade will be equal to the total heat absorbed by the ice to come to 0°C from -10.2°C along with the latent heat absorbed in the melting of ice at 0°C and the heat absorbed by the ice water of 0°C to reach a temperature of 11.8°C.
Now, mathematically:
[tex]Q_l=Q_i+Q_m+Q_w[/tex]
[tex]m_l.c_w.\Delta T_l=m_i.c_i.\Delta T_i_i+m_i.L+m_i.c_w.\Delta T_w[/tex]
[tex]m_l.c_w.(T_{il}-T_f)=m_i(c_i.\Delta T_i_i+L+c_w.\Delta T_w)[/tex]
[tex]m_l\times 4186\times (20.5-11.8)=0.055(2000\times (0-(-10.2))+340000+4186\times (11.8-0))[/tex]
[tex]m_l=0.619\ kg[/tex] (mass of lemonade)
The escape velocity at the surface of Earth is approximately 11 km/s. What is the mass, in units of ME (the mass of the Earth), of a planet with three times the radius of Earth for which the escape speed is three times that for Earth?
Answers
Answer:11 km/s
Explanation:
Given
Escape velocity at the surface of earth is 11 km/s
Escape velocity is given by
[tex]V_e=\sqrt{\frac{2GM}{R}}[/tex]
Escape velocity at the surface of earth
[tex]11=\sqrt{\frac{2GM}{R}}[/tex]--------------------1
If Escape velocity is three times and the radius is also the three times
[tex]V_e_2=\sqrt{\frac{2G(3M)}{3R}}[/tex]
[tex]V_e_2=\sqrt{\frac{2GM}{R}}=V_e[/tex]
i.e. [tex]V_e_2=11 km/s[/tex]
A police officer in hot pursuit of a criminal drives her car through an unbanked circular (horizontal) turn of radius 300 m at a constant speed of 22.2 m/s. Her mass is 55.0 kg. To the nearest degree, what is the angle (relative to vertical) of the net force of the car seat on the officer?
Answers
Answer:
The angle (relative to vertical) of the net force of the car seat on the officer to the nearest degree is 10°.
Explanation:
Given:
Mass of the driver is, [tex]m=55\ kg[/tex]
Radius of circular turn is, [tex]R=300\ m[/tex]
Linear speed of the car is, [tex]v=22.2\ m/s[/tex]
Since, the car makes a circular turn, the driver experiences a centripetal force radially inward towards the center of the circular turn. Also, the driver experiences a downward force due to her weight. Therefore, two forces act on the driver which are at right angles to each other.
The forces are:
1. Weight = [tex]mg=55\times 9.8=539\ N[/tex]
2. Centripetal force, 'F', which is given as:
[tex]F=\frac{mv^2}{R}\\F=\frac{55\times (22.2)^2}{300}\\\\F=\frac{55\times 492.84}{300}\\\\F=\frac{27106.2}{300}=90.354\ N[/tex]
Now, the angle of the net force acting on the driver with respect to the vertical is given by the tan ratio of the centripetal force (Horizontal force) and the weight (Vertical force) and is shown in the triangle below. Thus,
[tex]\tan \theta=\frac{90.354}{539}\\\tan \theta=0.1676\\\theta=\tan^{-1}(0.1676)=9.52\approx 10[/tex]°
Therefore, the angle (relative to vertical) of the net force of the car seat on the officer to the nearest degree is 10°.
The net force angle relative to the vertical is found by calculating the centripetal and vertical forces. The angle is approximately 10°. This involves understanding forces in circular motion.
To find the angle of the net force of the car seat on the police officer, we need to consider the forces involved and the direction of the net force.
The car is executing horizontal circular motion, which means there is a centripetal force acting horizontally towards the center of the circle. This force is provided by the friction between the car tires and the road. The centripetal force Fc can be calculated by:
Fc = m * ac
where m is the mass of the officer and ac is the centripetal acceleration, which is given by:
ac = v2 / r
Given:
Radius of the turn, r = 300 m
Speed of the car, v = 22.2 m/s
Mass of the officer, m = 55.0 kg
First, calculate the centripetal acceleration:
ac = (22.2 m/s)2 / 300 m = 1.6444 m/s2
Now, calculate the centripetal force:
Fc = 55.0 kg * 1.6444 m/s2 = 90.442 N
The vertical force acting on the officer is her weight:
Fw = m * g = 55.0 kg * 9.8 m/s2 = 539 N
Now, to find the angle θ relative to the vertical, use the tangent function, as the net force's components are the horizontal centripetal force and the vertical weight:
θ = tan-1(Fc / Fw) = tan-1(90.442 / 539)
θ ≈ 9.5°
Thus, the angle of the net force relative to the vertical is approximately 10° (to the nearest degree).
The coil in a generator has 100 windings and a cross sectional area of 0.01 m^2. If the coil rotates with constant speed in the magnetic field of the earth (B = 0.5x10^-4 T), how many revolutions per second must it complete to generate a maximum e.m.f. of one volt?
Using this result, is it practical to create a generator that uses only the earths magnetic field?
Answers
To develop the problem it is necessary to apply the concepts related to the induced voltage and its definition according to the magnetic field and angular velocity.
By definition the induced voltage or electromotive force is given by
[tex]\epsilon = BAN\omega[/tex]
Where,
B = Magnetic Field
A = Cross-sectional area
N = Number of turns
[tex]\omega[/tex]= Angular velocity
For the given case in the problem we will look for the angular velocity,
[tex]\omega = \frac{\epsilon}{BAN}\\\omega = \frac{1}{(0.5*10^{-4})(0.01)(100)}\\\omega = 2*10^4rad/s \\\omega = 2*10^4rad/s (\frac{1rev}{2\pi rad})\\\omega = 3183.098rev/s[/tex]
Therefore the number of revolutions is 3183.1rev/s.
Although the number of revolutions the magnetic field of the earth is discarded for use as a generator because the magnetic field of the earth compared to other current devices is considered weak. Very little power would be obtained from the generator
Identify the procedure to determine a formula for self-inductance, or inductance for short. Using the formula derived in the text, find the inductance in henries for a solenoid with 1500 loops of wire wound on a rod 13 cm long with radius 4 cm.
Answers
Answer:
L = 0.109 H
Explanation:
Given that,
Number of loops in the solenoid, N = 1500
Radius of the wire, r = 4 cm = 0.04 m
Length of the rod, l = 13 cm = 0.13 m
To find,
Self inductance in the solenoid
Solution,
The expression for the self inductance of the solenoid is given by :
[tex]L=\dfrac{\mu_o N^2 A}{l}[/tex]
[tex]L=\dfrac{4\pi \times 10^{-7}\times (1500)^2\times \pi (0.04)^2}{0.13}[/tex]
L = 0.109 H
So, the self inductance of the solenoid is 0.109 henries.
A hammer taps on the end of a 3.4-m-long metal bar at room temperature. A microphone at the other end of the bar picks up two pulses of sound, one that travels through the metal and one that travels through the air. The pulses are separated in time by 8.40 ms .What is the speed of sound in this metal?
Answers
Answer:
S = 2266.67 m/s
Explanation:
Given,
length of the metal = 3.4 m
pulses are separated in time = 8.4 ms
speed of sound in air= 343 m/s
speed of sound in this metal = ?
time taken
[tex]t = \dfrac{distance}{speed}[/tex]
[tex]t = \dfrac{3.4}{343}[/tex]
t = 9.9 ms
speed of sound in the metal is fast
t = 9.9 - 8.4 = 1.5 ms
time for which sound is in metal is equal to 1.5 ms
speed of sound in metal
[tex]speed= \dfrac{distance}{time}[/tex]
[tex]S = \dfrac{3.4}{1.5 \times 10^{-3}}[/tex]
S = 2266.67 m/s
Speed of sound in metal is equal to S = 2266.67 m/s
Final answer:
The speed of sound in the metal is approximately 404.76 m/s.
Explanation:
To determine the speed of sound in the metal, we need to use the time difference between the two pulses and the length of the metal bar. The time difference between the pulses is 8.40 ms and the length of the bar is 3.4 m. The speed of sound in the metal can be calculated using the formula: speed = distance / time. In this case, the distance is the length of the bar and the time is the time difference between the pulses.
So, by substituting the values into the formula, we get: speed = 3.4 m / (8.40 * 10^-3 s).
Simplifying the equation, we get: speed = 404.76 m/s.
Therefore, the speed of sound in the metal is approximately 404.76 m/s.
Jack and Jill have made up since the previous HW assignment, and are now playing on a 10 meter seesaw. Jill is sitting on one end of the seesaw, while Jack is some distance (d) from the fulcrum on the other side. If Jack and Jill have masses of 100 kg and 60 kg, respectively, where should Jack sit in order to balance the seesaw?
Answers
Answer: 3 m.
Explanation:
Neglecting the mass of the seesaw, in order the seesaw to be balanced, the sum of the torques created by gravity acting on both children must be 0.
As we are asked to locate Jack at some distance from the fulcrum, we can take torques regarding the fulcrum, which is located at just in the middle of the length of the seesaw.
If we choose the counterclockwise direction as positive, we can write the torque equation as follows (assuming that Jill sits at the left end of the seesaw):
mJill* 5m -mJack* d = 0
60 kg*5 m -100 kg* d =0
Solving for d:
d = 3 m.
The aorta carries blood away from the heart at a speed of about40 cm/s and has a radius of approximately 1.1 cm.
The aortabranches eventually into a large number of tiny capillaries thatdistribute the blood to the various body organs.
In a capillary,the blood speed is approximately 0.007 cm/s, and the radius isabout 6.0 10-4 cm.
Treat the blood as an incompressiblefluid, and use these data to determine the approximate number ofcapillaries in the human body.
Answers
Answer:
The no. of capillaries are [tex]1.92\times 10^{10}[/tex]
Solution:
As per the question:
Speed of the blood carried by the aorta, [tex]v_{a} = 40\ cm/s[/tex]
Radius of the aorta, [tex]R_{a} = 1.1 cm[/tex]
Speed of the blood in the capillaries, [tex]v_{c} = 0.007\ cm/s[/tex]
Radius of the capillaries, [tex]R_{c} = 6.0\times 10^{- 4} cm[/tex]
Now,
To determine the no. of capillaries:
Cross sectional Area of the Aorta, [tex]A_{a} = \pi R_{a}^{2} = \pi \times (1.1)^{2} = 1.21\pi \ m^{2}[/tex]
Cross sectional Area of the Capillary, [tex]A_{c} = \pi R_{c}^{2} = \pi \times (6.0\times 10^{- 4})^{2} = (3.6\times 10^{- 7})\pi \ m^{2}[/tex]
Let the no. of capillaries be 'n'
Also, the volume rate of flow in the aorta equals the sum total flow in the 'n' capillaries:
[tex]A_{a}v_{a} = nA_{c}v_{c}[/tex]
[tex]1.21\pi\times 40 = n\times 3.6\times 10^{- 7}\pi\times 0.007[\tex]
[tex]n = 1.92\times 10^{10}[/tex]
Using the principle of continuity, which states the constant volume flow rate of an incompressible fluid, we calculate that the human body has approximately ten billion capillaries.
Explanation:To determine the approximate number of capillaries in the human body based on the given data, we can use the principle of continuity, which states that the product of the cross-sectional area of a tube and the fluid speed through the tube is constant.
This implies that the blood flow (volume rate of flow) is the same in the aorta as in the capillaries, i.e., Aorta's cross-sectional area × speed of blood flow in Aorta = Capillary's cross-sectional area × speed of blood flow in capillaries × number of capillaries. From this, we can solve for the number of capillaries:
(π(1.1 cm)^2 × 40 cm/s) = (π(6.0 × 10^-4 cm)^2 × 0.007 cm/s) × number of capillaries.
When we do the calculations, we find that there are approximately ten billion capillaries in the human body, a vast network to ensure blood is delivered to every part of the body.
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A rectangular box with a volume of 784ft^3 is to be constructed with a square base and top. The cost per square foot for the bottom is 20 cents, for the top is 16 cents, and for the sides is 1.5 cents. What dimensions will minimize the cost? The length of one side of the base is ? The height of the box is?
Answers
Let's dimension the horizontal length with the name X and the vertical dimensions as Y.
In this way the total volume will be given under the function
[tex]V = x^2 y[/tex]
The cost for the bottom is given by [tex](x^2)(20)=20x^2[/tex]
While the cost for performing the top by [tex](x^2)(16) = 16x^2[/tex]
The cost for performing the sides would be given by [tex](1.5)(xy)(4) = 6xy[/tex]
Therefore the total cost would be
[tex]c_{total}= 36x^2 +6xy[/tex]
The total volume is equivalent to
[tex]784 = x^2y[/tex]
[tex]xy = \frac{784}{x}[/tex]
Replacing in our cost function
[tex]c_{total}= 36x^2 +6xy[/tex]
[tex]c_{total}= 36x^2 +6(\frac{784}{x})[/tex]
Obtaining the first derivative and equalizing to zero we will obtain the ideal measure, therefore
[tex]c' = 0[/tex]
[tex]c' = 72x-\frac{4707}{x^2}[/tex]
[tex]0= 72x-\frac{4707}{x^2}[/tex]
[tex]x = (\frac{523}{2})^{1/3}[/tex]
[tex]x = 6.3947ft[/tex]
Then,
[tex]784 = x^2y[/tex]
[tex]y = \frac{784}{x^2}[/tex]
[tex]y = \frac{784}{6.3947^2}[/tex]
[tex]y = 19.17ft[/tex]
In this way the measures of the base should be 6.3947ft (width and length) and the height of 19.17ft.
A barge floating in fresh water (rho = 1000 kg/m3) is shaped like a hollow rectangular prism with base area A = 550 m2 and height H = 2.0 m. When empty the bottom of the barge is located H0 = 0.55 m below the surface of the water. When fully loaded with coal the bottom of the barge is located H1 = 1.7 m below the surface.
Answers
Answer:
The buoyant force is 2964500 N.
Explanation:
Given that,
Density of water = 1000 kg/m³
Area = 550 m²
Height = 2.0 m
Depth = 0.55 m
Suppose we need to write an equation for the buoyant force on the empty barge in terms of the known data and find the value of force
We need to write the buoyant force equation
[tex]F_{b}=\rho\times V\times g[/tex]
The volume is
[tex]V=A\times h_{0}[/tex]
Put the value of volume into the buoyant force
[tex]F_{b}=\rho\times A\times h_{0}\times g[/tex]
We need to calculate the buoyant force
Put the value into the formula
[tex]F_{b}=1000\times550\times0.55\times9.8[/tex]
[tex]F_{b}=2964500\ N[/tex]
Hence, The buoyant force is 2964500 N.
The question describes a scenario involving Archimedes' Principle and the concept of density. It compares the submerged portions of a barge when it's empty versus when it's loaded. The difference in the submerged parts describes how the barge displaces more water to balance the increased weight.
Explanation:The phenomenon described in the question involves Archimedes' Principle and the concept of density, both of which are topics in Physics. The principle states that the buoyant force (upward force) that acts on an object submerged in a fluid is equal to the weight of the fluid the object displaces.
Initially, when the barge is empty, only a small part (H0 = 0.55 m) of it is submerged in the water, meaning it displaces a smaller volume of water equivalent to 0.55 m * 550 m². When it is loaded with coal, a larger part (H1 = 1.7 m) is submerged, and it displaces a larger volume of water (1.7 m * 550 m²).
The change in the submerged height of the barge when it's loaded compared to when it's empty can be attributed to increased weight which, to maintain equilibrium, is matched by the displacement of greater volume of water and hence increasing water's buoyant force.
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A glider of mass 5.0 kg hits the end of a horizontal rail and bounces off with the same speed, in the opposite direction. The collision is elastic and takes place in a time interval of 0.2s, with an average force of 100N. What was the speed, in m/s, of the glider
Answers
To solve the exercise it is necessary to apply the concepts given in Newton's second law and the equations of motion description.
Let's start by defining acceleration based on speed and time, that is
[tex]a = \frac{v}{t}[/tex]
On the other hand according to Newton's second law we have to
F=ma
where
m= Mass
a = Acceleration
Replacing the value of acceleration in this equation we have
[tex]F=m(\frac{v}{t})[/tex]
Substituting with our values we have
[tex]100N=(5Kg)\frac{v}{0.2}[/tex]
Re-arrange to find v
[tex]v=\frac{100*0.2}{5}[/tex]
[tex]v = 2m/s[/tex]
Therefore the speed of the glider is 2m/s
Diffraction effects become significant when the width of an aperture is comparable to the wavelength of the waves being diffracted. (a) At what speed will the de Broglie wavelength of a 65-kg student be equal to the 0.76-m width of a doorway? (b) At this speed, how much time will it take the student to travel a distance of 1.0 mm? (For comparison, the age of the universe is approximately 4 * 1017 s.)
Answers
To develop this problem it is necessary to apply the concepts related to Broglie hypothesis.
The hypothesis defines that
[tex]\lambda = \frac{h}{p}[/tex]
Where,
P = momentum
h = Planck's constant
The momentum is also defined as,
P = mv
Where,
m = mass
v = Velocity
PART A) Replacing at the first equation
[tex]\lambda = \frac{h}{mv}[/tex]
Our values are given as,
[tex]h = 6.626*10^{-34}Js[/tex]
[tex]m = 65Kg[/tex]
[tex]\lambda = 0.76m[/tex]
Re-arrange to find v, we have:
[tex]v = \frac{h}{m\lambda}[/tex]
[tex]v = \frac{6.626*10^{-34}}{65*0.76}[/tex]
[tex]v = 1.341*10^{-35}m/s[/tex]
PART B) From the kinematic equations of movement description we know that velocity is defined as displacement over a period of time, that is
[tex]v = \frac{x}{t}[/tex]
Re-arrange to find t,
[tex]t = \frac{d}{v}[/tex]
[tex]t = \frac{0.001}{ 1.341*10^{-35}}[/tex]
[tex]t = 7.455*10^{31}s[/tex]
[tex]7.455*10^{31} > 4*10^{17} \rightarrow[/tex]the age of the universe.
A tall glass cylinder is filled with a layer of water 17.0 cm deep, and floating on top of the water, a layer of oil 34.0 cm thick. The oil has a specific gravity of 0.900. What is the absolute pressure (in Pa) at the bottom of the cylinder? (Assume the atmospheric pressure is 1.013 ✕ 105 Pa. Round your answer to at least three significant figures.)
Answers
To develop this problem it is necessary to apply the law of Pascal. The pressure exerted on an incompressible and equilibrium fluid within a container of non-deformable walls is transmitted with equal intensity in all directions and at all points of the fluid. For which the pressure is defined as,
[tex]P = P_{atm}+P_{oil}+P_{water}[/tex]
The pressure of an object can be expressed by means of density, gravity and height
[tex]P = \rho*g*h[/tex]
Our values are given as,
[tex]g=9.8m/s^2\\h_w = 0.17m\\h_o = 0.34\\\gamma_g = 0.9\rightarrow \rho=0.9*10^{-3}[/tex]
Replacing we have to,
[tex]P = P_{atm}+P_{oil}+P_{water}[/tex]
[tex]P = P_{atm}+\rho_{oil}gh_{oil}+\rho_{water}gh_{water}[/tex]
[tex]P = 1.013*10^5+(0.9*10^3)(9.8)(0.34)+(10^3)(9.8)(0.17)[/tex]
[tex]P = 105964.8Pa[/tex]
Oil at 150°C flows slowly through a long, thin-walled pipe of 25-mm inner diameter. The pipe is suspended in a room for which the air temperature is 20°C and the convection coefficient at the outer tube surface is 11 W/m2·K. Estimate the heat loss per unit length of tube, in W/m.
Answers
Final answer:
The heat loss per unit length of the tube due to convection is approximately 141.7 W/m, calculated using the formula for heat transfer by convection with the provided temperatures and convection coefficient.
Explanation:
To estimate the heat loss per unit length of tube in W/m due to convection, we use the formula for heat transfer by convection which is Q = hA(Tsur - Tair). The convection coefficient 'h' is given as 11 W/m²·K, the temperature of the oil 'Tsur' is 150°C, and the air temperature 'Tair' is 20°C.
We need to find the surface area 'A' which for a pipe is calculated using A = π x d x L, where 'd' is the diameter and 'L' is the length of the pipe. Since we need the heat loss per unit length, L will be 1 meter.
Firstly, calculate the surface area per meter of the pipe: A = π x 0.025 m x 1 m = 0.0785 m². Then, plug these values into the convection formula to find heat loss per meter: Q = 11 W/m²·K x 0.0785 m² x (150°C - 20°C) = 141.7 W/m. Therefore, the heat loss per unit length of the tube is approximately 141.7 W/m.
A thin film of oil of thickness t is floating on water. The oil has index of refraction no = 1.4. There is air above the oil. When viewed from the air in the direction normal to the surface, there is constructive interference of reflected red light with l = 675 nm. Find the minimum thickness t of the oil film that will give constructive interference of reflected light with l = 675 nm.
Answers
Answer:
t = 120.5 nm
Explanation:
given,
refractive index of the oil = 1.4
wavelength of the red light = 675 nm
minimum thickness of film = ?
formula used for the constructive interference
[tex]2 n t = (m+\dfrac{1}{2})\lambda[/tex]
where n is the refractive index of oil
t is thickness of film
for minimum thickness
m = 0
[tex]2 \times 1.4 \times t = (0+\dfrac{1}{2})\times 675[/tex]
[tex]t = \dfrac{0.5\times 675}{2\times 1.4}[/tex]
t = 120.5 nm
hence, the thickness of the oil is t = 120.5 nm
Final answer:
The minimum thickness t of the oil film that will give constructive interference of reflected red light with a wavelength of 675 nm is 202.5 nm.
Explanation:
Constructive interference of reflected light occurs when the path difference between the two waves is equal to an integer multiple of the wavelength. In this case, we have constructive interference for red light with a wavelength of 675 nm.
The path difference can be represented as 2nt, where n is the index of refraction of the oil and t is the thickness of the oil film. Since we are viewing the film from air, we need to consider the difference in indices of refraction.
For constructive interference, the path difference must be equal to mλ, where m is the order of the interference. In this case, we have m = 1.
Using the formula for the path difference, we can set up the following equation:
2nt = (m + 1/2)λ
Plugging in the values, we have:
2(1.4)(t) = (1 + 1/2)(675 nm)
Simplifying, we get:
t = 202.5 nm
Therefore, the minimum thickness t of the oil film that will give constructive interference of reflected red light with a wavelength of 675 nm is 202.5 nm.
A 7.1 kg watermelon is placed at one end of a 4.8 m, 260 N scaffolding supported by two cables. One supporting cable is at the opposite end of the scaffolding, and the other is 0.53 m from the watermelon. How much tension is in the cable at the end of the scaffolding? The acceleration of gravity is 9.8 m/s 2 . Answer in units of N
Answers
The calculated tension is in the cable at the end of the scaffolding is 208.94 N.
First, let's calculate the weights acting on the system: the weight of the watermelon (Wm) and the weight of the scaffolding (Ws). Wm = mass of the watermelon × gravity = 7.1 kg × 9.8 m/s2 = 69.58 N, and Ws = 260 N given. Next, we choose the pivot point at the end of the scaffolding opposite the watermelon to find the tension in the cable at that end. The distance from the watermelon to the supporting cable is (4.8 m - 0.53 m) = 4.27 m, and the distance from the pivot to the scaffold's center of mass (assuming it's uniform) is half its length, or 2.4 m.
Applying the equilibrium condition for torques (Τclockwise = Τcounterclockwise), we have:
(69.58 N × 4.27 m) + (260 N × 2.4 m) = T × 4.8 m
⇒ T = ((69.58 N × 4.27 m) + (260 N × 2.4 m)) / 4.8 m
⇒ T = 208.94 N.
The tension in the cable at the end of the scaffolding where the watermelon is placed is [tex]{34.79 \text{ N}} \)[/tex].
Given:
- Mass of the watermelon, [tex]\( m = 7.1 \)[/tex] kg
- Acceleration due to gravity, [tex]\( g = 9.8 \)[/tex] m/s²
- Length of the scaffolding, [tex]\( L = 4.8 \)[/tex] m
- Distance of the watermelon from one end, [tex]\( d = 0.53 \)[/tex] m
1. Calculate the weight of the watermelon:
[tex]\[ W = mg = 7.1 \times 9.8 = 69.58 \text{ N} \][/tex]
2. Determine the tensions in the cables:
Since the system is in equilibrium, the tension [tex]\( T_1 \)[/tex] at the end where the watermelon is placed and the tension [tex]\( T_2 \)[/tex] at the opposite end satisfy:
[tex]\[ T_1 + T_2 = W \][/tex]
3. Find [tex]\( T_2 \)[/tex]:
From the equilibrium condition:
[tex]\[ T_2 = \frac{W}{2} = \frac{69.58}{2} = 34.79 \text{ N} \][/tex]
4. Calculate [tex]\( T_1 \)[/tex]:
Since [tex]\( T_1 + T_2 = W \)[/tex] and [tex]\( T_1 = -T_2 \)[/tex] (because [tex]\( T_1 \)[/tex] and [tex]\( T_2 \)[/tex] act in opposite directions):
[tex]\[ T_1 = W - T_2 = 69.58 - 34.79 = 34.79 \text{ N} \][/tex]
Therefore, the tension in the cable at the end of the scaffolding where the watermelon is placed is [tex]{34.79 \text{ N}} \)[/tex].
You (50 kg) are standing on a floating log (200 kg). Both are floating down a river at 1 m/s. The log points in the direction along the river. You walk in a direction that is down river for 5 seconds and your speed is 1.5 m/s as measured by a ground observer at the end of the 5 seconds.
a. What is the speed of the log at the end of the 5 seconds?
b. What is the average force between you and the log during those 5 seconds?
Answers
Answer:
0.875 m/s
5 N
Explanation:
[tex]m_1[/tex] = Mass of person = 50 kg
[tex]m_2[/tex] = Mass of log = 200 kg
[tex]v_1[/tex] = Velocity of person = 1.5 m/s
[tex]v_2[/tex] = Velocity of log
v = Velocity of log with respect to shore = 1 m/s
t = Time taken = 5 seconds
As the momentum of system is conserved we have
[tex](m_1+m_2)v=m_1v_1+m_2v_2\\\Rightarrow v_2=\frac{(m_1+m_2)v-m_1v_1}{m_2}\\\Rightarrow v_2=\frac{(50+200)1-50\times 1.5}{200}\\\Rightarrow v_2=0.875\ m/s[/tex]
Velocity of the log at the end of the 5 seconds is 0.875 m/s
Force is given by
[tex]F=\frac{m(v-u)}{t}\\\Rightarrow F=\frac{50(1.5-1)}{5}\\\Rightarrow F=5\ N[/tex]
The average force between you and the log during those 5 seconds is 5 N