Answer:
(A) object B has the largest mass because it has the least acceleration
(B) object C has the smallest mass because it has the largest acceleration
(C) mass A : mass B = 3 : 5
Explanation:
Given;
acceleration of object A = 5 m/s²
acceleration of object B = 3 m/s²
acceleration of object C = 7 m/s²
A constant force, F
According to Newton's second law of motion;
F = ma
m = F / a
Mass of object A:
m = F / 5
Mass of object B:
m = F / 3
Mass of object C:
m = F / 7
(A). Which object has the largest mass:
object B has the largest mass because it has the least acceleration
(B). Which object has the smallest mass:
object C has the smallest mass because it has the largest acceleration
(C). What is the ratio of mass A to mass B;
mass A = F / 5
mass B = F / 3
[tex]mass \ A : \ mass \ B = \frac{F}{5} : \frac{F}{3} \\\\\frac{mass \ A}{mass \ B} = \frac{F}{5} * \frac{3}{F}= \frac{3}{5} \\\\mass \ A : \ mass \ B = 3: 5[/tex]
A. The Object B has largest mass.
B. The Object A has smallest mass.
C. The ratio of mass A to mass B is, [tex]\frac{3}{5}[/tex]
Newton second law of motion:The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object.
[tex]F=ma\\\\m=\frac{F}{a}[/tex]
For constant force, mass is inversely proportional to acceleration of object.Given that, acceleration of object A is [tex]5m/s^{2}[/tex] and object B is [tex]3m/s^{2}[/tex]Thus, Object B has largest mass.Object A has smallest mass.the ratio of mass A to mass B is,[tex]\frac{m_{A}}{m_{B}} =\frac{a_{B}}{a_{A}} =\frac{3}{5}[/tex]
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Your lab instructor has asked you to measure a spring constant using a dynamic method—letting it oscillate—rather than a static method of stretching it. You and your lab partner suspend the spring from a hook, hang different masses, m, on the lower end, and start them oscillating. One of you uses a meter stick to measure the amplitude, A, and the other uses a stopwatch to time 10 oscillations, t. Your data are as follows:Mass, m(g) Amplitude, A(cm) Time, T(s) 100 6.5 7.8150 5.5 9.8200 6.0 10.9250 3.5 12.4Use the best-fit line of an appropriate graph to determine the spring constant.
Answer:
k = 6,547 N / m
Explanation:
This laboratory experiment is a simple harmonic motion experiment, where the angular velocity of the oscillation is
w = √ (k / m)
angular velocity and rel period are related
w = 2π / T
substitution
T = 2π √(m / K)
in Experimental measurements give us the following data
m (g) A (cm) t (s) T (s)
100 6.5 7.8 0.78
150 5.5 9.8 0.98
200 6.0 10.9 1.09
250 3.5 12.4 1.24
we look for the period that is the time it takes to give a series of oscillations, the results are in the last column
T = t / 10
To find the spring constant we linearize the equation
T² = (4π²/K) m
therefore we see that if we make a graph of T² against the mass, we obtain a line, whose slope is
m ’= 4π² / k
where m’ is the slope
k = 4π² / m'
the equation of the line of the attached graph is
T² = 0.00603 m + 0.0183
therefore the slope
m ’= 0.00603 s²/g
we calculate
k = 4 π² / 0.00603
k = 6547 g / s²
we reduce the mass to the SI system
k = 6547 g / s² (1kg / 1000 g)
k = 6,547 kg / s² =
k = 6,547 N / m
let's reduce the uniqueness
[N / m] = [(kg m / s²) m] = [kg / s²]
The spring-mass system forms a linear graph between the time period and mass. And the value of spring-constant from the given data is 6.46 N/m.
Given data:
Mass suspended by spring is, [tex]m=100 \;\rm g =0.1 \;\rm kg[/tex].
Number of oscillations is, [tex]n =10\;\rm oscillations[/tex].
Time period of oscillation is, [tex]T=7.8 \;\rm s[/tex].
The expression for the angular frequency of spring-mass system is,
[tex]\omega =\drac \sqrt{\dfrac{k}{m} }[/tex] ......................................................(1)
Here, k is the spring constant.
Angular frequency is also expressed as,
[tex]\omega = 2 \pi f[/tex] .........................................................(2)
here, f is the linear frequency of spring-mass system.
And linear frequency is,
[tex]f=\dfrac{n}{T}\\f=\dfrac{10}{7.81}\\f=1.28 \;\rm cycles/sec[/tex]
Then substitute equation (2) in equation (1) as,
[tex]2 \pi f=\drac \sqrt{\dfrac{k}{m} }\\2 \pi \times 1.28=\drac \sqrt{\dfrac{k}{0.1} }\\(2 \pi \times 1.28)^{2}= \dfrac{k}{0.1}\\k = 6.46 \;\rm N/m[/tex]
Thus, the value of spring constant is 6.46 N/m. And the suitable graph for the spring-mass system is given below.
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To practice Problem-Solving Strategy 6.1: Circular motion A highway curve with radius R = 274 m is to be banked so that a car traveling v = 25.0 m/s will not skid sideways even in the absence of friction. At what angle should the curve be banked?
Answer:
The curve should be banked at an angle of 13 degrees.
Explanation:
We have,
Radius of a highway curve is 274 m
Speed of car on this curve is 25 m/s
Let [tex]\theta[/tex] is the banking angle. On a banked curve, the angle of safe diving is given by following expression.
[tex]\tan\theta=\dfrac{v^2}{Rg}[/tex]
g = 10 m/s²
Plugging all the values in above formula,
[tex]\tan\theta=\dfrac{(25)^2}{274\times 9.8}\\\\\theta=\tan^{-1}\left(\dfrac{(25)^{2}}{274\times9.8}\right)\\\\\theta=13^{\circ}[/tex]
So, the curve should be banked at an angle of 13 degrees.
A heavy copper ball of mass 2 kg is dropped from a fiftieth-floor apartment window. Another one with mass 1 kg is dropped immediately after 1 second. Air resistance is negligible. The difference between the speeds of the two balls:__________.
a. increases over time at first, but then stays constant.
b. decreases over time.
c. remains constant over time.
d. increases over time.
Answer:
C
Explanation:
Because everything on Earth falls at the same speed, the masses of the balls do not matter. Since the acceleration due to gravity is constant, their speeds will both be increasing at the same rate, and therefore the difference in speeds would remain constant until they hit the ground. Hope this helps!
The friends now feel ready to try a problem. Suppose an Atwood machine has a mass of m1 = 2.5 kg and another mass of m2 = 8.5 kg hanging on opposite sides of the pulley. Assume the pulley is massless and frictionless, and the cord is ideal. Determine the magnitude of the acceleration of the two objects and the tension in the cord.
Answer:
a = 5.34 m/s²
T = 37.86 N
Explanation:
This is the case where two masses are hanging vertically on sides of the pulley. In such case, the formula for acceleration of objects is derived to be:
a = g(m₂ - m₁)/(m₂ + m₁)
where,
a = acceleration of both masses = ?
g = 9.8 m/s²
m₂ = heavier mass = 8.5 kg
m₁ = lighter mass = 2.5 kg
Therefore,
a = (9.8 m/s²)(8.5 kg - 2.5 kg)/(8.5 kg + 2.5 kg)
a = (9.8 m/s²)(6 kg)/(11 kg)
a = 5.34 m/s²
The formula for tension in cable is derived to be:
T = 2m₁m₂g/(m₁ + m₂)
T = (2)(2.5 kg)(8.5 kg)(9.8 m/s²)/(2.5 kg + 8.5 kg)
T = 37.86 N
Volume of an block is 5 cm3. If the density of the block is 250 g/cm3, what is the mass of the block ?
Answer:
1.25kg
Explanation:
Simply multiply volume and density together
250cm3 of fres
er of density 1000kgm-3 is mixed with 100cm3 of sea water of density 1030kgm-3. Calculate the density of the mixture. *
Answer:
1008.57kg/m3
Explanation:
Now the mass of fresh water is 250×1000 /1000000 = 0.25kg
Now the mass of salt water is
100×1030 /1000000 = 0.103kg
Note Density = mass / volume
Mass = volume × density
Note that converting from cm3 to m3 we divide by 1000000
Total mass = 0.25kg +0.103kg= 0.353kg.
Total volume also is (250 +100 )/1000000= 35 × 10^{-5}m3
Hence the density of the mixture= total mass / total volume
0.353kg/35 × 10^{-5}m3=1008.57kg/m3
What is the answer for this question
If the velocity of a runner changes from -2 m/s to -4 m/s over a period of time, the
runner's kinetic energy will become:
(a) four times as great as it was.
(b) half the magnitude it was.
(c) energy is conserved.
(d) twice as great as it was.
(e) four times less than it was.
Answer:
It will be A. So since its 2 times more the kinetic energy. But then you have to square it 2^2 = 4
wha is amplitde in sound
Answer:
The number of molecules displaced in a vibration makes the amplitude of a sound.
An airplane flies 2500 miles east in 245 seconds what is the velocity of the plane?
Speed = (distance) / (time)
Speed = (
Velocity = speed, and its direction
The velocity of the plane is 10.2 miles per second East.
(about 48 times the speed of sound)
To practice Problem-Solving Strategy 11.1 Equilibrium of a Rigid Body. A horizontal uniform bar of mass 2.7 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to the end of the bar, and string 2 is attached a distance 0.6 m from the other end. A monkey of mass 1.35 kg walks from one end of the bar to the other. Find the tension T1 in string 1 at the moment that the monkey is halfway between the ends of the bar.
Answer:
[tex]T_{1}[/tex] = 14.88 N
Explanation:
Let's begin by listing out the given variables:
M = 2.7 kg, L = 3 m, m = 1.35 kg, d = 0.6 m,
g = 9.8 m/s²
At equilibrium, the sum of all external torque acting on an object equals zero
τ(net) = 0
Taking moment about [tex]T_{1}[/tex] we have:
(M + m) g * 0.5L - [tex]T_{2}[/tex](L - d) = 0
⇒ [tex]T_{2}[/tex] = [(M + m) g * 0.5L] ÷ (L - d)
[tex]T_{2}[/tex] = [(2.7 + 1.35) * 9.8 * 0.5(3)] ÷ (3 - 0.6)
[tex]T_{2}[/tex]= 59.535 ÷ 2.4
[tex]T_{2}[/tex] = 24.80625 N ≈ 24.81 N
Weight of bar(W) = M * g = 2.7 * 9.8 = 26.46 N
Weight of monkey(w) = m * g = 1.35 * 9.8 = 13.23 N
Using sum of equilibrium in the vertical direction, we have:
[tex]T_{1}[/tex] + [tex]T_{2}[/tex] = W + w ------- Eqn 1
Substituting T2, W & w into the Eqn 1
[tex]T_{1}[/tex] + 24.81 = 26.46 + 13.23
[tex]T_{1}[/tex] = 14.88 N
ASK YOUR TEACHER A meter stick is found to balance at the 49.7-cm mark when placed on a fulcrum. When a 51.5-gram mass is attached at the 16.0-cm mark, the fulcrum must be moved to the 39.2-cm mark for balance. What is the mass of the meter stick
Answer:
0.114 kg or 114 g
Explanation:
From the diagram attaches,
Taking the moment about the fulcrum,
sum of clockwise moment = sum of anticlockwise moment.
Wd = W'd'
Where W = weight of the mass, W' = weight of the meter rule, d = distance of the mass from the fulcrum, d' = distance of the meter rule.
make W' the subject of the equation
W' = Wd/d'................ Equation 1
Given: W = mg = 0.0515(9.8) = 0.5047 N, d = (39.2-16) = 23.2 cm, d' = (49.7-39.2) = 10.5 cm
Substitute these values into equation 1
W' = 0.5047(23.2)/10.5
W' = 1.115 N.
But,
m' = W'/g
m' = 1.115/9.8
m' = 0.114 kg
m' = 114 g
You're driving a vehicle of mass 850 kg and you need to make a turn on a flat road. The radius of curvature of the turn is 80 m. The maximum horizontal component of the force that the road can exert on the tires is only 0.22 times the vertical component of the force of the road on the tires (in this case the vertical component of the force of the road on the tires is , the weight of the car, where as usual 9.8 N/kg, the magnitude of the gravitational field near the surface of the Earth). The factor 0.22 is called the "coefficient of friction" (usually written "", Greek "mu") and is large for surfaces with high friction, small for surfaces with low friction.(a) What is the fastest speed you can drive and still make it around the turn? Invent symbols for the various quantities and solve algebraically before plugging in numbers.
maximum speed = ____ m/s
(b) Which of the following statements are true about this situation?
The net force is nonzero and points away from the center of the kissing circle.The momentum points toward the center of the kissing circle.The net force is nonzero and points toward the center of the kissing circle.The rate of change of the momentum is nonzero and points toward the center of the kissing circle.The centrifugal force balances the force of the road, so the net force is zero.The rate of change of the momentum is nonzero and points away from the center of the kissing circle.
(c) Look at your algebraic analysis and answer the following question. Suppose your vehicle had a mass 3 times as big (5250 kg). Now what is the fastest speed you can drive and still make it around the turn?
maximum speed = ____ m/s
(d) Look at your algebraic analysis and answer the following question. Suppose you have the original 1750 kg vehicle but the turn has a radius twice as large (166 m). What is the fastest speed you can drive and still make it around the turn?
maximum speed = ____m/s
Answer:
(a) v = 13.13 m/s
(b) The centrifugal force balances the force of the road, so net force is zero.
(c) v = 13.13 m/s
(d) v = 18.92 m/s
Explanation:
(a)
To make it around the turn without skidding the frictional force on cat must balance the centrifugal force. Therefore:
Frictional Force = Centrifugal Force
μR = mv²/r
where,
R = Normal Reaction = Weight of Car = mg
Therefore,
μmg = mv²/r
μg = v²/r
v = √μgr
where,
v = maximum possible velocity of car = ?
μ = coefficient of friction = 0.22
g = 9.8 m/s²
r = radius of curvature = 80 m
Therefore,
v = √[(0.22)(9.8 m/s²)(80 m)
v = 13.13 m/s
(b)
In order for the car to move without skidding around the turn, all the forces in horizontal direction must be equal. Hence, the centrifugal force and the frictional force (force of the road) must balance each other. So the true statement is:
The centrifugal force balances the force of the road, so net force is zero.
(c)
v = √μgr
Since the formula for speed is independent of mass. Therefore, the speed will remain same.
v = 13.13 m/s
(d)
v = √μgr
v = √[(0.22)(9.8 m/s²)(166 m)
v = 18.92 m/s
Suppose a stone is thrown vertically upward from the edge of a cliff on Mars (where the acceleration due to gravity is only about 12 ft/s2 with an initial velocity of 64 ft/s from a height of 192 ft above the ground. The height s of the stone above the ground after t seconds is given by
s=−6t2+64t+192
a. Determine the velocity v of the stone after t seconds. b. When does the stone reach its highest point? c. What is the height of the stone at the highest point? d. When does the stone strike the ground? e. With what velocity does the stone strike the ground?
Answer:
a) v = -12t + 64
b) t = 5.33s
c) s = 362.66ft
d) t = 13.10s
e) v = 93.2ft/s
Explanation:
You have the following equation for the height of a stone thrown in Mars:
[tex]s(t)=-6t^2+64t+192[/tex] (1)
a) The velocity of the stone after t seconds is obtained with the derivative of s in time:
[tex]v=\frac{ds}{st}=-12t+64[/tex] (2)
The equation for the speed of the stone is v = -12t + 64
b) The highest point is obtained when the speed of the stone is zero. Then, from the equation (2) equal to zero, you can obtain the time when the stone is at its maximum height:
[tex]-12t+64=0\\\\t=5.33s[/tex]
The time in which the stone is at the maximum height is 5.33s
c) For this time the stone is at the maximum height. Then, you replace t in the equation (1):
[tex]s(1)=-6(5.33)^2+64(5.33)+192=362.66ft[/tex]
the maximum height is 362.66 ft
d) To find the time when the stone arrive to the ground you equal the equation (1) to zero and you solve for t:
[tex]0=-6t^2+64t+192[/tex]
you use the quadratic formula:
[tex]t_{1,2}=\frac{-64\pm\sqrt{64^2-4(-6)(192)}}{2(-6)}\\\\t_{1,2}=\frac{-64\pm 93.29}{-12}\\\\t_1=13.10s\\\\t_2=-2.44s[/tex]
You use the result with positive values because is the onlyone with physical meaning.
The time for the stone hits the ground is 13.10 s
e) You replace 13.10s in the equation (2) to obtain the velocity of the stone when it strike the ground:
[tex]v=-12t+64=-12(13.10)+64=-93.2\frac{ft}{s}[/tex]
The minus sign is because the stone's direction is downward.
The speed of the stone just when it strikes the ground is 93.2ft/s
You have just landed on Planet X. You take out a ball of mass 100 gg , release it from rest from a height of 16.0 mm and measure that it takes a time of 2.90 ss to reach the ground. You can ignore any force on the ball from the atmosphere of the planet. How much does the 100-g ball weigh on the surface of Planet X?
Answer:
0.173 N.
Explanation:
We will calculate the mass and then use the following calculations on the surface of planet X that is :
[tex]W=mg[/tex]
We would use the following equation to get the value of g for planet X that is :
[tex]y_f-y_i=v_{yi}t+\frac{1}{2}gt^2[/tex]
Then, put the values in the above equation.
[tex]16=0+\frac{1}{2}\times g\times(2.90)^2[/tex]
[tex]\bf\mathit{g=3.80\;m/s^2}[/tex]
Now, we will measure the ball weight on planet X's surface:
[tex]m=\frac{100}{1000} \;\;\;\;\;\;\;\;\;\;[1kg=1000g][/tex]
Then, we have to put the value in the above equation.
[tex]W=0.1\times 1.73=0.173\:N[/tex]
A cylindrical shell of radius 7.00 cm and length 2.44 m has its charge uniformly distributed on its curved surface. The magnitude of the electric field at a point 21.9 cm radially outward from its axis (measured from the midpoint of the shell) is 36.0 kN/C.(a) Use approximate relationships to find thenet charge on the shell.
(b) Use approximate relationships to find theelectric field at a point 4.00 cm from theaxis, measured radiallyoutward from the midpoint of the shell.
Answer:
(b) Use approximate relationships to find theelectric field at a point 4.00 cm from the axis, measured radiallyoutward from the midpoint of the shell.
Shrinking Loop. A circular loop of flexible iron wire has an initial circumference of 165 cmcm , but its circumference is decreasing at a constant rate of 14.0 cm/scm/s due to a tangential pull on the wire. The loop is in a constant uniform magnetic field of magnitude 0.800 TT , which is oriented perpendicular to the plane of the loop. Assume that you are facing the loop and that the magnetic field points into the loop.
(a) Find the emf induced in the loop at the instant when 9.0 s have passed.
(b) Find the direction of the induced current in the loop as viewed looking along the direction of the magnetic field.
Answer:
(a) emf = 1.18 mV
(b) counter-clockwise sense
Explanation:
(a) The induced emf is given by the following formula:
[tex]emf=-\frac{d\Phi_B}{dt}[/tex] (1)
where:
ФB: magnetic flux = AB = (area of the loop)*(magnitude of the magnetic field)
A = πr^2
B = 0.800 T
You replace the expression for the magnetic flux in the equation (1):
[tex]emf=-B\frac{\Delta A}{\Delta t}=-B\frac{A_2-A_1}{t_2-t_1}[/tex]
A1: initial area
A2: final area
t2-t1: time interval = 9.0s
Then you have to calculate the change in the area of the loop, by using the information about the circumference of the loop. First you calculate the radius of the loop for a circumference of 165 cm = 1.65m
[tex]s=1.65m=2\pi r\\\\r=\frac{1.65m}{2\pi}=0.262m[/tex]
You calculate the initial area A1:
[tex]A_1=\pi (0.262m)^2=0.215m^2[/tex]
After 9.0 second the circumference will be:
[tex]s'=1.65m-0.14\frac{m}{s}(9.0s)=0.39m[/tex]
the new radius and the final area is:
[tex]r=\frac{0.39m}{2\pi}=0.062m[/tex]
[tex]A_2=\pi(0.062m)^2=0.012m^2[/tex]
Finally, you replace in the equation (1):
[tex]emf=-(0.800T)\frac{0.012m^2-0.215m^2}{9.0s}=1.8*10^{-3}V=1.8mV[/tex]
The induced emf in the circular loop is 1.18mV
(b) The induced emf generates an electric current, which produces a magnetic field that is opposite to the direction of the constant magnetic field of 0.800T. Due to this magnetic field point into the loop. The current has to have a direction in a counter-clockwise sense.
A politician holds a press conference that is televised live. The sound picked up by the microphone of a TV news network is broadcast via electromagnetic waves and heard by a television viewer. This viewer is seated 2.7 m from his television set. A reporter at the press conference is located 5.5 m from the politician, and the sound of the words travels directly from the celebrity's mouth, through the air, and into the reporter's ears. The reporter hears the words exactly at the same instant that the television viewer hears them. Using a value of 343 m/s for the speed of sound, determine the maximum distance between the television set and the politician. Ignore the small distance between the politician and the microphone. In addition, assume that the only delay between what the microphone picks up and the sound being emitted by the television set is that due to the travel time of the electromagnetic waves used by the network.
Answer:
Explanation:
Time taken by sound waves to cover distance between politician and reporter = time taken by em waves to travel distance between politician and the television viewer.
5.5 / 343 = d / 3 x 10⁸ + 2.7 / 343
d is distance between politician and television set + time taken by sound to travel distance between television and its viewer.
.0160349 = d / 3 x 10⁸ + .0078717
d / 3 x 10⁸ = .0081632
d = 2448960 m
= 2448.96 km
= 2449 km .
A 2 kg object is subjected to three forces that give it an acceleration −→a = −(8.00m/s^2)ˆi + (6.00m/s^2)ˆj. If two of the three forces, are −→F1 = (30.0N)ˆi + (16.0N)ˆj and −→F2 = −(12.0N)ˆi + (8.00N)ˆj, find the third force.
Answer:
[tex]\vec{F_3}=(-34.0N)\hat{i}+(-12.0N)\hat{j}[/tex]
Explanation:
You have three forces F1, F2 an F3 that produce the following acceleration:
a = −(8.00m/s^2)ˆi + (6.00m/s^2)ˆj
you know that force F1 and F2 are:
F1 = (30.0N)ˆi + (16.0N)ˆj
F2 = −(12.0N)ˆi + (8.00N)ˆj
and the force F3 is unknown:
F3 = F3x ˆi + F3y ˆj
The second Newton law is given by the following equation:
[tex]\vec{F}=m\vec{a}[/tex]
F: the total force = F1 +F2 + F3
m: mass of the object = 2 kg
By the properties of vectors you have:
[tex]\vec{F_1}+\vec{F_2}+\vec{F_3}=m\vec{a}\\\\(30.0-12.0+F_{3x})N\hat{i}+(16.0+8.00+F_{3y})N\hat{j}=(2.0kg)[(-8.00m/s^2)\hat{i}+(6.00m/s^2)\hat{j}]\\\\(18.0+F_{3x})N\hat{i}+(24.0+F_{3y})\hat{j}=(-16.00N)\hat{i}+(12.0N)\hat{j}[/tex]
Both x and y component must be equal in the previous equality, then you have:
[tex]18.0N+F_{3x}=-16.00N\\\\F_{3x}=-34.00N\\\\24.0N+F_{3y}=12.0N\\\\F_{3y}=-12.00N[/tex]
Hence, the vector F3 is:
[tex]\vec{F_3}=(-34.0N)\hat{i}+(-12.0N)\hat{j}[/tex]
If the radius of curvature of the cornea is 0.75 cm when the eye is focusing on an object 36.0 cm from the cornea vertex and the indexes of refraction are as described before, what is the distance from the cornea vertex to the retina
Answer:
The distance from the cornea vertex to the retina is 2.36 cm
Explanation:
The question is incomplete.
The complete question is as follows;
A Nearsighted Eye. A certain very nearsighted person cannot focus on anything farther than 36.0 cm from the eye. Consider the simplified model of the eye. In a simplified model of the human eye, the aqueous and vitreous humors and the lens all have a refractive index of 1.40, and all the refraction occurs at the cornea, whose vertex is 2.60 cm from the retina.
If the radius of curvature of the cornea is 0.65 cm when the eye is focusing on an object 36.0 cm from the cornea vertex and the indexes of refraction are as described before, what is the distance from the cornea vertex to the retina?
Solution.
We use image-object reaction to calculate the distance from the cornea vertex to the retina.
Mathematically;
n1/s + n2/s’ = n2-n1/R
From the question, we identify the following;
n1 ; Refractive index of air = 1
n2 ; Refractive index of lens = 1.4
S ; Object Distance = 36 cm
S’ = ?
R ; Radius of curvature of the cornea = 0.65
Substituting these values into the equation above;
1/36 + 1.4/S’ = (1.4-1)/0.65
{S’+ 36(1.4)}/36S’ = 0.4/0.65
{S’ + 50.4}/36S’ = 0.62
S’ + 50.4 = 22.32S’
50.4 = 22.32S’ -S’
21.32S’ = 50.4
S’ = 50.4/21.32
S’ = 2.36 cm
A hydraulic lift is made by sealing an ideal fluid inside a container with an input piston of cross-sectional area 0.004 m2 , and an output piston of cross-sectional area 1.2 m2 . The pistons can slide up or down without friction while keeping the fluid sealed inside. What is the maximum weight that can be lifted when a force of 60 N is applied to the input piston
Answer:
Maximum weight that can be lifted = 18,000 N
Explanation:
Given:
Cross-sectional area of input (A1) = 0.004 m²
Cross-sectional area of the output (A2) = 1.2 m ²
Force (F) = 60 N
Computation:
Pressure on input piston (P1) = F / A1
Assume,
Maximum weight lifted by piston = W
Pressure on output piston (P2) = W / A2
We, know that
P1 = P2
[F / A1] = [W / A2]
[60 / 0.004] = [W / 1.2]
150,00 = W / 1.2
Weight = 18,000 N
Maximum weight that can be lifted = 18,000 N
A skater wearing in – line skates (no friction) is standing in the middle of the aisle inside a bus and is not holding on to anything. Which way would the skater move in reaction to the bus as it pulls away from the bus stopA skater wearing in – line skates (no friction) is standing in the middle of the aisle inside a bus and is not holding on to anything. Which way would the skater move in reaction to the bus as it pulls away from the bus stop
Before the bus starts moving, the bus and the skater are both standing still.
When the bus starts moving and pulls away from the bus-stop, the skater stays right where she is.
The people outside on the sidewalk see her standing still, and they see the bus moving out from under her.
The other passengers on the bus see her rolling backwards down the aisle, toward the back of the bus.
The driver of a train moving at 23m/s applies the breaks when it pases an amber signal. The next signal is 1km down the track and the train reaches it 76s later. The acceleration is -0.26s^2. Find its speed at the next signal.
Answer:
3.2 m/s
Explanation:
Given:
Δx = 1000 m
v₀ = 23 m/s
a = -0.26 m/s²
t = 76 s
Find: v
This problem is over-defined. We only need 3 pieces of information, and we're given 4. There are several equations we can use. For example:
v = at + v₀
v = (-0.26 m/s²) (76 s) + (23 m/s)
v = 3.2 m/s
Or:
Δx = ½ (v + v₀) t
(1000 m) = ½ (v + 23 m/s) (76 s)
v = 3.3 m/s
Or:
v² = v₀² + 2aΔx
v² = (23 m/s)² + 2(-0.26 m/s²)(1000 m)
v = 3.0 m/s
Or:
Δx = vt − ½ at²
(1000 m) = v (76 s) − ½ (-0.26 m/s²) (76 s)²
v = 3.3 m/s
As you can see, you get slightly different answers depending on which variables you use. Since 1000 m has 1 significant figure, compared to the other variables which have 2 significant figures, I recommend using the first equation.
b. A locomotive of a train exerts a constant force of 280KN on a train while pulling
it at 50 km/h along a level track. What is:
[4 marks)
i. Workdone in quarter an hour and
[4 marks]
Answer:
Work-done in quarter an hour = 3.5 × 10⁶ J
Explanation:
Given:
Force (F) = 280 KN = 280,000 N
Velocity (V) = 50 km / h
Time (t) = 1 / 4 = 0.25 hour
Find:
Work-done in quarter an hour
Computation:
⇒ Displacement = Velocity (V) × Time
⇒ Displacement = 50 × 0.25
⇒ Displacement = 12.5 km
⇒ Work-done = Force (F) × Displacement
⇒ Work-done in quarter an hour = 280,000 × 12.5
⇒ Work-done in quarter an hour = 3,500,000
Work-done in quarter an hour = 3.5 × 10⁶ J
An airplane takes off a runway at a constant speed of 49m/s at constant angle 30 to the horizontal
Complete Question
An airplane takes off a runway at a constant speed of 49 m/s at constant angle 30 to the horizontal.How high (in meters ) is the airplane above the ground 13 seconds after takeoff?
Answer:
The height is [tex]H = 318.5 \ m[/tex]
Explanation:
From the question we are told that
The speed at which the plane takes off is [tex]u = 49 \ m/s[/tex]
The angle at which it takes off is [tex]\theta = 30 ^o[/tex]
The time taken is [tex]t = 13 s[/tex]
The vertical distance traveled is mathematically represented as
[tex]H = u sin \theta t[/tex]
Substituting values
[tex]H = (49) * sin (30) *13[/tex]
[tex]H = 318.5 \ m[/tex]
g Tether ball is a game children play in which a ball hangs from a rope attached to the top of a tall pole. The children hit the ball, causing it to swing around the pole. What is the total initial acceleration of a tether ball on a 2.0 m rope whose angular velocity changes from 13 rad/s to 7.0 rad/s in 15 s
Answer:
a = -0.8 m/s²
Here, negative sign indicates that the acceleration has opposite direction to the direction of motion.
Explanation:
First we find the angular acceleration of the ball from the following formula:
α = (ωf - ωi)/t
where,
α = angular acceleration = ?
ωf = final angular velocity = 7 rad/s
ωi = initial angular velocity = 13 rad/s
t = Time taken = 15 s
Therefore,
α = (7 rad/s - 13 rad/s)/15 s
α = - 0.4 rad/s
negative sign shows that acceleration is in opposite direction to the direction of motion.
Now, for the linear acceleration, we use the formula:
a = rα
where,
a = linear acceleration = ?
r = radius of circular path = length of rope = 2 m
therefore,
a = (2 m)(- 0.4 rad/s²)
a = -0.8 m/s²
Here, negative sign indicates that the acceleration has opposite direction to the direction of motion.
Convert from standard form to scientific notation:
0.00000013
A)1.3 x 10-7
B)13 x 108
C)1.3 x 107
D)13 x 10-8
In which situation is chemical energy being converted to another form of energy?
Answer:
A burning candle. (chemical energy into energy of heat and light, i.e. thermal and wave)
Explanation:
You rub a balloon on your head and it becomes negatively charged. The balloon will be most attracted to what?
Answer:
To things that are positive charged
A 148 g ball is dropped from a tree 11.0 m above the ground. With what speed would it hit the ground
Answer:
14.68m/s
Explanation:
As per the question, the data provided is as follows
Mass = M = 0.148 kg
Height = h = 11 m
Initial velocity = U = 0 m/s
Final velocity = V
Gravitational force = F
Mass = M
Based on the above information, the speed that hit to the ground is
As we know that
Work to be done = Change in kinetic energy
[tex]F ( S) = (\frac{1}{2} ) M ( V^2 - U^2 )[/tex]
[tex]M g h = (\frac{1}{2} ) M ( V^2 - U^2 )[/tex]
[tex]g h = (\frac{1}{2} ) ( V^2 - U^2 )[/tex]
[tex]V^2 - U^2 = 2gh[/tex]
[tex]V^2 - 0 = 2gh[/tex]
[tex]V = \sqrt{2 g h}[/tex]
[tex]= \sqrt{2\times9.8\times11}[/tex]
= 14.68m/s