Answer:
8 months needed to grow one inch in diameter by water oak and it grows 24 inches of height in a year.
Explanation:
In 2 different ways the trunk of a tree grows first the height than second in diameter usually in diameter tree grows one ring per year. We can say on counting the number of rings we can determine the age of a tree both of the growth does not occur in the same rate.
Tree grows more higher as compared to the diameter growth and mature trees usually grows 1 inch in diameter every year.
Water oak gets 24 inches height growth every year and 1.5 inch growth in diameter annually means if we divide 1.5 inches by 12 months we gets 0.125 inches growth monthly.so we can say 8 months are needed to grow the diameter of water oak 1 inch.
Two resistors of 5.0 and 9.0 ohms are connected inparallel. A
4.0 Ohm resistor is then connected in series withthe parallel
combination. A 6.0V battery is then connected tothe series-parallel
combination. What is the current throughthe 9.0 ohm resistor?
Answer:
The current through [tex]9 \Omega[/tex] is 0.297 A
Solution:
As per the question:
[tex]R_{5} = 5.0 \Omega[/tex]
[tex]R_{9} = 9.0 \Omega[/tex]
[tex]R_{4} = 5.0 \Omega[/tex]
V = 6.0 V
Now, from the given circuit:
[tex]R_{5}[/tex] and [tex]R_{9}[/tex] are in parallel
Thus
[tex]\frac{1}{R_{eq}} = \frac{1}{R_{5}} + \frac{1}{R_{9}}[/tex]
[tex]R_{eq} = \frac{R_{5}R_{9}}{R_{5} + R_{9}}[/tex]
[tex]R_{eq} = \frac{5.0\times 9.0}{5.0 + 9.0} = 3.2143 \Omega[/tex]
Now, the [tex]R_{eq}[/tex] is in series with [tex]R_{4}[/tex]:
[tex]R'_{eq} = R_{eq} + R_{4} = 3.2143 + 4.0 = 7.2413 \Omega[/tex]
Now, to calculate the current through [tex]R_{9}[/tex]:
[tex]V = I\times R'_{eq}[/tex]
[tex]I = {6}{7.2143} = 0.8317 A[/tex]
where
I = circuit current
Now,
Voltage across [tex]R_{eq}[/tex], V':
[tex]V' = I\times R_{eq}[/tex]
[tex]V' = 0.8317\times 3.2143 = 2.6734 V[/tex]
Now, current through [tex]R_{9}[/tex], I' :
[tex]I' = \frac{V'}{R_{9}}[/tex]
[tex]I' = \frac{2.6734}{9.0} = 0.297 A[/tex]
Final answer:
To find the current through the 9.0 ohm resistor in a series-parallel circuit, we can calculate the equivalent resistance of the parallel combination.
Explanation:
To find the current through the 9.0 ohm resistor, we need to first determine the equivalent resistance of the circuit. The two resistors of 5.0 and 9.0 ohms that are connected in parallel have an equivalent resistance given by the formula:
1/Req = 1/R1 + 1/R2
1/Req = 1/5.0 + 1/9.0
1/Req = (9.0 + 5.0)/(5.0 * 9.0)
1/Req = 14.0/45.0
Req = 45.0/14.0 ≈ 3.21 ohms
The equivalent resistance of the parallel combination is approximately 3.21 ohms.
On a distant planet, golf is just as popular as it is on earth. A golfer tees off and drives the ball 3.00 times as far as he would have on earth, given the same initial velocities on both planets. The ball is launched at a speed of 44.4 m/s at an angle of 25° above the horizontal. When the ball lands, it is at the same level as the tee. On the distant planet, what are (a) the maximum height and (b) the range of the ball?
Answer:
(a) Maximum height = 53.88 meters
(b) Range of the ball = 924.36 meters
Explanation:
The ball has been launched at a speed = 44.4 meters per second
Angle of the ball with the horizontal = 25°
Horizontal component of the speed of the ball = 44.4cos25° = 40.24 meters per second
Vertical component = 44.4sin25° = 18.76 meters per second
We know vertical component of the speed decides the height of the ball so by the law of motion,
v² = u² - 2gh
where v = velocity at the maximum height = 0
u = initial velocity = 18.76 meter per second
g = gravitational force = 9.8 meter per second²
Now we plug in the values in the given equation
0 = (18.76)² - 2(9.8)(h)
19.6h = 352.10
h = [tex]\frac{352.10}{19.6}[/tex]
h = 17.96 meters
By another equation,
[tex]v=ut-\frac{1}{2}gt^{2}[/tex]
Now we plug in the values again
[tex]0=(18.76)t-\frac{1}{2}(9.8)t^{2}[/tex]
18.76t = 4.9t²
18.76 = 4.9t
t = [tex]\frac{18.76}{4.9}=3.83[/tex]seconds
Since time t is the time to cover half of the range.
Therefore, time taken by the ball to cover the complete range = 2×3.83 = 7.66 seconds
So the range of the ball = Horizontal component of the velocity × time
= 40.24 × 7.66
= 308.12 meters
This we have calculated all for our planet.
Now we take other planet.
(a) Since the golfer drives the ball 3 times as far as he would have on earth then maximum height achieved by the ball = 17.96 × 3 = 53.88 meters
(b) Range of the ball = 3×308.12 = 924.36 meters
An oil bath maintained at 50.5°C loses heat to its surroundings at the rate of 4.68 kJ/min. Its temperature is maintained by an electrically heated coil with a resistance of 60 operated from a 110 V line. A thermoregulator switches the current on and off. What fraction of the time will the current be turned on?
Answer:
The fraction of time for turn on is 0.3852
Solution:
As per the question:
Temperature at which oil bath is maintained, [tex]T_{o} = 50.5^{\circ}[/tex]
Heat loss at rate, q = 4.68 kJ/min
Resistance, R = [tex]60\Omega[/tex]
Operating Voltage, [tex]V_{o} = 110 V[/tex]
Now,
Power that the resistor releases, [tex]P_{R} = \frac{V_{o}^{2}}{R}[/tex]
[tex]P_{R} = \frac{110^{2}}{60} = 201.67 W = 12.148 J/min[/tex]
The fraction of time for the current to be turned on:
[tex]P_{R} = \frac{q}{t}[/tex]
[tex]12.148 = \frac{4.68}{t}[/tex]
t = 0.3852
You have two square metal plates with side length of 16.50 cm. You want to make a parallel-plate capacitor that will hold a charge of 18.5 nC when connected to a 37.8 V potential difference. Determine the necessary separation in mm. Round your answer to three significant figures.
Answer:
d = 3.44 *10^{-7} m
Explanation:
given data:
length of metal plates = 16.50 cm
capacitor charge = 18.5 nC
potential difference = 37.8 V
capacitance of parallel plate capacitor
[tex]C = \frac{A\epsilon _{0}}{d}[/tex]
area of the individual plate
A=[tex] a^2 = (16.5*10^{-2})^2 = 272.25 *10^{-4}[/tex] m2
capacitance
[tex]C = QV = 18.5 *10^{-9} *37.8 = 699.3 * 10^{-9} C[/tex]
separation between plates d is given as[tex] = \frac{A\epsilon _{0}}{C }[/tex]
[tex]d = \frac{272.25 *10^{-4} *8.85*10^{-12}}{699.3 *10^-9}[/tex]
d = 3.44 *10^{-7} m
A vector has an x-component of -26.5 and a y-component of 43 units. Find magnitude and direction of vector
Answer:
The vector has a magnitude of 33.86 units and a direction of 121.64°.
Explanation:
To find the magnitude, you use the Pitagorean theorem:
[tex]||V|| = \sqrt{x^2 + y^2}= \sqrt{(-26.5)^2 + (43)^2} = 33.86 units[/tex]
In order to find the direction, you can use trigonometry. You have to keep in mind, that as the y component of the vector is positive and the x component is negative, the vector must have an angle between 90 and 180°, or in the second quadrant of the plane.:
[tex]tan(\alpha) = \frac{y}{x} \\\alpha = tan^{-1}(\frac{43}{-26.5})= 180 - tan^{-1}(\frac{43}{26.5}) = 121.64[/tex]° or, 58.36° in the second quadrant
A plane traveling north at 100.0 km/h through the air gets caught in a 40.0 km/h crosswind blowing west. This turbulence caused a beverage cart to brake free and begin rolling at 20.0 km/h toward the tail of the plane. What is the velocity of the cart relative to the ground? (you do not have to convert these since they are all the same unit)
The velocity of the cart relative to the ground is approximately 89.44 km/h in a direction about 63.43 degrees north of west.
To find the velocity of the cart relative to the ground when a plane is traveling north at 100.0 km/h with a 40.0 km/h crosswind blowing west and the cart is rolling at 20.0 km/h towards the tail of the plane, we can use vector addition to determine the resultant velocity.
1. First, break down the velocities into their horizontal (west-east) and vertical (north-south) components:
- Plane's velocity (north): 100.0 km/h
- Crosswind velocity (west): 40.0 km/h
- Cart's velocity (towards tail): 20.0 km/h
2. The horizontal component of the cart's velocity is the crosswind velocity (40.0 km/h), and the vertical component is its velocity towards the tail of the plane (20.0 km/h).
3. To find the resultant velocity, we can use vector addition by adding the horizontal and vertical components of the velocities separately:
Horizontal component: 40.0 km/h (west) - 0 km/h (east) = 40.0 km/h (west)
Vertical component: 100.0 km/h (north) - 20.0 km/h (south) = 80.0 km/h (north)
4. Now, we can use the Pythagorean theorem to find the magnitude of the resultant velocity:
Resultant velocity = √(40.0^2 + 80.0^2)
Resultant velocity = √(1600 + 6400)
Resultant velocity = √8000
Resultant velocity ≈ 89.44 km/h
5. To find the direction of the resultant velocity, we can use trigonometry:
Direction = arctan(vertical component / horizontal component)
Direction = arctan(80.0 / 40.0)
Direction = arctan(2)
Direction ≈ 63.43 degrees north of west
6. Therefore, the velocity of the cart relative to the ground is approximately 89.44 km/h in a direction about 63.43 degrees north of west.
A 1 kg particle moves upward from the origin to (23) m. Wit is work done by the force of gravity which is in - y direction s B. 19,6 D. 29.4) A.-19.6 J C. -29.4 J
Answer:
Explanation:
mass, m = 1 kg
Position (2, 3 ) m
height, h = 2 m
acceleration due to gravity, g = 9.8 m/s^2
Here, no force is acting in horizontal direction, the force of gravity is acting in vertical direction, so the work done by the gravitational force is to be calculated.
Force mass x acceleration due to gravity
F = 1 x 9.8 = 9.8 N
Work = force x displacement x CosФ
Where, Ф be the angle between force vector and the displacement vector.
Here the value of Ф is 180° as the force acting vertically downward and the displacement is upward
So, W = 9.8 x 2 x Cos 180°
W = - 19.6 J
Thus, option (A) is correct.
Three point charges are arranged on a line. Charge q3 = +5.00 nC and is at the origin. Charge q2 = -2.00 nC and is at x = 5.00 cm . Charge q1 is at x = 2.50 cm .What is q1 (magnitude and sign) if the net force on q3 is zero?
Answer:
q₁= +0.5nC
Explanation:
Theory of electrical forces
Because the particle q3 is close to three other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.
To solve this problem we apply Coulomb's law:
Two point charges (q1, q2) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:
o solve this problem we apply Coulomb's law:
Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:
F=K*q₁*q₂/d² Formula (1)
F: Electric force in Newtons (N)
K : Coulomb constant in N*m²/C²
q₁,q₂:Charges in Coulombs (C)
d: distance between the charges in meters
Data:
Equivalences
1nC= 10⁻⁹ C
1cm= 10⁻² m
Data
q₃=+5.00 nC =+5* 10⁻⁹ C
q₂= -2.00 nC =-2* 10⁻⁹ C
d₂= 5.00 cm= 5*10⁻² m
d₁= 2.50 cm= 2.5*10⁻² m
k = 8.99*10⁹ N*m²/C²
Calculation of magnitude and sign of q1
Fn₃=0 : net force on q3 equals zero
F₂₃:The force F₂₃ that exerts q₂ on q₃ is attractive because the charges have opposite signs,in direction +x.
F₁₃:The force F₂₃ that exerts q₂ on q₃ must go in the -x direction so that Fn₃ is zero, therefore q₁ must be positive and F₂₃ is repulsive.
We propose the algebraic sum of the forces on q₃
F₂₃ - F₁₃=0
[tex]\frac{k*q_{2} *q_{3} }{d_{2}^{2} } -\frac{k*q_{1} *q_{3} }{d_{1}^{2} }=0[/tex]
We eliminate k*q₃ of the equation
[tex]\frac{q_{1} }{d_{1}^{2} } = \frac{q_{2} }{d_{2}^{2} }[/tex]
[tex]q_{1} =\frac{q_{2} *d_{1} ^{2} }{d_{2}^{2} }[/tex]
[tex]q_{1} =\frac{2*10^{-9}*2.5^{2}*10^{-4} }{5^{2}*10^{-4} }[/tex]
q₁= +0.5*10⁻⁹ C
q₁= +0.5nC
Final answer:
The magnitude and sign of charge q1 can be found by setting the electric force by q1 on q3 equal and opposite to the force exerted by q2 on q3 using Coulomb's Law and solving for q1 given the known distances.
Explanation:
The question is asking to find the magnitude and sign of charge q1 such that the net electrostatic force on charge q3 is zero when placed in a line with charges q2 and q3. To solve this, we can apply Coulomb's law, which states that the electrostatic force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.
To have the net force on q3 be zero, the force exerted on q3 by q1 must be equal in magnitude and opposite in direction to the force exerted on q3 by q2. We can set up equations based on Coulomb's law and solve for q1. First, determine the force F31 between q3 and q1, and the force F32 between q3 and q2. Since these forces must be equal and opposite to cancel each other out, we set F31 = F32 and solve for q1. Since the distance between q3 and q1 is 2.50 cm and between q3 and q2 is 5.00 cm, and accounting for the sign of q2, we can calculate the magnitude and sign of q1 that makes the net force on q3 zero.
How fast does a 2 MeV fission neutron travel through a reactor core?
Answer:
The answer is [tex] 1.956 \times 10^7\ m/s[/tex]
Explanation:
The amount of energy is not enough to apply the relativistic formula of energy [tex]E = mc^2[/tex], so the definition of energy in this case is
[tex]E = \frac{1}{2}m v^2[/tex].
From the last equation,
[tex]v= \sqrt{2E/m}[/tex]
where
[tex]E = 2 MeV = 3.204 \times 10^{-13} J[/tex]
and the mass of the neutron is
[tex]m = 1.675\times 10^{-27}\ Kg[/tex].
Then
[tex]v = 1.956 \times 10^7\ m/s[/tex]
the equivalent of [tex]0.065[/tex] the speed of light.
Consider two force vectors in the xy-horizontal plane. Suppose a force of 12.7 N pointing along the +x-axis is added to a second force of 18.1 N directed at 30 degrees to the +x-axis , also in the horizontal plane. Find the resultant vector for this sum. magnitude direction degrees above the +x-axis in the horizontal plane
Answer:
[tex]F_1+F_2= (28.26, 9.05) N[/tex]
[tex]\alpha = 17.7\º[/tex]
[tex]F = 29.67 N[/tex]
Explanation:
Hi!
In a (x, y) coordinate representation, the two forces are:
[tex]F_1=(12.7N, 0)\\F_2=(18.1N\cos(30\º), 18.1N \sin(30\º) )\\\cos(\º30)=0.86\\\sin(\º30)= 0.5[/tex]
The sum of the two forces is:
[tex]F_1 + F_2 = ( 12.7 + 0.86*18.1, 18.1*0.5) N[/tex]
[tex]F_1+F_2= (28.26, 9.05) N[/tex]
The angle to x-axis is calculated using arctan:
[tex]\alpha = \arctan(\frac{F_y}{F_x}) = \arctan(\frac{9.05}{28.26} = 17.7\º[/tex]
The magnitude is:
[tex]F = \sqrt {F_x^2 + F_y^2}= \sqrt{798.6 + 81.9} = 29.67 N[/tex]
A walker covers a disatance of 4.0 km in a time of 53 minutes. (a) What is the average speed of the walker for this distance in km/hr?
Answer:
Average speed, v = 48.01 km/h
Explanation:
Given that,
Distance covered by the walker, d = 4 km
Tim taken to cover that distance, t = 53 min = 0.0833 hours
We need to find the average speed of the walker for this distance. It is given by :
[tex]speed=\dfrac{distance}{time}[/tex]
[tex]speed=\dfrac{4\ km}{0.0833\ h}[/tex]
Speed, v = 48.01 km/h
So, the average speed of the walker is 48.01 km/h. Hence, this is the required solution.
The lighting needs of a storage room are being met by six fluorescent light fixtures, each fixture containing four lamps rated at 60 W each. All the lamps are on during operating hours of the facility, which is 6 a.m. to 6 p.m., 365 days a year. The storage room is actually used for an average of 3 h a day. If the price of electricity is $0.11/kWh, determine the amount of energy and money that will be saved as a result of installing motion sensors. Also, determine the simple payback period if the purchase price of the sensor is $32 and it takes 1 h to install it at a cost of $66.
By installing motion sensors, the storage room would save $1.27 per day in energy costs. The simple payback period for installing motion sensors is approximately 77.17 days.
Explanation:To determine the amount of energy and money that will be saved as a result of installing motion sensors, we need to compare the energy consumption and cost of the current lighting system with that of the motion sensor system. Currently, the storage room uses six fluorescent light fixtures, each containing four lamps rated at 60 W each. The lamps are on for 12 hours a day. So, the total energy consumed by the current system per day is 6 * 4 * 60 W * 12 hours = 17,280 W. With an electricity price of $0.11/kWh, the cost of energy consumed per day is 17,280 W / 1000 * $0.11 = $1.90.
Now, let's consider the motion sensor system. If the room is only used for an average of 3 hours a day, the total energy consumed by the motion sensor system per day would be 6 * 4 * 60 W * 3 hours = 5,760 W. This results in a cost of 5,760 W / 1000 * $0.11 = $0.63 per day. Therefore, by installing motion sensors, the storage room would save $1.90 - $0.63 = $1.27 per day in energy costs.
To determine the simple payback period, we need to calculate the cost of the motion sensor system and the savings achieved per day. The total cost of the motion sensor system is $32 for the purchase price + $66 for installation, which equals $98. With daily savings of $1.27, the payback period can be calculated as $98 / $1.27 = 77.17 days. Therefore, the simple payback period for installing motion sensors in the storage room is approximately 77.17 days.
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A person travels by car from Tucson to Phoenix at a constant speed of 75 km/hr. They then return from Phoenix to Tucson at a constant speed of 65 km/hr. What was their average velocity?
Answer:
[tex]v=0[/tex]
Explanation:
Knowing that the formula for average velocity is:
[tex]v=\frac{x_{2}-x_{1}}{t_{2}-t_{1}}[/tex]
Being said that, we know that the person's displacement is zero because it returns to its starting point
[tex]x_{2}=x_{1}[/tex]
That means [tex]x_{2}-x_{1}=0[/tex]
[tex]v=\frac{0}{t_{2}-t_{1}}=0[/tex]
A golfer takes three putts to get the ball into the hole. The first putt displaces the ball 3.45 m north, the second 1.53 m southeast, and the third 0.877 m southwest. What are (a) the magnitude and (b) the angle between the direction of the displacement needed to get the ball into the hole in just one putt and the direction due east?
Answer:
magnitude is 2.52 m with 66.15° north east
Explanation:
given data
displace 1 = 3.45 m north = 3.45 j
displace 2 = 1.53 m southeast = 1.53 cos(315) i + sin(315) j
displace 3 = 0.877 m southwest = 0.877 cos(225) i- sin(225) j
to find out
displacement and direction
solution
we consider here direction i as east and direction j as north
so here
displacement = displace 1 + displace 2 + displace 3
displacement = 3.45 j + 1.53 cos(315) i + sin(315) j + 0.877 cos(225) i- sin(225) j
displacement = 3.45 j + 1.081 i -1.0818 j - 0.062 i -0.062 j
displacement = 1.019 i + 2.306 j
so magnitude
magnitude = [tex]\sqrt{1.019^{2} + 2.306^{2}}[/tex]
magnitude = 2.52
and
angle will be = arctan(2.306/1.019)
angle = 1.15470651 rad
angle is 66.15 degree
so magnitude is 2.52 m with 66.15° north east
An arrow is shot from a height of 1.7 m toward a cliff of height H. It is shot with a velocity of 26 m/s at an angle of 60° above the horizontal. It lands on the top edge of the cliff 3.4 s later. What is the height of the cliff?
what is the maximum height (in m) reached by the arrow along it trajectory?
what is the arrows impact speed (in m/s) just before hitting the cliff?
The height of the cliff is 16.97 m, the maximum height reached by the arrow is 24.47 m, and the impact speed of the arrow just before hitting the cliff is 16.47 m/s.
Explanation:In this scenario, we can apply the equations of motion to calculate the height of the cliff, the maximum height reached by the arrow, and its impact speed.
Firstly, the height of the cliff can be calculated using the equation Y = Yo + Vy*t - 0.5*g*t^2, where g is the gravity, t is the time, Vy is the initial vertical speed, and Yo is the initial height. Given Yo = 1.7m, Vy = 26sin(60°), t = 3.4s, and g = 9.8 m/s^2, the height H of the cliff is 16.97 m.
Secondly, the maximum height reached by the arrow can be calculated by the equation Hmax = Yo + Vy*t - 0.5*g*(t)^2, where t is the time it takes to reach the maximum height, which can be Ve/g. Ve is the initial vertical velocity whose value is Vy = 26sin(60°). Hence the maximum height Hmax is 24.47 m.
Finally, the arrow’s impact speed can be calculated by using Pythagoras' theorem. The impact speed V = sqrt((Vx)^2 + (Vy)^2), where Vx is the horizontal velocity and Vy is the final vertical velocity. Given Vx = 26cos(60°) and Vy = Ve - g*t, with Ve = 26sin(60°) and t = 3.4s, the impact speed V of the arrow just before hitting the cliff is 16.47 m/s.
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The height of the cliff is 21.6 m, and the arrow’s impact speed just before hitting the cliff is approximately 16.8 m/s. These calculations use projectile motion equations for both components.
A. To determine the height of the cliff, we can use the vertical motion equation:
[tex]y = y_0 + v_{0y}t - 0.5gt^2[/tex]
Where:
y₀ = initial height = 1.7 m[tex]v_{0y}[/tex] = initial vertical velocity = v₀sin(θ) = 26sin(60°) = 22.5 m/sg = acceleration due to gravity = 9.8 m/s²t = time = 3.4 sSubstituting these values into the equation:
y = 1.7 + 22.5(3.4) - 0.5(9.8)(3.4)2
y = 1.7 + 76.5 - 56.6 = 21.6 m
Therefore, the height of the cliff is 21.6 m.
B. To find the impact speed, we need to calculate both the final vertical and horizontal components of velocity:
Horizontal component (vₓ): It remains constant:The total impact speed is found using the Pythagorean theorem:
[tex]v_f = \sqrt{(v_x^2 + v_{fy}^2)} = \sqrt{(132 + (-10.8)2)} \approx 16.8 m/s[/tex]
Thus, the arrow's impact speed is approximately 16.8 m/s.
A certain car takes 30m to stop when it is traveling at 25m/s. If a pedestrian is 28m in front of this car when the driver starts braking (starting at 25m/s), how long does the pedestrian have to get out of the way?
Answer:
It take 1.78033 second get away
Explanation:
We have given that a car takes 30 m to stop when its speed is 25 m/sec
As the car stops its final speed v = 0 m/sec
Initial speed u = 25 m/sec
Distance s = 30 m
From third law of motion [tex]v^2=u^2+2as[/tex]
So [tex]0^2=25^2+2\times a\times 30[/tex]
[tex]a=-10.4166m/sec^2[/tex]
Now in second case distance s = 28 m
So [tex]v^2=25^2+2\times -10.4166\times 28[/tex]
[tex]v^2=41.666[/tex]
v = 6.4549 m/sec
Now from first equation of motion v=u+at
So [tex]6.4549=25-10.4166\times t[/tex]
t = 1.78033 sec
A ball is thrown vertically upward with a speed of 18.0 m/s. (a) How high does it rise? (b) How long does it take to reach its highest point? (o) How long does the bali take to hit the ground after it reaches its highest point? (d) what is its velocity when it returns to the level from which it started?
Answer:
a) [tex]y=16.53m[/tex]
b) [tex]t_{up}=1.83s[/tex]
c)[tex]t_{down}=1.84s[/tex]
d) [tex]v=-18m/s[/tex]
Explanation:
a) To find the highest point of the ball we need to know that at that point the ball stops going up and its velocity become 0
[tex]v^{2} =v^{2} _{o} +2g(y-y_{o})[/tex]
[tex]0=(18)^{2} -2(9.8)(y-0)[/tex]
Solving for y
[tex]y=\frac{(18)^{2} }{2(9.8)}=16.53m[/tex]
b) To find how long does it take to reach that point:
[tex]v=v_{o}+at[/tex]
[tex]0=18-9.8t[/tex]
Solving for t
[tex]t_{up} =\frac{18m/s}{9.8m/s^{2} }= 1.83s[/tex]
c) To find how long does it take to hit the ground after it reaches its highest point we need to find how long does it take to do the whole motion and then subtract the time that takes to go up
[tex]y=y_{o}+v_{o}t+\frac{1}{2}gt^{2}[/tex]
[tex]0=0+18t-\frac{1}{2}(9.8)t^{2}[/tex]
Solving for t
[tex]t=0 s[/tex] or [tex]t=3.67s[/tex]
Since time can not be negative, we choose the second option
[tex]t_{down}=t-t_{up}=3.67s-1.83s=1.84s[/tex]
d) To find the velocity when it returns to the level from which it started we need to use the following formula:
[tex]v=v_{o}+at[/tex]
[tex]v=18m/s-(9.8m/s^{2} )(3.67s)=-18m/s[/tex]
The sign means the ball is going down
An object of mass m = 4.0 kg, starting from rest, slides down an inclined plane of length l = 3.0 m. The plane is inclined by an angle of θ = 30◦ to the ground. The coefficient of kinetic friction μk = 0.2. At the bottom of the plane, the mass slides along a rough surface with a coefficient of kinetic friction μk = 0.3 until it comes to rest. The goal of this problem is to find out how far the object slides along the rough surface. What is the work done by the friction force while the mass is sliding down the in- clined plane? (Is it positive or negative?) (b) What is the work done by the gravitational force while the mass is sliding down the inclined plane? (Is it positive or negative?)
Answer:
(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface
(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:
Wf= -20.4 J is negative
(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane:
Wg= 58.8 J is positive
Explanation:
Nomenclature
vf: final velocity
v₀ :initial velocity
a: acceleleration
d: distance
Ff: Friction force
W: weight
m:mass
g: acceleration due to gravity
Graphic attached
The attached graph describes the variables related to the kinetics of the object (forces and accelerations)
Calculation de of the components of W in the inclined plane
W=m*g
Wx₁ = m*g*sin30°
Wy₁= m*g*cos30°
Object kinematics on the inclined plane
vf₁²=v₀₁²+2*a₁*d₁
v₀₁=0
vf₁²=2*a₁*d₁
[tex]v_{f1} = \sqrt{2*a_{1}*d_{1} }[/tex] Equation (1)
Object kinetics on the inclined plane (μ= 0.2)
∑Fx₁=ma₁ :Newton's second law
-Ff₁+Wx₁ = ma₁ , Ff₁=μN₁
-μ₁N₁+Wx₁ = ma₁ Equation (2)
∑Fy₁=0 : Newton's first law
N₁-Wy₁= 0
N₁- m*g*cos30°=0
N₁ = m*g*cos30°
We replace N₁ = m*g*cos30 and Wx₁ = m*g*sin30° in the equation (2)
-μ₁m*g*cos30₁+m*g*sin30° = ma₁ : We divide by m
-μ₁*g*cos30°+g*sin30° = a₁
g*(-μ₁*cos30°+sin30°) = a₁
a₁ =9.8(-0.2*cos30°+sin30°)=3.2 m/s²
We replace a₁ =3.2 m/s² and d₁= 3m in the equation (1)
[tex]v_{f1} = \sqrt{2*3.2*3} }[/tex]
[tex]v_{f1} =\sqrt{2*3.2*3}[/tex]
[tex]v_{f1} = 4.38 m/s[/tex]
Rough surface kinematics
vf₂²=v₀₂²+2*a₂*d₂ v₀₂=vf₁=4.38 m/s
0 =4.38²+2*a₂*d₂ Equation (3)
Rough surface kinetics (μ= 0.3)
∑Fx₂=ma₂ :Newton's second law
-Ff₂=ma₂
--μ₂*N₂ = ma₂ Equation (4)
∑Fy₂= 0 :Newton's first law
N₂-W=0
N₂=W=m*g
We replace N₂=m*g inthe equation (4)
--μ₂*m*g = ma₂ We divide by m
--μ₂*g = a₂
a₂ =-0.2*9.8= -1.96m/s²
We replace a₂ = -1.96m/s² in the equation (3)
0 =4.38²+2*-1.96*d₂
3.92*d₂ = 4.38²
d₂=4.38²/3.92
d₂=4.38²/3.92
(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface
(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:
Wf = - Ff₁*d₁
Ff₁= μ₁N₁= μ₁*m*g*cos30°= -0.2*4*9.8*cos30° = 6,79 N
Wf= - 6.79*3 = 20.4 N*m
Wf= -20.4 J is negative
(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane
Wg=W₁x*d= m*g*sin30*3=4*9.8*0.5*3= 58.8 N*m
Wg= 58.8 J is positive
The work done by the friction force is negative, and the work done by the gravitational force is positive as the object slides down the inclined plane.
Explanation:The work done by the friction force while the object slides down the inclined plane is negative. The work done by the gravitational force while the object slides down the inclined plane is positive.
When the object slides down the inclined plane, the friction force acts in the opposite direction to its motion. Since friction always opposes the motion, the work done by friction is negative.
The gravitational force, on the other hand, acts in the same direction as the object's motion. Therefore, the work done by the gravitational force is positive.
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A movie star catches a paparazzi reporter snapping pictures of her at home and claims that he was trespassing. He, of course denies the allegations. To prove her point, she submits as evidence the film that she confiscated. Her height of 1.75 m appears as an 8.25 mm high image on the film. Additionally, she submits that the camera that was used has a focal length of 210 mm. How far away was the reporter when he took the picture? (All the information that is given).
Answer:
44.755 m
Explanation:
Given:
Height of the movie star, H = 1.75 m = 1750 mm
Height of the image, h = - 8.25 mm
Focal length of the camera = 210 mm
Let the distance of the object i.e the distance between camera and the movie star be 'u'
and
distance between the camera focus and image be 'v'
thus,
magnification, m = [tex]\frac{\textup{h}}{\textup{H}}[/tex]
also,
m = [tex]\frac{\textup{-v}}{\textup{u}}[/tex]
thus,
[tex]\frac{\textup{-v}}{\textup{u}}=\frac{\textup{h}}{\textup{H}}[/tex]
or
[tex]\frac{\textup{-v}}{\textup{u}}=\frac{\textup{-8.25}}{\textup{1750}}[/tex]
or
[tex]\frac{\textup{1}}{\textup{v}}=-\frac{\textup{1750}}{\textup{-8.25}}\times\frac{1}{\textup{u}}[/tex] ....................(1)
now, from the lens formula
[tex]\frac{\textup{1}}{\textup{f}}=\frac{\textup{1}}{\textup{u}}+\frac{1}{\textup{v}}[/tex]
on substituting value from (1)
[tex]\frac{\textup{1}}{\textup{210}}=\frac{\textup{1}}{\textup{u}}+-\frac{\textup{1750}}{\textup{-8.25}}\times\frac{1}{\textup{u}}[/tex]
or
[tex]\frac{\textup{1}}{\textup{210}}=\frac{\textup{1}}{\textup{u}}(1 -\frac{\textup{1750}}{\textup{-8.25}})[/tex]
or
u = 210 × ( 1 + 212.12 )
or
u = 44755.45 mm
or
u = 44.755 m
A 2.0-mm-diameter copper ball is charged to 40 nC . What fraction of its electrons have been removed? The density of copper is 8900 kg/m^3.
Answer:
0.02442 × 10⁻⁹
Explanation:
Given:
Diameter of copper ball = 2.00 mm = 0.002 m
Charge on ball = 40 nC = 40 × 10⁻⁹ C
Density of copper = 8900 Kg/m³
Now,
The number of electrons removed, n = [tex]\frac{\textup{Charge on ball}}{\textup{Charge of an electron}}[/tex]
also, charge on electron = 1.6 × 10⁻¹⁹ C
Thus,
n = [tex]\frac{40\times10^{-9}}{1.6\times10^{-19}}[/tex]
or
n = 25 × 10¹⁰ Electrons
Now,
Mass of copper ball = volume × density
Or
Mass of copper ball = [tex]\frac{4}{3}\pi(\frac{d}{2})^3[/tex] × 8900
or
Mass of copper ball = [tex]\frac{4}{3}\pi(\frac{0.002}{2})^3[/tex] × 8900
or
Mass of copper ball = 0.03726 grams
Also,
molar mass of copper = 63.546 g/mol
Therefore,
Number of mol of copper in 0.03726 grams = [tex]\frac{ 0.03726}{63.546}[/tex]
or
Number of mol of copper in 0.03726 grams = 5.86 × 10⁻⁴ mol
and,
1 mol of a substance contains = 6.022 × 10²³ atoms
Therefore,
5.86 × 10⁻⁴ mol of copper contains = 5.86 × 10⁻⁴ × 6.022 × 10²³ atoms.
or
5.86 × 10⁻⁴ mol of copper contains = 35.88 × 10¹⁹ atoms
Now,
A neutral copper atom has 29 electrons.
Therefore,
Number of electrons in ball = 29 × 35.88 × 10¹⁹ = 1023.37 × 10¹⁹ electrons.
Hence,
The fraction of electrons removed = [tex]\frac{25\times10^{10}}{1023.37\times10^{19}}[/tex]
or
The fraction of electrons removed = 0.02442 × 10⁻⁹
The fraction of electrons removed from the copper ball can be calculated by comparing its net charge to the charge of a single electron. Using the provided information, we can find that approximately 2.5 x 10^10 electrons have been removed from the ball. To determine the fraction, we need to compare this number to the total number of electrons in the ball, which can be calculated using the mass, density, and atomic mass of copper. By plugging in the values, we can find the fraction of electrons removed.
Explanation:To determine the fraction of electrons that have been removed from the copper ball, we need to compare the net charge of the ball to the charge of a single electron. The net charge of the ball is 40 nC, which is equivalent to 40 x 10^-9 C. The charge of a single electron is 1.60 x 10^-19 C.
We can calculate the number of electrons that have been removed using the formula:
Number of electrons removed = Net charge of the ball / Charge of a single electron
Number of electrons removed = (40 x 10^-9 C) / (1.60 x 10^-19 C) = 2.5 x 10^10 electrons
To find the fraction of electrons removed, we need to compare the number of electrons removed to the total number of electrons in the ball. The total number of electrons in the ball can be calculated using the formula:
Total number of electrons = Number of copper atoms x Number of electrons per copper atom
The number of copper atoms can be calculated using the formula:
Number of copper atoms = Mass of the ball / Atomic mass of copper
The mass of the ball can be calculated using the formula:
Mass of the ball = Volume of the ball x Density of copper
Given that the diameter of the ball is 2.0 mm, the volume of the ball can be calculated using the formula for the volume of a sphere:
Volume of the ball = (4/3) x pi x (radius)^3
As the ball is a sphere, the radius is half the diameter, so the radius is 1.0 mm or 1 x 10^-3 m.
Using the given density of copper (8900 kg/m^3) and atomic mass of copper (63.5 g/mol), we can now calculate the fraction of electrons removed:
Fraction of electrons removed = Number of electrons removed / Total number of electrons = (2.5 x 10^10 electrons) / (Number of copper atoms x Number of electrons per copper atom)
A person is riding on a Ferris Wheel. When the wheel makes one complete turn, is the net work done by the gravitational force posiitive, negative, or zero?
Answer:
Zero
Explanation:
The overall work done by gravitational force on completion of a complete turn is zero.
Since the work done by gravitaional force is conservative and depends only on the initial and end position here height and the path followed does not matter.
Since, in a complete turn the wheel return to its final position as a result of which displacement is zero and as work is the dot product of Force exerted and displacement, the work done is zero.
Also the work done in half cycle by gravity is counter balance by the work which is done against the gravity in the other half cycle.
The net work done by the gravitational force when a person is riding on a Ferris Wheel and it makes one complete turn is zero.
To understand why the net work done by the gravitational force is zero, we need to consider the definition of work and the nature of the motion on a Ferris Wheel. Work done by a force is defined as the product of the force and the displacement in the direction of the force. Mathematically, this is expressed as:
[tex]\[ W = F \cdot d \cdot \cos(\theta) \][/tex]
where ( W ) is the work, ( F ) is the force, ( d ) is the displacement, and [tex]\( \theta \)[/tex] is the angle between the force and the displacement.
In the case of the Ferris Wheel, the gravitational force acts vertically downward towards the center of the Earth, while the displacement of the person on the Ferris Wheel is along the circumference of the wheel, which is horizontal at any given point. Since the force and displacement are perpendicular to each other at every point in the circle, the angle \( \theta \) between them is always 90 degrees. Therefore, the cosine of 90 degrees is zero, which means that the work done by the gravitational force at each point is zero:
[tex]\[ W = F \cdot d \cdot \cos(90^\circ) = F \cdot d \cdot 0 = 0 \][/tex]
Moreover, over one complete turn of the Ferris Wheel, the initial and final positions of the person are the same. This means that the total displacement over one complete cycle is zero. Even if we consider the components of the gravitational force along the direction of displacement during different parts of the cycle, the net effect is zero because the person is raised to a certain height and then lowered back to the starting point. The work done against gravity to raise the person is equal in magnitude and opposite in sign to the work done by gravity as the person descends, resulting in a net work of zero for the entire cycle.
Therefore, the gravitational force does no net work on the person over one complete turn of the Ferris Wheel.
Three charges, each of magnitude 10 nC, are at separate corners of a square of edge length 3 cm. The two charges at opposite corners are positive, and the other charge is negative. Find the force exerted by these charges on a fourth charge q = +3 nC at the remaining (upper right) corner. (Assume the +x axis is directed to the right and the +y axis is directed upward.)
Answer:
The force exerted by three charges on the fourth is [tex]F_{resultant}=2.74\times10^{-5}\ \rm N[/tex]
Explanation:
Given:
The magnitude of three identical charges, [tex]q=10\ \rm nC[/tex]Length of the edge of the square a=3 cmMagnitude of fourth charge ,Q=3 nCAccording to coulombs Law the force F between any two charge particles is given by
[tex]F=\dfrac{kQq}{r^2}[/tex]
where r is the radial distance between them.
Since the force acting on the charge particle will be in different directions so according to triangle law of vector addition
[tex]F_{resultant}=\sqrt ((\dfrac{kQq}{L^2})^2+(\dfrac{kQq}{L^2} })^2)+\dfrac{kQq}{(\sqrt{2}L)^2}\\F_{resultant}=\dfrac{kQq}{L^2}(\sqrt{2}-\dfrac{1}{2})\\F_{resultant}=\dfrac{9\times10^9\times10\times10^{-10}\times3\times10^{-9}}{0.03^2}(\sqrt{2}-\dfrac{1}{2})\\F_{resultant}=2.74\times 10^{-5}\ \rm N[/tex]
The force on the fourth charge is calculated by first determining the individual forces exerted by each of the three other charges separately using Coulomb's Law and then adding these forces as vectors. This involves resolving each force into its x and y components, combining them separately, and then determining the resultant force's magnitude and direction.
Explanation:The problem here involves Coulomb's Law and the superposition principle in physics. Coulomb's Law defines the force between two point charges as directly proportional to the product of their charges, and inversely proportional to the square of the distance between them.
First, you need to calculate the forces exerted on the fourth charge by each of the three other charges separately. This involves calculating the distance from each existing charge to the fourth charge, then subbing these distances, along with the relevant charge values, into the Coulomb's Law formula. Remember that if the charge is positive (like in the case of charge +q), the force vector points directly from the charge, while if the charge is negative, the force vector points towards the charge.
After calculating the force vectors resulting from each charge, you add these vectors together to get the resultant force vector which is the force exerted on the fourth charge. This problem also involves trigonometry as when you add the force vectors, you have to take into account the direction which each force vector is pointing.
Force due to the positive charge at the lower left: F1 is in the first quadrant
Force due to the positive charge at the lower right: F2 is in the fourth quadrant
Force due to the negative charge at the upper left: F3 is in the third quadrant
In each case, you'll need to resolve each force into its x and y components, and then add up all the x and y components separately to get the x and y components of the total force. Finally, calculate the magnitude of the total force using the Pythagorean theorem.
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A small company manufactures a certain product. The price-production relationship for this product is P = -0.7*D + 300, where P is the unit sales price of the product and D is the annual production (number of units produced). Suppose variable costs are $25 per unit produced and fixed costs are $10,287 . Find the break even point (minimum number of units that must be produced for a business to become profitable). Note: there are 2 roots in the breakeven equation, choose the smallest root.
Answer:42 units
Explanation:
Given
Price-production relationship=-0.7D+300
Total cost=Fixed cost+ variable cost
Total cost=10,287+25D
where D is the units produced
Total revenue[tex]=\left ( -0.7D+300\right )D[/tex]
Total revenue[tex]=-0.7D^2+300D[/tex]
For Break even point
Total revenue=Total cost
[tex]10,287+25D=-0.7D^2+300D[/tex]
[tex]7D^2-2750D+102870=0[/tex]
[tex]D=\frac{2750\pm \sqrt{2750^2-4\times 7\times 102870}}{2\times 7}[/tex]
[tex]D=41.869\approx 42[/tex] units
The potential difference between A and B is 5.0 V. A proton starts from rest at A. When it reaches B what is its kinetic energy? (e = 1.60 x 10^-19 C)
Answer:
total kinetic energy is 8 × [tex]10^{-19}[/tex] J
Explanation:
given data
potential difference = 5 V
e = 1.60 × [tex]10^{-19}[/tex] C
to find out
what is kinetic energy
solution
we will apply here conservation of energy that is
change in potential energy is equal to change in kinetic energy
so
change potential energy is e × potential difference
change potential energy = 1.60 × [tex]10^{-19}[/tex] × 5
change potential energy = 8 × [tex]10^{-19}[/tex] J
so change in kinetic energy = 8 × [tex]10^{-19}[/tex] J
and we know proton start from rest that mean ( kinetic energy is 0 ) so
change in KE is total KE
total kinetic energy is 8 × [tex]10^{-19}[/tex] J
A particular Bohr orbit in a hydrogen atom has a total energy of -0.28 eV . What is the kinetic energy of the electron in this orbit? What is the electric potential energy of the system?
Answer:
Explanation:
The total energy of an electron in an orbit consists of two components
1 ) Potential energy which is - ve because the field is attractive
2) Kinetic energy which represents moving electron having some velocity.
Kinetic is always positive.
3 ) In an orbit , The magnitude of potential energy is twice that of kinetic energy. So if -2E is the value of potential energy E wil be the value of kinetic energy.
4 ) Total energy will become some of potential energy and kinetic energy
-2E + E = -E
5 ) So total energy becomes equal to kinetic energy with only sign reversed.
In the given case total energy is -0.28 eV . Hence kinetic energy will be +0.28 eV.
When kinetic energy is calculated as +.28 eV , the potential energy will be
- 2 x .28 or - 0.56 eV .
The electric field in a region is uniform (constant in space) and given by E-( 148.0 1 -110.03)N/C. An additional charge 10.4 nC (nano-coulombs) is placed into this region. What is the y component of the electric force on this charge?
Answer:
[tex]-1.144\ \mu C[/tex]
Explanation:
Given:
[tex]\vec{E}[/tex] = uniform electric field in the space = [tex](148.0\ \hat{i}-110.0\ \hat{j})\ N/C[/tex]Q = Charge placed in the region = [tex]10.4 nC\ = 1.04\times 10^{-8}\ nC[/tex]Assume:
[tex]\vec{F}[/tex] = Electric force on the charge due to electric fieldWe know that the electric field is the electric force applied on a unit positive charge i.e.,
[tex]\vec{E}=\dfrac{\vec{F}}{Q}[/tex]
This means the electric force applied on this additional charge placed in the field is given by:
[tex]\vec{F}=Q\vec{E}\\\Rightarrow \vec{F} = 1.04\times 10^{-8}\ n C\times (148.0\ \hat{i}-110.0\ \hat{j})\ N/C\\\Rightarrow \vec{F} = (1.539\ \hat{i}-1.144\ \hat{j})\ \mu N\\[/tex]
From the above expression of force, we have the following y-component of force on this additional charge.
[tex]F_y = -1.144\ \mu N[/tex]
Hence, the y-component of the electric force on the this charge is [tex]-1.144\ \mu N[/tex].
Electric fields are vector quantities whose magnitudes are measured in units of volts/meter (V/m). Find the resultant electric field when there are two fields, E1 and E2, where E1 is directed vertically upward and has magnitude 99 V/m and E2 is directed 48° to the left of E1 and has magnitude 164 V/m.
Answer:
The resultant field will have a magnitude of 241.71 V/m, 30.28° to the left of E1.
Explanation:
To find the resultant electric fields, you simply need to add the vectors representing both electric field E1 and electric field E2. You can do this by using the component method, where you add the x-component and y-component of each vector:
E1 = 99 V/m, 0° from the y-axis
E1x = 0 V/m
E1y = 99 V/m, up
E2 = 164 V/m, 48° from y-axis
E2x = 164*sin(48°) V/m, to the left
E2y = 164*cos(48°) V/m, up
[tex]Ex: E_{1_{x}} + E_{2_{x}} = 0 V/m - 164 *sin(48) V/m= -121.875 V/m\\Ey: E_{1_{y}} + E_{2_{y}} = 99 V/m + 164 *cos(48) V/m = 208.74 V/m\\[/tex]
To find the magnitude of the resultant vector, we use the pythagorean theorem. To find the direction, we use trigonometry.
[tex]E_r = \sqrt{E_x^2 + E_y^2}= \sqrt{(-121.875V/m)^2 + (208.74V/m)^2} = 241.71 V/m[/tex]
The direction from the y-axis will be:
[tex]\beta = arctan(\frac{-121.875 V/m}{208.74 V/m}) = 30.28[/tex]° to the left of E1.
the resultant electric field has a magnitude of approximately 247.8 V/m and is directed at an angle of approximately 63.5 degrees above the horizontal, not the vertical.
To find the resultant electric field, you should add the horizontal and vertical components of E1 and E2 separately:
Vertical Component:
E1y = 99 V/m (vertical component of E1)
E2y = 164 V/m * sin(48°) ≈ 123.6 V/m (vertical component of E2)
Horizontal Component:
E2x = 164 V/m * cos(48°) ≈ 109.8 V/m (horizontal component of E2)
Add the vertical components:
Ey = E1y + E2y
Ey = 99 V/m + 123.6 V/m
Ey ≈ 222.6 V/m
Add the horizontal components:
Ex = E2x
Ex ≈ 109.8 V/m
Calculate the magnitude of the resultant electric field (E) using the Pythagorean theorem:
E = √(Ex² + Ey²)
E = √((109.8 V/m)² + (222.6 V/m)²)
E ≈ 247.8 V/m
So, the resultant electric field has a magnitude of approximately 247.8 V/m and is directed at an angle of approximately 63.5 degrees above the horizontal, not the vertical, as previously stated.
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David Scott's experiment shows that all objects, regardless of their weight, fall equally in the absence of: a. The gravity
b. The air
c. The pressure
d. The gravitational force
Answer:
b. the air
Explanation:
David Scott's experiment was performed on the moon, where there is gravity but there is no air. This experiment consisted of letting a hammer and a feather fall at the same time. The result was that the two objects touch the ground simultaneously.
Since these two objects obviously have a different mass, the experiment shows that in a vacuum, objects fall with the same acceleration regardless of their mass.
A bicyclist starts at rest and speeds up to 30 m/s while accelerating at 4 m/s^2. Determine the distance traveled.
Answer:
Distance, d = 112.5 meters
Explanation:
Initially, the bicyclist is at rest, u = 0
Final speed of the bicyclist, v = 30 m/s
Acceleration of the bicycle, [tex]a=4\ m/s^2[/tex]
Let s is the distance travelled by the bicyclist. The third equation of motion is given as :
[tex]v^2-u^2=2as[/tex]
[tex]s=\dfrac{v^2-u^2}{2a}[/tex]
[tex]s=\dfrac{(30)^2}{2\times 4}[/tex]
s = 112.5 meters
So, the distance travelled by the bicyclist is 112.5 meters. Hence, this is the required solution.
A 3-cm high object is in front of a thin lens. The object distance is 4 cm and the image distance is –8 cm. (a) What is the focal length of the lens and the kind of lens is used? (b) What is the magnification and height of the image? (c) Describe the image in terms of its type, orientation and size relative to the object?
Answer:
a) Focal length of the lens is 8 cm which is a convex lens
b) 6 cm
c) The lens is a convex lens and produces a virtual image which is upright and two times larger than the object.
Explanation:
u = Object distance = 4 cm
v = Image distance = -8 cm
f = Focal length
Lens Equation
[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}=\frac{1}{4}+\frac{1}{-8}\\\Rightarrow \frac{1}{f}=\frac{1}{8}\\\Rightarrow f=\frac{8}{1}=-8\ cm[/tex]
a) Focal length of the lens is 8 cm which is a convex lens
Magnification
[tex]m=-\frac{v}{u}\\\Rightarrow m=-\frac{-8}{4}\\\Rightarrow m=2[/tex]
b) Height of image is 2×3 = 6 cm
Since magnification is positive the image upright
c) The lens is a convex lens and produces a virtual image which is upright and two times larger than the object.
(a) The focal length of the lens is -2.67 cm.
(b) The magnification of the image is 2 and the height is 6 cm.
(c) The imaged formed by the lens is upright, virtual and magnified.
Focal length of the lensThe focal length of the lens is determined by using lens formulas as given below;
[tex]\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \\\\\frac{1}{f} = \frac{-1}{8} - \frac{1}{4} \\\\\frac{1}{f} = \frac{-1 -2}{8} = \frac{-3}{8} \\\\f = -2.67 \ cm[/tex]
Magnification of the imageThe magnification of the image is calculated as follows;
[tex]m = \frac{v}{u} \\\\m = \frac{8}{4} \\\\m = 2\\\\[/tex]
Height of the imageThe height of the image is calculated as follows;
[tex]H = mu\\\\H = 2(3cm) = 6\ cm[/tex]
Properties of the imageThe imaged formed by the lens is;
UprightVirtualTwo times larger than the object (magnified)Learn more about lens here: https://brainly.com/question/25779311