Answer:
No, how do you tell that?
Telescope is an example of technology that use on seeing the object on far place try to do that, you acn see a planet like small pieceof like an tungaw
Accomplishment of SPTA Officers and all parents this 1st Quarter.. Enjoy watching..Mabbalo
Explanation:
HOPE IT HELP TO YOUWhat is the normal force of a 0.0037 kg tennis ball rolling at a constant speed of 3 m/s
across a desk?
The normal force of the rolling ball is 0.037 N
Mass of tennis ball= 0.0037 kg
Constant speed= 3m/s
we need to apply the concept of laws of motion
Since the ball is rolling at a constant speed, it is an example of uniform motion.
So,
Normal force=weight of the body
= 0.0037 x 10 ( acceleration of gravity= 10 m/s²)
Normal force= 0.037 N
Therefore the normal force on the ball is 0.037 N
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What is the density of a 45.87 g golf ball with a diameter of 4.287 cm?
We are asked to determine the density of a gulf ball given its mass and volume. To do that, we will use the formula for density:
[tex]D=\frac{m}{V}[/tex]Where:
[tex]\begin{gathered} D=\text{ density} \\ m=\text{ mass} \\ V=\text{ volume} \end{gathered}[/tex]To determine the volume we will use the fact that the gulf ball can be approximated to a sphere and the volume of a sphere is given by:
[tex]V=\frac{4}{3}\pi r^3[/tex]Where:
[tex]r=\text{ radius}[/tex]We are given the diameter. We know that the diameter is twice the radius, therefore:
[tex]r=\frac{D}{2}[/tex]Substituting the value of the diameter we get:
[tex]r=\frac{4.287\operatorname{cm}}{2}[/tex]Solving the operations:
[tex]r=2.144\operatorname{cm}[/tex]Now, we substitute the value of the radius in the formula of the volume:
[tex]V=\frac{4}{3}\pi(2.144\operatorname{cm})^3[/tex]Solving the operation we get:
[tex]V=41.282\operatorname{cm}^3[/tex]Now, we substitute the value of the volume and the mass in the formula for density:
[tex]D=\frac{45.87g}{41.282\operatorname{cm}^3}[/tex]Solving the operation:
[tex]D=1.11\frac{g}{\operatorname{cm}^3}[/tex]Therefore, the density of the ball is 1.11 g/cm^3.
A car is being tested for safety by colliding it with a brick wall.The 1300 kg car is initially driving towards the wall at a speedof 15 m/s, and after colliding with the wall the car movesaway from the wall at 2 m/s. If the car is in contact with thewall for 0.5 s, calculate the average force exerted on the carby the wall.
Answer:
44200 N
Explanation:
To calculate the average force exerted on the car, we will use the following equation
[tex]\begin{gathered} Ft=\Delta p \\ Ft=m(v_f-v_i) \end{gathered}[/tex]Where F is the average force, t is the time, m is the mass, vf is the final velocity and vi is the initial velocity of the car.
Replacing t = 0.5s, m = 1300 kg, vf = -2 m/s, and vi = 15 m/s and solving for F, we get
[tex]\begin{gathered} F(0.5s)=(1300\text{ kg\rparen\lparen-2 m/s - 15 m/s\rparen} \\ F(0.5s)=(1300\text{ kg\rparen\lparen-17 m/s\rparen} \\ F(0.5s)=-22100\text{ kg m/s} \\ F=\frac{-22100\text{ kg m/s}}{0.5\text{ s}} \\ F=-44200\text{ N} \end{gathered}[/tex]Therefore, the average force exerted on the car by the wall was 44200 N
A man can crack a nut by applying a force of 100 N a lever of length 0.5 m. What should be the length of the lever if he wants to use a force of 75 N to crack the nut?
Answer:
So, to apply the force of 75N length of nut cracker should be 66.6 cm
Answer:
Explanation:
Given:
F₁ = 100 N
L₁ = 0.5 m
F₂ = 75 N
__________
L₂ - ?
According to the rule of moments:
M₁ = M₂
F₁·L₁ = F₂·L₂
The length of the lever:
L₂ = F₁·L₁ / F₂
L₂ = 100·0.5 / 75 ≈ 0.67 m
Marc is sitting at his desk with good posture. What MOST likely is Marc doing?
It is most likely that Marc is studying while assuming a good sitting posture.
What is posture?Posture is defined as the the various ways an individual carry their body in order to assume a particular position which may be while sitting of standing up.
There different types of postures which are classified under good or bad posture. They include the following:
Healthy Posture.Kyphosis Posture.Flat Back Posture.Swayback Posture.Forward Head Posture.To sit correctly at a dest with a good posture is an example of a healthy posture which was assumed by Marc.
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How is kinetic energy and pontential energy alike ?
The kinetic energy of the particle is associated with the velocity of the particle where as the potential energy of the particle depneds upon the position ofthe particle.
Both the energies can transform into each other.
For example,
When a ball is dropped from the building's top floor, the potential energy of the ball is maximum at the top of the builiding and transform into the kinetic energy while moving in the downward direction.
The kinetic energy of the ball is maximum just before hitting the ground.
This shows the tranformation of the potential energy to the kinetic energy.
Thus, the kinetic energy and potential can transform into each other is one the similarity in behaviour.
Students conducted an experiment in which a ball was thrown into the air and then caught at the same height from which it was released. They determined that the ball had 100 J of kinetic energy when it was released. They calculated the ball's energy just before it was caught and determined that it's kinetic energy was less than 100 J.Which of the following statements MOST LIKELY accounts for the difference in the ball's initial and final kinetic energy?
A ball was thrown into the air and then caught at the same height from which it was released.
The initial kinetic energy of the ball was 100 J and the ball's kinetic energy just before it was caught was less than 100 J.
QUESTION 22As the rocket moves from position "b" to posisition "c", its speed is:constant.O continuously increasing.O continuously decreasing.increasing for a while and constant thereafter.constant for a while and decreasing thereafter.
The rocket movement due to the direction between a and b, and turning on the engine, movement is described by E image
Is continuously increasing, the force is constant and the acceleration too, which means that the speed continues increasing
this is a 3 part question20) A 9.50-g bullet has a speed of 1.30 km>s. (a) What is its kinetic energy in joules? (b) What is the bullet’s kinetic energy if its speed is halved? (c) If its speed is doubled?
Given that mass of bullet, m = 9.50 g = 0.0095 kg
speed of bullet, v is 1.30 km/s
(a) Kinetic energy is given by the formula
[tex]K\mathrm{}E\text{. = }\frac{1}{2}mv^2[/tex]Substituting the values in the above formula, we get
[tex]K\mathrm{}E\mathrm{}=\frac{1}{2}(0.0095)(1.30)^2=8.0275\times10^{-3}\text{ J}[/tex](b) Speed of bullet is v/2
Sustituting this value in the formula of kinetic energy, we get
[tex]\begin{gathered} K.E._1=\text{ }\frac{1}{2}m(\frac{v}{2})^2 \\ =\frac{1}{2}\times(0.0095)\times(\frac{1.30}{2})^2 \\ =2.0068\times10^{-3\text{ }}J \end{gathered}[/tex](c) Speed of bulllet becomes 2v
Sustituting this value in the formula of kinetic energy, we get
[tex]\begin{gathered} K.E._2=\text{ }\frac{1}{2}m(2v)^2 \\ =\frac{1}{2}\times(0.0095)\times(2\times1.30)^2 \\ =0.03211J \end{gathered}[/tex]12. How could extreme heat (resulting from Climate Change) affect human andanimal life?
ANSWER:
The answer is given in the step by step of the question
STEP-BY-STEP EXPLANATION:
Extreme heat can affect human and animal life in the following ways:
• Animals are sensitive to changes in temperature, as much or more than humans. Think how bad you feel when you have a fever. They also suffer this type of discomfort as a consequence of climate change.
,• Climate change is pushing many species to the limit. They lack water to drink or suffer from temperatures in which they are not comfortable.
,• Provocation of forest fires damaging the habitat of animals and humans
Two cars travel in the same direction along a straight highway, one at a constant speed of 55 mi/h and the other at 60 mi/h.(a) Assuming they start at the same point, how much sooner does the faster car arrive at a destination 14 mi away?_____ min(b) How far must the faster car travel before it has a 15-min lead on the slower car?_____ mi
We will have the following:
a) We first determine the time it takes to travel the distance to both vehicles:
*
[tex]t_1=\frac{14mi\ast1h}{55}\Rightarrow t_1=\frac{14}{55}h[/tex]*
[tex]t_2=\frac{14mi\ast1h}{60mi}\Rightarrow t_2=\frac{11}{12}h[/tex]So, we determine now the difference in time:
[tex]\frac{11}{12}h-\frac{14}{55}h=\frac{437}{660}h\approx0.66h[/tex]So, the fastest car will arrive approximately 0.66 hours sooner.
b) We determmine the distance it must travel the fastest car to have a 15 minute lead on the other one as follows:
First, we determine the time difference required:
[tex]t=\frac{15min\ast1h}{60min}\Rightarrow t=0.25h[/tex]Then, since both vehicles will move relative to each other, we will have that:
[tex]d_{c1}=(60mi/h)(0.25h)\Rightarrow d_{c1}=15mi[/tex]So, the fastest car must be 15 miles ahead of the other car in order to have a 15 minute lead with respect to the second car.
An Olympic long jumper is capable of jumping 8.0 m. Assuming his horizontal speed is 9.1 m/s as he leaves the ground, how long is he in the air and how high does he go?
Given data:
* The initial velocity of the jumper is u = 9.1 m/s.
* The horizontal range in the given case is 8 m.
Solution:
(a). By the kinematic equation, the time taken by the jumper in terms of the initial velocity and the horizontal range is,
[tex]R=ut+\frac{1}{2}at^2[/tex]where a is the acceleration of the jumper in the horizontal direction,
As there is no force acting on the jumper in the horizontal direction, thus, the value of acceleration is zero.
Substituting the known values,
[tex]\begin{gathered} 8=9.1\times t \\ t=\frac{8}{9.1} \\ t=0.88\text{ s} \end{gathered}[/tex]Thus, the time for which the jumper remains in the air is 0.88 seconds.
(b). By the kinematics equation, the initial velocity of the jumper in the upward direction is,
[tex]v_y-u_y=gt^{\prime}_{}\ldots\ldots\ldots(1)[/tex]where u_y is the initial velocity, v_y is the final velocity of the jumper at the top of vertical displacement, g is the acceleration due to gravity, and t' is the time taken by the jumper to reach the top of vertical displacement,
The jumper will come to rest at the higher position, thus, the final velocity of the jumper at the highest position is zero.
The time taken by the jumper to reach the maximum height is equal to the time taken by the jumper to reach the ground from the maximum height.
As the total time for the complete motion of the jumper is t, thus, the time taken by the jumper to reach the maximum height from the ground is,
[tex]\begin{gathered} t^{\prime}=\frac{t}{2} \\ t^{\prime}=\frac{0.88}{2} \\ t^{\prime}=0.44\text{ s} \end{gathered}[/tex]Substituting the known values in the equation (1),
[tex]\begin{gathered} 0-u_y=-9.8\times0.44_{} \\ u_y=4.312\text{ m/s} \end{gathered}[/tex]By the kinematics equation, the maximum height reached by the jumper is,
[tex]h=u_yt^{\prime}+\frac{1}{2}gt^{\prime}^2[/tex]Substituting the known values,
[tex]\begin{gathered} h=4.312\times0.44+\frac{1}{2}\times(-9.8)\times(0.44)^2 \\ h=1.9-0.95 \\ h=0.95\text{ m} \end{gathered}[/tex]Thus, the maximum height reached by the jumper is 0.95 meters.
a piece of steel expands 10 cm when it is heated from 20 to 40 deg C. How much would it expand if it was heated from 20 to 60 deg.C?
ANSWER
20 cm
EXPLANATION
Given:
• The change in length of a piece of steel, ΔL = 10 cm, when its temperature change is ΔT = 20°C
Find:
• The change in length of the same piece of steel, ΔL, when its temperature is changed ΔT = 40°C
The change in length of a solid material is,
[tex]\Delta L=\alpha\cdot L_o\cdot\Delta T[/tex]Where L₀ is the original length, α is the linear expansion coefficient and ΔT is the change in temperature.
In this case, we know how much is the piece of steel expanded when the temperature change is 20°C, so we can find the product αL₀,
[tex]\alpha L_o=\frac{\Delta L}{\Delta T}[/tex]Replace the known values and solve,
[tex]\alpha L_o=\frac{10\text{ }cm}{20\degree C}=0.5cm/\degree C[/tex]If the temperature change now is 40°C,
[tex]\Delta L=\alpha L_o\Delta T=0.5cm/\degree C\cdot40\degree C=20cm[/tex]Hence, when the piece of steel is heated from 20°C to 60°C it will expand 20 centimeters.
A traffic light hangs from a pole as shown in (Figure 1). The uniform aluminum pole AB is 7.20 m long and has a mass of 11.0 kg . The mass of the traffic light is 22.0 kg .
-Determine the tension in the horizontal massless cable CD.
-Determine the vertical component of the force exerted by the pivot A on the aluminum pole.
-Determine the horizontal component of the force exerted by the pivot A on the aluminum pole.
There is a 409.30 N force in the horizontal massless cable CD.
The vertical component of the pivot A's force on the aluminum pole, which is 323.73N.
The force that the pivot A applies to the aluminum pole has a horizontal component of 409.30 N.
Force is any influence to the motion of the body .
Basically the product of mass and acceleration.Length of the pole = 7.20 m
Mass of the pole = m = 11.0 kg
Mass of the traffic light = M = 22.0 kg
Let the length of the rod AD be L
Also [tex]Lsin\alpha = 3.8[/tex]
[tex]\alpha = 37[/tex]°
[tex]Sin\alpha = 0.602[/tex]
L = [tex]\frac{3.8}{0.602}[/tex]
L = 6.312 m
In order to determine the tension (T) in the cable, the free body diagram will give the detailed information.
On balancing all the forces in both the 'x' and 'y' direction, i.e. summation of all the forces in 'x' and 'y' direction must equal to zero.∑[tex]F_{x}[/tex] = 0
[tex]R_{x}[/tex] - T = 0
[tex]R_{x}[/tex] = T
∑[tex]F_{y}[/tex] = 0
[tex]R_y- W_p - W_l = 0[/tex]
[tex]W_p[/tex] = 9.81 * 11 = 107.91 kg
[tex]W_l\\[/tex] = 9.81 * 22 = 215.82 kg
[tex]R_y\\[/tex] = [tex]W_p + W_l[/tex]
[tex]R_y\\[/tex] = 107.91 + 215.82
[tex]R_y\\[/tex] = 323.73 N
The vertical component of the force exerted by the pivot A on the aluminum pole is 323.73 N
Taking moment along x-axis
M = 0
[tex]Th-mgCos\alpha* \frac{1}{2} - mglCos\alpha[/tex] = 0
T = [tex]\frac{glCos\alpha }{h} *\frac{m}{2}*M[/tex]
T = 409.30 N
The tension in the horizontal massless cable CD is 409.30 N
[tex]R_{x}[/tex] = T = 409.30 N
The horizontal component of the force exerted by the pivot A on the aluminum pole is 409.30 N.
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Could someone please help me ASAP?
As shown in the picture a person is swinging a yo-yo in the circle then the direction of the velocity vector is given by the vector D.
What is Velocity?
The total displacement covered by any object per unit of time is known as velocity. It depends on the magnitude as well as the direction of the moving object.
As given in the problem shown in the picture a person is swinging a yo-yo in a circle, we have to find the direction of the velocity vector,
As seen in the image, a person is swinging a yo-yo in a circle, and the vector D indicates the direction of the velocity vector.
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What is the average velocity of a car that travels 48 km north in 2.0 h?
The average velocity of the car would be 24 kilometers per hour in the north direction if the car travels 48 kilometers in the north for 2 hours.
What is Velocity?The total displacement covered by any object per unit of time is known as velocity. It depends on the magnitude as well as the direction of the moving object.
As given in the problem, we have to find the average velocity of the car if the car travels 48 kilometers in the north for 2 hours.
The average velocity of the car = 48 kilometers / 2 hours
= 24 kilometers per hour
Thus, the average velocity of the car would be 24 kilometers per hour
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the answer to this question
The car and the delivery truck both start from rest and accelerate at the same rate. So, the final speed of the car as compared to the truck is four times as much. So, option D is correct.
What is acceleration?Acceleration is the term used in mechanics to describe how quickly an object's velocity changes in relation to time. The magnitude of accelerations as a vector. An object will accelerate in the direction that it is being pulled in by the net force.
Acceleration is a vector quantity since it has a magnitude and a direction. Velocity is another aspect of vessel quantities. The change in the vector of velocity over a time interval divided by the time interval is the definition of acceleration.
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The potential energy of a system can be changed by varying the position of objects in the system. At which point do the coaster cars have the most potential energy?
To find:
At which point the potential energy of the coaster car is the highest?
Explanation:
The potential energy of an object is the energy possessed by the object due to the position of the object. The gravitational potential energy of the object is directly proportional to the height of the object.
Thus an object will have the highest potential energy when it is at the highest point.
Final answer:
Thus the coaster cars will have the highest potential energy when it is at the highest point on the roller coaster.
Therefore, the coaster cars will have the most potential energy at point A.
The displacement (in meters) of a particle moving in a straight line is given by S= t^2 - 7t + 17 ii)iii)iv)
Given displacement of a particle moving in a straight line,
[tex]S=t^2-7t+17[/tex](i) Calculate the average velocity in the interval [3,4]
[tex]\begin{gathered} \text{Average velocity = }\frac{S(4)-S(3)}{4-3} \\ \text{Average velocity = }\frac{\lbrack(4)^2-7\times4+17\rbrack-\lbrack(3)^2-7\times3+17\rbrack}{4-3} \\ \text{Average velocity =}\frac{5-5}{1} \\ \text{Average velocity = 0 m/s} \end{gathered}[/tex](iii) Calculate the average velocity in the time interval [4,5].
[tex]\begin{gathered} \text{Average velocity = }\frac{S(5)-S(4)}{5-4} \\ \text{Average velocity = }\frac{\lbrack(5)^2-7\times5+17\rbrack-\lbrack(4)^2-7\times4+17\rbrack}{5-4} \\ \text{Average velocity = }\frac{7-5}{1} \\ \text{Average velocity = }2 \end{gathered}[/tex](iv) Calculate the average velocity in the time interval [4,4.5].
[tex]\begin{gathered} \text{Average velocity = }\frac{S(4.5)-S(4)}{4.5-4} \\ \text{Average velocity = }\frac{\lbrack(4.5)^2-7\times4.5+17\rbrack-\lbrack(4)^2-7\times4+17\rbrack}{4.5-4} \\ \text{Average velocity = }\frac{5.75-5}{0.5} \\ \text{Average velocity = }1.5 \end{gathered}[/tex]Point charges create equipotential lines that are circular around the charge (in the plane of the paper). What is the potential energy, in nJ, of a 1 nC charge located 1.99 m from a 2 nC charge ?
The potential energy between two charges can be written as:
[tex]U_e=\frac{kq_1q_2}{r}[/tex]In our case, it'll be equal to:
[tex]U_e=\frac{9*10^9*1*10^{-9}*2*10^{-9}}{1.99}=9.045nJ[/tex]Then, our answer is PE=9.045nJ
A 3 kg ball is dropped from a height of 100 m above the surface of Planet Z. If the ball reaches a velocity of 45 m/s in 7 s, what is the ball’s weight on Planet Z? What is the gravitational field strength on Planet Z?
We are given the following information
Mass of ball = 3 kg
Height = 100 m
Final velocity = 45 m/s
Time = 7 s
Recall from the equations of motion
[tex]s=u\cdot t+\frac{1}{2}\cdot g\cdot t^2[/tex]Where u is the initial velocity of the ball that is zero.
[tex]\begin{gathered} s=u\cdot t+\frac{1}{2}\cdot g\cdot t^2 \\ 100=0\cdot7+\frac{1}{2}\cdot g\cdot7^2 \\ 100=\frac{1}{2}\cdot g\cdot49 \\ g=\frac{2\cdot100}{49} \\ g=4.08\; \frac{m}{s^2} \end{gathered}[/tex]So, the gravitational acceleration of the planet Z is 4.08 m/s^2
The weight o the ball is given by
[tex]\begin{gathered} W=m\cdot g \\ W=3\cdot4.08 \\ W=12.24\; N \end{gathered}[/tex]Therefore, the weight of the ball is 12.24 N
When a 6.0-F capacitor is connected to a generator whose rms output is 25 V, the current in the circuit is observed to be 0.40 A. What is the frequency of the source?
The frequency of the source is 1.66 Hz.
What is the impedance?Let us recall that the impedance of the circuit is the opposition that is offered to the flow of current by a circuit component that is not a resistor. Now let us find the impedance.
I = V/Z
I = current
V = voltage
Z = impedance
Z = V/I
Z = 25/0.4
Z = 62.5 ohm
Z^2 = R^2 + Xc^2
Z^2 = Xc^2
Xc= Z
Xc = 2πfC
f = frequency
C = capacitance
f= Xc/2πC
f = 62.5/2 * 3.142 * 6
f = 1.66 Hz
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Use the acceleration vs time graph to answer this question. The graph shows the motion with an initial velocity of -4 m/s. Each tick mark on the x-axis represents 1 second. Calculate the velocity at t = 8 seconds.
Answer:
4 m/s
Explanation:
To find the velocity at t = 8 seconds, we will use the following equation:
[tex]v_f=v_i+at[/tex]Where vf is the final velocity, vi is the initial velocity, a is the acceleration and t is the time.
From t = 0 seconds to t = 3 seconds, we have an acceleration of 6 m/s², so we can calculate the velocity at t = 3 seconds as:
[tex]\begin{gathered} v_f=-4m/s+6m/s^2(3\text{ s)} \\ v_f=-4\text{ m/s + 18 m/s} \\ v_f=14\text{ m/s} \end{gathered}[/tex]Now, from t = 3 seconds to t = 8 seconds, the acceleration is equal to -2 m/s². So we need to use the same equation but this time, the initial velocity will be 14 m/s and the time will be 5 seconds because t = 8 s - 3 s = 5s. Then, we get:
[tex]\begin{gathered} v_f=14m/s-2m/s^2(5s) \\ v_f=14\text{ m/s - 10 m/s} \\ v_f=4\text{ m/s} \end{gathered}[/tex]Therefore, the velocity at t = 8 seconds is 4 m/s
The density of copper is 8.94 g/cm^3. What is the mass of a rectangular sheet of copper: 10 cm wide, 45 cm long, and 0.2 cm thick? I have no idea how to solve for mass or where to start. :(
Mass = 804.6 g
Explanation:The dimension of the rectangular copper = 10 cm wide, 45 cm long, and 0.2 cm thick
The volume of the rectangular copper = 0.1m x 0.45m x 0.002m
The volume of the rectangular copper = 0.00009 m³
The density of copper = 8.94 g/cm³ = 8.94 x 1000 kg/m³
The density of copper = 8940 kg/m³
Mass = Density x Volume
Mass = 8940 x 0.00009
Mass = 0.8046 kg
Mass = 804.6 g
help, Asap!!!!!!!!!!!!!!!!!!
Answer:
0
Explanation:
all of the movement and opposite movement are the same
A helicopter takes off and travels forward at an angle of 59.4 above horizontal. After following this path for 294 meters, the pilot changes the angle of flight to 10.5 degrees above horizontal and follows this path for 849 meters. After these two legs, what is the helicopter’s horizontal distance from the point of take off?
985 m
447 m
408 m
964 m
After the two legs, the helicopter’s horizontal distance from the point of take off is 979 m
For the first leg,
d = 294 m
θ = 59.4°
[tex]d_{x}[/tex] = d cos θ
[tex]d_{x}[/tex] = 294 * cos 59.4°
[tex]d_{x}[/tex] = 147 m
For the second leg,
d = 849 m
θ = 10.5°
[tex]d_{x}[/tex] = d cos θ
[tex]d_{x}[/tex] = 849 * cos 10.5°
[tex]d_{x}[/tex] = 832 m
Total horizontal distance = [tex]d_{x}[/tex] ( 1st leg ) + [tex]d_{x}[/tex] ( 2nd leg )
Total horizontal distance = 147 + 832
Total horizontal distance = 979 m
Therefore, after the two legs, the helicopter’s horizontal distance from the point of take off is 979 m
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Which of these uses digital signals to store, send, or receive information?
Answer:
here you go
Explanation:
For storing information we use digital signals since that’s what our digital computer storage uses.
For transmitting signals in a computer we use digital signals by simply changing the voltage on the data route or bus.
For wirelessly transmitting signals we use analog signals since electromagnetic radiation is analog itself. Although, the information can be analog modulated OR digitally modulated, depending on the application.
The inductor in the RLC tuning circuit of an AM radio has a value of 9 mH. What should be the value of the variable capacitor, in picofarads, in the circuit to tune the radio to 94 kHz
We are asked to determine the capacitance of an RLC circuit given the frequency. To do that we will use the following formula:
[tex]C=\frac{1}{4\pi^2f^2L}[/tex]Where:
[tex]\begin{gathered} C=\text{ capacitance} \\ f=\text{ frequency} \\ L=\text{ inductance} \end{gathered}[/tex]Now, We plug in the values:
[tex]C=\frac{1}{4\pi^2(94\times10^3Hz)^2(9\times10^{-3}H)}[/tex]Now, we solve the operations:
[tex]C=3.19\times10^{-10}F[/tex]The capacitance is 3.19x10^-10 farads. In Picofarads this is equivalent to:
[tex]C=0.0319pF[/tex]A block of known mass hanging from an ideal spring of known spring constant is oscillating vertically. A motion detector records the position, velocity, and acceleration of the block as a function of time. Which of the following indicates the measured quantities that are sufficient to determine whether the net force exerted on the block equals the vector sum of the individual forces?A. Acceleration only B. Acceleration and position only C. Acceleration and velocity onlyD. Acceleration, position, and velocityPart 1. “Whether The net force exerted on the block equals the vector sum of the individual forces” really means “Newton’s Second Law”. The problem wants you to make measurements to show that the net force equals mass times acceleration. How would you find the force exerted by the spring? How do you find the force exerted by gravity? Part 2. Connect your answer to the previous question with the right answer. Clearly explain which quantities must be measured (between acceleration, velocity, and position) and explain what each quantity is used for to show Newton’s second law.
Answer:
B. Acceleration and position only
Explanation:
We need to identify the measurements that show that the net force is equal to the sum of the force exerted by the spring and the force of gravity, so we want to know if the following equation is satisfied
[tex]\begin{gathered} F_{net}=F_s-mg \\ ma=k(\Delta x)-mg \end{gathered}[/tex]Where m is the mass, a is the acceleration, k is the spring constant, Δx is the stretched, and g is the gravity. The mass m, the spring constant k, and the gravity g are known. So, the measurement quantities that we need are the acceleration and the position.
So, the answer is
B. Acceleration and position only
Part 1.
How would you find the force exerted by the spring?
The force exerted by the spring is equal to k(Δx ), so to find Δx, we need to identify the position.
How do you find the force exerted by gravity?
The force exerted by gravity is calculated as mass times gravity, so it is known.
Part 2.
We need to measure Acceleration and position.
The acceleration to calculate the net force because by the second law of newton Fnet = ma
The position to calculate the force exerted by the spring.
Ellipses have only one focus. Is this true or false?
In any ellipse there are two foci. This two points are fixed and are fundamental for the construction opf the ellipse. Therefore the statement is false.