Given:
the height of the object is
[tex]h_0=5\text{ cm}[/tex]The distance of the object is
[tex]d_0=-20\text{ cm}[/tex]The focal length of the lens is
[tex]f=15\text{ cm}[/tex]Required: the distance of the image and height of the image.
Explanation:
the lens formula is given by
[tex]\frac{1}{f}=\frac{1}{d_i}-\frac{1}{d_0}[/tex]Plugging all the values in the above relation, we get:
[tex]\begin{gathered} \frac{1}{15\text{ cm}}=\frac{1}{d_i}-\frac{1}{-20\text{ cm}} \\ \frac{1}{d_i}=\frac{1}{15\text{ cm}}-\frac{1}{20\text{ cm}} \\ \frac{1}{d_i}=\frac{4-3}{60\text{ cm}} \\ d_i=60\text{ cm} \end{gathered}[/tex]Thus, the distance of the image is 60 cm.
now calculate the height of the image
we know that
[tex]\frac{h_i}{h_0}=\frac{d_i}{d_0}[/tex]substitute all the values in the above relation, we get:
[tex]\begin{gathered} h_i=5\text{ cm}\times\frac{60}{-20} \\ h_i=-15\text{ cm} \end{gathered}[/tex]Thus, the height of the image is 15 cm.
A 150 N crate is being pulled up a perfectly smooth ramp that slopes upward at 15 degrees by a pull that is directly at 30 degrees above the surface of the ramp. What is the magnitude of the pull required to make the crate move up the ramp at a constant velocity of 1.75 m/s?
The magnitude of the pull required to make the crate move up the ramp at a constant velocity of 1.75 m/s is 44.83 N.
What is the magnitude of the force required?
The magnitude of the force pull required to make the crate move up the ramp at a constant velocity of 1.75 m/s is calculated by applying Newton's second law of motion.
F(net) = ma
where;
F(net) is the net force on the cratem is the mass of the cratea is the acceleration of the crateat constant velocity, the acceleration of the crate = 0
F(net) = 0
Fx - Fgx = 0
where;
Fx is the horizontal component of the applied forceFgx is the horizontal component of the weight of the crateFcos(30) - 150sin(15) = 0
F = 150 sin(15) / cos30
F = 44.83 N
Learn more about applied force here: https://brainly.com/question/24255032
#SPJ1
Find the power output of the engine of a 1200 kg car while the car accelerates from 30 km/h to 100 km/h in 10 s.
According to work energy thereom, the work done on the car to accelerate is,
[tex]W=\frac{1}{2}mv^2-\frac{1}{2}mu^2[/tex]Plug in the known values,
[tex]\begin{gathered} W=\frac{1}{2}(1200kg)(100km/h)^2(\frac{1000\text{ m}}{1\text{ km}})^2(\frac{1\text{ h}}{3600\text{ s}})^2(\frac{1\text{ J}}{1kgm^2s^{-2}})-_{}\frac{1}{2}(1200kg)(30km/h)^2(\frac{1000\text{ m}}{1\text{ km}})^2(\frac{1\text{ h}}{3600\text{ s}})^2(\frac{1\text{ J}}{1kgm^2s^{-2}}) \\ =462963\text{ J-}41667\text{ J} \\ =421296\text{ J} \end{gathered}[/tex]The power output of the engine can be given as,
[tex]P=\frac{W}{t}[/tex]Substitute the values,
[tex]\begin{gathered} P=\frac{421296\text{ J}}{10\text{ s}}(\frac{1\text{ W}}{1\text{ J/s}}) \\ =42129.6\text{ W} \end{gathered}[/tex]Therefore, the power output of the engine is 42129.6 W.
URGENT!! ILL GIVE
BRAINLIEST!!!! AND 100 POINTS!!!!!!
Answer:
Its the fourth/last answer
A 5 kg mass, hung onto a spring, causes the spring to stretch 7.0 cm. What is the spring constant? What is the potential energy of the spring?
Given,
The mass is m=5 kg.
The extensionis d=7.0 cm
The force is
F=mg
F=5x0.07=0.35N
Thus the spring constant is:
[tex]k=\frac{F}{d}=\frac{0.35}{0.07}=\frac{5N}{m}[/tex]The potential energy is:
[tex]U=\frac{1}{2}kd^2=\frac{1}{2}\times5\times(0.07)^2=0.012Nm^2[/tex]A particle moves in a straight line, and you are told that the torque acting on it is zero about some unspecified origin. Does this necessarily imply that the total force on the particle is zero? Can you conclude that its angular velocity is constant?
Horizontal force will be there but angular force is zero because it is dependent on angle and angular velocity is constant.
[tex]\tau = rF\sin\theta[/tex]
T= torque
r= radius
In mechanics, every action that tries to preserve, alter, or change a body's motion is referred to as a force. Isaac Newton's Principia Mathematica has three principles of motion that are usually used to explain the concept of force (1687).
Newton's first law states that in the absence of an external force, a body will continue to be in either its resting or evenly moving condition along a straight path. According to the second law, any time an outside force acts on a body, the body accelerates (changing velocity) in the force's direction.
To know more about force visit : brainly.com/question/13191643
#SPJ9
If you drop a 2.3kg ball from the top of a 44 m high building, how much potential energy will it have just before it hits the ground? Round your answer to the nearest whole number and include an appropriate unit
Answer: Potential energy = 0
Explanation:
The formula for calculating potential energy is expressed as
Potential energy = mgh
where
m = mass of object
g = acceleration due to gravity = 9.8m/s^2
h = height of object above the ground
From the information given,
m = 2.3
h = 44
Just before the object hits the ground, all potential energy is converted to kinetic energy. Thus, the answer is
Potential energy = 0
An air-filled pipe is found to have successive harmonics at 800 Hz , 1120 Hz , and 1440 Hz . It is unknown whether harmonics below 800 Hz and above 1440 Hz exist in the pipe. The length is 53.5 cm. Identify the correct pressure variation graph for the 1120 Hz standing wave in the pipe. Note that the closed end of the pipe is on the right.
Using the concept of Harmonic-wave, we got the desired graph which is shown in the image.
It must be an open-closed pipe.
Open-closed pipe
Let the fundamental frequency of f be
The second harmonic is 3f.
3rd harmonic = 5f
Therefore, the difference between harmonics is 3f-f = 2f
So, 2f = (1120-800) = 1440-1120 = 320
Also, f = 160Hz
800Hz = 5th Harmonic
1120 Hz = 7th
1440 = 9th
For an open- or closed-pipe,
f = v/4L
where v is the speed of sound in the air = 343 M/s
So, 160 = 343/4L
Also, L = 0.5336 m = 53.6cm
Hence for the frequency of 1120Hz, we got the desired graph of a wave which is shown in the below graph.
To know more about the mechanical wave, visit here:
https://brainly.com/question/8885243
#SPJ9
rock that is attached to a 2.6 m rope is whirled around and covers 3 Radians every 1.2seconds. What is the Tangential velocity of the rock when the rope is released?
Answer: Tangential velocity = 6.5 m/s
Explanation:
The formula for calculating tangential velocity is expressed as
v = rw
where
v represents tangential velocity
r represents the radius of the circular path
w is the angular velocity
From the information given,
r = 2.6
If 3 radians is covered in 1.2 seconds, then
angular velocity, w = distance /time = 3/1.2
w = 2.5 rad/s
Thus,
v = 2.6 x 2.5
v = 6.5 m/s
Tangential velocity = 6.5 m/s
A simple machine does 30 J of work with an efficiency of 28%. How much energy was put into the machine?
Take into account that efficiency is given by the following expression:
[tex]\text{ efficiency=(output work/input work)}\cdot100[/tex]In this case, you have:
efficieny = 28%
output work = 30J
input work = ?
Replace the previous values of the parameters into the formula for efficieny, solve for input work and simplify:
28% = (30J/input work)*100%
input work = 30J*(100% / 28%)
input work = 107.14 J
Hence,about 107.14 J was put into the machine
The cores of the terrestrial worlds are made mostly of metal because ______.
A. metals sunk to the centers a long time ago when the interiors were molten throughout
B. the terrestrial worlds as a whole are made mostly of metal
C. the core contained lots of radioactive elements that decayed into metals
D. over billions of years, convection gradually brought dense metals downward to the core
A yo‑yo with a mass of 0.0600 kg and a rolling radius of =1.60 cm rolls down a string with a linear acceleration of 5.20 m/s2.
Calculate the tension magnitude in the string and the angular acceleration magnitude of the yo‑yo.
What is the moment of inertia of this yo‑yo?
The tension magnitude in the string is 0.312 N.
The angular acceleration magnitude of the yo‑yo is 325 rad/s².
The moment of inertia of the yo-yo is 7.68 x 10⁻⁶ kgm².
What is the moment of inertia of the yo-yo?
The moment of inertia of the yo-yo is calculated by applying the following equation as shown below;
I = ¹/₂MR²
where;
M is the mass of the yo-yoR is the radius of the yo-yoI = ¹/₂(0.06 kg)(0.016)²
I = 7.68 x 10⁻⁶ kgm²
The angular acceleration of the yo-yo is calculated as follows;
α = a/R
where;
a is the linear accelerationα = 5.2/0.016
α = 325 rad/s²
The tension in the string is calculated as follows;
T = ma
where;
m is the mass of the yo-yoa is the linear acceleration of the yo-yoT = 0.06 kg x 5.2 m/s²
T = 0.312 N
Learn more about moment of inertia here: https://brainly.com/question/14460640
#SPJ1
Which pan of water shows molecules that have received the most heat from a stove?Select one:a. Ab. Bc. Cd. D
Given:
The water in the pan is heated the most.
To find:
Pan of water that has received the most heat from the stove.
Explanation:
When water is heated, the water molecules in the water acquire kinetic energy due to a rise in water temperature. The more the water is heated, the higher the temperature of the water. Thus the kinetic energy of molecules of water will be more.
The higher kinetic energy of water molecules corresponds to the higher speed of the water molecules.
Thus, in option (a), the water molecules are moving very fast, meaning they have high kinetic energy due to high temperature. And the temperature will be maximum if we provide maximum heat to the water.
Thus, option (a) is correct.
Conclusion: The correct option is option (a).
How do I solve this problem? String one is 5kg String 2 is 11 kg
ANSWERS
• Forces on the 5kg block: ,3
,• Forces on the 11kg block: ,2
,• Forces balanced? ,Yes, the system is at rest.
,• Tension in string 1:
,• Tension in string 2: ,107.91 N
EXPLANATION
First we have to draw the forces on each block:
Hence, block 1 has 3 forces acting on it, while block 2 has 2 forces acting on it.
It is said that the system is at rest, which means that the forces on the system are balanced.
By Newton's second law we know that,
[tex]F_{t1}-F_{t2}-F_{g1}=0[/tex]And,
[tex]F_{t2}-F_{g2}=0[/tex]Both equal zero because the blocks are not moving, and therefore there's no acceleration.
From the second equation we can find the tension in string 2,
[tex]\begin{gathered} F_{t2}=F_{g2} \\ F_{t2}=m_2\cdot g \end{gathered}[/tex]m2 = 11kg and g = 9.81m/s²,
[tex]F_{t2}=11\operatorname{kg}\cdot9.81m/s^2=107.91N[/tex]Now that we know that the tension in string 2 is 107.91N, we can find the tension in string 1 replacing this into the first equation and solving for Ft1,
[tex]F_{t1}=F_{t2}+F_{g1}[/tex][tex]undefined[/tex]If a planet was located approximately 25 thousand light-years from the center of a galaxy and orbits that center once every 247 million years, how fast is the planet traveling around the galaxy in km/hr? If needed, use 3.0 × 10^8 m/s for the speed of light.
The distance of the planet from the centre i.e. radius,
[tex]r=25000\text{ light years}[/tex]The time period is
[tex]T=247\text{ million years}[/tex]The speed of the planet is given by the formula
[tex]v=\frac{2\pi r}{T}[/tex]Substituting the values, the speed will be
[tex]\begin{gathered} v=\frac{2\times\pi\times25000\text{ }\times9.461\times10^{12}}{247\times10^6\times876} \\ =6.866\times10^5\text{ km/hr} \end{gathered}[/tex]From the list of world record track events in the table, calculate the average speeds for each race. Assume the length of each event is known to the nearest 0.1 m. Calculate only the last one, for 10,000 m.
Answer:
0.095 m/s
Explanation:
The average speed can be calculated as
Avg speed = distance/time
For the last one, we get that distance is 10000 m and the time 29 hours, 17 min, and 45 seconds.
So, first, we need to convert 29:17.45 to seconds as follows
29 hours x 3600 s/ 1 hour = 104400 s
17 min x 60 s / 1 min = 1020 s
45 s = 45 s
Total time = 104400 s + 1020 s + 45 s
Total time = 105,465 s
Then, the average speed is equal to
Avg speed = 10,000 m/105,465 s
Avg speed = 0.095 m/s
Therefore, the answer is 0.095 m/s
An object weighing 297 N in air is immersedin water after being tied to a string connectedto a balance. The scale now reads 263 N.Immersed in oil, the object appears to weigh270 N.Find the density of the object.Answer in units of kg/m^3Do not round.
Given
Wair = 297 N
Wwater = 263 N
Density of water = 1000 kg/m^3
Acceleration due to gravity = 9.8
Procedure
We can calculate the thrust of the water with the difference of weights
[tex]\begin{gathered} W_{\text{air}}-W_{\text{water}}=\rho_wgV \\ V=\frac{W_{\text{air}}-W_{\text{water}}}{\rho_wg} \\ V=\frac{297N-263N}{1000\cdot9.8} \\ V=0.00347\text{ m}^3 \end{gathered}[/tex]Now for the object
[tex]\begin{gathered} \rho_{\text{obj}}=\frac{W_{\text{air}}}{gV} \\ \rho_{\text{obj}}=\frac{297}{9.8\cdot0.00347} \\ \rho_{\text{obj}}=8735.29kg/m^3 \end{gathered}[/tex]
Find the net force on the 4th part. Given, the value of velocity remain same at every point.
In the fourth case, when the weight and the tension are acting vertically downward.
The net force acting on the bucket is,
[tex]F_{\text{net}}=\frac{mv^2}{r}[/tex]where v is the velocity, r is the radius, and m is the mass,
The net force acting on the bucket will remain the same if the magnitude of the velocity remains the same.
As the direction of the bucket changes thus, the value of net force is not equal to zero.
As the velocity of the bucket is 3 m/s.
Thus, the value of net force acting on the bucket is,
[tex]\begin{gathered} F_{\text{net}}=\frac{0.5\times3^2}{0.6} \\ F_{\text{net}}=7.5\text{ N} \end{gathered}[/tex]Thus, (at the constant velocity at every point case) the net force acting is 7.5 N.
A block weighing 200N is pushed along a surface. If it takes 80N toget the block moving and 40N to keep the block moving at a constantvelocity, what are the coefficient of friction us and uk ?
Answer:
see below
Explanation:
coefficient of static F = 80 / 200 = .4
kinetic F = 40 /200 = .2
What is the potential difference across a curling iron if there is a current of 2.5 A transferring 9360 J in 32s?
Power is expressed as energy/time
Thus means that
power = energy/time
From the information given,
energy = 9360 J
time = 32 s
Thus,
power = 9360/32 = 292.5 J/s = 292.5 watts
The formula relating potential difference which is the same as voltage, current and power is expressed as
P = IV
where
P = power = 292.5
I = current = 2.5
V = potential difference = ?
Thus, we have
292.5 = 2.5V
V = 292.5/2.5
V = 117 V
The potential difference is 117 V
An object is dropped from the top of a building. How fast is it moving after 5s? What is the acceleration at this time?
Any object under free fall accelerates at a constant rate given by the gravitational acceleration:
[tex]g=9.81\frac{m}{s^2}[/tex]On the other hand, the speed v of an object under free fall after t seconds, if it starts from rest, is given by the formula:
[tex]v=gt[/tex]Replace g=9.81m/s^2 and t=5s to find the speed of the object 5 seconds after the object is dropped:
[tex]v=(9.81\frac{m}{s^2})(5s)=49.05\frac{m}{s}[/tex]Therefore, the speed of the object after 5 seconds is approximately 49 meters per second. Its acceleration is always the same and it is equal to 9.81 meters per second squared.
b. Calculate the electric force that exists between two objects that are 0.500 m apart and carrycharges of 0.00450 C and 0.00240
Given
The two charges,
[tex]\begin{gathered} q_1=0.00450\text{ C} \\ q_2=0.00240\text{ C} \end{gathered}[/tex]Distance between them,
[tex]r=0.5\text{ m}[/tex]To find
The electric force
Explanation
We know the force is given by
[tex]F=\frac{kq_1q_2}{r^2}[/tex]Putting the values,
[tex]\begin{gathered} F=9\times10^9\frac{0.00450\times0.00240}{(0.500)^2} \\ \Rightarrow F=388,800\text{ N} \end{gathered}[/tex]Conclusion
The electric force is 388,800 N
A certain violet light has a wavelength of 413 nm. Calculate the frequency of the light. Calculate the energy content of one quantum of the light.
We will have the following:
First, we transform from nm to m,t that is:
[tex]\lambda=413nm\cdot\frac{m}{1\cdot10^9nm}\Rightarrow\lambda=4.13\cdot10^{-7}m[/tex]Then:
[tex]4.13\cdot10^{-7}m=\frac{3.0\cdot10^8m/s}{f}\Rightarrow f=\frac{3.0\cdot10^8m/s}{4.13\cdot10^{-7}m}[/tex][tex]\Rightarrow f\approx7.26\cdot10^{14}Hz[/tex]So, the frequency of the ligth is approximately 7.26*10^14 Hz.
Now, the energy will be:
[tex]E=(6.63\cdot10^{-34}m^2Kg/s)(7.26\cdot10^{14}/s)\Rightarrow E\approx4.81\cdot10^{-19}m^2Kg/s^2[/tex][tex]\Rightarrow E\approx4.81\cdot10^{-19}m^2Kg/s^2\cdot(J/m^2Kg/s^2))\Rightarrow E\approx4.81\cdot10^{-19}J[/tex]And the energy is approximately 4.81*10^-19 J.
A person is pushing on a car with a force of <-14,4> and another person is pushing on a car with a force of -6,1> What is the net force in the vertical direction?
We are given the following vector forces:
[tex]\begin{gathered} F_1=(-14,4) \\ F_2=(-6,1) \end{gathered}[/tex]These vector forces are expressed as:
[tex]F=(x,y)[/tex]Where the first component "x" is the horizontal component, and "y" is the vertical component. Since we are asked about the net force in the vertical component we need to add the components "y" of both vectors, we get:
[tex]\Sigma F_y=4+1=5[/tex]Therefore, the net force in the vertical direction is 5.
A pitcher throws a 0.140 kg baseball with a speed of 42.3 m/s the batter strikes it with an average force of 5120 N which result in the ball traveling with an initial speed of 31.0 m/s toward the pitcher for how long were the bat and ball in contact
The impulse exerted over an object is equal to the change in its linear momentum:
[tex]I=\Delta p[/tex]On the other hand, the impulse is equal to the force exerted over the object multiplied by the time during which the force was exerted:
[tex]I=F\cdot\Delta t[/tex]First, find the impulse by finding the change in linear momentum of the baseball. The linear momentum is given by the product of the speed of the baseball times its mass:
[tex]p=mv[/tex]Assume that the negative direction is towards the batter and the positive direction is towards the pitcher. Then, the initial velocity of the ball is -42.3 m/s and the final velocity of the ball is 31.0 m/s. Then:
[tex]\begin{gathered} p_i=mv_i=(0.140\operatorname{kg})(-42.3\frac{m}{s})=-5.922\operatorname{kg}\cdot\frac{m}{s} \\ \\ p_f=mv_f=(0.140\operatorname{kg})(31.0\cdot\frac{m}{s})=4.34\operatorname{kg}\cdot\frac{m}{s} \end{gathered}[/tex]Use the initial and final linear momentum to find the change in linar momentum, which is equal to the impulse:
[tex]\begin{gathered} I=\Delta p \\ =p_f-p_i \\ =(4.34\operatorname{kg}\cdot\frac{m}{s})-(-5.922\operatorname{kg}\cdot\frac{m}{s}) \\ =10.262\operatorname{kg}\cdot\frac{m}{s} \end{gathered}[/tex]Isolate Δt from the equation that relates force and impulse and substitute the corresponding values for I and F to find the time during which the bat and the ball were in contact:
[tex]\begin{gathered} \Delta t=\frac{I}{F} \\ =\frac{10.262\operatorname{kg}\cdot\frac{m}{s}}{5120N} \\ =0.00200429687\ldots s \\ \approx0.00200s=2.00ms \end{gathered}[/tex]Therefore, the bat and the ball were in contact during a time interval of 2 miliseconds.
3. 1.819 m4. 5.291 m5. 6.321 m39. Answer: B40. Work done by the non conservative forces actingon an object is equal1.to the change in the mechanical energy of the object2. to the change in the kinetic energy of the object3. to the work done by the conservative forces4. to the change in the potential energy of the object5. to the net work done on the object41. Answer: A
We are asked to determine the distance that spring will stretch when a given mass is attached to it.
To do that we will use Hook's law:
[tex]F=kx[/tex]where:
[tex]\begin{gathered} F=\text{ force } \\ k=\text{ spring constant} \\ x=\text{ distance that the spring is stretched} \end{gathered}[/tex]We will determine the constant of the spring "k" first using the fact that the spring is stretched 0.8 meters when a mass of 3kg hangs from it.
Since the only force acting on the spring is the weight of the object we have:
[tex]F=mg[/tex]Where:
[tex]\begin{gathered} m=\text{ mass} \\ g=\text{ acceleration of gravity} \end{gathered}[/tex]Now, we substitute and we get:
[tex]mg=kx[/tex]Now, we divide both sides by "x":
[tex]\frac{mg}{x}=k[/tex]Now, we plug in the values:
[tex]\frac{(3kg)(9.8\frac{m}{s^2})}{0.8m}=k[/tex]Solving the operations:
[tex]36.75\text{ N/m}=k[/tex]Now, we substitute the value of "k":
[tex]F=(36.75\text{ N/m\rparen}x[/tex]Now, we solve for "x":
[tex]\frac{F}{36.75\text{ N/m}}=x[/tex]Now, we substitute the value of the weight of the second object:
[tex]\frac{(14kg)(9.8\frac{m}{s^2})}{36.75\text{ N/m}}=x[/tex]Solving the operations:
[tex]3.733m=x[/tex]Therefore, the spring will stretch by 3.733 meters.
instantaneous velocity for a displacement function dt=2-2t at any given time
a -2t
b-2
c 2
d 2t
Instantaneous velocity for a displacement function [tex]d_{t}[/tex] = 2-2t is -2 m/s. So the correct option is (b)
What is instantaneous velocity?Instantaneous velocity is defined as the rate of change of position over a very short (near zero) time interval. Measured in SI units m/s. The instantaneous velocity is the magnitude of the instantaneous velocity. Same value as instantaneous velocity, but without direction. Simply put, the velocity of an object at that point in time is called the instantaneous velocity. Hence the definition is given as "velocity of a moving object at a given point in time". It can also be determined by taking the slope of the displacement-time plot or the x-t plot.
If the object's velocity is constant, the instantaneous velocity can be the same as the default velocity.
For the given case,
Displacement function [tex]d_{t}[/tex] = 2-2t
Assume x = 2-2t
Time = t
Since [tex]v_{t}[/tex] = [tex]\frac{d_{} _{x} }{d_{t} }[/tex]
[tex]v_{t}[/tex] = [tex]\int\limits^t_0[/tex]2 [tex]d_{t}[/tex] - [tex]\int\limits^t_0[/tex]-2t [tex]d_{t}[/tex]
[tex]v_{t}[/tex] = -2 m/s
To know more about instantaneous velocity visit:
https://brainly.com/question/28837697
#SPJ4
relationship between force of failure and diameter if ultimate tensile strength is same for all material tested i need equation
The relationship between the force and area of the material is,
[tex]F_f=\frac{P}{A}[/tex]where P is the load and A is the area.
As the tensile strength of the material is same for all the material tested.
As the area of the material is directly proportional to the diameter.
Thus, the force of failure is inversely proportional to the diameter.
A set of charged plates have anarea of 5.10*10^-3 m^2 andseparation 1.42*10^-5 m. Howmuch charge must be placed onthe plates to create a potentialdifference of 125 V across them?(The answer is *10^-7 C. Just fill inthe number, not the power.)
Given data:
Area of plates:
[tex]A=5.10\times10^{-3}\text{ m}^2[/tex]Separation between the plates:
[tex]d=1.42\times10^{-5}\text{ m}[/tex]Potential difference:
[tex]V=125\text{ V}[/tex]The capacitance of the capacitor is given as,
[tex]C=\frac{A\epsilon_{\circ}}{d}[/tex]Here, ε_o is the permittivity of the free space.
Substituting all known values,
[tex]\begin{gathered} C=\frac{5.10\times10^{-3}\times8.85\times10^{-12}}{1.42\times10^{-5}} \\ =3.178\times10^{-9}\text{ F} \end{gathered}[/tex]The charge on the capacitor is given as,
[tex]Q=CV[/tex]Substituting all known values,
[tex]\begin{gathered} Q=3.178\times10^{-9}\times125 \\ =3.9725\times10^{-7}\text{ C} \end{gathered}[/tex]Therefore, the charge on the plates is 3.9725×10^-7 C.
if a rocket travles 5600km in 3 hours, what is its speed?
Given the distance traveled by rocket is
[tex]D=5600[/tex]The time taken by the rocket is 3 hours.
[tex]T=3\text{ hours}[/tex]Calculate the speed of the rocket.
[tex]\begin{gathered} \text{Speed = }\frac{5600km}{3\text{ hours}} \\ \text{Speed = 1866.66 }\frac{km}{hr} \end{gathered}[/tex]Thus, the time taken by the rocket is 1866.66 km/hr.
Which of the following choices correctly ranks the colors of visible light from the lowest to the highest frequency?Select one:a. Ab. Bc. Cd. D
d.D
Explanation
Visible light is the small part of the electromagnetic spectrum that we can see.
Colors exist at different wavelengths from lowest energy to highest energy:
The more energy a wave has, the higher its frequency, and vice versa. When it comes to visible light, the highest frequency color, which is violet, also has the most energy. The lowest frequency of visible light, which is red, has the least energy
therefore, the visible colors from the lowest to the highest frequency are
therefore, the answer is
d.D
I hope this helps you