The electric field charges, q₁, and q₂, which are each -6×10⁻⁶ μC, gives:
First part:
The magnitude of the electric field at x = 4 cm is -3.456×10⁷ N/CThe direction of the electric field is towards the origin, along the x-axisSecond part:
The force experienced by the charge is 69.12 NWhat is an electric field?An electric field is the field around a particle that is electrically charged and which exerts a force on charged particles within the field.
The given information are:
The electric charges, q₁ = q₂ = -6 μC
The location of the charge q₁ = y₁ = 3 cm on the y-axis
Location of the charge q₂ = y₂ = -3 cm
First part:
The required location of the point where the electric field magnitude and direction is required is x = 4 cm
The electric field formula is: [tex]\displaystyle{E = \frac{k\cdot q}{r^2}[/tex]
Where:
k = The electrostatic constant ≈ 9 × 10⁹ N·m²/C²
The distances, r, of the charges from the required point are therefore obtained using Pythagorean theorem as follows:
r = √(3² + 4²) = 5
r = 5 cm = 0.05 m
Which gives;
[tex]\displaystyle{E = \frac{9 \times 10^9\times (-6) \times 10^{-6}}{(0.05)^2} = -2.16\times 10^{7}[/tex]
Given that the magnitude of the electric field along the y-axis cancel out, the magnitude of the electric field along the x-axis is found as follows:
[tex]E_x = 2 \times -2.16\times 10^{7}\times \dfrac{4}{5} = -3.456 \times 10^7[/tex]
The magnitude of the electric field at x = 4 is -3.456 × 10⁷ N/C
Second part: The magnitude of the test charge is q₀ = 2 × 10⁻⁶ μC
The force of an electric field, F = E × q
The force experienced by the test charge is therefore:
F = -3.456 × 10⁷ × 2 × 10⁻⁶ = -69.12
The force the test charge experiences is 69.12 N acting towards the origin from the point x = 4 cm.
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PLEASE HELP
A planet's distance from _____ and its _____ both determine its overall gravity.
A) the sun; mass
B) the Kuiper belt; diameter
C) Mars; temperature
D) the Milky Way; perimeter
A cat chases a mouse across a 0.66 m high table. The mouse steps out of the way, and the cat slides off the table and strikes the floor taylor (jdt3899) – Homework 3, 2d motion 22-23 – tejeda – (LermaHPHY1 1) 3 2.4 m from the edge of the table. The acceleration of gravity is 9.81 m/s 2 . What was the cat’s speed when it slid off the table?
The cat’s speed when it slid off the table will be 6.552 m/s
The branch of physics that defines motion with respect to space and time, ignoring the cause of that motion, is known as kinematics. Kinematics equations are a set of equations that can derive an unknown aspect of a body’s motion if the other aspects are provided.
a = -g = 9.8 m[tex]/s^{2}[/tex]
using equation of motion
x = u(horizontal )*t + 1/2 * a (horizontal) * [tex]t^{2}[/tex]
since , a (horizontal) = 0
x = u(horizontal )*t
u = x / t equation 1
similarly
y = u(vertical)*t + 1/2 * a (vertical) * [tex]t^{2}[/tex]
u(vertical) = 0
t = [tex]\sqrt{2y / a}[/tex] equation 2
substituting the value of equation 2 in equation 1
u = x / [tex]\sqrt{2y / a}[/tex]
= [tex]\sqrt{\frac{-9.81}{2*-0.66} } * 2.4[/tex]
= 6.552 m/s
The cat’s speed when it slid off the table will be 6.552 m/s
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A baseball is rolling along a tabletop with avelocity of 3.9 m/s to the right. The tabletopis 1.1 m above the ground. The ball rolls offthe edge of the table and falls to theground.A.) What is the ball's final vertical Velocity?B.) How long does the ball take to fall?C.) how far from the table does the ball land?
To answer this question we need to notice that once the ball starts falling we have a projectile motion; which means that horizontally we have a rectilinear motion and vertically we have an uniformly accelerated motion.
Then we can use the following equations for each direction:
[tex]\begin{gathered} \text{ Horizontal motion:} \\ x=x_0+v_{0x}t \\ \text{ Vertical motion:} \\ a=\frac{v_f-v_0}{t} \\ y=y_0+v_0t+\frac{1}{2}at^2 \\ v_f^2-v_0^2=2a(y-y_0) \end{gathered}[/tex]Since the ball is moving down in the vertical direction we will think that down is the positive direction vertically.
a)
We know that the ball is rolling to the right when it rolls off the edge of the table, this means that vertically the initial velocity is zero; we also know that the ball will fall for 1.1 m and that the acceleration is the gravitational acceleration. Then we can use the third vertical motion equation to find the final velocity, plugging the values we know we have that:
[tex]\begin{gathered} v_f^2-0^2=2(9.8)(1.1) \\ v_f=\sqrt{2(9.8)(1.1)} \\ v_f=4.64 \end{gathered}[/tex]Therefore, the final vertical velocity is 4.64 m/s.
b)
To determine the time we can use the second vertical equation with the values we know:
[tex]\begin{gathered} 1.1=0+0t+\frac{1}{2}(9.8)t^2 \\ 4.9t^2=1.1 \\ t^2=\frac{1.1}{4.9} \\ t=\sqrt{\frac{1.1}{4.9}} \\ t=0.474 \end{gathered}[/tex]Therefore, it takes 0.474 s for the ball to fall.
c)
While the ball is falling it is also moving horizontally, in this direction we know the initial velocity is 3.9 m/s; using the horizontal equations we have:
[tex]\begin{gathered} x=0+(3.9)(0.474) \\ x=1.85 \end{gathered}[/tex]Therefore, the ball lads 1.85 m from the table.
The wiring in a house must be thick enough so it does not become so hot as to start a fire.part aWhat diameter must a copper wire be if it is to carry a maximum current of 34 A and produce no more than 1.6 W of heat per meter of length?
Given:
The maximum current in the circuit is,
[tex]i=34\text{ A}[/tex]The power per length is,
[tex]\frac{P}{l}=1.6\text{ W/m}[/tex]To find:
The diameter of the copper wire
Explanation:
The power (P) produced by current i, through a copper wire of resistance R and length l is given by,
[tex]\begin{gathered} Pl=i^2R \\ \frac{R}{l}=\frac{P}{i^2} \\ \frac{R}{l}=\frac{1.6}{34\times34} \end{gathered}[/tex]Now,
[tex]\begin{gathered} R=\frac{\rho l}{A} \\ R=\frac{\rho l}{\pi r^2} \end{gathered}[/tex]The resistivity of copper is,
[tex]\rho=1.72\times10^{-8}\text{ ohm.m}[/tex]So, we can write,
[tex]\begin{gathered} \frac{R}{l}=\frac{\rho}{\pi r^2} \\ \frac{1.6}{34\times34}=\frac{1.72\times10^{-8}}{\pi r^2} \\ r^2=\frac{1.72\times10^{-8}\times34\times34}{1.6} \\ r=3.5\times10^{-3}\text{ m} \\ diamer\text{ is,} \\ 2r=7.0\times10^{-3}\text{ m} \end{gathered}[/tex]Hence, the diameter is,
[tex]7.0\times10^{-3}\text{ m}[/tex]The wave shown below is headed towards the end to the right. What will happen to the wave when it reaches the end of the string?-The wave will be absorbed by the support.-The wave will be reflected but inverted.-The wave will stop at the end of the string. -The wave will be reflected but not inverted.
Answer:
-The wave will be reflected but inverted
Explanation:
This is a transverse wave because the wave is moving to the right and the particles are moving up and down. When a transverse wave reaches the end, it is reflected and inverted so the crest becomes through and the through becomes valleys. So, the answer is
-The wave will be reflected but inverted.
Because when a wave finds a fixed end, the wave is reflected, which means that there will be a wave with the same speed and amplitude but in the opposite direction.
Alnico is _____.an alloy of metals with strong magnetic propertiesa brittle mixture of substances containing ferromagnetic elementsany material containing ironan element found in nature that behaves like a magnet
Alnico is an alloy made of iron combined with other metals, aluminum, nickel, and cobalt.
The alnico is a permanent magnet
A 6.5 kg lump of clay is sliding to the right on a fricitonless surface with a speed of 23 m/s. It collides head-on and sticks to a 2 kg metal sphere that is sliding to the left with a speed of -7 m/s. What is the kinetic energy of the combined objects after the collision?
The kinetic energy of the combined objects after the collision = 1768.25 Joules
Explanation:The mass of the lump of clay, m₁ = 6.5 kg
The speed of the lump of clay, v₁ = 23 m/s
The mass of the metal sphere, m₂ = 2 kg
The speed of the metal sphere, v₂ = -7 m/s
The Kinetic Energy (KE) of the combined objects after collision is calculated as shown below:
[tex]KE=\frac{1}{2}m_1v^2_1+\frac{1}{2}m_2v^2_2^{}[/tex][tex]\begin{gathered} KE=\frac{1}{2}(6.5)(23^2)+\frac{1}{2}(2)(-7)^2 \\ KE=1719.25+49 \\ KE\text{ = }1768.25J \end{gathered}[/tex]The kinetic energy of the combined objects after the collision = 1768.25 Joules
Which of the following scientists discovered that atoms contain electrons?a. Daltonb. RutherfordC. Thomsond. Bohr
Until 1897, atom was thought as the fundemental particle. But in 1897 J.J Thompson discovered that the atoms contains electrons. He discovered during his experiment with cathod ray tube.
Thus the correct answer is option C.
The fact that light can be polarized and sound cannot be polarized can be best explained by which?a) Light is a transverse wave and sound is not.b) Light travels much faster than sound.c) Light is a longitudinal wave, and sound is not.d) Light travels much slower than sound.
a) Light is a transverse wave and sound is not.
Explanation:
Light waves are transverse waves, their direction of propagation is perpendicular to the direction of vibration. Transverse waves can be polarized.
In
A 3000-kg satellite orbits the Earth in a circular orbit 11797 km above the Earth's surface (Earth radius = 6380 km, Earth Mass = 5.97x10^24 kg). Reminders:Distance should be in meters, not kilometers. 1000 m = 1 km.The total radius needed for the problem is r=r earth + hightWhat is the gravitational force (in newtons, N) between the satellite and the Earth?Hint: The radius of the Earth + the height of the orbit = the center-to-center distance needed for the equation. You also need the universal gravitational constant (G), which is not 9.81 m/s^2. Be careful.Fg=Gm1m2/r2Answer: __________ N
We have:
m1 = mass 1 = 3000 kg
h = height = 11797 km = 11797000 m
r2 = 6380 km = 6380000
m2 = mass 2 = 5.97x10^24 kg
G = gravitational constant = 6.6743 × 10-11 Nm^2 /kg^2
r= distance = h + r2 = 11797000 m + 6390000 m = 18,177,000 m
Apply:
Fg = G m1m2/ r^2
Replacing:
Fg = 6.6743 × 10-11 Nm^2/kg^2 ( 3000 kg * 5.97x10^24 kg ) / (18,177,000 m)^2
Fg= 3,617.9 N
A student makes the following claim, "Acceleration is when an object changes speed, so it can be discussed as a scalar quantity." Explain the error in the student's claim. Provide an example of each quantity to support your answer.
A student makes the following claim, "Acceleration is when an object changes speed, so it can be discussed as a scalar quantity." The error here is that acceleration is said to be done when either speed of that object changes or direction of that object changes. Hence , acceleration is not a scaler quantity.
Scalar quantities are quantities that only have a magnitude and do not have any direction
A vector quantity is defined as the physical quantity that has magnitude as well as directions associated to it.
Acceleration is said to be occurred in two cases :
when the object changes its speed
or when the object changes its direction
since , acceleration depends upon both direction as well as magnitude ,hence it is a vector quantity not a scaler quantity.
for example : a stone attached to a string moving in a circular motion at a constant speed will be considered in accelerated motion because it is constantly changing its direction. Here we can see speed is constant hence magnitude (value of speed) is not changing but direction of the stone is changing . since , direction is changing the object is said to be in accelerated motion.
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Question 24 of 25What disadvantage of analog signals is overcome by sending digital signals?A. The waves used to transmit analog signals carry more energy.B. The waves used to transmit analog signals are more dangerous.dC. Noise decreases the quality of analog signals.O0D. Noise decreases the loudness of analog signals.SUBMIT
The correct answer is option C, "Noise decreases the quality of the analog signals."
The anlog signals q
Vector A= 30 m/s towards East and vector B= 80 m/s towards south. Find A- B [Perform the subtraction of the vector].
ANSWER
[tex]\begin{equation*} 85.44\text{ }m\/s \end{equation*}[/tex]EXPLANATION
First, let us make a sketch of the two vectors:
The vector A - B is represented by line BA in the figure above.
To evaluate A - B, we will apply the Pythagoras theorem:
[tex]\begin{gathered} A-B=BA=\sqrt{(30)^2+(80)^2} \\ A-B=BA=\sqrt{900+6400} \\ A-B=BA=\sqrt{7300} \\ A-B=BA=85.44\text{ }m\/s \end{gathered}[/tex]That is the answer.
an ice has a volume of 8975 ft^3. what is the mass in kilograms of the iceberg? the density of ice 0.917 g/cm^3
The density is given by:
[tex]\rho=\frac{m}{V}[/tex]where V is the volume and m is the mass.
To determine the mass we have to solve the equation for m:
[tex]m=\rho V[/tex]Now, before we can calculate the mass we have to convert the volume given to cubic meter, this comes from the fact that the density is given in g/cm^3 units. We have to remember that a ft is equal to 30.48 cm, then we have:
[tex]8975ft^3(\frac{30.48\text{ cm}}{1\text{ ft}})(\frac{30.48\text{ cm}}{1\text{ ft}})(\frac{30.48\text{ cm}}{1\text{ ft}})=2.54\times10^8[/tex]Hence the volume of the iceberg is:
[tex]2.54\times10^8cm^3[/tex]Now that we have the volume in the correct units we plug its value and the density in the equation for the mass above:
[tex]\begin{gathered} m=2.54\times10^8(0.917) \\ m=2.32\times10^8 \end{gathered}[/tex]Hence the mass of the iceber is 2.32x10^8 g.
Therefore the mass of the iceberg in kilograms is:
[tex]2.32\times10^5\text{ kg}[/tex]Imagine that someone is looking out of the top floor window of a skyscraper with a brick in his hand at the same instant someone else is looking out if the window on the floor below also holding a brick if both bricks were dropped at the same instant would the distance between them increase decrease or remain the same over time why
The distance between them remain the same over time both of them are accelerating - because of gravity
What is acceleration ?
acceleration: the rate at which the speed and direction of a moving object vary over time. A point or object going straight ahead is accelerated when it accelerates or decelerates. Even if the speed is constant, motion on a circle accelerates because the direction is always shifting. Both effects contribute to the acceleration for all other motions.
Both of them are accelerating - because of gravity - and the one that you dropped first has been accelerating for longer - and is therefore going faster than they one that you dropped later.
Until the two objects both reach their terminal velocities - assuming the objects are identical - their speeds will eventually be the same - and from that point onwards - the distance between them won’t change
The distance between them remain the same over time both of them are accelerating - because of gravity
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What is the voltage drop across point A and B?
We are asked to find the voltage drop at point A and B
Notice that point A and B have 3 resistors connected in parallel so the voltage across these 3 resistors will be the same.
First, we have to find the equivalent resistance of these 3 parallel resistors.
[tex]\begin{gathered} R_{AB}=\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}} \\ R_{AB}=\frac{1}{\frac{1}{120}+\frac{1}{60}+\frac{1}{30}} \\ R_{AB}=17.14\; \Omega \end{gathered}[/tex]So, the resistance of the parallel resistors is 17.14
Now, we can simply use the voltage drop formula to find the voltage drop at point A and B
[tex]\begin{gathered} V_{AB}=\frac{R_{AB}}{R_{total}}\times V_{\text{in}} \\ V_{AB}=\frac{R_{AB}}{R_{AB}+R_{CD}}\times V_{\text{in}} \end{gathered}[/tex]Where Vin is the input voltage that is 100 V
[tex]\begin{gathered} V_{AB}=\frac{17.14}{17.14+100}\times100 \\ V_{AB}=14.63\; V \end{gathered}[/tex]Therefore, there is a 14.63 V drop at point A and B
A raindrop has a mass of 7.7 × 10-7 kg and is falling near the surface of the earth. Calculate the magnitude of the gravitational force exerted (a) on the raindrop by the earth and (b) on the earth by the raindrop.(a)Fraindrop= _________________ units ________(b)Fearth= _________________ units_____________
ANSWER:
a) Fraindrop
[tex]F=7.546\cdot10^{-6}N[/tex](b) Fearth
[tex]F=-7.546\cdot10^{-6}N[/tex]STEP-BY-STEP EXPLANATION:
(a)
We calculate the force, multiplying the value of the mass by gravity, just like this:
[tex]\begin{gathered} F=m\cdot a \\ F=7.7\cdot10^{-7}\cdot9.8 \\ F=7.546\cdot10^{-6}N \end{gathered}[/tex](b)
by newton's 3rd law they are are equal and opposite so:
[tex]F=-7.546\cdot10^{-6}N[/tex]The amount of work done by two boys who apply 300 N of force in an unsuccessful attempt to move a stalled car is:1. 600 N · m.2. 300 N.3. 300 N · m.4. 600 N.5. 0.
The work done by a force can be calculated with the formula below:
[tex]W=F\cdot d[/tex]Where W is the work in J, F is the force in N and d is the distance in meters.
Since in this case, the attempt was unsuccessful, the car didn't move, so the distance is zero.
Therefore the work is zero, and the correct option is 5 (work is zero).
A sea wave propagates with a speed of 12 cm/s and a length of 0.4 meters,find its period.A:0,1 minB:10sC:1sD:3,33s
Given:
The speed of the wave is v = 12 cm/s = 0.12 m/s
The wavelength of the wave is
[tex]\lambda\text{ = 0.4 m}[/tex]To find the period.
Explanation:
The time period can be calculated by the formula
[tex]T=\frac{\lambda}{v}[/tex]On substituting the values, the time period will be
[tex]\begin{gathered} T=\frac{0.4}{0.12} \\ =3.33\text{ s} \end{gathered}[/tex]Thus, the time period of the sea wave is 3.33 s
3. An object of mass 8 kg is sliding down a friction lessinclined plane of length 11 m that makes an angleof 70 deg with the horizontal. Calculate thework done by gravitational force as the objectslides from the top of the inclined plane to theground. (1 point)A. 099.045 JB. O1135.872 JC.810.391 JD.499.917
Given
Mass of the object, m=8 kg
Length of the inclined plane, l=11 m
Angle of inclination,
[tex]\theta=70^o[/tex]To find
Calculate the work done by gravitational force as the object slides from the top of the inclined plane to the
ground.
Explanation
The height of the ramp
[tex]h=lsin\theta=11sin70^o[/tex]The work done by gravitational force,
[tex]\begin{gathered} W=mgh \\ \Rightarrow W=8\times9.8\times11sin70^o \\ \Rightarrow W=810.390J \end{gathered}[/tex]Conclusion
The work done is C.810.391
a 2403 kg racecar has a total momentum of 9.912*10^4kgm/s at one point in the race. calculate the speed of the racecar at that point
In order to calculate the speed, we can use the formula for the momentum:
[tex]p=m\cdot v[/tex]Where p is the momentum (in kg m/s), m is the mass (in kg) and v is the speed (in m/s).
So, using p = 99120 kg m/s and m = 2403 kg, we have:
[tex]\begin{gathered} 99120=2403\cdot v\\ \\ v=\frac{99120}{2403}\\ \\ v=41.25\text{ m/s} \end{gathered}[/tex]When the buoyant force on an object is equal to or greater than its weight, the object __
When the buoyant force on an object is equal to or greater than its weight, the object accelerates upwards and floats.
What is buoyant force?
Buoyant force is the upward force exerted on an object that is fully or partly immersed in a fluid.
This upward force is also called Upthrust.
According to Archimedes' principle which states that the buoyant force on an object is equal to the weight of the fluid displaced by the object.
An object will accelerate if its upthrust is greater than its weight, but will reach an upward terminal velocity when upthrust is equal to weight plus drag force.
Thus, when the buoyant force on an object is equal to or greater than its weight, the object accelerates upwards and floats.
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Atmospheric pressure is about 1.00 × 105 Pa on average.A. What is the downward force of the air on a desktop with surface area 2.59 m2?B. Convert the downward force of the air on a desktop with surface area 2.59 m2 to pounds to help others understand how large it is.
A)
The formula for calculating pressure is expressed as
pressure = force/area
From the information given,
pressure = 1.0 x 10^5 pa
Recall, 1 pa = 1 N/m^2
This means that
pressure = 1.0 x 10^5 N/m^2
surface area = 2.59 m^2
Force = pressure x area
Force = 1.0 x 10^5 x 2.59 = 259000 N
Recall,
B)
1 newton = 0.224808943 pounds
259000 newtons = 259000 x 0.224808943
= 58226 pounds
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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A 244 kg motorcycle is travelling with aspeed of 14.7 m-s-1A) Calculate the kinetic energy (in J) of themotorcycle.B) If the speed of the motorcycle is increasedby a factor of 1.6, by what factor does itskinetic energy change?C) Calculate the speed (in m-s-1) of themotorcycle if its kinetic energy is 1/3 of thevaluefound in (a).
Given data:
* The mass of the motorcycle is m = 244 kg.
* The speed of the motorcycle is u = 14.7 m/s.
Solution:
(A). The kinetic energy of the motorcycle is,
[tex]K_1=\frac{1}{2}mu^2[/tex]Substituting the known values,
[tex]\begin{gathered} K_1=\frac{1}{2}\times244\times(14.7)^2_{} \\ K_1=26362.98\text{ J} \end{gathered}[/tex]Thus, the value of kinetic energy is 26362.98 J.
(B). If the speed of the motorcycle is increased by a factor of 1.6,
[tex]\begin{gathered} v=14.7\times1.6 \\ v=23.52\text{ m/s} \end{gathered}[/tex]Thus, the kinetic energy of the motorcycle becomes,
[tex]\begin{gathered} K_2=\frac{1}{2}mv^2 \\ K_2=\frac{1}{2}\times244\times(23.52)^2 \\ K_2=67489.23\text{ m/s} \end{gathered}[/tex]Dividing K_2 by K_1,
[tex]\begin{gathered} \frac{K_2}{K_1}=\frac{67489.23}{26362.98} \\ \frac{K_2}{K_1}=2.56 \end{gathered}[/tex]Thus, the kinetic energy is increased by the factor of 2.56.
(C). The 1/3 of the kinetic energy in the first part is,
[tex]\begin{gathered} K=\frac{1}{3}\times K_1 \\ K=\frac{1}{3}\times26362.98 \\ K=8787.66\text{ J} \end{gathered}[/tex]Thus, the speed of the motorcycle with the kinetic energy K is,
[tex]\begin{gathered} K=\frac{1}{2}mv^2_{}_{} \\ 8787.66=\frac{1}{2}\times244\times v^2 \\ 8787.66=122\times v^2 \end{gathered}[/tex]By simplifying,
[tex]\begin{gathered} v^2=\frac{8787.66}{122} \\ v^2=72.03 \\ v\approx8.5\text{ m/s} \end{gathered}[/tex]Thus, the speed of the motorcycle is 8.5 m/s.
An installation consists of a 30-kVA, 3-phase transformer, a 480-volt primary, and a 240-volt secondary. Calculate the largest standard size circuit breaker permitted for primary-only protection to be used without applying Note 1 of Table 450.3(B).
Answer: 45 A
Explanation:
Primary only protection 3-phase
I = 3 phase kVA / ( 1.723 * V)
I = 30000 / ( 1.732 * 480 ) = 36.085 A
Table 450.3(B)
Currents of 9A or more column
primary only protection = 125%
Max OCPD pri = 125% of I = 1.25 * 36.085 = 45.11 A
Table 450.3(B) Note 1 does not apply, use next smaller Table 240.6(A)
Next smaller = 45 A
What is the image distance if a 5.00 cm tall object is placed 2.33 cm from a converging lens with a focal length of 5.75 cm?0.603cm1.66cm-0.255cm-3.92cm
We will have the following:
First, we will recall that:
[tex]\frac{1}{f}=\frac{1}{v}+\frac{1}{u}[/tex]That is:
[tex]\begin{gathered} \frac{1}{5.75}=\frac{1}{2.33}+\frac{1}{u}\Rightarrow\frac{1}{u}=-\frac{1368}{5359} \\ \\ \Rightarrow u=-\frac{5359}{1368}\Rightarrow u\approx-3.92 \end{gathered}[/tex]So, the image distance is approximately -3.92 cm.
A truck covers 40.0 m in 9.00 s while uniformly slowing down to a final velocity of 2.20 m/s.(a) Find the truck's original speed. m/s(b) Find its acceleration. m/s2
Given:
The distance covered by truck: d = 40.0 m
The time taken to cover the distance is: t = 9.00 s
The final velocity of the truck is: v2 = 2.20 m/s
To find:
a) the speed of the truck.
b) the acceleration
Explanation:
a)
The speed of the truck before it slows down can be calculated as:
[tex]d=\frac{1}{2}(v_2+v_1)t[/tex]Substituting the values in the above equation, we get
[tex]\begin{gathered} 40=\frac{1}{2}(2.20+v_1)\times9 \\ \\ \frac{40\times2}{9}-2.20=v_1 \\ \\ v_1=6.69\text{ m/s} \end{gathered}[/tex]b)
The truck is initially moving at a speed of 6.69 m/s. It then slows down to the final velocity of 2.20 m/s. The acceleration of the truck can be determined as:
[tex]d=v_1t+\frac{1}{2}at^2[/tex]Substituting the values in the above equation, we get:
[tex]\begin{gathered} 40=6.69\times9+\frac{1}{2}\times a\times9^2 \\ \\ 40=60.21+40.5a \\ \\ a=\frac{40-60.21}{40.5} \\ \\ a=-0.499 \\ \\ a\approx-0.5\text{ m/s}^2 \end{gathered}[/tex]Final answer:
a) The original speed of the truck is 6.69 m/s.
b) The acceleration of the truck is - 0.5 m/s^2.
A car initially at rest travels with a uniform acceleration of 8 m / s^2 . Calculate the distance covered by the car in 3 s .
We know that
• The initial velocity is zero because it starts from rest.
,• The acceleration is 8 m/s^2.
,• The time elapsed is 3 seconds.
Use a formula that relates initial velocity, acceleration, time, and distance.
[tex]d=v_0t+\frac{1}{2}at^2[/tex]Use the given magnitudes to find d.
[tex]\begin{gathered} d=0\cdot3\sec +\frac{1}{2}\cdot(8\cdot\frac{m}{s^2})(3\sec )^2 \\ d=4\cdot\frac{m}{s^2}\cdot9s^2 \\ d=36m \end{gathered}[/tex]Therefore, the distance covered is 36 meters.7. What is the velocity of a 850kg car after starting at rest when 13,000J of work is done to it.
Answer:
5.53 m/s
Explanation:
The work is equal to the change in the kinetic energy, so
[tex]\begin{gathered} W=\Delta KE \\ W=\frac{1}{2}m(v^2_f-v^2_i)^{}^{} \end{gathered}[/tex]Since the car starts at rest, the initial velocity vi = 0 m/s, so we can solve for the final velocity vf as follows
[tex]\begin{gathered} W=\frac{1}{2}mv^2_f \\ 2W=mv^2_f \\ \frac{2W}{m}=v^2_f \\ v_f=\sqrt[]{\frac{2W}{m}} \end{gathered}[/tex]So, replacing the work W = 13,000J and the mass m = 850kg, we get:
[tex]\begin{gathered} v_f=\sqrt[]{\frac{2(13,000J)}{850\operatorname{kg}}} \\ v_f=5.53\text{ m/s} \end{gathered}[/tex]Therefore, the velocity is 5.53 m/s
