We are given that the height of a rock in terms of the time is given by the following equation:
[tex]h\mleft(t\mright)=120+20t-5t^2[/tex]We are asked to determine the height after two seconds. To do that we will substitute in the equation the value of "t = 2s", like this:
[tex]h(2)=120+20(2)-5(2)^2[/tex]Solving the operations we get:
[tex]h(2)=140[/tex]Therefore, the height after 2 seconds is 140 ft.
Now, to determine an equation for the velocity we will determine the derivative with respect to the time of the equation for the height.
[tex]\frac{dh}{dt}=\frac{d}{dt}(120+20t-5t^2)[/tex]Now, we distribute the derivative:
[tex]\frac{dh}{dt}=\frac{d}{dt}(120)+\frac{d}{dt}(20t)-\frac{d}{dt}(5t^2)[/tex]For the first derivative we will use the following rule:
[tex]\frac{d}{dt}(a)=0[/tex]Where "a" is a constant. Applying the rule we get:
[tex]\frac{dh}{dt}=\frac{d}{dt}(20t)-\frac{d}{dt}(5t^2)[/tex]For the second derivative we will use the following rule:
[tex]\frac{d}{dt}(at)=a[/tex]Where "a" is a constant. Applying the rule we get:
[tex]\frac{dh}{dt}=20-\frac{d}{dt}(5t^2)[/tex]For the last derivative we will use the following rule:
[tex]\frac{d}{dt}(at^n)=\text{nat}^{n-1}[/tex]Applying the rule we get:
[tex]\frac{dh}{dt}=20-10t[/tex]Since the derivative of the position with respect to time is the velocity we have:
[tex]\frac{dh}{dt}=v=20-10t[/tex]Now, we substitute the value of "t = 2s":
[tex]v=20-10(2)[/tex]Now, we solve the operations:
[tex]\begin{gathered} v=20-20 \\ v=0 \end{gathered}[/tex]Therefore, the velocity after 2 seconds is 0.
if i kicked a empty soda can would it travel further than a filled up soda can, if so why? & what newton law would this be ?
If we kick both sodas with the same force, the soda that has a higher weight will have a lower acceleration. This is explained by Newton's second law of motion
The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object.
F = ma
where
m = mass
a = accelaration
F = Force
Scientists might make a computer model of volcanic eruptions. What is the
biggest benefit of this model?
Answer:
Computer Model May Help to More Accurately Predict Volcano Eruptions. Scientists at the GFZ German Research Center in Potsdam, Germany, have developed a computer model which they say boosts the accuracy of volcanic eruption prediction.
Pls Mark As Brainliest
how far will you go (km) in 3 min traveling 60 km/hr?
We will have the following:
First, we transform minutes to hours, that is:
[tex]3\min \cdot\frac{1h}{60\min }=0.05h[/tex]Now, we determine the distance traveled:
[tex]d=(60km/h)(0.05h)\Rightarrow d=3km[/tex]So, you will go 3km.
A __________ is described as a device with specific resistance and is used to control current.batteryresistorparallel connectionseries connection
ANSWER:
resistor
STEP-BY-STEP EXPLANATION:
A device that has a specific resistance and is also used to control current is the resistor.
Therefore:
A resistor is described as a device with specific resistance and is used to control current.
Explain why clothes stick together when they are removed from a drier. What is static electricity?
ANSWER:
What happens in clothes is a phenomenon called static cling is a phenomenon caused by static electricity. When dry materials rub against each other, they can exchange electrons, creating an electrical charge. This charge can build up in the form of static electricity and cause two objects, in this case clothing, to stick or stick together.
When the substance that loses electrons becomes positively charged and the substance that gains electrons becomes negatively charged. These charges are stationary and remain on the surface of the material. Since there is no flow of electrons, this is called static electricity.
A spring of length 9.7 meters stretches to 9.8 meters when a 0.4 kg mass is hung vertically from one end. What is the spring constant?
Given,
The initial length of the spring, l=9.7 m
The length of the spring after stretching, L=9.8 m
The mass, m=0.4 kg
The magnitude of the restoring force of the spring due to the stretching from the mass will be equal to the force applied by the mass, which is nothing but the weight of the mass.
Thus,
[tex]\begin{gathered} mg=k\Delta x \\ =k(L-l) \end{gathered}[/tex]Where g is the acceleration due to gravity, k is the spring constant, and Δx is the stretch in the length of the spring.
On substituting the known values,
[tex]\begin{gathered} k=\frac{mg}{(L-l)_{}} \\ =\frac{0.4\times9.8}{9.8-9.7} \\ =\frac{3.92}{0.1} \\ =39.2\text{ N/m} \end{gathered}[/tex]Thus the spring constant is 39.2 m
Sound is an example of a longitudinal wave because the wave travels by compressions and rarefactions of air particles?True or false
To find
The given statement is true or false.
Explanation
The sound waves propag5ate th6roug5h6 a medium by the compression and rarefraction of the m
The distance and direction in which an object travels per unit of time. A. velocity B. magnitude C. speed
Answer:
Answer is C. Speed
Which of the following is an appropriate measure of electric power on a toaster label?220 W55 Ω110 V2.0 A
A Watt is the unit of electrical power
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
what is the smallest amount of time in which the person can accelerate the car from rest to 23 m/s and still keep the coffee cup on the roof. The coefficient of the static friction is 0.21. The maximum acceleration of the car that is allowed so that the cup does not fall is 2.1 m/s^2
Given:
The coefficient of the static friction, μ=0.21
The maximum acceleration of the car so that the cup does not fall, a=2.1 m/s²
The initial velocity of the car, u=0 m/s
The final velocity of the car, v=23 m/s
To find:
The smallest amount of the time in which the car can accelerate so that the coffee cup will still be on the roof.
Explanation:
From the equation of motion,
[tex]v=u+at[/tex]Where t is the smallest amount of time in which the person can accelerate and still keep the cup on the car.
On rearranging the above equation,
[tex]t=\frac{v-u}{a}[/tex]On substituting the known values,
[tex]\begin{gathered} t=\frac{23-0}{2.1} \\ =10.95\text{ s} \end{gathered}[/tex]Final answer:
Thus the smallest amount of the time in which the person can accelerate the car at the given rate and still keep the cup on the roof of the car is 10.95 s
An air compressor has a volume of 100.L What volume of gas is pumped into the tank if the pressure goes from 750 torr to a pressure of 145 psi?Remember to convert pressure to atm. Refer to picture for conversions if needed.
ANSWER:
1000 L
STEP-BY-STEP EXPLANATION:
The first thing is to convert the unit of both pressures to atm, with the help of the conversion equivalences shown, just like this:
[tex]\begin{gathered} P_1=750\text{ torr }\cdot\frac{1\text{ atm}}{760\text{ torr}}=0.987\text{ atm} \\ P_2=145\text{ psi }\cdot\frac{1\text{ atm}}{14.7\text{ psi}}=9.87\text{ atm} \end{gathered}[/tex]Now, applying Boyle's Law, we calculate the value of the volume, like this:
[tex]\begin{gathered} P_1\cdot V_1=P_2\cdot V_2 \\ V_1=\frac{P_2\cdot V_2}{P_1} \\ \text{ replacing} \\ V_1=\frac{9.87\cdot100}{0.987}_{} \\ V_1=1000\text{ L} \end{gathered}[/tex]The volume of gas is 1000 L
100 POINTS
A man pulls a crate with a rope. The crate slides along the floor in the horizontal direction (x direction). The man exerts a force of 50 N on the rope, and the rope is at an angle . Describe how the force components change as the angle increases from 0° to 90° and use your graph to explain your answer. Give a detailed explanation of the forces at . Show a sample calculation at one angle for both components.
The exerted 50 N force at an angle on the crate can be resolved into a horizontal and vertical force component in which the horizontal component, Fₓ, decreases, while the vertical force component, [tex]F_y[/tex], increases as the value of the angle formed by the rope, θ, increases from 0° to 90°.
What is a component of a force?The components of a force are the force parts acting in perpendicular directions, horizontal and vertical, that combine to give the specified force.
The direction the crate is sliding = The horizontal, x-direction
The force exerted by the man = 50 N
The angle of the rope = θ
The components of the force are therefore:
Horizontal component, Fₓ = 50 × cos(θ)
Vertical component, [tex]F_y[/tex] = 50 × sin(θ)
The value of cos(θ) and sin(θ) as the angle the rope makes with the horizontal, θ, increases from 0° to 90° are as follows:
[tex]\begin{center} \begin{tabular}{ |c|c |c |} \theta & cos(\theta) & sin(\theta) \\ 0^{\circ} & 1 & 0 \\ 30^{\circ} & \sqrt{3} /2 & 0.5 \\ 45^{\circ}&\sqrt{2}/2 &\sqrt{2}/2 \\ 60^{\circ}&0.5&\sqrt{3}/2 \\ 90^{\circ} &0&1\\ \end{tabular}[/tex]
Therefore, the horizontal component of the force exerted by the man, Fₓ has a maximum value at θ = 0, and decreases to 0, as θ increases from 0° to 90°.
The vertical component of the force exerted, [tex]F_y[/tex], has a minimum value of 0 at θ = 0°, and the value of sin(θ) and therefore [tex]F_y[/tex], increases to a maximum of (sin(90°) = 1) 50 N as increases to 90°.
Please find attached the graph showing the components of the force, Fₓ, and [tex]F_y[/tex], exerted by the man as the angle formed by the rope increases from 0° to 90°
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Use the momentum equation for photons found in this week's notes, the wavelength you found in #3, and Planck’s constant (6.63E-34) to calculate the momentum of this photon:
The wavelength is divided by Plank's constant to get the momentum equation for photons: p = h /λ.
What is photon?The electromagnetic force is carried by a photon, a basic particle that is a quantum of the electromagnetic field and includes electromagnetic radiation like light and radio waves. Since photons have no mass, they constantly move at the 299792458 m/s speed of light in a vacuum.
Assuming the wavelength determined in a prior issue, = 656 nm = 656 * 10 - 9 m, you get:
p = (6.63 * 10 ^-34) / (656 * 10 ^ -9) kg * m/s
P, which must be rounded to three significant numbers, is equal to 1.01067 * 10 - 27 kg*m/s.
Consequently, p = 1.01 * 10 -27 kg*m/s
Our number, rounded to two significant figures, is 1.0 * 10 - 27 kg*m/s because the answers are only rounded to two significant values.
Therefore, given that the wavelength is 656 nm, the first option—1.0*10-27 kg*m/s—is the correct response.
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Calculate the potential energy of a 2 kg ball that is about to be dropped of a height of 25 m
Given:
The mass of the ball, m=2 kg
The height from which the ball is about to be dropped, h=25 m
To find:
The potential energy of the ball.
Explanation:
The type of potential energy that is stored in this ball is called gravitational potential energy. The gravitational potential energy is the energy that an object possesses due to its height. The gravitational potential energy is directly proportional to the height at which the object is situated.
The potential energy of the ball is given by,
[tex]E=\text{mgh}[/tex]Where g is the acceleration due to gravity.
On substituting the known values,
[tex]\begin{gathered} E=2\times9.8\times25 \\ =490\text{ J} \end{gathered}[/tex]Final answer:
The potential energy of the ball is 490 J
A boy walks 30m [E25°S] then 60m [E40°N]. Determine his net displacement.
Given,
The distances; a=30 m
b=60 m
Angles; θ=E25°S
α=E40°N
From the diagram, ∠A is given by,
[tex]\angle A=180\degree-\theta-\alpha[/tex]On substituting the known values,
[tex]\begin{gathered} \angle A=180\degree-25\degree-40\degree \\ =115\degree^{} \end{gathered}[/tex]From the cos rule,
[tex]d^2=a^2+b^2-2ab\cos A[/tex]On substituting the known values,
[tex]\begin{gathered} d^2=30^2+60^2-2\times30\times60\times\cos 115\degree \\ \Rightarrow d=\sqrt[]{30^2+60^2-2\times30\times60\times\cos 115\degree} \\ =77.6\text{ m} \end{gathered}[/tex]Thus the total displacement of the boy is 77.6 m
a) your one friend and their bumper boat (m = 210 kg) are traveling to the left at 3.5 m/s and your other friend and their bumper boat (m = 221 kg) are traveling in the opposite direction at 1.8 m/s. The two boats then collide in a perfectly elastic one-dimensional collision. How fast is your first friend and their boat moving after the collision?b) after the collision, a constant drag force between the water and the boat causes your first friends boat to come to a stop. If the boat travels 2.7 m before stopping, what is the magnitude of the constant drag force?
ANSWER:
a) 1.757 m/s
b) 119.91 N
STEP-BY-STEP EXPLANATION:
Given:
Mass 1 (m1) = 210 kg
Initial speed 1 (u1) = 3.5 m/s
Mass 2 (m2) = 221 kg
Initial speed (u2) = 1.8 m/s
We make a sketch of the situation:
a)
We make a momentum balance by taking into account the conservation of momentum:
[tex]\begin{gathered} m_1u_1-m_2u_2=m_1v_1+m_2v_2 \\ \\ v_2=\frac{m_1u_1-m_2u_2-m_1v_1}{m_2}\rightarrow(1) \end{gathered}[/tex]Now an energy balance taking into account the conservation of energy, as follows:
[tex]\begin{gathered} \frac{1}{2}m_1(u_1)^2+\frac{1}{2}m_2(u_2)^2=\frac{1}{2}m_1(v_1)^2+\frac{1}{2}m_2(v_2)^2 \\ \\ m_1\left(u_1\right)^2+m_2\left(u_2\right)^2=m_1\left(v_1\right)^2+m_2\lparen v_2)^2\rightarrow(2) \end{gathered}[/tex]Now, we substitute equation (1) in (2) and we get the following:
[tex]\begin{gathered} m_1\left(u_1\right)^2+m_2\left(u_2\right)^2=m_1\left(v_1\right)^2+m_2\left(\frac{m_1u_1-m_2u_2-m_1v_1}{m_2}\right)^2 \\ \\ m_1\left(u_1\right)^2+m_2\left(u_2\right)^2=m_1\left(v_1\right)^2+m_2\left(\frac{\left(m_1u_1\right)^2-2\left(m_1u_1\right)\left(m_2u_2\right)-2\left(m_1u_1\right)\left(m_1v_1\right)+\left(m_2u_2\right)^2+2\left(m_2u_2\right)\left(m_1v_1\right)+\left(m_1v_1\right)^2}{(m_2)^2}\right) \\ \\ m_1\left(u_1\right)^2+m_2\left(u_2\right)^2=m_1\left(v_1\right)^2+\left(\frac{\left(m_1u_1\right)^2-2\left(m_1u_1\right)\left(m_2u_2\right)-2\left(m_1u_1\right)\left(m_1v_1\right)+\left(m_2u_2\right)^2+2\left(m_2u_2\right)\left(m_1v_1\right)+\left(m_1v_1\right)^2}{m_2}\right) \\ \\ v_1=u_1\frac{m_1-m_2}{m_1+m_2}+u_2\frac{2m_2}{m_1+m_2} \\ \\ \text{ Now, we substitute each value, like so:} \\ \\ v_1=3.5\cdot\frac{210-221}{210+221}+1.8\cdot\frac{2\cdot221}{210+221} \\ \\ v_1=1.757\text{ m/s} \end{gathered}[/tex]b)
We use the following formula to determine the force:
[tex]\begin{gathered} v^2=u^2+2as \\ \\ \text{ Wee replacing:} \\ \\ (1.756)^2=0^2+2\cdot a\cdot2.7 \\ \\ a=0.003375\text{ m/s}^2 \\ \\ \text{ Therfore:} \\ \\ F=m\cdot a=210\cdot0.571=119.91\text{ N} \end{gathered}[/tex]A cheetah is running at a velocity of 8 m/s when it accelerptes at 1.5 m/s^2 for 6 seconds. If the
cheetah started at a position of 20m what is the final Position of the cheetah?
Answer:
[tex]95[/tex]
Explanation:
[tex]x= x. + vt+\frac{1}{2} at^{2}[/tex]
[tex]s=20 + (8*6)+ \frac{1}{2} (1.5 * 6^{2} )[/tex]
[tex]s= 95[/tex]
A car starts from rest and travels for 9.0 s with a uniform acceleration of +2.4 m/s²?. The driver then applies the brakes, causing a uniform acceleration of -2.5 m/s². If the brakes areapplied for 2.0 s, determine each of the following(a) How fast is the car going at the end of the braking period?m/s(b) How far has the car gone?m
a) At the end of the braking period, the car is moving at a speed of 16.6m/s
b) The car has traveled a total distance of 135.4m
Explanations:The car starts from rest
The initial velocity, u = 0 m/s
Uniform acceleration, a = 2.4 m/s²
time, t = 9.0 seconds
Find the final velocity when the car accelerates at 2.4m/s² using the equation
v = u + at
v = 0 + 2.4(9)
v = 21.6 m/s
The distance covered when the car accelerates at 2.4m/s²
s = ut + 0.5at²
s = 0(9) + 0.5(2.4)(9²)
s = 97.2 m
The distance covered when the car accelerates at 2.4m/s² is 97.2 m
The driver then applied a brake for 2.0 s and accelerates at -2.5m/s²
The initial velocity now becomes 21.6 m/s
That is, u = 21.6 m/s
t = 2 seconds
a = -2.5m/s²
The final speed v is calculated as:
v = u + at
v = 21.6 + (-2.5)(2)
v = 21.6 - 5
v = 16.6m/s
At the end of the braking period, the car is moving at a speed of 16.6m/s
The distance covered during the braking period is calculated as:
[tex]\begin{gathered} s\text{ = (}\frac{u+v}{2})t \\ s\text{ = }\frac{21.6+16.6}{2}\times2 \\ s\text{ = }\frac{38.2}{2}\times2 \\ s\text{ = 38.2 m} \end{gathered}[/tex]The car traveled a distance of 38.2 m during the braking period
Total distance covered = 97.2m + 38.2m
Total distance covered = 135.4m
The car has gone a distance of 135.4m
When considering the Law of Universal Gravitation, the graph of force v. distance is _____.1)linear2)parabolic3)circular4)none of the above
According to Law of Universal Gravitation, the graph between Force and distance is hyperbolic as the force depends sqaure of the distance. Therefore, the correct option is (4).
How much kinetic energy does Usain Bolt (m=94kg) have when he hits his top
speed of 12 m/s?
Answer:
6768 Joules (J)
Explanation:
kinetic energy = 1/2mv^2
1/2 (94x12^2) = 6768
How many cubic inches are there in 3.25 yd3?Express the volume in cubic inches to three significant figures.What is the mass in grams of 16.86 mL of acetone?Express your answer to four significant figures and include the appropriate units.What is the volume in milliliters of 7.06 g of acetone?Express your answer to three significant figures and include the appropriate units.
Answer:
[tex]\text{ 3.25 yd}^3=151,632in^3[/tex]Explanation: We need to convert cubic-yards into cubic inches, this can be simply done as follows:
[tex]\frac{46656\text{ Cubic inches}}{1\text{ Cubic Yard}}^{}[/tex]Therefore we have:
[tex]\begin{gathered} 3.25\text{ Cubic yards }\times\text{ }\frac{46656\text{ Cubic Inches}}{1\text{ Cubic Yard}}=151,632\text{ Cubic Inches} \\ \therefore\rightarrow \\ \text{ 3.25 yd}^3=151,632in^3 \end{gathered}[/tex]A car travels at an average speed of 60 km/h for 15 minutes.How far does the car travel in 15 minutes?D900 kmC 240 kmA4.0 km B 15 km
Notice that the speed is written using units of km/h and the time is written using units of minutes.
Convert the time interval to hours. Remember that 1 hour equals 60 minutes:
[tex]15\min =15\min \times\frac{1h}{60\min }=0.25h[/tex]The distance d that a particle travels during a time t if it moves at an average speed v is given by the formula:
[tex]d=v\cdot t[/tex]Replace v=60km/h and t=0.25h to find the distance traveled by the car:
[tex]d=(60\frac{km}{h})\times(0.25h)=15km[/tex]Therefore, the car travels 15km in 15 minutes.
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.
10. ABC.Per=1200 NNet Force:Pit=600 NEngen-SONFrid=20 NPry=800 NPrax=800 NF50NWhich situation above would best describe free fall velocity?Which situation above would best describe a crane lifting an object?If situation C had a Fapp of 40N to the right, the net force on the object would beIf situation Chad Fapp of 20N to the right, the forces would be (balanced, unbalanced) and thehorizontal velocity would be (constant, + accelerating, - accelerating). Circle the correct terms
When the body is under free fall its apperaent weight will be zero.
Therefore
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
carts, bricks, and bands
2. Which of the following conclusions are specifically supported by the data in Table 1?
a. A constant mass causes the acceleration value to increase.
b. An increase in the number of bricks causes the acceleration to decrease.
c. An increase in the length of the rubber band causes the acceleration to increase.
d. An increase in the number of rubber bands causes an increase in the acceleration.
The conclusions that are specifically supported by the data in Table 1 is that An increase in the number of rubber bands causes an increase in the acceleration. That is option D.
What is acceleration?Acceleration is defined as the rate at which the velocity of a moving object changes with respect to time which is measured in meter per second per second (m/s²).
From the table given,
Trial 1 ----> 1 band = 0.24m/s²
Trial 2 ----> 2 bands = 0.51 m/s²
Trial 3 ----> 3 bands = 0.73 m/s²
Trial 4 -----> 4 bands = 1.00 m/s²
This clearly shows that increase in the number of bands increases the acceleration of one brick that was placed on the cart.
This is because increasing the number of rubber bands has the effect of doubling the force leading to an effective increase in velocity of the moving cart.
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A +10.31 nC charge is located at (0,8.47) cm and a -2.09 nC charge is located (3.91, 0) cm. Where would a -14.84 nC charge need to be located in order that the electric field at the origin be zero? Express your answer, in cm, as the magnitude of the distance of q3 from the origin.
To make the E-field at the origin become 0, we need to find the E-field at the origin before
E = kq/r^2
E1 = 12934 V/m
E2 = 12303.69 V/m
Etotal = 17851.34 V/m
17851.34 = kq/r^2
r = 8.6526 cm
Suppose that the Towngas supply pressure is 8.5 kPa (gauge pressure) andthe total volume is 2.4 m? enters a building from outside where thetemperature is 10 °C and passes into a building where the temperature is 35°C, if the pressure was reduced to 2 kPa. What would be the new volume ofthe gas?
Given,
The initial pressure of the gas, P₁=8.5 kPa
The initial volume of the gas, V₁=2.4 m³
The initial temperature of the gas, T₁=10 °C=282.15 K
The temperature of the building, i.e., the temperature of the gas after entering the building, T₂=35 °C=308.15 K
The pressure of the gas after entering the building, P₂=2 kPa
Let us assume the new volume of the gas is V₂
From the combined gas law,
[tex]\begin{gathered} \frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} \\ \Rightarrow V_2=\frac{P_1V_1T_2}{T_1P_2} \end{gathered}[/tex]On substituting the known values,
[tex]undefined[/tex]Directions: Solve the following problems. Show your solutions.2. A circuit has three resistors connected in parallel. Their resistances are 11 Ω, 17 Ω, and 12 Ω as shown on the figure below. Find for: a. Voltage in R1 (V1)b. Voltage in R2 (V2)c. Voltage in R3 (V3)d. Total Resistance (RT)e. Total Current (IT)f. Current in R1 (I1)g. Current in R2 (I2)h. Current in R3 (I3)
Since the resistances are in parallel, the voltage in each one is the same, so:
a. V1 = 60 V
b. V2 = 60 V
c. V3 = 60 V
d.
The total resistance of parallel resistances can be calculated with the formula below:
[tex]\begin{gathered} \frac{1}{RT}=\frac{1}{R1}+\frac{1}{R2}+\frac{1}{R3}\\ \\ RT=\frac{R1\cdot R2\cdot R3}{R1R2+R2R3+R1R3}\\ \\ RT=\frac{11\cdot17\cdot12}{11\cdot17+17\operatorname{\cdot}12+11\operatorname{\cdot}12}\\ \\ RT=\frac{2244}{523}\\ \\ RT=4.29\text{ ohms} \end{gathered}[/tex]e.
The total current is given by the voltage divided by the total resistance:
[tex]IT=\frac{V}{RT}=\frac{60}{4.29}=13.99\text{ A}[/tex]The current in each resistor is given by the voltage divided by the resistance:
f.
[tex]I1=\frac{V1}{R1}=\frac{60}{11}=5.45\text{ A}[/tex]g.
[tex]I2=\frac{V2}{R2}=\frac{60}{17}=3.53\text{ A}[/tex]h.
[tex]I3=\frac{V3}{R3}=\frac{60}{12}=5\text{ A}[/tex]