Calculate the total capacitance of three capacitors 30µF, 20µF & 12µF connected in parallel across a d.c supply The answer is :Consider that the equivalent capacitance of three capacitors C1, C2 and C3 in parallel is given by:C=C1+C2+C3In this case:C1 = 30µFC2 = 20µFC3 = 12µFReplace the previous values into the formula for C and simplify:C=30μF+20μF+12μF=62μFHence, the total capacitance is 62µFQuestion 5 : Calculate the total charge on the capacitors connected in parallel if the supply voltage is 500V. Sketch a circuit diagram and label this to show how the charges are located
Answers
The circuit diagram is shown below:
From the diagram we notice that the same voltage will flow in every capacitor, this will be helpful later.
We know that this three capacitors are equivalent to a single equivalent capacitor with 62µF capacitance. The charge in this equivalent capacitor is:
[tex]Q_{eq}=(62\times10^{-6})(500)=0.031[/tex]Now, as we mentioned, the voltage is the same in each capacitor then the charge in each of them is:
[tex]\begin{gathered} Q_1=(30\times10^{-6})(500)=0.015 \\ Q_2=(20\times10^{-6})(500)=0.01 \\ Q_3=(12\times10^{-6})(500)=0.006 \end{gathered}[/tex]To check if this is correct we need to remember that the charge in the equivalent capacitor is equal to the sum of the charge in each capactior; for this case this conditon is fulfil; therefore we conclude that:
• The charge in the first capacitor is 0.015 C
,• The charge in the second capacitor is 0.01 C
,• The charge in the third capacitor is 0.006 C
The diagram with the labels is shown below:
Related Questions
Suppose you stand with one foot on ceramic flooring and one foot on a wool carpet,
making contact over an area of 80.0cm2 with each foot. Both the ceramic and the
carpet are 2.00 cm thick and are 10.0ºC on their bottom sides. At what rate must
heat transfer occurs from each foot to keep the top of the ceramic and carpet at
33.0ºC?
Answers
The rate of heat transfer from each foot to keep the top of the ceramic and carpet at the desired temperature is 27.6 J/s.
What is thermal heat transfer?
Thermal heat transfer is defined as the movement of heat across the border of the system due to a difference in temperature between the system and its surroundings.
The rate at which heat transfer occurs from each foot to keep the top of the ceramic and carpet at the given temperature is calculated as follows;
Q/t = kA(T₂ - T₁)/d
where;
Q/t is the rate of heat transfer
k is heat transfer coefficientT₂ is the final temperatureT₁ is the initial temperatured is the distance between the wool and ceramics
The heat transfer coefficient of human skin = 0.3 W / m⁰C
Area = 80 cm² = 0.008 m²
thickness, d = 2 cm = 0.02 m
change in temperature, = 33⁰ C - 10 ⁰C = 23 ⁰C
Q/t = (0.3 x 0.008 x 23) / (0.002)
Q/t = 27.6 W = 27.6 J/s
Thus, the rate of heat transfer from each foot to keep the top of the ceramic and carpet at the desired temperature is 27.6 J/s.
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If John Glenn weighed 640 N on Earth's surface, a) how much would he haveweighed if his Mercury spacecraft had (hypothetically) remained at twice thedistance from the center of Earth? b) Why is it said that an astronaut is nevertruly "weightless?"
Answers
Given:
The weight of John Glenn, w=640 N
To find:
a) The weight if the distance was twice that of the initial value.
b) Why is an astronaut never weightless.
Explanation:
a)
Let the distance between the spacecraft and the earth be r.
If it becomes twice, then the distance is 2r.
The initial gravitational force on John Glenn is,
[tex]F=w=\frac{GMm}{r^2}[/tex]Where G is the gravitational constant, M is the mass of the earth and m is the mass of John Glenn.
The force when the distance is twice,
[tex]\begin{gathered} w_n=\frac{GMm}{(2r)^2} \\ =\frac{GMm}{4r^2} \\ =\frac{w}{4} \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} w_n=\frac{640}{4} \\ =160\text{ N} \end{gathered}[/tex]b)
Even when the astronaut is in space they still have the mass and so does the earth. Thus there will always be a gravitational force of attraction between the earth and the astronaut. The astronaut does not feel the weight because there will be nothing in space that pushes them back. That is why an astronaut is never truly weightless.
Final answer:
a) Thus the weight of John Glenn will be 160 N
50 POINTS
A velocity vs time graph is shown. What is the acceleration of the object?
Answers
Answer:
Explanation:
Given:
V₀ = 0 m/s
V = 20m/s
t = 5 s
___________
a - ?
The acceleration:
[tex]a = \frac{(V -V_{0} )}{t} \\\\[/tex]
[tex]a=\frac{(20 - 0)}{5} = 4 \frac{m}{s^{2} } \\[/tex]
Car A is traveling with a constant velocity of 18 [m/s]. Car B speeds up from 0 [m/s] to 10 [m/s] in 4 seconds. Which car has a greater acceleration?Car ACar BCar A and Car B have the same accelerationNeither car is accelerating
Answers
Given:
Car A is traveling with a constant velocity of,
[tex]18\text{ m/s}[/tex]The initial speed of car B is,
[tex]v_i=0\text{ m/s}[/tex]After t=4 s, carB's speed is,
[tex]v_f=10\text{ m/s}[/tex]To find:
Which car has a greater acceleration
Explanation:
The acceleration of Car A is Zero as there is no change in velocity with time.
The acceleration of car B is,
[tex]\begin{gathered} a_B=\frac{v_f-v_i}{t} \\ =\frac{10-0}{4} \\ =2.5\text{ m/s}^2 \end{gathered}[/tex]So, Car B has greater acceleration.
Hence, Car B has greater acceleration.
my choice1A child pushes a toy truck up an inclined plane. The lifting force is...Select one:ut ofO a. the weight of the truck.O b. the length of the inclined plane.O c. the force the child uses to push the truck.uestionO d. equal to the total amount of work done.
Answers
When pushing a toy truck up an inclined plane, the force that makes the truck goes up is the force that the child uses to push the truck. (The force that pushes
Therefore the correct option is C.
Why is the (k) a negative value in hooks law.
Answers
Well, we will hav that neither "k" or "x" are negative, the negative sign is external to both values and the reason is that this is written in that way to represent the "direction" of the movement of the system. Form example, when the spring is extended then it will represent that it will pull back, and when it is compacted it represents that it will pull outwards.
[tex]F_s=-kx[/tex]It is simply a technicallity.
A box is standing on a conveyor belt that is not in motion. At one point the belt starts moving with some acceleration. At that point the box starts moving too (without slipping). Which force is responsible for the acceleration of the box. a. The air resistance force. b. The force of the pull. c. The force of friction. d. The normal force.
Answers
Given that a box is standing on a conveyor belt that is not in motion.
When the belt starts moving with some acceleration, the box starts moving too without slipping.
Let's determine the force that is responsible for the acceleration of the box.
Here, since the box starts moving without slipping when the belt starts moving, there will be static friction between the box and the belt since the belt was fixed.
Now, the force which is responsible for the acceleration of the box will be the force of gravity and the normal force.
Applying the Newton's second law, if the there is only force of gravity and the normal force acting on the box, there will be zero horizontal acceleration.
In order for the box to accelerate without slipping, the force responsible will be the static frictional force.
ANSWER:
c. The force of friction.
The period of a simple harmonic oscillator is the time it takes for one complete cycle of oscillation to be completed. Is this true or false?
Answers
Yes, the given statement is true.
The period of a simple harmonic oscillator is the time it takes for one complete cycle of oscillation to be completed.
In the setup of a cart pulled by a hanging mass, without friction, we have the following information: the mass of the cart is 0.53 Kg., and the hanging mass is 0.077 Kg.
Determine the acceleration of the cart (in m/s2).
Answers
The acceleration of the cart
a=1.2431m/s^2
What is acceleration?Generally, The equation for Newton's second law of motion is
2nd law of motion,
Fnet=m a
on hanging mass,
m_1 g-T=m_1 a
m_1 g-m_2 a=m_1 a
Acceleration, [tex]$a=\frac{m_1 g}{m_1+m_2}$[/tex]
[tex]&a=\frac{0.077 \mathrm{~kg} \times 9.8 \mathrm{~m} / \mathrm{s}^2}{0.077 \mathrm{~kg}+0.53 \mathrm{~kg}} \\[/tex]
a=1.2431m/s^2
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Calculate the depth in the ocean at which thepressure is three times atmospheric pressure.Atmospheric pressure is 1.013 x 10^5 Pa. Theacceleration of gravity is 9.81 m/s^2and them/sdensity of sea water is 1025 kg/m^3Answer in units of m.
Answers
In order to determine the depth in the ocean, use the following equation:
[tex]h=\frac{P-P_o}{\rho g}[/tex]h: depth
P: pressure = 3*Po
Po: atmospheric pressure = 1.013*10^5Pa
g: gravitational acceleration constant = 9.8m/s^2
p: density of water = 1025 kg/m^3
Replace the previoua values into the formula for h and simplify:
[tex]\begin{gathered} h=\frac{3P_o-P_o}{\rho g}=\frac{2P_o}{\rho g} \\ h=\frac{2(1.013\cdot10^5Pa)}{(1025\frac{kg}{m^3})(9.8\frac{m}{s^2})}\approx20.17m \end{gathered}[/tex]Hence, the depth in the ocean is approximately 20.17m
The weight of a proton is 1.64×10−26 N. The charge on a proton is +1.60×10−19 C. If a proton is placed in a uniform electric field so that the electric force on the proton just balances its weight, what is the magnitude and direction of the field?
Answers
Given:
The weight of the proton is: W = 1.6 × 10^(-26) N.
The charge on a proton is: q = 1.60 × 10^(-19) C
To find:
The magnitude and the direction of the electric field.
Explanation:
The weight of the proton is the force that the proton experiences due to its mass and acceleration. The electric force balances the weight of the proton. Thus we have,
F = W
Here, F is the electric force a proton experiences when it is placed in an electric field and W is the weight of the proton,
The force experienced by a photon when it is placed in an electric field is given as,
[tex]F=Eq[/tex]Here, E is the electric field.
Rearranging the above equation and substituting the values, we get:
[tex]\begin{gathered} E=\frac{F}{q} \\ \\ E=\frac{1.64\times10^{-26}\text{ N}}{1.60\times10^{-19}\text{ C}} \\ \\ E=1.025\times10^{-7}\text{ N/C} \end{gathered}[/tex]Thus, the magnitude of the electric field is 1.025 × 10^(-7) N/C.
The charge on the proton is positive and when it is placed in the electric field, the electric force on the proton is balanced by the weight of the proton. Thus, The direction of the electric force is opposite to the direction of the weight of the proton which is radially outward.
Final answer:
The magnitude of the electric field is 1.025 × 10^(-7) N/C and it has a radially outward direction that is opposite to the wight of the proton.
A PVC pipe has a length of 45.132 centimeters.a. What are the frequencies of the first three harmonics when the pipe is open at both ends? Include units in your answers.b. What are the frequencies of the first three harmonics when the pipe is closed at one end and open at the other? Include units in your answers.
Answers
ANSWERS
a. f₁ = 380 Hz; f₂ = 760 Hz; f₃ = 1140 Hz
b. f₁ = 190 Hz; f₃ = 570 Hz; f₅ = 950 Hz
EXPLANATION
a. For a pipe of length L open at both ends, the frequencies of the first three harmonics are:
[tex]\begin{cases}f_1=\frac{v}{2L} \\ \\ f_2=2f_1=\frac{v}{L} \\ \\ f_3=3f_1=\frac{3v}{2L}\end{cases}[/tex]Assuming that the speed of the wave is the speed of sound: 343 m/s and knowing that the length of the pipe is L = 45.132 cm = 0.45132 m we can find the frequencies of the first three harmonics:
[tex]\begin{cases}f_1=\frac{343m/s}{2\cdot0.45132m}\approx380Hz \\ \\ f_2=2f_1=2\cdot380Hz\approx760Hz \\ \\ f_3=3f_1=3\cdot380Hz\approx1140Hz\end{cases}[/tex]b. For a pipe of length L closed at one end and open at the other, the frequencies of the first three harmonics are:
[tex]\begin{cases}f_1=\frac{v}{4L} \\ \\ f_2=DNE \\ \\ f_3=3f_1=\frac{3v}{4L}\end{cases}[/tex]In a closed pipe, there can only be odd harmonics (1, 3, 5...). Therefore, the second harmonic does not exist and the "third harmonic" would be the 5th,
[tex]\begin{cases}f_1=\frac{v}{4L} \\ \\ f_3=3f_1=\frac{3v}{4L} \\ \\ f_5=5f_1=\frac{5v}{4L}\end{cases}[/tex]Again, the length of the pipe is 45.132 cm = 0.45132 m, so the first three harmonics are:
[tex]\begin{cases}f_1=\frac{343m/s}{4\cdot0.45132m}\approx190Hz \\ \\ f_3=3f_1=3\cdot190Hz=570Hz \\ \\ f_5=5f_1=5\cdot190Hz=950Hz\end{cases}[/tex]Which picture below correctly identifies the effort length and lifting length of a lever?Select one:a. Ab. Bc. Cd. D
Answers
d.D
Explanation
A lever is a simple machine made of a rigid beam and a fulcrum.
where
the Load is the object which we are lifting,Fulcrum is the point at which the lever is pivoted. and Effort is he force applied to make the object move
Step 1
check for the graph that:
the lifting length goes from the fulcrum to the load
and
the effort length goes from the fulcrum to the applied force,
therefore,
the answer is
d.D
I hope this helps you
Find the volume of a cuboid with the following measurments
Length:4.5 cm Breadth: 2 cm Height: 50 cm
ans is :45,000
But I am not getting the correct ans
please send working
Answers
Answer:
The answer should be 450 cm³
Cannot be 4500 unless one of the measurements is in meters. See detailed explanation below
Explanation:
This is pretty much straightforward
V = l x w x h
where
l = length, w = width, h = height
Given the values for l, w, h
V = 4.5 x 2 x 50
= 450 cm³
The only way you can get 45,000 is if one of the measurements is in meters and others in cm. Please check your question carefully to see what units were used and what units the answer should be in
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
Answers
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]Determine the order of magnitude of the following small number by entering the appropriate exponent n below: 4,870×1021kg≈10nkg .(PLEASE LOOK AT THE SCREENSHOT FOR CORRECT NUMBERING)
Answers
To find the order of magnitude we move the decimal point until we have only one interger before it:
[tex]4870\times10^{21}=4.870\times10^{24}[/tex]Now that we have in this form we conclude that the order of magnitude of the number is 24
A student has a cannon that can fire a cannonball at speeds up to 97.0mph. The students wants to determine the maximum range of the cannon and if she could hit a target on the ground as shown. Neglect drag and the initial height of the cannonball.
Answers
Answer:
88 ft / sec = 60 mph
Thus 97/60 * 88 = 142 ft/sec maximum speed of cannonball
R = V^2 sin 2 θ / g = 142^2 / 32 = 630 ft
Using 3.28 ft / m
630 ft / 3.28 f/m = 192 m is the maximum range of the cannonball
Vy = 142 ft / sec * sin 45 = 100 ft/sec vertical speed at 45 deg
Tup = 100 ft/sec / 32 ft/sec^2 = 3.12 sec time to reach height
T = 2 * 3.12 = 6.24 total time in air when fired at 45 deg
An aluminum rod and a nickel rodare both 5.00 m long at 20.0°C.The temperature of each is raisedto 70.0°C. What is the differencein length between the two rods?AluminumNickela = 23.10-6C+ B = 69.10-6 0-1a = 13.10-6C1 B = 39.10-6-1(Unit = m)Enter
Answers
The difference in length between the two rods = 0.0025m
Explanations:Linear expansivity of a material is given by the formula:
[tex]\alpha\text{ = }\frac{l_2-l_1}{l_1(\theta_2-\theta_1)}[/tex]For the Aluminium rod:
[tex]\begin{gathered} l_{A1}\text{ = 5.0m} \\ \theta_{A1}=20^0C \\ \theta_{A2}=70^0C \\ \alpha_A\text{ = }23\times10^{-6}C^{-1} \\ \alpha_A\text{ = }\frac{l_{A2}-l_{A1}}{l_{A1}(\theta_{A2}-\theta_{A1})} \\ \text{ }23\times10^{-6}\text{ = }\frac{l_{A2}-5}{5(70-20)} \\ 5\times50\times\text{ }23\times10^{-6}=\text{ }l_{A2}-5 \\ l_{A2}=\text{ (}5750\text{ }\times10^{-6})\text{ + 5} \\ l_{A2}=\text{ 0.00575+5} \\ l_{A2}=\text{ 5.00575m} \end{gathered}[/tex]For the Nickel rod:
[tex]\begin{gathered} l_{N1}\text{ = 5.0m} \\ \theta_{N1}=20^0C \\ \theta_{N2}=70^0C \\ \alpha_N=\text{ 13}\times10^{-6}C^{-1} \\ \alpha_N\text{ = }\frac{l_{N2}-l_{N1}}{l_{N1}(\theta_{N2}-\theta_{N1})} \\ \text{ 1}3\times10^{-6}\text{ = }\frac{l_{N2}-5}{5(70-20)} \\ 5\times50\times\text{ 1}3\times10^{-6}=\text{ }l_{A2}-5 \\ l_{N2}=\text{ (32}50\text{ }\times10^{-6})\text{ + 5} \\ l_{N2}=\text{ }0.00325+5 \\ l_{N2}=\text{ 5.00325m} \end{gathered}[/tex]The difference in length between the two rods will be given as:
[tex]\begin{gathered} l_{A2}-l_{N2}=\text{ 5.00575-5.00325} \\ l_{A2}-l_{N2}=0.0025m \end{gathered}[/tex]The difference in length between the two rods = 0.0025m
Acceleration problems use the 4 Step method to solve each of the following. a wagon has a beginning speed of 10 km/hr and it reaches 75 km/hr in 7 sec
Answers
Vi = initial speed = 10 km/h
Vf= final speed = 75 km/h
t = time = 7 sec
a = acceleration
First, express km/h in m/s
Vi = 2.78 m/s
Vf= 20.83 m/s
Apply
a = (vf-vi) / t
replace:
a = (20.83 - 2.78 ) / 7 = 2.58 m/s^2
An intrepid hiker reaches a large crevasse in his hiking route. He sees a nice landing ledge 60.0 cm below his position but it is across a 2.3 m gap. He spends 1.2 s accelerating horizontally at 5.92 m/s2 [right] in an attempt to launch himself to the safe landing on the far side of the gap. Does he make it?
Answers
The hiker made it to a safe landing on the other side of the gap after travelling horizontally at 2.49 m.
What is the time motion from the vertical height?
The time taken for the hiker to fall from the given height is calculated as follows;
h = vt + ¹/₂gt²
where;
v is the vertical velocity = 0t is the time of motiong is acceleration due to gravityh is the height of fallh = ¹/₂gt²
t = √(2h/g)
t = √[(2 x 0.6) / (9.8)]
t = 0.35 seconds
The horizontal velocity of the hiker during the period of acceleration is calculated as follows;
Vₓ = at
Vₓ = (5.92 m/s²) x (1.2 s)
Vₓ = 7.104 m/s
The horizontal distance travelled during the time period of 0.35 seconds;
X = Vₓt
X = 7.104 x 0.35
X = 2.49 m
Thus, the hiker made it to a safe landing on the other side of the gap which is 2.3 m wide and smaller to his horizontal displacement of 2.49 m.
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At what point, if any, during a dive is a skydiver experiencing complete freefal? Explain. (1 point)•skydiver will experience complete freefall the moment right before they jump out of the plane because they are free to start fallingat any moment.•A skydiver will experience complete freefall when they first jump out of the plane because they only experience air resistance oncethey deploy their parachute.•A skydiver will never experience complete freefall until after they have deployed their parachute because they are now falling at asafe speed for their landingA skydiver will never experience complete freefall because as soon as they start their dive, they will experience air resistance.
Answers
To find:
Which of the given statements is true.
Explanation:
The free fall is defined as the motion of an object under the influence of only gravitational force. In free fall, there will be no forces, except gravitational force acting on the object.
When the skydiver jumps, gravity will pull the diver downwards. And the air resistance due to the air in the atmosphere will oppose this motion of the diver. Thus there will be two forces acting on the diver, gravity and the air resistance. Thus the skydiver will never experience free fall.
Final answer:
Thus the correct answer is option D.
How much power is used by a contact lens heating unit that draws 0.05 A of current from a 197 V line?
Answers
Given,
The current drawn by the contact lens heating unit, I=0.05 A
The supply voltage, V=197 V
The electric power is given by the product of the current drawn and the supply voltage.
Thus the power used by the given device is given by,
[tex]P=VI[/tex]On substituting the known values,
[tex]\begin{gathered} P=197\times0.05 \\ =9.85\text{ W} \end{gathered}[/tex]Thus the power used by the contact lens heating unit is 9.85 W
A driver of a car going 90km/hr suddenly sees the lights of a barrier 40.0m ahead. It take the driver 0.75s before he applies the brakes (this is known as reaction time). Once he does begin to brake, he decelerates at a rate of 10m/s^2. Does he hit the barrier?
Answers
First, consider that the distance traveled by the car in 0.75s is:
[tex]x=v\cdot t[/tex]Convert 90km/h to m/s as follow:
[tex]\frac{90\operatorname{km}}{h}\cdot\frac{1h}{3600}\cdot\frac{1000m}{1\operatorname{km}}=\frac{25m}{s}[/tex]Then, the distance x is:
[tex]x=(\frac{25m}{s})(0.75s)=18.75m[/tex]Then, when the driver start to apply the brakes, the distance to the barrier is:
x' = 40.0 m - 18.75 m = 21.25 m
Next, calculate the distance that the car need to stop completely, by using the following formula:
[tex]v^2=v^2_o-2ad[/tex]where,
v: final velocity = 0m/s (the car stops)
vo: initial velocity = 25m/s
a: acceleration = 10m/s^2
d: distance = ?
Solve the previous equation for d and replace the values of the other parameters:
[tex]d=\frac{v^2_0-v^2}{2a}=\frac{(\frac{25m}{s})^2-(\frac{0m}{s})^2}{2(\frac{10m}{s})^{}}=31.25m[/tex]Then, the drive needs 31.25 m to stop. If you compare the previous result with the distance of the car related to the barrier when the driver applies the brakes
(x' = 18.75 m), you can notice that d is greater than x'.
Hence, the car does hit the barrier.
Given v = 520 sin (30t - 5π/4), what is the phase angle?Question 4 options:-225 degrees90 degrees-135 degrees-90 degrees
Answers
Given data:
The voltage can be expressed as,
v = 520 sin (30t - 5π/4) ...... (1)
Now, the general equation of the sine wave can be given as,
[tex]v=V_m\text{ sin}(\omega t+\phi)\ldots\ldots\text{ }(2)[/tex]Here,
[tex]\phi[/tex]is the phase angle,
[tex]V_m[/tex]is the maximum voltage, and
[tex]\omega[/tex]is the angular frequency.
Compare equations (1) and (2), we get:
[tex]\begin{gathered} \phi=-\frac{5\pi\text{ rad}}{4}(\frac{180\degree}{\pi}) \\ =-225\degree \end{gathered}[/tex]Thus, the phase angle is
[tex]-225\degree[/tex]and the first option (-225 degrees) is correct.
find the pressure increase in a fluid when a force of 25 N is exerted on a closed syringe where the piston radius is 2 cm.
Answers
Given:
• Force, F = 25 N
,• Radius, r = 2 cm
Let's find the pressure increase.
To find the pressure increase, apply the formula:
[tex]\Delta P=\frac{Force}{Area}=\frac{F}{\pi r^2}[/tex]Where:
F = 25 N
r is the radius in meters = 2 cm
P is the pressure
Convert the radius from cm to meters.
Where:
100 cm = 1 m
2 cm = 2 /100 = 0.02 m
Hence, we have:
[tex]\begin{gathered} \Delta P=\frac{25}{\pi(0.02)^2} \\ \\ \Delta P=\frac{25}{\pi(0.0004)} \\ \\ \Delta P=\frac{25}{0.00125664} \\ \\ \Delta P=19894.4\text{ Pa}\approx1.99\times10^4\text{ Pa} \end{gathered}[/tex]Therefore, the pressure increase is 1.99 x 10⁴ Pa.
ANSWER:
1.99 x 10⁴ Pa
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
Answers
Why isn't geothermal energy used more often?Question 12 options:It's hard to find geothermal energy sources that are close to the surface.It produces too much pollution.It adds a large amount of water into the water cycle through evaporation.It is not a renewable energy source.
Answers
Geothermal energy can be used only if an energy source is present.
Mining is required to use geothermal energy.
The depth of mining increases earthquake risk.
Thus, it is difficult to find geothermal energy sources that are close to the surface.
having been given a newly discovered mineral, carry out a method in the physics lab on how to determine its specific gravity
Answers
ANSWER
Weigh a piece of the material, drop it in water and weigh the amount of water displaced. Specific gravity is the quotient between the mass of the mineral and the mass of water.
EXPLANATION
The specific gravity is the ratio of the mineral's density to the density of water at 23°C,
[tex]specific\text{ }gravity=\frac{density\text{ }of\text{ }mineral}{density\text{ }of\text{ }water}[/tex]Density, denoted with the greek letter ρ, is the ratio between the mass and volume of the substance,
[tex]\rho=\frac{m}{V}[/tex]If the mineral is a solid piece, we can weigh it - so we know the mass, and can find its volume by dropping it into water. Then, if we weigh the amount of water displaced, we have the mass of water as well - by Archimedes principle, the volume of water will be the volume of the object,
[tex]specific\text{ }gravity=\frac{\frac{m_{mineral}}{V}}{\frac{m_{water}}{V}}=\frac{m_{mineral}}{m_{water}}[/tex]How much heat in kcal must be added to 0.68 kg of water at room temperature (20°C) to raise its temperature to 45°C?answer:____ kcal
Answers
Given:
Mass, m = 0.68 kg
Initial temperature, T1 = 20°C
Final temperature, T2 = 45°C
Let's find the amount of heat needed.
Apply the Specific Heat Capacity formula:
[tex]\begin{gathered} Q=mc\Delta T \\ \\ Q=mc(T_2-T_1) \end{gathered}[/tex]Where:
m is the mass = 0.68 kg
c is the specific heat of water = 4.187 J/kg °C−1
T1 = 20°C
T2 = 45°C
Plug in the values and solve for Q:
[tex]\begin{gathered} Q=0.68*4.187*(45-20) \\ \\ Q=2.84716(25) \\ \\ Q=71.179\text{ kJ} \end{gathered}[/tex]Where:
1 kJ = 0.239 kCal
Since the answer is to be in kCal, we have:
[tex]Q=71.179*0.239=17.01\text{ kCal}[/tex]Therefore, the amount if heat added is 17.01 kCal.
ANSWER:
17.01 kCal
8.1 kg of copper sits at a temperature of 64 oF. How much heat is required to raise its temperature to 743 oF? The specific heat of copper is 385 J/kg-oC. Submit your answer in exponential form.
Answers
ANSWER:
STEP-BY-STEP EXPLANATION:
Given:
Mass (m) = 8.1 kg
T1 = 64 °F
T2 = 743 °F
Specific heat (C) = 385 J/kg*°C
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