Given:
The current drawn by the heating element is: I = 18.014 A.
Mass of the water is: m = 176.251 g.
The temperature difference of water is: T = 87.714 deg C - 20 deg C = 67.714 deg C.
Time for which the change in temperature occurs: t = 7.167 minutes
The specific heat of water is: c = 4.186 J/g degC.
To find:
The resistance of the heating element.
Explanation:
The expression for the heat is given as:
[tex]Q=mcT[/tex]Substitute the values in the above equation, we get:
[tex]\begin{gathered} Q=176.251\text{ g}\times4.186\text{ J/g deg C}\times67.714\text{ deg C} \\ \\ Q=49958.4876\text{ J} \end{gathered}[/tex]Joule's equation for electric heating is given as:
[tex]Q=I^2Rt[/tex]Here, R is the resistance of the heating element.
Rearranging the above equation, we get:
[tex]R=\frac{Q}{I^2t}[/tex]Substituting the values in the above equation, we get:
[tex]\begin{gathered} R=\frac{49958.4876\text{ J}}{18.014^2\text{ A}^2\times7.167\text{ m}} \\ \\ R=\frac{49958.4876\text{ J}}{324.5042\text{ A}^2\times7.167\times60\text{ s}} \\ \\ R=0.3580\text{ }\Omega \end{gathered}[/tex]Final Answer:
The resistance of the heating element is 0.3580 Ω.
Resistances of 2.0 Ω, 4.0 Ω, and 6.0 Ω and a 24-V emf device are all in parallel. The current in
the 2.0-Ω resistor is
Answer:
The current in the 2.0 Ω resistor is 12 A
Explanation:
Given:
R₁ = 2.0 Ω
R₂ = 4.0 Ω
R₃ = 6.0 Ω
ξ = 24 V
______________
I₁ - ?
With parallel connection:
U₁ = U₂ = U₃ = ξ
Ohm's law:
I₁ = U₁ / R₁ = 24 / 2,0 = 12 A
Resistances of 2.0 Ω, 4.0 Ω, and 6.0 Ω and a 24-V emf device are all in parallel, the current in the 2.0-Ω resistor is 6 A.
The voltage across each resistor in a parallel circuit is the same, while the current divides among the different branches.
The sum of the currents passing through each resistor equals the total current flowing into the parallel circuit.
1/RTotal = 1/R1 + 1/R2 + 1/R3
1/RTotal = 1/2.0 Ω + 1/4.0 Ω + 1/6.0 Ω
1/RTotal = 3/12 Ω
1/RTotal = 1/4 Ω
RTotal = 4 Ω
Now,
I = V / RTotal
I = 24 V / 4 Ω
I = 6 A
Thus, the current in the 2.0-Ω resistor is 6 A.
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Which of the following X-Y tables agrees withthe information in this problem?A puck moves 2.35 m/s in a -22.0° direction. A hockeystick pushes it for 0.215 s, changing its velocity to 6.42m/s in a 50.0° direction. What was the acceleration?A)хYYYC)хV 0.8802.18-0.8802.35-2.18ViVE4.134.92B)XVi2.35VE6.42a?Ax 0.2156.42> > 04.924.13a???Ax0.215ΔΧ??t0.2150.215tt 0.2150.215H
let's find the components for the initial velocity
[tex]v_ix=2.35\cdot\cos (-22)=2.18[/tex][tex]v_iy=2.35\cdot\sin (-22)=-0.88_{}[/tex]then for the final velocity
[tex]v_fx=6.42\cdot\cos (50)=4.13[/tex][tex]v_fy=6.42\cdot\sin (50)=4.92[/tex]The table that agrees with the problem is A)
Then for the acceleration, we will use the next formula
[tex]a=\frac{v_f-v_i_{}}{t}[/tex]for the acceleration in x
[tex]a_x=\frac{v_fx-v_ix}{t}=\frac{4.13-2.18}{0.215}=9.06m/s^2[/tex]then for the acceleration in y
[tex]a_y=\frac{v_fy-v_iy}{t}=\frac{4.92-(-0.88)}{0.215}=26.97m/s^2[/tex]then we calculate the magnitude
[tex]a=\sqrt[]{9.06^2+26.97^2}=28.45\text{ m/s}^2[/tex]A 5.0n force is applied to a 3.0kg ball to change its velocity from +9.”m/s to +3.0 m/s. The impulse on the ball is _I think it’s -18
Given,
The force is F=5 N.
The mass is m=3.0 kg
The velocities are 9 m/s to 3 m/s
The impulse is:
[tex]\begin{gathered} I=m(v_f-v_i) \\ \Rightarrow I=3(9-3) \\ \Rightarrow I=\frac{18kgm}{s} \end{gathered}[/tex]The answer is 18 kgm/s
Please answer the questions
Answer:
Where's the question? So that i can answer it!
Anyone available for teaching me simple pendulum in physics
Simple pendulum is a device that has a periodic motion.
In periodic motion, the object repeats its path after an interval of time.
The simple pendulum can be drawn as
It has a simple bob connected to a fixed end through a massless string.
During a hockey game, two hockey players (m1 = 82kg and m2 = 76kg) collide head on in a 1 dimensional perfectly instant collision. If the first hockey player is moving to the left at a velocity 4.2 m/s and the second hockey player is moving in the opposite direction at a velocity of 3.4 m/s, how fast are they both moving after they collide, assuming they stick together after they collide? How much kinetic energy is lost as a result of the collision?
Given:
The mass of player 1 is m1 = 82 kg
The initial velocity of player 1 is
[tex]v_{i1}=\text{ 4.2 m/s}[/tex]towards left.
The mass of player is m2 = 76 kg
The initial velocity of player 2 is
[tex]v_{i2}=\text{ -3.4 m/s}[/tex]in opposite direction.
Required:
The final velocity after the collision assuming they stick together.
The loss of kinetic energy after collision.
Explanation:
According to the conservation of momentum, the final velocity will be
[tex]\begin{gathered} m1v_{i1}+m2v_{i2}=(m1+m2)v_f \\ v_f=\frac{m1v_{i1}+m2v_{i2}}{m1+m2} \\ =\text{ }\frac{(82\times4.2)-(76\times3.4)}{82+76} \\ =0.544\text{ m/s} \end{gathered}[/tex]The loss in kinetic energy will be
[tex]\begin{gathered} \Delta KE=KE_{after}-KE_{before} \\ =\frac{1}{2}(m1+m2)(v_f)^2-\frac{1}{2}m1(v_{i1})^2-\frac{1}{2}m2(v_{i2})^2 \\ =\frac{1}{2}(82+76)(0.544)^2-\frac{1}{2}\times82\times(4.2)^2-\frac{1}{2}\times76\times(3.4)^2 \\ =23.38-723.24-439.28 \\ =-1139.14\text{ J} \end{gathered}[/tex]Final Answer:
The final velocity after the collision assuming they stick together is 0.544 m/s.
The loss of kinetic energy after the collision is 1139.14 J.
MULTIPLE CHOICE QUESTION Which word is synonymous to tension? Push Pull
Pull
Explanation:Tension is a pulling force that is transmitted in a string, rope or cable.
Therefore, tension is synonymous to pull
The diagram below shows an example of an object on which tension is transmitted.
It is a pull because it is tranmitted towards the source rather than away from the source
The Taipei Tower is a 508 meter, 101 story skyscraper. If you were to toss a orange of 0.13 kg off the top, how much kinetic energy would it have when it hits the sidewalk? Ignore air resistance .
Answer:
647.19 J
Explanation:
By the conservation of energy, the potential energy when you toss the orange is converted to kinetic energy when it hits the sidewalk, so
Ef = Ei
KE = PE
KE = mgh
Where m is the mass, g is 9.8 m/s², and h is the height of the Tower. Replacing m by 0.13 kg and h by 508 meters, we get
KE = (0.13kg)(9.8 m/s²)(508 m)
KE = 647.19 J
So, the orange would have 647.19 J of kinetic energy when it hits the sidewalk.
What is the resistance of a lamp which draws 250 milliamperes when connected toa 12.6 volt battery
V = I*R
R = V / I
250 ma = 0.25 A
R = 12.6 / 0.25
R =50.4 ohm
The example of a book falling off of a table shows a(n) _____.1) contact force2) scalar quantity3) absence of acceleration4) field force
anotherGravity is the force of attraction between two objects. A gravitational force field is modeled as space around a massive body in which other bodies experiences force.
In Newtonian gravity, a particle of mass M creates a gravitational force field around itself given as,
[tex]g=\frac{GM}{r^2}[/tex]Here, G is the universal gravitational constant, and r is the separation between the bodies.
Hence, the example of a book falling off of a table shows a(n) field force. Therefore, option (4) is the correct choice.
a ray of light hits a flat mirror with an incident angle of 39.20 degrees. What angle does it reflect at? (do not put in units)
Given
A ray of light hits a flat mirror with an incident angle of 39.20 degrees.
To find
What is the angle of reflection?
Explanation
According to the law of reflection,
The angle of incidence is equal to the angle of reflection.
Here the angle of incidence is 39.20 deg
So the angle of reflection is 39.20 deg
Conclusion
The angle of reflection is 39.20 deg
a clay ball (0.65kg) is thrown at a wall and sticks. The speed of the ball before hitting the wall was 15 m/s. It took 45 miliseconds for the clay to come to a stop and make contact with the wall. What was the average force the wall exerted on the wall during the collision?
In order to determine the average force the wall exerted on the ball, use the following formula:
[tex]F=\frac{\Delta p}{\Delta t}[/tex]where
Δp: change in momentum
Δt: change in time = 45 ms = 0.045 s
F: force = ?
calculate the change in momentum as follow:
[tex]\Delta p=m\Delta v=(0.65kg)(15\frac{m}{s}-0\frac{m}{s})=9.75\text{ kg}\cdot\text{m/s}[/tex]next, replace Δp and Δt into the formula for F:
[tex]F=\frac{9.75\text{ kgm/s}}{0.045s}=216.67N[/tex]Hence, the average force exterded by the wall on the ball was approximately 216.67N
Question 2
An object has a momentum of 500 kg mls. If its mass is 20 kg, its speed must be
O 25 m/sO 20 m/s50 m/sO 500
Given:
The momentum of the object is p = 500 kg m/s
The mass of the body is m = 20 kg
To find the speed of the object.
Explanation:
Speed can be calculated by the formula
[tex]\begin{gathered} p=mv \\ v=\frac{p}{m} \end{gathered}[/tex]On substituting the values, the speed of the object will be
[tex]\begin{gathered} v=\frac{500}{20} \\ =25\text{ m/s} \end{gathered}[/tex]Thus, the speed of the object is 25 m/s
What is the escape speed (in km/s) from an Earth-like planet with mass 4.9e+24 kg and radius 70.0 × 105 m? Use the gravitational constant G = 6.67 × 10-11 m3 kg-1 s-2.
The escape velocity = 9.66 km/s
Explanation:The mass of the planet is represented by m
[tex]m=4.9\times10^{24}\operatorname{kg}[/tex]The radius is represented by r
[tex]r=70.0\times10^5m[/tex]The gravitational constant is represented by G
[tex]G=6.67\times10^{-11}m^3kg^{-1}s^{-2}[/tex]The escape velocity (v) is given by the formula:
[tex]\begin{gathered} v=\sqrt[]{\frac{2GM}{r}} \\ v=\sqrt[]{\frac{2\times6.67\times10^{-11}\times4.9\times10^{24}}{70\times10^5}} \\ v=\sqrt[]{\frac{65.366\times10^{13}}{70\times10^5}} \\ v=\sqrt[]{0.9338\times10^8} \\ v=\sqrt[]{93380000} \\ v=9663.33\text{ m/s} \\ v=9.66\text{ km/s} \end{gathered}[/tex]The escape velocity = 9.66 km/s
Which of the following is NOT a characteristic of average speed?elloa. Average speed includes both the rate of motion and its direction.b. Average speed is the rate of motion.C. Average speed is always a positive value.d. The greater the speed of an object the faster it moves.
The average speed is defined as the total distance traveled by an object divided into the time that it takes for the object to travel that distance.
Average speed is a scalar quantity, which means that it can be represented using a number with no need of specifying a direction. Since the total distance is a positive value, then average speed is always also a positive value.
From the options, the statement "aveage speed includes both rate of motion and its direction" is false.
Therefore, the answer is: Option A.
How far did a runner travel if they ran 3 mph for 30 minutes?
Given:
Speed = 3 mph
Time = 30 minutes
Let's determine how far(distance) the runner travelled.
To find the distance, apply the formula:
[tex]Dis\tan ce=speed\ast time[/tex]Since the time is in minutes, let's convert to hours by dividing the time by 60 minutes.
Thus, we have:
[tex]\begin{gathered} D=3\ast\frac{30}{60} \\ \\ D=3\ast\frac{1}{2} \\ \\ D=1.5\text{ m} \end{gathered}[/tex]Therefore, the runner travelled 1.5 miles
What will be the effect of multiplying negative scalar number with a vector.. Explain with the help of drawing.
When we multiply a vector by a scalar number, each dimension of the vector will be multiplied by the number.
For example, multiplying the vector <3, 4> by 2 would create the vector <6, 8>.
But if this scalar is negative, the dimensions of the vector will change the signal:
vector <3, 4> multiplied by -2 creates the vector <-6, -8>.
This causes the original vector to change its direction to the opposite direction, that is, it flips 180°:
an ocean moves at 22.9 m/s and has a wavelength of 27.0. with what frequency do the waves pass you?
The zombies are closing in on him and Mr. Mangan sees a ramp to his left. Hedrives at it and hits the ramp with a velocity of 30 m/s. If it took him 7 s toget to the ramp, what was his acceleration?
Given data
*The given velocity is v = 30 m/s
*The given time is t = 7 s
The formula for the acceleration is given as
[tex]a=\frac{v}{t}[/tex]Substitute the values in the above expression as
[tex]\begin{gathered} a=\frac{30}{7} \\ =4.28m/s^2 \end{gathered}[/tex]Hence, the acceleration is a = 4.28 m/s^2
37.Why is hot water used to loosen a sealed jar lid?Select one:a. The jar contracts when heated.b. The lid expands when heated.c. The lid contracts when heated.d. The jar and the lid both contract.
Answer:
Explanation:
Heat causes expansion of the lid at a faster rate th
Our galaxy, the Milky Way, contains approximately 4.0 x 1011 stars with anaverage mass of 2.0 X 1030 kg each. How far away is the Milky Way from ournearest neighbor, the Andromeda Galaxy, if Andromeda contains roughly thesame number of stars and attracts the Milky Way with a gravitational force of2.4 x 1030 N?
We will have the following:
First, we remember:
[tex]F=\frac{Gm_1m_2}{r^2}[/tex]Then, from the problem we will have that:
[tex]2.4\ast10^{30}N=\frac{G(2.0\ast10^{30}kg)(2.0\ast10^{30}kg)}{r^2}[/tex][tex]\Rightarrow r^2=\frac{G(2.0\ast10^{30}kg)^2}{2.4\ast10^{30}N}\Rightarrow r\approx1.1116666667\ast10^{20}m[/tex]So, the Andromeda galaxy is approximately 1.1*10^20 meters from the milky way.
The acceleration of the Andromeda galaxy towards the milky way is:
[tex]2.4\ast10^{30}N=(2.0\ast10^{30}kg)a\Rightarrow a=1.2m/s^2[/tex]So, the acceleration towards the milky ways is 1.2m/s^2.
27. Write a number in scientific notation with4 sig figs, and 2 zeros
Answer:
1.600 x 10³
Explanation:
The numbers in scientific notation are multiplied by a power of 10, additionally, if there is a decimal point, all the numbers including zeros are significant figures. Therefore, a number in scientific notation with 4 significant figures and 2 zeros is:
1.600 x 10³
The largest single publication in the world is the 1112-volume set of British Parliamentary Papers for 1968 through 1972. The complete set has a mass of 3,548 kg. Suppose the entire publication is placed on a cart that can move without friction. The cart is at rest, and a librarian is sitting on top of it, just having loaded the last volume. The librarian jumps off the cart with a horizontal velocity relative to the floor of 2.65 m/s to the right. The cart begins to roll to the left at a speed of 0.03 m/s. Assuming the cart’s mass is negligible, what is the librarian’s mass? Round to the hundredths.
ANSWER:
40.17 kg
STEP-BY-STEP EXPLANATION:
Given:
m1 = 3548 kg
V1 = -0.03 m/s
V2 = 2.65 m/s
We apply law of conservation of linear momentum:
[tex]\begin{gathered} P_i=P_f \\ P_i=0 \\ P_f=m_1\cdot V_1+m_2\cdot V_2 \end{gathered}[/tex]We replace and calculate the mass as follows:
[tex]\begin{gathered} m_1\cdot V_1+m_2\cdot V_2=0 \\ 3548\cdot-0.03+2.65\cdot m_2=0 \\ -106.44+2.65\cdot m_2=0 \\ m_2=\frac{106.44}{2.65} \\ m_2=40.17\text{ kg} \end{gathered}[/tex]Therefore the mass is equal to 40.17 kg
Find the work W W done by the 30-newton force.
The work done by the 30 newton force is 4156.8 J
How do I determine the work done?We know that work done is defined according to the following formula:
Work done (Wd) = force (F) × distance (d)
Wd = fd
Where angle projection is involved, the work done is defined as follow::
Wd = FdCosθ
Haven know the formula for calculating work done, we shall thus, determine the work done by the 30 newton force as follow:
Force (F) = 30 newtonAngle (θ) = 30 degreesDistance (d) = 160 mWorkdone (Wd) =?Wd = FdCosθ
Wd = 30 × 160 × Cos 30
Wd = 4800 × 0.8660
Wd = 4156.8 J
Thus, from the calculation made above, we can conclude that the work done is 4156.8 J
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The radioactive isotope 239Pu has a half-life of approximately 24100 years. After 2000 years, there are 5g of 239Pu.(1) What was the initial quantity? (Round your answer to three decimal places.) g Tries 0/99(2) How much of it remains after 20000years? (Round your answer to three decimal places.) g Tries 0/99
Using the following formula:
[tex]\begin{gathered} N(t)=N_o(0.5)^{\frac{t}{t_{1/2}}} \\ where: \\ N(t)=Remaining_{\text{ }}quantity_{\text{ }}after_{\text{ }}time_{\text{ }}t \\ N_o=Initial_{\text{ }}quantity \\ t=time_{\text{ }}in_{\text{ }}years \\ t_{1/2}=half-life=24100 \end{gathered}[/tex](1)
[tex]\begin{gathered} t=2000 \\ t_{1/2}=24100 \\ N(2000)=5g \\ so: \\ 5=N_o(0.5)^{\frac{2000}{24100}} \\ N_o=\frac{5}{(0.5)^{\frac{2000}{24100}}} \\ N_o\approx5.296 \end{gathered}[/tex](2)
Using the initial quantity calculated previously:
[tex]\begin{gathered} t=20000 \\ N(20000)=5.296(0.5)^{\frac{20000}{24100}} \\ N(20000)=2.979 \end{gathered}[/tex]Answers:
For (1): 5.296
For (2): 2.979
Which of the following is the type of force that pulls objects towards the center of motion?A. Inertia B. MomentumC. AccelerationD. Centripetal
When an object is in a circular motion, the acceleration responsible for changing the direction of the movement is known as the centripetal acceleration:
This acceleration points towards the center of the circular motion.
Using the second law of newton, we can find the centripetal force Fc:
[tex]F_c=m*a_c[/tex]Therefore the force that pulls objects towards the center of motion is the centripetal force.
Correct option: D.
i would like some help with this problem, i already tried to complete it on my own but would like to check my answer
Given:
Two circuits with multiple resistors
To find:
The net resistance of the circuits
Explanation:
For the first circuit
The equivalent series resistance of the 1 and 2 is,
[tex]\begin{gathered} R_1+R_2 \\ =20+16 \\ =36\text{ ohm} \end{gathered}[/tex]The equivalent series resistance of the 4 and 5 is,
[tex]\begin{gathered} 10+14 \\ =24\text{ ohm} \end{gathered}[/tex]The net resistance of the resistances is,
[tex]\begin{gathered} \frac{1}{R}=\frac{1}{36}+\frac{1}{24}+\frac{1}{12} \\ \frac{1}{R}=\frac{11}{72} \\ R=\frac{72}{11}\text{ohm} \\ R=6.54\text{ ohm} \end{gathered}[/tex]Hence, the net resistance of the upper circuit is 6.54 ohms.
For the circuit below:
[tex]R_2,\text{ R}_3,\text{ R}_4\text{ are in series}[/tex]So, the equivalent resistance is,
[tex]\begin{gathered} R_2+R_3+R_4 \\ =2+2+2 \\ =6\text{ ohm} \end{gathered}[/tex]This equivalent resistance in parallel with
[tex]R_5[/tex]So, the equivalent resistance is,
[tex]\begin{gathered} \frac{6\times2}{6+2} \\ =1.5\text{ ohm} \end{gathered}[/tex]Now 1.5 ohm is in series with the rest two resistances.
So, the net resistance is,
[tex]\begin{gathered} R_1+R_6+1.5 \\ =2+2+1.5 \\ =5.5\text{ ohm} \end{gathered}[/tex]Hence, the net resistance of the circuit below is 5.5 ohms.
A block of ice is sliding down a ramp of slope 40 to the horizontal. What is the rate of acceleration of the block? Assume the force of friction is not significant.
The given problem can be exemplified in the following diagram:
Since there is no friction or any other external force, the only force acting in the direction of the movement is the component of the weight of the block, therefore, applying Newton's second law:
[tex]\Sigma F=ma[/tex]Replacing the values:
[tex]mg\sin 40=ma[/tex]We may cancel out the mass:
[tex]g\sin 40=a[/tex]Using the gravity constant as 9.8 meters per square second:
[tex]9.8\frac{m}{s^2}\sin 40=a[/tex]Solving the operations:
[tex]6.3\frac{m}{s^2}=a[/tex]Therefore, the acceleration is 6.3 meters per square second.
One closed organ pipe has a length of 2.24 meters. When a second pipe is played at the same time, a beat note with a frequency of 1.1 hertz is heard. By how much is the second pipe too long? Include units in your answer.
ANSWER
0.07 m
EXPLANATION
First, we have to find the frequency that the first pipe would play,
[tex]\lambda=4L=4\cdot2.24m=8.96m[/tex]The frequency is,
[tex]f=\frac{v}{\lambda}=\frac{343m/s}{8.96m}=38.28125Hz[/tex]The frequency played by the second pipe is,
[tex]f=38.25125Hz-1.1Hz=37.18125Hz[/tex]The wavelength of this note is,
[tex]\lambda=\frac{v}{f}=\frac{343m/s}{37.18125s^{-1}}\approx9.2251m[/tex]So the length of the second pipe is,
[tex]L=\frac{\lambda}{4}=\frac{9.2251m}{4}\approx2.31m[/tex]The difference between the pipes' length is,
[tex]2.31m-2.24m=0.07m[/tex]Hence, the second pipe is 0.07 meters too long
Calculate the Mach number for sound given the temperature and the speeda. 332 m/s at 30°Cb. 340 m/s at -10°Cc. 6000 km/h at 13°Cd. 6000 km/h at -13°0
The Mach number indicates how many times a speed is greater than the speed of sound.
So, to find the Mach number, we just need to divide the speed by the speed of sound.
a) at 30°C, the speed of sound is 349 m/s, so we have:
[tex]\frac{332}{349}=0.95[/tex]b) at -10°C, the speed of sound is 325 m/s, so we have:
[tex]\frac{340}{325}=1.05[/tex]c) at 13°C, the speed of sound is 339 m/s. First, let's convert the speed from km/h to m/s:
[tex]\begin{gathered} 6000\text{ km/h}=\frac{6000}{3.6}\text{ m/s}=1666.67\text{ m/s} \\ \frac{1666.67}{339}=4.92 \end{gathered}[/tex]d) at -13°C, the speed of sound is 323 m/s, so we have:
[tex]\frac{1666.67}{323}=5.16[/tex]