Plum pudding
Explanation:Note these points:
• J.J Thompson performed the Cathose ray experiment
,• From this experiment, he discovered that all atoms contain tiny negatively charged subatomic particles called electrons
,• J.J Thompson proposed the plum pudding model of the atom
,• This model contained negatively-charged electrons embedded within a positively charged soup
,• It was on April 30, 1897 that J.J Thompson announced the presence of the electrons in the atom, a result of his cathode ray experiment
From the points highlighted above, we can conclude that J.J Thompson gained recognition fpr discovering the electron and also won a noble prize for his plum pudding model of the atom
An ocean A. Open system B. Closed system C . Isolated
ANSWER:
A. Open system
STEP-BY-STEP EXPLANATION:
The ocean is a component of the hydrosphere, and the ocean surface represents the interface between the hydrosphere and the atmosphere above it. Which means that it generates an exchange of matter and energy, so the ocean is an example of an open system.
Christie and Daisy measured the applied forces while pulling on opposite ends of a rope, but they forgot to fill in their table ofresults shown below. Predict the most likely values that belong in Box C and Box D.Scale 1 (Christie) Scale 2 (Daisy) Observed Effect on Motion2 kg2 kgBox A3 kgBox BRope moved toward DaisyBox CBox DRope moved toward Christie(1 point)O Box C 4 kg; Box D: 4 kgO Box C: 3 kg, Box D: 3 kgO Box C: 4 kg, Box D: 3 kgO Box C: 3 kg; Box D: 4 kg
To find:
Most likely values that belong in box C and box D.
Explanation:
From Newton's second law of motion, the net force acting on an object is equal to the product of the mass of the object and the acceleration of the object produced due to the net force.
That is,
[tex]\vec{F}=m\vec{a}[/tex]Thus the object will accelerate in the direction of the net force.
It is given that in the third trail, the box moves towards Christie. Thus the net force is directed towards the Christie. Thus the force applied by Christie must be greater than the force applied by Daisy.
Final answer:
Thus the correct answer is option C.
The net torque of a bod about certain axis of rotation is 150 Nxm. You recalculate the net torque about a different axis of rotation. If your work is correctyou have to find the same number.is this tstemnt true or false?
Every individual torque acting on the body depends on the axis of rotation, however the net torque is independent of the axis we choose. Therefore, the statement is true.
At the bottom of the first hill, the 300 kg car is traveling 39.6 m/s. What is the kinetic energy of the car at this point? (ignore friction forces)
ANSWER:
1st option: 235,000 J
STEP-BY-STEP EXPLANATION:
Given:
Mass (m) = 300 kg
Speed (v) = 39.6 m/s
We can calculate the kinetic energy using the following formula:
[tex]\begin{gathered} E_K=\frac{1}{2}mv^2 \\ \text{ We replacing:} \\ E_K=\frac{1}{2}\cdot300\cdot39.6^2 \\ E_K=235224\cong235000\text{ J} \end{gathered}[/tex]The kinetic energy is 235000 J.
Hello there, to start off. I am BLIND literely and can't read the graphics on the website so the information about how to answer these problems needs to be written out. I am sorry but just how it is. Now the question. You are camping with two friends, Joe and Karl. Since all three of you like your privacy, you don't pitch your tents close together. Joe's tent is 21.5 m from yours, in the direction 21.5 ∘ north of east. Karl's tent is 44.5 m from yours, in the direction 40.5 ∘ south of east.Part AWhat is the distance between Karl's tent and Joe's tent?
The three tents form a triangle with two sides given by the distances between your tent and the tents of your friends; the remaining side is the distance between Joe's and Karl's tents.
To determine this side (and hence this distance) we need to notice that the angle between the two known sides is the addition of the angles between those sides and your position, hence the angle between the sides is 62°.
Now that we know this we can use the cosine law to determine the distance between the tents. The cosine law states that:
[tex]c^2=a^2+b^2-2ab\cos \gamma[/tex]this means that the squared of the distance we are looking for is equal to the sum of the squares of the other two distances minus twice the product of the other two sides and the cosine of the angle between them.
Then we have that:
[tex]\begin{gathered} c=\sqrt[]{21.5^2+44.5^2-2(21.5)(44.5)\cos (62)} \\ c=39.3 \end{gathered}[/tex]the distance we are looking for is the square root of twenty one point five squared plus forty four point five squared minus two multiplied by twenty one point five by forty four point five by the cosine of 62.
By making the operations we conclude that the distance between Joe's and Karl's tents is thirty nine point three meters
Please answer the questions
Answer:
Where's the question? So that i can answer it!
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.
Two forces act on an object as shown. Calculate the magnitude of the resultant force.[ Hint: Use the Pythagorean Theorem]
Using law of vector addition , the magnitude (R) of resultant force is given by
[tex]\begin{gathered} R=\sqrt{30^2+40^2+2\times30\times40\sin90\degree;} \\ R=\text{ }\sqrt{4900}=\text{ 70N}\begin{cases}sin90\degree={1} \\ \end{cases} \end{gathered}[/tex]Final answer is 70 N.
Note:- we need to use law of vector addition to get resultant of two forces but we cant use Pythagorean theorem for vector .
When 2 identical charged particles get closer to each other the strength of theelectrical force between them.A. increasesB. stays the sameC. decreasesD. you have to know the amount of the charge on the particles to answer
Given
2 identical charged particles get closer to each other
To find
The strength of the electrical force between them.
Explanation
The electrostatic force is indirectly proportional to the square of the distance between the charges.
So as the distance decreases, the electric force increases.
Conclusion
The correct opttion is
A. increases
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
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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Missing part of question:
See attached photo
2. A Roborovski hamster attempts to cross a desert road in Mongolia. He first moves to the right 70 cm, then to the left 30 cm, the back to the right 90 cm and then finally back to the left for 370 cm. What is the total displacement of the hamster? What was the average speed of the little fellow if it took him 6's to bust those moves? What is the average velocity in the same time interval?
The total displacement of the hamster is 240 cm.
The average speed of the little fellow is 93.33 cm/seconds.
What is distance?Distance is a scalar quantity that indicates "how much ground an object has covered" while moving. Displacement is a vector quantity that describes "how far out of place an object is"; it is the overall change in the position of the object.
The average speed is calculated by dividing the total distance traveled by the total time it took to travel that distance. Speed is the rate at which something moves at any given time. Average speed is the average rate of speed over the course of a journey.
For total displacement:
taking right as +ve direction
Travel D = +70 - 30 + 90 - 370
Displacement = 240 cm
Total distance travelled will be:
= 70+30+90+370
= 560 cm
Average speed will be:
= Total distance travelled /time
= 560/6
= 93.33 cm/sec
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The spectral classification for spica is B1V. This means it is A. A red giant star since B1 means it is very large B. A whit dwarf star since it is B1,hence it lies along the left edge of the HR for blue C.a blue giant star since it is B1, whee B stands for blue D. A main sequence star as indicated by V
The given spectral classification for spica is B1V.
B1 means it is B type star which have very high temperature and high mass. Due to its high temperature it will emit the light of lower wavelength which corresponds to blue color.
Thus, B1V is a white dwarf star since it is B1, hence it lies along the left edge of HR for blue which means option (B) is correct.
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
URGENT!! ILL GIVE
BRAINLIEST!!!! AND 100 POINTS!!!!!!
Answer:
the answer is B
Explanation:
because the hot air is transferred through heating being irrational from a thermal burning, because how the oven works
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
A car has a velocity of 21.3 m/s. It then accelerates at a uniform rate of 3.6 m/s per second for the next 5.0 seconds. What distance does the car cover during this time? Round to 4 decimal places if necessary
Explanation
U uniformly accelerated motion is the one in which the acceleration of the particle throughout the motion is uniform,the formula to find the distance is as follows:
[tex]\begin{gathered} x=v_ot+\frac{1}{2}at^2 \\ where \\ v_o\text{ is the initial velocity} \\ t\text{ is the time} \\ a\text{ is the acceleration} \end{gathered}[/tex]so
Step 1
a)let
[tex]\begin{gathered} v_o=21.3\text{ }\frac{m}{s} \\ a=3.6\frac{m}{s^2} \\ t=5\text{ s} \end{gathered}[/tex]b) now,replace in the formula and calculate
[tex]\begin{gathered} x=v_{o}t+\frac{1}{2}at^{2} \\ x=21.3\frac{m}{s}*5s+\frac{1}{2}3.6\frac{m}{s^2}*(5\text{ s\rparen}^2 \\ x=106.5\text{ m}+45\text{ m} \\ x=151.5\text{ m} \end{gathered}[/tex]therefore, the answer is 151.5 meters
I hope this helps you
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
Which element is a good material for use as a semiconductor?copperoxygensiliconsodium
To find:
Which of the given elements is a good material for semiconductor?
Explanation:
The silicon is widely used for the manufacturing of semiconductors. The stable structure of the silicon makes it a suitable candidate for the manufacture of semiconductors. The silicon is widely abundant in nature. This makes it economical.
a student who weighs 556 Newtons climbs the stairway (vertical height of 4.0 m) in 25 s (a) how much work is done (b) what is the power output of the student
Given data:
Weight of the student;
[tex]\begin{gathered} F=mg \\ =556\text{ N} \end{gathered}[/tex]Height;
[tex]h=4.0\text{ m}[/tex]Time;
[tex]t=25\text{ s}[/tex]Part (a)
The work done by the student is given as;
[tex]W=Fh[/tex]Substituting all known values,
[tex]\begin{gathered} W=(556\text{ N})\times(4.0\text{ m}) \\ =2224\text{ J} \end{gathered}[/tex]Therefore, the work done by the student is 2224 J.
Part (b)
The power is defined as the rate of doing work. Mathematically,
[tex]P=\frac{W}{t}[/tex]Substituting alll known values,
[tex]\begin{gathered} P=\frac{2224\text{ J}}{25\text{ s}} \\ =88.96\text{ W} \end{gathered}[/tex]Therefore, the power output of the student is 88.96 W.
How much heat transfer is required to completely boil 3500 g of water (already at its boiling point of 100°C) into a gas?answer in joules
Answer:
7910000 J
Explanation:
The heat transfer to phase change is calculated as:
[tex]Q=mH_v[/tex]Where m is the mass and Hv is the heat of vaporization.
For water, Hv = 2260 J/g.
So, replacing m = 3500 g and Hv = 2260 J/g, we get:
Q = 3500 g (2260 J/g)
Q = 7910000 J
Therefore, the answer is:
7910000 J
Consider a container filled with coconut oil having a density of 0.903 g/ml, What is the gauge pressure (Pa) 6.35 cm below the surface? 1ml = 1cm^3, 1000g = 1kg
The gauge pressure is 562.5 Pa.
Given data:
The density of oil is ρ=0.903 g/ml.
The distance below the surface is h=6.35 cm.
The density in kg/cm³ will be,
[tex]\begin{gathered} \rho=0.903\text{ g/ml }\times\frac{1\text{kg}}{1000\text{ g}}\times\frac{1\text{ml}}{1\text{ }cm^3}\times\frac{10^6\text{ }cm^3}{1m^3} \\ \rho=\frac{903kg}{m^3} \end{gathered}[/tex]The gauge pressure can be calculated as,
[tex]\begin{gathered} p=\rho gh \\ p=(\frac{903kg}{m^3})(\frac{9.81m}{s^2})(6.35cm\times\frac{1\text{ m}}{100cm}) \\ p=\frac{562.5kg}{ms^2}\times\frac{1\text{ Pa}}{\frac{kg}{ms^2}} \\ p=562.5\text{ Pa} \end{gathered}[/tex]Thus, the gauge pressure is 562.5 Pa.
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
Find the final temperature of the thermometer assuming no heat flows to the surroundings
ANSWER
[tex]\begin{equation*} 71.51\degree C \end{equation*}[/tex]EXPLANATION
Parameters given:
Mass of thermometer, m = 300 g
Initial temperature of thermometer, t1 = 35°C
Volume of water, V = 258 cm³
Initial temperature of water, T1 = 80°C
First, let us find the mass of the water using the formula for density:
[tex]\begin{gathered} \rho=\frac{M}{V} \\ \Rightarrow M=\rho *V \end{gathered}[/tex]where ρ = density of water = 1 g/cm³
Therefore, the mass of the water is:
[tex]M=1*258=258g[/tex]According to the conservation of energy, the total heat flow (the sum of the heat energy of the thermometer and water) must be equal to 0 since no heat flows to the surroundings:
[tex]\begin{gathered} Q_g+Q_w=0 \\ mc(T-t_1)+MC(T-T_1)=0 \end{gathered}[/tex]where c = specific heat capacity of glass thermometer = 0.2 cal/g°C
C = specific heat capacity of water = 1 cal/g°C
T = final temperature of thermometer and water
Hence, solving for T, we have that:
[tex]\begin{gathered} T=\frac{mct_1+MCT_1}{mc+MC} \\ T=\frac{(300*0.2*35)+(258*1*80)}{(300*0.2)+(258*1)} \\ T=\frac{2100+20640}{60+258}=\frac{22740}{318} \\ T=71.51\degree C \end{gathered}[/tex]That is the final temperature of the thermometer.
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
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
Part AA space vehicle accelerates uniformly from 80m/s att 0 to 167 m/s att 10.0 sHow far did it move between t = 2.0 s and t - 6.08 ?Express your answer to two significant figures and include the appropriate units.
Given:
Velocity at t = 0: 80 m/s
Velocity at t = 10.0 s: 167 m/s
Let's find the distance it covered between t = 2.0s and t = 6.0 s
Apply the kinematics formula:
[tex]v=u+at[/tex]Where:
v is the final velocity ==> 167 m/s
u is the initial velocity ==> 80 m/s
a is the acceleration
t is the time ==> 10.0 - 0 = 10.0 s
Now, let's find the acceleration.
Rewrite the formula for a:
[tex]\begin{gathered} a=\frac{v-u}{t} \\ \\ a=\frac{167-80}{10} \\ \\ a=\frac{87}{10} \\ \\ a=8.7m/s^2 \end{gathered}[/tex]The acceleration is 8.7 m/s²
To find the distance, apply the formula:
[tex]s=ut+\frac{1}{2}at^2[/tex]Thus, we have:
• At t = 2.0 s:
[tex]\begin{gathered} d_1=80(2)+\frac{1}{2}(8.7)(2)^2 \\ \\ d_1=160+17.4 \\ \\ d_1=177.4\text{ m} \end{gathered}[/tex]• At t = 6.0 s:
[tex]\begin{gathered} d_2=ut+\frac{1}{2}at^2 \\ \\ d_2=80(6)+\frac{1}{2}(8.7)(6)^2 \\ \\ d_2=480+156.6 \\ \\ d_2=636.6\text{ m} \end{gathered}[/tex]To find the distance traveled between t = 2.0 s and t = 6.0 s, we have:
d = d2 - d1 = 636.6 m - 177.4 m
d = 459.2 m
Therefore, the distance traveled betwen t = 2.0 s and t = 6.0 s is 459.2 m
ANSWER:
459.2 m
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
E. O-27.7 m/s9. An object of mass 2 kg moving with a speedof 26 m/s to the right collides with an objectof mass 13 kg moving with a speed of 2 m/sto the left. If the collision is completely inelastic,calculate their speed after collision. (1 point)A. O-0.831 m/sB. O 1.733 m/sC. 06.089 m/sD. O5.39 m/sE. O 0.388 m/s10. An object of mass 16 kg moving with a speed
Given
Mass of an object, m=2 kg
Velocity of the object, u=26 m/s towards right
Mass of the other object, m'=13 kg
Speed of the object, u'=2 m/s towards left
To find
If the collision is completely inelastic,calculate their speed after collision.
Explanation
Since the collision is perfectly inelastic, so after collision both the body would combine together and move as one
Let the velocity after collision be v.
Considering the direction towards right as positive.
Thus,
By conservation of momentum
[tex]\begin{gathered} mu+m^{\prime}u^{\prime}=(m+m^{\prime})v \\ \Rightarrow2\times26-13\times2=(2+13)v \\ \Rightarrow v=1.733\text{ m/s} \end{gathered}[/tex]Conclusion
The velocity is B. 1.733 m/s