We know that
• The sphere diameter is 8.55 cm.
,• The temperature change is from 30 C to 155 C.
First, we have to find the radius of the sphere. The radius is the half diameter.
[tex]\begin{gathered} r=\frac{d}{2}=\frac{8.55\operatorname{cm}}{2} \\ r=4.275\operatorname{cm} \end{gathered}[/tex]Now we have to find the volume of the sphere using the following formula.
[tex]V=\frac{4}{3}\pi(r)^3[/tex]Where r = 4.275 cm.
[tex]\begin{gathered} V=\frac{4}{3}\pi(4.275\operatorname{cm})^3 \\ V\approx327.26(cm)^3 \end{gathered}[/tex]Then, we use the following formula
[tex]V_2-V_1=BV_0(T_2-T_1)[/tex]Where the initial volume is 327.26 cubic cm, B is a constant about thermal expansion for aluminum, and we have to find the final volume to then calculate the percentage change.
[tex]\begin{gathered} \Delta V=75\times10^{-6}\cdot327.26(155-30) \\ \Delta V=75\times10^{-6}\cdot327.26(125) \\ \Delta V\approx3.07(cm)^3 \end{gathered}[/tex]This means that the volume change is 3.07 cubic centimeters.
At last, we have to divide the volume change by the initial volume, and then we have to multiply it by 100% to express it as a percentage.
[tex]\frac{\Delta V}{V_1}\times100=\frac{3.07}{327.26}\times100\approx0.938[/tex]Therefore, the percentage change is 0.938%.I think this is true I’m not really sure can you explain it?
Assuming there was gravitational force, an object flying upwards would decelerate due its weight. This would cause the velocity to decrease. In this scenario, the area is free of gravitational forces. The velocity is likely to remain the same. Therefore, the engine must not run at some maximum level in order to maintain the ships velocity. Thus, the answer is False
A 5.2 × 105 kg subway train is brought to a stop from a speed of 0.55 m/s in 0.51 m by a large spring bumper at the end of its track. What is the force constant k of the spring in N/m?
Given:
The mass of the subway train is,
[tex]m=5.2\times10^5\text{ kg}[/tex]The initial speed of the train is,
[tex]v=0.55\text{ m/s}[/tex]The distance moved by train is,
[tex]x=0.51\text{ m}[/tex]To find:
the spring constant
Explanation:
The kinetic energy of the train converts into the potential energy of the spring. So we can write,
[tex]\begin{gathered} \frac{1}{2}kx^2=\frac{1}{2}mv^2 \\ k=\frac{mv^2}{x^2} \end{gathered}[/tex]Substituting the values we get,
[tex]\begin{gathered} k=\frac{5.2\times10^5\times0.55^2}{0.51^2} \\ =0.60\times10^6\text{ N/m} \end{gathered}[/tex]Hence, the spring constant is
[tex]0.60\times10^6\text{ N/m}[/tex]A ball is sliding from the top to the bottom of a plank without rolling (e.g. imagine the surface is covered in ice, so very slippery). The ball is returned to the top and released again, but this time the ball is rolling (without slipping) down the plank (imagine the ice has melted). Compare the speeds of the ball at the bottom.a.The final speed is the same in both cases.b.The final speed is larger in the second case (with rolling).c.The final speed is larger in the first case (without rolling).d.The final speed is larger in the first case (without rolling) if the the plank is at an angle bigger than 45o and smaller if the angle is less than that.
To find:
Compare the speeds of the ball at the bottom of the plank.
Explanation:
From the law of conservation of energy, the total energy of the system always remains constant. Thus the total energy of the ball at the bottom of the plank must be equal to its total energy at the top of the plank.
When the ball is at the top of the plank, the ball has only potential energy. When the ball slides down to the bottom, this potential energy is converted into translational kinetic energy.
The translational kinetic energy is directly proportional to the square of the velocity of the object. Thus when the translational kinetic energy is high the velocity of the ball will be high.
When the ball rolls down the bottom of the plank, the initial potential energy of the ball is converted into translational and rotational kinetic energy.
The rotational kinetic energy of an object is proportional to the square of the angular velocity of the object.
Thus in the first case, the translational kinetic energy and hence the speed of the ball will be larger compared to that in the second case.
Final answer:
Thus the correct answer is option C.
Standing on a bridge, you throw a stone straight upward. The stone hits a stream, 32.5 m below the point atwhich you release it, 3.10 s later. What is the speed of the stone (in m/s) just after it leaves your hand? Pleasedo not include any units in your answer below. Type in only the numerical result. If you include units, youranswer will be marked as incorrect.
Given:
Distance the stone hits the stream = 32.5m below the released point
Time = 3.10 seconds
Let's find the speed of the stone just after it leaves your hand.
To find the speed of the stone, apply the kinematic formula:
[tex]\Delta y=v_{iy}\ast t-\frac{1}{2}g\ast t^2[/tex]Since the point the stone hits the stream is below the released point is, the change in distance is:
[tex]\Delta y=0-32.5=-32.5m[/tex]Where:
a = -g = -9.8 m/s^2
t = 3.10 s
Substituet values into the formula and solve for the speed of the stone (vy).
We have:
[tex]\begin{gathered} -32.5=v_{iy}\ast3.10-\frac{1}{2}(9.8)\ast3.10^2 \\ \\ -32.5=v_{iy}\ast3.10-4.9\ast9.61 \\ \\ -32.5=v_{iy}\ast3.10-47.089 \end{gathered}[/tex]Solving further:
[tex]\begin{gathered} v_{iy}\ast3.10=-32.5+47.089 \\ \\ v_{iy}\ast3.10=14.589 \end{gathered}[/tex]Divide both sides by 3.10:
[tex]\begin{gathered} \frac{v_{iy}\ast3.10}{3.10}=\frac{14.589}{3.10} \\ \\ v_{iy}=4.706\text{ m/s} \end{gathered}[/tex]Therefore, the speed of the stone just after it leaves your hand is 4.706 m/s
ANSWER:
4.706 m/s
-4, 0, -2/3, 4.11111…, 2, π, √6 'which members of this set are irrartional?
An irrational number is a real number that cannot be written as a simple fraction.
4 = 4/1 = rational
0 = 0/1 = rational
-2/3 = rational
4.11111 = irrational
2 = 2/1 = rational
π = 3.141592... = irrational
√6= irrational
Answer:
4.1111, π , √6
Which picture correctly shows the path of the reflected light rays given an object outside the focal point?Select one:a. Ab. Bc. Cd. D
We will have that the graph that describes the scenario is given by graph B.
Can you explain the second law of Kepler? Simple terms please
According to Kepler's second law.
The planets move in an elliptical orbit. An elliptical orbit is non-circular which means its radius is not uniform as Sun is present in one of the foci.
According to Kepler's law
Areal speed is constant. This means that when the planet is near the sun, its speed increases and when it is farther from the sun its speed decreases.
When speed is more, the time taken is less and when the speed is less, the time taken will be more.
Thus, areal speed always remains constant.
What does it mean that energy is quantized?
Electrons can posses only certain values of energy.
Explanation:
because of where the particle lies it. may be able to access more than that but it will be held at that position due to size. at least I think that's what this is. I took chemistry last year.
A neutral cloth and a neutral rod are rubbed together. Electrons are transferred from the cloth to the rod. What is the final charge on the overall cloth-rod system?A. PositiveB. NeutralC. NegativeD. Not enough info
Given that a neutral cloth and a neutral rod are rubbed together and electrons are transferred from the cloth to the rod.
Let's find the final charge on the overall cloth-rod system.
Here, since a neutral cloth and a neural rod are rubbed together, there is a transfer of charges between both materials.
The metal rod is a conductor while the cloth is an insulator.
As both materials are rubbed together, there is a transfer of electrons. This is due to friction on the materials.
Now, we know that electrons are negatively charged.
As electrons transfer, the rod becomes positively charged while the cloth becomes negatively charged.
Since both materials have opposite charges, the final charge on the overall cloth-rod system will be a NEUTRAL charge.
When a system is neutral, it means that the system has an equal amount of positive and negative cha
• ANSWER:
B. Neutral
What is the index of refraction of a substance if the critical angle for a ray going into Water (n=1.33) is 69.30?1.422.2652.11.24
First, remembering Snell's law, we know that:
[tex]sin(\theta)=\frac{n_{entering}}{n_{leaving}}[/tex]So, we know that n_entering is 1.33, as the ray is entering water, and we have the angle, thus:
[tex]sin(69.30°)=\frac{1.33}{n_{leaving}}[/tex]And therefore:
[tex]n_{leaving}=\frac{1.33}{sin(69.30°)}=1.42[/tex]A toy that Is 0.5 ft long is used to model the actions of an actual car that is 15 ft long. Which ratio shows the relationship between the sizes of the model and the actual car????
The length of the toy is,
[tex]l=0.5\text{ ft}[/tex]The actual length is given as,
[tex]l^{\prime}=15\text{ ft}[/tex]The ratio of these two values is,
[tex]\begin{gathered} \frac{l}{l^{\prime}}=\frac{0.5}{15} \\ \frac{l}{l^{\prime}}=\frac{1}{30} \end{gathered}[/tex]Thus, the model is 1/30 times size of an actual car.
Based on the Mohs Hardness Scale, which mineral could be scratched by a penny but not by a fingernail?Question options:A) FluoriteB) TalcC) CalciteD) Gypsum
Given:
Scale number of a penny = 3.5
Scale number of a fingernail = 2.5
Let's determine the mineral from the options which could be scratched by a penny but not a fingernail.
A mineral with a scale number can scratch any material with equal or lower scale number.
Since a penny has a scale number of 3.5, it can scratch any material with a scale number of 3.5 or lower.
Also, since a fingernail could not scratch the material here, this means the scale number if the material must be greater than 2.5.
Thus, the mineral that could be scratched by a penny but not a fingernail must have a scale number equal to or less than 3.5 but greater than 2.5.
Hence, from the MOHS Hardness Scale, the mineral with a scale number between 3.5 and 2.5 is the Calcite with a scale number of 3.
Therefore, the mineral which could be scratched by a penny but not a fingernail is Calcite.
• ANSWER:
C) Calcite.
A 2-newton force applied to a 2-kilogram object caused it to move in a straight line 2 meters during an interval of 2 seconds. The object gains kinetic energy K during this interval. In which of the following cases will the object gain the same kinetic energy K? A. The same force is applied to a 4-kilogram object for the same time. B. The same force is applied to a 4-kilogram object for the same distance.C. A 4-Newton force is applied to a 4-kilogram object for the same time.D. A 4-Newton force is applied to a 4-kilogram object for the same distance.E. The same force is applied to a 4-kilogram object for 4 seconds.
By the kinetic energy theorem we know that:
[tex]W=\Delta K[/tex]where W is the work done by the force and delta K is the change in kinetic energy.
In the original scenario the work done by the force is:
[tex]W=Fd=(2)(2)=4[/tex]Which means that the object gained 4 J in kinetic energy.
Now we need to determine in which scenario given the work done is the same, from the options we notice that in option B we apply the same force for the same distance, which means that the work will be the same and hence the object will gain the same kinetic energy (note that this does not mean that the final velocity will be the same, just that the object will gain the same amount of kinetic energy). In all the other cases we can't conclude the same.
Therefore, the answer is option B.
A box with a mass of 2kg slides across a surface at the velocity of 10 m/s. A force of 32 N is applied. What force will cause the box to continue with a velocity of 10 m/s?
Answers:
F2 = 32N
a = 10 m/s²
Explanation:
The object will continue with the same velocity if the net force is equal to 0. So, to make the net force equal to 0, F2 should be equal and opposite to F1. It means that if F2 = 32 N, the object will continue with a velocity of 10 m/s
On the other hand, If F2 = 12 N, the net force will be equal to
Net Force = F1 - F2
Net Force = 32N - 12N
Net Force = 20N
Then, by the second law of Newton, the acceleration is equal to the net force divided by the mass. Since the mass m = 2kg, the acceleration is
a = F/m
a = 20N/2kg
a = 10 m/s²
Therefore, the answers are:
F2 = 32N
a = 10 m/s²
A girl ties a rope securely to the windowsill of a tower and climbs down to the base of it. She is with a group of friends and wants them to come down too, but easier. She ties the other end of the rope to a tree 30m down and 60m away horizontally, creating a zipline. The first person down the zip line weighs 80 kg. How quickly does he accelerate, and how long does it take for him to reach the end of the zipline?
Answer:
Acceleration = 4.38 m/s²
Time = 5.54 s
Explanation:
We can represent the situation as follows:
So, first, we need to find the angle θ. Using trigonometric functions, we get:
[tex]\begin{gathered} \tan \theta=\frac{opposite}{adjacent} \\ \text{tan}\theta=\frac{60}{20} \\ \tan \theta=2 \\ \theta=tan^{-1}(2)=63.43 \end{gathered}[/tex]Then, the net force in the direction of the rope is equal to:
[tex]\begin{gathered} F_{\text{net}}=mg\cos \theta \\ F_{\text{net}}=(80\operatorname{kg})(9.8m/s^2)\cos (63.43) \\ F_{\text{net}}=350.62N \end{gathered}[/tex]By the second law of Newton, this force is equal to mass times acceleration, so we can solve for acceleration as follows:
[tex]\begin{gathered} F_{\text{net}}=ma \\ a=\frac{F_{net}}{m}=\frac{350.62N}{80\operatorname{kg}}=4.38m/s^2 \end{gathered}[/tex]So, the first person accelerates at 4.38 m/s².
Now, we need to find the length of the rope. Using the Pythagorean theorem, we get:
[tex]\begin{gathered} L=\sqrt[]{60^2+30^2} \\ L=\sqrt[]{3600+900} \\ L=\sqrt[]{4500}=67.08\text{ m} \end{gathered}[/tex]Then, using a kinetic equation, we get:
[tex]\begin{gathered} x=v_it+\frac{1}{2}at^2 \\ x=\frac{1}{2}at^2 \\ 2x=at^2 \\ \frac{2x}{a}=t^2 \\ t=\sqrt[]{\frac{2x}{a}} \end{gathered}[/tex]Where x is the distance traveled, vi is the initial velocity, which is 0 m/s, a is the acceleration and t is the time.
Now, we can replace x by the length of the rope 67.08m and a by 4.38 to get:
[tex]t=\sqrt[]{\frac{2(67.08)}{4.38}}=5.54\text{ s}[/tex]So, the first person takes 5.54 s to reach the end of the zipline.
Therefore, the answers are
Acceleration = 4.38 m/s²
Time = 5.54 s
URGENT!! ILL GIVE
BRAINLIEST!!!! AND 100 POINTS!!!!!!
Answer:
a - high thermal conductivity
Answer:
A. high thermal conductivity
Explanation:
high thermal conductivity means a material has the ability to easily take up heat from their environment, in this case the metal pan is taking in heat from it's environment and therefore the mitt is needed.
hope this helps
Explain numericals of electricity chapter of class 10 and also diagrams.What is ammeter?What is voltmeter?potential difference?Ohm's law?Resistance?Electric circuitElectric current
Let's explain the following electrical terms.
• Electric current, can be said to be the flow of electric charge. These charges are through a conductor by moving electrons.
• Ammeter, is an instrument used to measure the amount of electric current in a circuit.
The name ammeter was derived from the unit of electric current (Amperes).
• Voltmeter, can be said to be an instrument used to measure the voltage(potential difference) in a circuit. It measures the electric potential between two points in an electrical circuit. It can also be called voltage meter.
• Electric circuit, can be defined as the conductive path for the flow of electric current.
It allows electric charge carriers to flow continuously.
• Resistance ,can be said to be the property of an electrical conductor which resists the flow of electric current. It is measured in ohms.
• Ohms law, states that the potential difference (V) between two points is directly proportional to the electric current across two points.
It is deonted as: V = I x R
Voltage = Current x Resistance
• Electric potential ,can be said to be the amount of electric potential energy at a point.
The change in angular velocity divided by time.Also equal to a/r
The angular acceleration of the body in terms of the angular velocity is,
[tex]\alpha=\frac{\text{angular velocity}}{\text{time}}[/tex]The angular velocity in terms of the linear velocity is,
[tex]\text{angular velocity=}\frac{v}{r}[/tex]where v is the linear velocity and r is the radius,
Thus, the angular acceleration becomes,
[tex]\begin{gathered} \alpha=\frac{1}{r}\times\frac{v}{t} \\ \alpha=\frac{a}{r} \end{gathered}[/tex]where a is the linear acceleration of the body,
Explain how the energy of a rubber ball is transformed as it rolls down a ramp. Give evidence that the energy of the ball remains the same at all points on the ramp.
At the top of the ramp:
The height, h, is maximum, hence the ball has a maximum potential energy
Since the ball is not moving, the speed = 0 m/s. Hence the ball has zero kinetic energy
As the rubber ball moves down the ramp:
The ball will be in motion and have a certain amount of speed
Therefore, some of the Potential Energy is transformed to Kinetic Energy. That is, the rubber ball has both Kinetic and Potential energy
The Kinetic energy increases while the Potential energy decreases
At the bottom of the ramp:
The height = 0
The Potential energy of the rubber ball will be zero
The law of conservation of energy states that energy can neither be created nor destroyed, but can be transformed from one form to another. This is an evidence that the energy of the ball remains constant at all the points on the ramp.
[tex]\text{Total Energy = Kinetic Energy + Potential Energy}[/tex]A proton is accelerated by a constant electric field of the magnitude 600. N/C. Find: a) the force acting on the proton by the electric field;b) the acceleration of the particle due to the electric force.
ANSWER
[tex]\begin{gathered} (a)9.61\cdot10^{-17}N \\ (b)5.75\cdot10^{10}m\/s^2 \end{gathered}[/tex]EXPLANATION
(a) To find the force acting on the proton, we have to apply the formula representing the relationship between electric field and electric force:
[tex]F=qE[/tex]where q = charge of the proton
E = electric field
The charge of a proton is:
[tex]q=1.602\cdot10^{-19}C[/tex]Hence, the force acting on the proton by the electric field is:
[tex]\begin{gathered} F=1.602\cdot10^{-19}\cdot600 \\ F=9.61\cdot10^{-17}N \end{gathered}[/tex](b) To find the acceleration of the particle, apply the relationship between force and acceleration:
[tex]F=ma[/tex]where m = mass; a = acceleration
The mass of a proton is:
[tex]m=1.67\cdot10^{-27}\operatorname{kg}[/tex]Hence, the acceleration of the proton is:
[tex]\begin{gathered} 9.61\cdot10^{-17}=1.67\cdot10^{-27}\cdot a \\ \Rightarrow a=\frac{9.61\cdot10^{-17}}{1.67\cdot10^{-27}} \\ a=5.75\cdot10^{10}m\/s^2 \end{gathered}[/tex]find the drag force on a car traveling at 15 m/s. assume that the cars' cross-sectional area is 3.5 m^2, and that air has a density of 1.2 kg/m^3
The drag force on a car traveling at 15 m/s. assume that the cars' cross-sectional area is 3.5 m^2, and that air has a density of 1.2 kg/m^3 will be 472.5 C(d) Newton
Drag force = 1/2 * density of fluid * [tex](speed)^{2}[/tex] * drag coefficient * cross sectional area
= 1/2 * 1.2 * [tex]15^{2}[/tex] * C (d) * 3.5
= 472.5 C(d) Newton
To learn more about drag force here :
https://brainly.com/question/12774964
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A 120-turn, 9.604-cm-diameter coil is at rest in a horizontal plane. A uniform magnetic field 12 degrees away from vertical increases from 0.316 T to 5.553 T in 98.254 s. Determine the emf induced in the coil.
Given:
The number of turns is,
[tex]n=120[/tex]The diameter of the coil is
[tex]\begin{gathered} d=9.604\text{ cm} \\ =9.604\times10^{-2}\text{ m} \end{gathered}[/tex]The angle of the magnetic field with the coil is
[tex]\theta=12\degree[/tex]The change in a magnetic field is from
[tex]0.316\text{ T to 5.553 T}[/tex]in
[tex]t=98.254\text{ s}[/tex]To find:
The induced emf
Explanation:
The induced emf in the coil is,
[tex]\xi=-nA\frac{dB}{dt}cos\theta[/tex]Here, the area of the coil is,
[tex]\begin{gathered} A=\pi\frac{d^2}{4} \\ =\pi\times\frac{(9.604\times10^{-2})^2}{4} \\ =7.244\times10^{-3}\text{ m}^2 \end{gathered}[/tex]The induced emf is,
[tex]\begin{gathered} \xi=-120\times7.244\times10^{-3}\times\frac{5.553-0.316}{98.254}cos12\degree \\ =-0.045\text{ V} \end{gathered}[/tex]Hence, the induced emf is 0.045 V.
No tutor is helping with this difficult question. Please someone help
The labelled diagram of the situation is shown below
Recall the kinetic energy formula,
work done = 1/2mv^2
where
m is the mass
v is the velocity
Recall, work done = force x distance
Thus,
force x distance = 1/2mv^2
From the information given,
Force = 2000
distance = 0.1
m = 5
Thus,
2000 x 0.1 = 1/2 x 5 x v^2
200 = 2.5v^2
v^2 = 200/2.5 = 80
v = √80
v = 8.94 m/s
The ball will collide with the the crate with a velocity of 8.94. The velocity at which the crate will move is v2
We would apply the law of momentum.
Initial momentum before collision = final momentum after collision
m1u1 + m2u2 = m1v1 + m2v2
where
m1 and m2 are the masses of the ball and crate respectively
u1 and u2 are the initial velocities of the ball and crate respectively
v1 and v2 are the final velocities of the ball and crate respectively
From the information given,
m1 = 5
m2 = 31
u1 = 8.94
u2 = 0 (because it was stationary)
v1 = - 1.44(because it moved in the opposite direction
Thus,
5 x 8.94 + 3.1 x 0 = 5 x - 1.44 + 31 x v2
44.7 = - 7.2 + 31v2
31v2 = 44.7 + 7.2 = 51.9
v2 = 51.9/31
v2 = 1.67 m/s
This means that the velocity with which the crate will start moving is 1.67 m/s
With respect to springs, conservation of energy is expressed as
1/2kx^2 = 1/2mv^2
where
k is the spring constant
x is the compression of the spring
m is the mass of the crate
v is the velocity of the crate
From the information given,
k = 250
Thus,
1/2 x 250 x x^2 = 1/2 x 31 x 1.67^2
125x^2 = 43.22795
x^2 = 43.22795/125 = 0.3458236
x = √0.3458236
x = 0.59 m
The spring will compress by 0.59 m
Extra credit
With the presence of friction, the ball will exert lesser force on the crate and this would cause the compression of the spring to be reduced. The spring will stretch less.
How much force would someone need to move a 200kg fridge to accelerate at 5m/s^2
Given that the mass of fridge, m = 200 kg and acceleration is, a = 5 m/s^2
Force can be calculated using, Newton's law of motion,
[tex]F=ma[/tex]Substituting the values, the force will be
[tex]\begin{gathered} F=200\times5 \\ =1000\text{ N} \end{gathered}[/tex]Thus, the force needed to apply to the fridge is 1000 N.
A green ball (ball 1) of mass M collides with an orange ball (ball 2) of mass 1.26. the initial speed of the green ball is 5.4 m/s the final speed of the green ball is 2.6 m/s and theta=36.9° A. find the magnitude of the final speed of the orange ball? B. what is the direction of the final speed of the orange ball?
Given data:
The mass of ball 1 is m.
The mass of ball 2 is 1.26m.
The initial speed of ball 1 is u=5.4 m/s.
The final speed of the ball 1 U=2.6 m/s.
The angle at which the ball 1 moves from x-axis is θ=36.9.
Applying the conservation of momentum in x-direction,
[tex]\begin{gathered} mu=mU\cos \theta+(1.26m)V\cos \alpha \\ u=U\cos \theta+(1.26)V\cos \alpha \\ 5.4=2.6\cos 36.9+(1.26)V\cos \alpha \\ V\cos \alpha=2.63\ldots\ldots\text{.}(1) \end{gathered}[/tex]Here, V is the final speed of ball 2, and α is the angle of ball 2 with x-axis after the collision.
Applying the conservation of momentum in y-direction,
[tex]\begin{gathered} 0=mU\sin \theta+(1.26m)V\sin \alpha \\ 0=U\sin \theta+(1.26)V\sin \alpha \\ 0=2.6\sin 36.9+(1.26)V\sin \alpha \\ V\sin \alpha=-1.56\ldots\ldots\text{.}(2) \end{gathered}[/tex]Dividing equation (2) and (1),
[tex]\begin{gathered} \frac{V\sin \alpha}{V\cos \alpha}=\frac{-1.56}{2.63} \\ \tan \alpha=0.593 \\ \alpha=30.6\degree \end{gathered}[/tex]Subsitute the value of α in equation (1),
[tex]\begin{gathered} V\cos \alpha=2.63 \\ V\cos 30.6\degree=2.63 \\ V=3.05\text{ m/s} \end{gathered}[/tex]Thus, the final speed of the ball 2 (orange ball) is 3.05 m/s, and the direction of the orange ball is 30.6⁰.
This week, you completed a simulation in which you tested conductivity of various materials to see if you could complete the circuit and light the bulb. The screenshot above shows you a simple circuit, along with an opening for different objects to be tested. Among the objects provided in the simulation are the following:A dollar billA coinA paper clipAn eraserQuestion to answer:Which two objects listed would act as conductors if clipped into the circuit, and which two objects listed would act as insulators? Please include a sentence to summarize how you'd know if each object is considered a conductor or insulator based on the way the circuit is arranged.hi, can someone help me with this? you don’t have to give me an answer or anything cause i know it’s a lot but maybe tell me the steps on how to find the answer?
We will have the following:
From the options provided we can see that the two objects that could act as conductors in order to complete the circuit are the two metallic objects: the coin and the paper clip; reason being that metals are most of the time much better conductors than other materials.
The two objects then that could potentially act as insulators are the dollar bill and the eraser; reason being in part process of elimination and the fact that paper and the material of the eraser are most of the times bad conductors and present a high resistance value due to the nature of the material.
A 12 inch-wide conveyor unloads 5000 bushels of wheat per hour. At this speed, how many bushels per hour will an 18-inch-wide conveyor unload?
12-inch wide conveyor unloads 5000 bushels of wheat per hour. At this same speed, the number of bushels that will be unloaded per hour by an 18-inch wide conveyor can be calculated below
[tex]\begin{gathered} 12\text{ = 5000} \\ 18=\text{?} \\ u\text{nload amount=}\frac{5000\times18}{12}=\frac{90000}{12}=7500\text{ bushels per hour } \end{gathered}[/tex]Hello could you please help me with this problem? Write an equation in standard form for the line that has an undefined slope and passes through (5,-3).
Answer:
[tex]-3=5m+b[/tex]Explanation: We have to write the equation od line in a standard form that has an unknown slope and passes through (5,-3), in general, the equation of a line is as follows:'
[tex]\begin{gathered} y(x)=mx+b\Rightarrow(1) \\ m=\frac{\Delta y}{\Delta x} \\ b\rightarrow\text{ y-intercept} \end{gathered}[/tex]Passing through a point implies the following:
[tex](x,y)\Rightarrow(5,-3)[/tex]Plugging the x and y in (1) finally gives the following:
[tex]-3=5m+b[/tex]What is the result of (3.60 m ✕ 1.50 m)/0.50 m with the proper number of significant figures? answer in: m
We will have the following:
[tex]\frac{3.60m\ast1.50m}{0.50m}=10.8\approx11[/tex]So, the solution with the proper number of significant figures is approximately 11. [2 significant figures]
Question 3 of 25Which of the following statements is true of the phases of a given substance?A. The particles of a gas are in a fixed location.B. The particles of liquids have more kinetic energy than the particles of gases.C. Gases have strong intermolecular forces between the atoms or molecules.D. The particles of solids have less kinetic energy than the particles of liquids or gases.
ANSWER
D. The particles of solids have less kinetic energy than the particles of liquids or gases.
EXPLANATION
We want to identify the statement that is true about the phases of a given substance.
A gas is a state of matter that expands freely so as to fill the container which contains it. This implies that gases have no fixed shape or location.
The particles of a liquid are more closely packed than gases, implying that motion is restricted for liquids. Since kinetic energy is dependent on motion, it implies that less motion equals less kinetic energy and so liquids have lesser kinetic energy than gases.
In gases, the intermolecular forces are very weak and can be said to be negligible since the particles move freely.
As stated above, kinetic energy is dependent on motion. This means that the more movement that occurs, the more the kinetic energy. The particles of solids are more closely packed than in liquids and gases. This implies that these particles possess less kinetic energy.
Therefore, the correct option is option D.