Object AB is moving with the velocity of 1.23 m/s.
Givne data:
The mass of object A is m=3 kg.
The velocity of object A is u=2 m/s.
The mass of object B is M=10 kg
The velocity of object B is v=1 m/s.
Applying the conservation of momentum before and after the collision,
[tex]\begin{gathered} mu+Mv=(m+M)V \\ (3)(2)+(10)(1)=(3+10)V \\ V=1.23\text{ m/s} \end{gathered}[/tex]Thus, the object AB moving with a velocity of 1.23 m/s after the collision.
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 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.
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]
A 0.45 kg spike is hammered into a railroad tie. The initial speed of the spike is equal to1.9 m/s.If the tie and spike together absorb 22 per- cent of the spike’s initial kinetic energy as internal energy, calculate the increase in in- ternal energy of the tie and spike.Answer in units of J.
Given:
mass of the spike is
[tex]0.45\text{ kg}[/tex]initial speed od the spike is
[tex]v_=\text{ 1.9 m/s}[/tex]if the tie and spike together absorb 22 percent of the spike's initial kinetic energy.
Required:
calculate the increase in internal energy of the tie and spike.
Explanation:
here we apply conservation of energy.
[tex]\Delta U+\Delta K+\Delta P=0[/tex]total change in energy is zero.
here potential energy is not given so neglect that part. we have only
[tex]\Delta U+\Delta K=0[/tex][tex]\Delta U=-\Delta K[/tex]Here,
[tex]\Delta U=-(K_2-K_1)[/tex]K1 is initial kinetic energy and K2 is final kinetic energy that is equal to zero.
now we have
[tex]\Delta U=K_1[/tex]we know that kinetic energy is
[tex]K=\frac{1}{2}mv^2[/tex]then
[tex]\Delta U=\frac{1}{2}mv^2[/tex]plugging all the values in the above relation. we get
[tex]\begin{gathered} \Delta U=\frac{1}{2}\times0.45\text{ kg }\times(1.9\text{ m/s})^2\times0.22 \\ \Delta U=0.18\text{ J} \end{gathered}[/tex]Thus, the change in internal energy is
[tex]0.18\text{ J}[/tex]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
An iron railroad railis 750 ft long when the temperature is 34°C. What is its length (in ft) when the temperature is -18°C? (Round your answer to at least 3 decimal placeFt
Answer: Final length = 749.998 ft
Explanation:
We would assume that the coefficient of linear expansion of iron is 36 x 10^- 6/C
We would apply the formula for calculating linear thermal expansion which is expressed as
L1 = aL(θ2 - θ1)
where
L1 = change in length
L = original length
a = coefficient of linear expansion of iron
θ2 = final temperature
θ1 = initial temperature
From the information given,
L = 750
θ2 = - 18
θ1 = 34
a = 12 x 10^- 6/C
By substituting these values into the formula,
L1 = 36 x 10^- 6(- 18 - 34)
L1 = 36 x 10^- 6(- 52)
L1 = - 0.001872
Thus,
Final length - initial length = - 0.001872
Final length - 750 = - 0.001872
Final length = - 0.001872 + 750
Final length = 749.998 ft
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]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
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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1) Which of the following is not true about vectors.a) Must have magnitude and direction.b) When doing math with vectors, we can usually treat the x and y components independently.c) If several vectors are added together, their order does not matter.d) The magnitude of a vector is the sum of the magnitude of its x and y components.2) Explain briefly your argument or reasoning.
a) A vector must have a magnitude and direction. A physical quantity with only the magnitude is called a scalar.
b) When doing the math we can treat the x and y components independently. We can use only x-component or only y-component for the necessary calculations. For example, the projectile motion. In projectile motion, we use the components of velocities independently.
c) We can add or subtract more than two vectors in any order. For example, If there are 3 vectors, A, B, and C then, from the associative law of vector addition,
[tex]\vec{A}+(\vec{B}+\vec{C})=(\vec{A}+\vec{B})+\vec{C}[/tex]d) The magnitude of the vectors is not the sum of their x and y components. Let vector A be represented as
[tex]\vec{A}=x\hat{i}+y\hat{j}[/tex]Where x and y are the x and y components of vector A respectively. And î and j cap are the unit vectors along the x and y direction respectively.
Then, the magnitude of vector A is given by,
[tex]A=\sqrt[]{x^2+y^2}[/tex]Thus the correct answer is option d. That is the statement in option d is not true.
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.
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]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 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.
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]E. O-8.237 kg m/s4. An object of mass 0.2 kg falling vertically downwardhits the ground with a speed of 13 m/s and bouncesback vertically upward with a speed 4 m/s. If the object was incontact with the ground for 0.5 seconds, calculate theaverage force exerted by the ground on the object. (1 point)A. 06.8 NB. 10.563 NC. 09.048 ND. 12.856 NE. 03.149 N5. An object was acted upon by a force of 24 for 0.8seconds. Calculate the change in its momentum. (1 point)
4)
We woud apply the formula for calculating rate of change of momentum which is expressed as
F = m(v2 - v1)/t
where
F is the force
m is the mass of the object
v2 is the final velocity
v1 is the initial velocity
t is the time
From the information given,
m = 0.2
v1 = -13
v2 = 4
t = 0.5
F = 0.2(4 - -13)/0.5
F = 6.8 N
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
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.
Line D represents movement that starts out toward the south, slows down, and stops. Why is the slope of the line positive?1) The velocity is positive.2) The velocity is negative.3) The acceleration is negative.4) The acceleration is positive.
Given that line D represents the movement that start out towards the south, slow down, and stops. Let's determine why the slope of the line is positive.
The slope of a velocity-time graph is said to represent the acceleration of the moving object.
In this case, the slope starts out toward the south, slows down and stops, this means the acceleration of the object that moved is positive.
Since the acceleration is positive, the slope of the line is aslo positive.
Therefore, we can say the slope of the line is positive because the acceleration is positive.
ANSWER:
4) The acceleration is positive
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
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.
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.
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.
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.
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.
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,
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.
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]the earth moves at a speed of 2.95*10^4m/s and has a mass of 6.0*10^24. calculate the momentum of the earth
Given:
the speed of the earth is
[tex]v=2.95\times10^4\text{ m/s}[/tex]mass of the earth is
[tex]m=6.0\times10^{24}\text{ kg}[/tex]Required:
momentum of the earth needs to be calculated.
Explanation:
To calculate the momentum of the earth we will use momentum formula that is given as
[tex]P=mv[/tex]here P is momentum, m is the mass of the earth and v is the velocity of the earth.
plugging all the values in the above relation. we get,
[tex]\begin{gathered} P=6.0\times10^{24}\text{ kg }\times2.95\times10^4\text{ m/s} \\ P=\text{ 17.7 }\times10^{28}\text{ kg m/s} \end{gathered}[/tex]Thus, the momentum of the earth is
[tex]P=\text{17.7}\times10^{28}\text{kg\frac{m}{s}}[/tex]-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