Answers
We will have the next diagram
Then we can sum the forces in x and sum the forces in y
Forces in x
[tex]\sum ^{}_{}F_x=-T_1\sin (\alpha)+T_2\cos (\beta)=0[/tex][tex]\sum ^{}_{}F_x=-T_1\sin (15)+T_2\cos (35)=0[/tex]Forces in y
[tex]\sum ^{}_{}F_y=T_1\cos (\alpha)+T_2\sin (\beta)=mg[/tex][tex]\sum ^{}_{}F_y=T_1\cos (15)+T_2\sin (35)=19.5(9.8)[/tex]We simplify the equations found and we found the next system of equation
[tex]\begin{gathered} -T_1\sin (15)+T_2\cos (35)=0 \\ T_1\cos (15)+T_2\sin (35)=191.1 \end{gathered}[/tex]then we isolate the T2 of the first equation
[tex]T_2=\frac{T_1\sin(15)}{\cos(35)}[/tex]We substitute the equation above in the second equation
[tex]T_1\cos (15)+(\frac{T_1\sin(15)}{\cos(35)})\sin (35)=191.1[/tex]we simplify
[tex]T_1(\cos (15)+\frac{\sin (15)\sin (35)}{\cos (35)})=191.1[/tex][tex]T_1(1.147)=191.1[/tex]We isolate the T1
[tex]T_1=\frac{191.1}{1.147^{}}=166.6N[/tex]then we can substitute the value we found in T1 in the euation with T2 isolate
[tex]T_2=\frac{(166.6)_{}\sin (15)}{\cos (35)}=52.54N[/tex]
Related Questions
Do you know the answer for these? [ showing the work] I'm trying to figure out the answers to the bullet points under your mission section.
Answers
Answer:
Explanation:
• Orbital Radius,:
We are already told that the altitude of the satellite is 300,000 meters. Having this information in hand, it is easy to find the radius of orbit using the following relation:
[tex]R_{\text{satellite}}=A+R_{\text{earth}}[/tex]where A is the altitude.
Now we know that A = 300,000 m and earth radius = 6.37 * 10^6 m; therefore, the orbital radius of the satellite is
[tex]R_{\text{satellite}}=3\cdot10^6+6.37\cdot10^6[/tex][tex]\boxed{R_{\text{satellite}}=9.37\times10^6m}[/tex]which is our answer!
• Velocity:
We are told that the velocity of the satellite is given by
[tex]v=\sqrt[]{G\frac{m_E}{R_{\text{satellite}}}}[/tex]where G is the gravitational constant and m_E is the mass of the earth.
Substituting the numerical values for these constants gives
[tex]v=\sqrt[]{(6.67\times10^{-11})\frac{5.98\times10^{24}}{9.37\times10^6}}[/tex]Using a calculator we evaluate the above to be:
[tex]\boxed{v=6.52\cdot10^3m/s}[/tex]which is around 6.5 km per second!
• Orbital Period:
The orbital period T of the satellite is given by
[tex]T=2\pi\sqrt[]{\frac{R^3_{satellite}}{Gm_E}}[/tex]putting in the numerical values for the constants gives
[tex]T=2\pi\sqrt[]{\frac{(9.37\times10^6)^3}{(6.67\times10^{-11})(5.98\times10^{24})}}[/tex][tex]\boxed{T=9024s\approx2.5hr}[/tex]Hence, the period of satellites orbit is only 2.5 hours! This means that we can see the same satellite multiple times in the night sky if it is observable!
• Orbital Path:
The problem with satellites is that since they are travelling so fast, they don't get to observe one location on earth for a long time. One solution to this is to place the satellites into something called the geosynchronous orbit. In such an orbit, the period of the satellite matches the earth's period of rotation. This way, when observed from the earth, the satellite looks stationary, but in fact, it is travelling with the earth in the same orbital period. Such a satellite can be launched to observe locations along the arctic and the antarctic circles to obtain substantial data.
Block 1 (2 kg) is sliding to the right on a level surface at aspeed of 3 m/s. Block 2 (5 kg) is initially at rest, and block 1collides with it. After the collision, block 2 is moving to theright with a speed of 1.5 m/s. Calculate the magnitude anddirection of the velocity of block 1 after the collision.
Answers
Given:
The mass of block 1, m₁=2 kg
The mass of the block 2, m₂=5 kg
The initial velocity of the block 1, u₁=3 m/s
The initial velocity of the block 2, u₂=0 m/s
The velocity of the block 2 after the collision, v₂=1.5 m/s
To find:
The magnitude and direction of the velocity of block 2 after the collision.
Explanation:
From the law of conservation of momentum, the total momentum of blocks before the collision must be equal to the total momentum of the blocks after the collision.
Thus,
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]Where v₁ is the velocity of block 1 after the collision.
On rearranging the above equation,
[tex]\begin{gathered} m_1v_1=m_1u_1+m_2u_2-m_2v_2 \\ \implies v_1=\frac{m_1u_1+m_2u_2-m_2v_2}{m_1} \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} v_1=\frac{2\times3+5\times0-5\times1.5}{2} \\ =-0.75\text{ m/s} \end{gathered}[/tex]The negative sign indicates that block 1 will be sliding to the left after the collision.
Final answer:
Thus the magnitude of the velocity of block 1 after the collision is 0.75 m/s.
And the direction of block 1 after the collision is to the left.
Derive Kinetic Energy and Potential Energy.
Answers
Answer:
[tex]\begin{gathered} \Delta K=\frac{1}{2}m(v_f)^2-\frac{1}{2}m(v_i)^2 \\ P=m\cdot g\cdot h \end{gathered}[/tex]Explanation: We need to derive the formula for both Kinetic energy and potential energy, the derivation of these formulas is as follows:
(i) Kinetic energy:
[tex]\begin{gathered} \Delta K=\Delta W=F\cdot\Delta d\cdot\cos (\theta) \\ \theta=0 \\ \therefore\Rightarrow \\ \Delta K=F\cdot\Delta d\Rightarrow(1) \end{gathered}[/tex]By using the Kinematic equations of motions, equation (1) can be changed to the kinetic energy formula as follows:
[tex]\begin{gathered} (1)\Rightarrow\Delta K=F\cdot\Delta d=m\cdot a\cdot\Delta d\Rightarrow(2) \\ (v_f)^2=(v_i)^2+2a\Delta d \\ \therefore\Rightarrow \\ a\Delta d=\frac{(v_f)^2-(v_i)^2}{2} \\ \text{ Substituting above in the }(2)\text{ gives the formula:} \\ \Delta K=m\cdot\frac{(v_f)^2-(v_i)^2}{2} \\ \Delta K=\frac{1}{2}m(v_f)^2-\frac{1}{2}m(v_i)^2 \\ \text{ This is the kinetic energy formula} \end{gathered}[/tex](ii) Potential-energy:
Potential energy is basically the work needed to lift an object to a certain height, mathematically, the derivation would be as follows:
[tex]\begin{gathered} \Delta P=\Delta W=F\cdot\Delta d \\ \Delta d=\Delta h \\ \therefore\Rightarrow \\ \Delta P=m\cdot a\cdot\Delta h \\ a=g \\ \therefore\Rightarrow \\ \Delta P=m\cdot g\cdot\Delta h \\ \text{ Simple version:} \\ P=m\cdot g\cdot h \end{gathered}[/tex]Given a material of specific heat c in Cal/gramC^o and mass 6 grams. If the material is heated by absorbing 7 calories of heat then which of these expressions yields the change in temperature of the material in Celsius degrees? A)7 divided by (c - 6) B)7 times (c - 6) C)7 divided by (6 times c)
Answers
Given:
The mass is m = 6 grams.
The heat absorbed is Q = 7 calories.
The unit of specific heat, c of the material is cal/gram degree Celsius
To find the change in temperature of the material in degrees Celsius.
Explanation:
The formula to calculate the temperature change is
[tex]\begin{gathered} Q\text{ = mc}\Delta T \\ \Delta T=\frac{Q}{mc} \end{gathered}[/tex]Substituting the values, the change in temperature will be
[tex]\Delta T\text{ = }\frac{7}{6\times c}[/tex]Final Answer: The change in temperature will be 7 divided by (6 times c).
12000 inches to yards
Answers
ANSWER
[tex]\begin{equation*} 333.33\text{ yds} \end{equation*}[/tex]EXPLANATION
We want to convert 12000 inches to yards.
To do this, divide the value in inches by 36:
[tex]\begin{gathered} 1\text{ in }=\frac{1}{36}\text{ yd} \\ \\ 12000\text{ in }=\frac{12000}{36}\text{ yds }=333.33\text{ yds} \end{gathered}[/tex]That is the answer.
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.
Answers
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.
How much time is required for a car engine to do 278 kJ of work, if its maximum power is 95 kW?
Answers
W= work = 278 kj
P = Power = 95 Kw
t= time
P = W / t
Isolate t
t = W /P
Replacing:
t= 278kj / 95Kw = 2.93 seconds
It is required 2.93 seconds
A coyote was holding a large bird cage a few meters above the ground. He lowers it at a slow constant speed onto the X he has drawn on the ground. Which statement best describes the change in the total mechanical energy of the Earth-bird cage system?Question 5 options:The total mechanical energy decreases, because the coyote does negative work on the bird cage by exerting a force in the direction opposite to its displacement.The total mechanical energy is unchanged, because there is no change in the bird cage’s kinetic energy as it is lowered to the table.The total mechanical energy is unchanged, because no work is done on the Earth-bird cage system while the book is lowered.The total mechanical energy decreases, because the coyote does positive work on the bird cage by exerting a force that opposes the gravitational force.
Answers
Total mechanical energy is conserved in a system.
Thus, the total mechanical energy is unchanged, because no work is done on the Earth-bird cage system while the book is lowered.
Find the equivalent resistance of thiscircuit.R₁www400 ΩIR₂600 ΩReq = [?] 2R3www500 Ω
Answers
We are given the following information.
Resistor: R₁ = 400 Ω
Resistor: R₂ = 600 Ω
Resistor: R₃ = 500 Ω
We are asked to find the equivalent resistance of the given circuit.
Notice that the resistors R₁ and R₂ are in parallel and this parallel combination is in series with resistor R₃.
First, let us find the parallel resistance of R₁ and R₂.
[tex]R_p=\frac{R_1\times R_2}{R_1+R_2}=\frac{400\times600}{400+600}=\frac{240000}{1000}=240\;\Omega[/tex]Finally, let us find the series resistance of Rp and R₃.
[tex]\begin{gathered} R_{eq}=R_p+R_3 \\ R_{eq}=240+500 \\ R_{eq}=740\;\Omega \end{gathered}[/tex]Therefore, the equivalent resistance of the given circuit is 740 Ω.
What is the main purpose for learning about significant figures in science and/or technology courses? Give a detailed answer without plagiarism.
Answers
The significant figure of a value or measurement is the number of important digits that it contains
The main purpose for learning about significant figures in science and/or technology courses is precision of measurements.
In science and technology experiments and reseaches, we always want to ensure that measured values are as close to the true values as possible. This is because any deviation from the true value can invalidate the result of the experiment. Also, errors due to repeated approximations can have cummulative effects on the experimental results.
Therefore, the knowledge of significant figures is useful in science and technology to be able to record values of experimental measurements as close as possible to the actual value.
Question 1001 10The device shown demonstrates which energy transformation?A Blectrical energy - electromagnetic energyB. Besteomaretic energy - Kinetic energyC. Yinetic energy-electrical energyD. Chemical energy-electromagnetic energyPlease answer this question!!
Answers
ANSWER
A. electrical energy → electromagnetic energy
EXPLANATION
The device shown is a lamp. To turn on the lamp we have to connect it to the electric power; there we have electrical energy. The lamp has a light bulb, which when it's on it emits light. Remember that light is an electromagnetic wave. Therefore the electrical energy is transformed into electromagnetic energy.
What is the power rating of a heating coil with a resis-tance of 12 12 that draws a current of 20 A?
Answers
According to the problem, we have the following
[tex]\begin{gathered} R=12 \\ I=20 \end{gathered}[/tex]We have to use Ohm's law
[tex]V=I\cdot R[/tex]Let's replace the given information
[tex]V=20\cdot12=240[/tex]Hence, the power is 20 Volts.A bowling ball of mass 7.29 kg and radius 11.0 cm rolls without slipping down a lane at 3.00 m/s. Calculate the total kinetic energy.
Answers
We have the next information
m=7.29 kg
v=3 m/s
r=11cm=0.11m
We can find the kinetic energy using the next formula
[tex]KE=\frac{1}{2}mv^2+\frac{1}{2}Iw^2[/tex]I is the moment of inertia
w is the angular velocity
First, we need to calculate the moment of inertia
[tex]I=\frac{2}{5}mr^2[/tex][tex]I=\frac{2}{5}(7.29)(0.11)^2=0.0353kgm^2[/tex]Then for the angular velocity
[tex]\omega=\frac{v}{r}[/tex][tex]\omega=\frac{3}{0.11}=27.28[/tex]then we will substitute the values in the kinetic energy formula
[tex]KE=\frac{1}{2}(7.29)(3)^2+\frac{1}{2}(0.0353)(27.28)^2[/tex][tex]KE=45.9J[/tex]The total kinetic energy is 45.9 J
Victor is driving south. He is traveling at 12 m/s, when he enters an area with a new speed limit. Aver a period of 6 seconds, his speed increases from 12 m/s to 29 m/s. What is Victor’s acceleration during this period?
Answers
Answer:
1.1417 m/s²
Explanation:
Acceleration = final velocity - initial velocity/ time
a= [tex]\frac{v-u}{t}[/tex]
a= [tex]\frac{29-12}{12}[/tex]
a= [tex]\frac{17}{12}[/tex]
a= 1.417 m/s² or [tex]\frac{17}{12}[/tex] m/s² in fraction form
I think this is all statements are true but I just want to make sure
Answers
Given that a bug flies into the windshield of a car going. Let's select the correct statements.
According the NEwton's third law, the force exerted on the bug by the car is equal to the force extered on the car by the bug.
To determine the acceleration, we have:
[tex]a=\frac{F}{m}[/tex]Where:
F is the force
m is the mass
The mass of the car will be greater than the mass of the bug.
Since the mass of the car is greater than the mass of the bug and they have the same force, we can say the acceleration of the bug is greater than the acceleration of the car.
Statement B is correct.
The force of impact is the same for both according to Newton's third Law.
Both the car and the bug deliver the same magnitude of impulse on each other.
Therefore, all statements are correct.
ANSWER:
All statements are true.
Find the y-component of thisvector:12.0 m73.3°Remember, angles are measured fromthe +X axis.y-component (m)
Answers
Given data:
The magnitude of the given vector is,
[tex]A=12.0\text{ m}[/tex]An angle between the given vector and +x axis is,
[tex]\theta=73.3^o[/tex]The formula of y-component is as follows:
[tex]A_y=A\sin \theta[/tex]Here,
[tex]A\text{ is the magnitude of the given vector-A}[/tex]Now, substitute known values in above equation;
[tex]\begin{gathered} A_y=12\text{ m}\times\sin 73.3^o \\ A_y=11.49\text{ m} \end{gathered}[/tex]Therefore, the y-component of the given vector is 11.49 m
Two equally charged, 4.982 g spheres are placed with 3.173 cm between their centers. When released, each begins to accelerate at 258.312 m/s2. What is the magnitude of the charge, in micro-Coulombs, on each sphere?
Answers
Given:
The mass of the spheres is m = 4.982 g
The distance between the center of spheres is d = 3.173 cm
The acceleration is a = 258.312 m/s^2.
To find the magnitude of charge in micro Coulomb on each sphere.
Explanation:
According to Newton's second law, the force will be
[tex]F\text{ =ma}[/tex]According to Coulomb's law, the force will be
[tex]F=\frac{kq^2}{r^2}[/tex]Here, k is the Coulomb's constant whose value is
[tex]k=9\times10^9\text{ N m}^2\text{ /C}^2[/tex]On equating the forces, the charge will be
[tex]\begin{gathered} ma=\frac{kq^2}{r^2} \\ q=\sqrt{\frac{mar^2}{k}} \end{gathered}[/tex]On substituting the values, the magnitude of charge will be
[tex]\begin{gathered} q=\sqrt{\frac{(4.982\times10^{-3})\times258.312\times(3.173\times10^{-2})^2}{9\times10^9}} \\ =3.79\text{ }\times10^{-7}\text{ C} \\ =0.379\text{ }\times10^{-6}\text{ C} \\ =0.379\text{ }\mu C \end{gathered}[/tex]The magnitude of the charge of each sphere is 0.379 microCoulomb
A 1,982-kg car starts from rest at the top of a driveway 6.74 m long that is sloped at an angle of 30 degrees with the horizontal. If an average friction force of 2,721 N impedes the motion of the car, find the speed (in m/s) of the car at the bottom of the driveway.Use the approximation that g ≈ 10 m/s2.
Answers
In this situation, we cannot apply the law of conservation of energy, as there is friction. For us to solve, let us start by writing the balance equations. We'll have:
[tex]\sum F_x=P*sin(30)-Fat=ma[/tex][tex]\sum F_y=N-P*cos(30)=0[/tex]In order to find out the acceleration, we can use the first equation:
[tex]a=\frac{P*sin(30)-Fat}{m}=\frac{1982*10*sin(30)-2721}{1982}=3.627\frac{m}{s^2}[/tex]The car will then suffer this acceleration on the sloped plane. With this, we can calculate its speed by the end using the equations for a uniformly accelerated movement:
[tex]S(t)=S_0+v_0t+\frac{at^2}{2}\Rightarrow6.74=\frac{3.627*t^2}{2}\Rightarrow t=1.928s[/tex]This is the time the car will take to reach the bottom. By replacing this on the equation for the velocity we get:
[tex]v(t)=v_0+at=0+3.627*1.928=7\frac{m}{s}[/tex]Then, our final answer is 7 m/s
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
Answers
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
Scenario 3: a clock attached to the wall.
1. State the object stationary.
2. State the outside, external, unbalanced force acting on the object.
Answers
answer: this object is stationary because it is in rest position until or unless the external force act on it we can say this by netwons 1st law
and the stationaty object has relatively zero velocity.
netwons 1st law: Newton's first law states that unless compelled to change its state by the action of an external force, every object will remain at rest or in uniform motion in a straight line.
to learn more about netwons 1st law link
https://brainly.in/question/4132747#:~:text=Expert%2Dverified%20answer,question&text=According%20to%20Newton%27s%20first%20law,moving%20with%20the%20same%20velocity.
#SPJ1
2. here the object has contact with the wall hence there is a normal force between clock and wall
the clock is at rest this means that the forcee is balanced and normal force is balanced by gravitational force .the normal force is always perpendicular to the surface of the object.
there is no external force is acting on the clock.
to learn more about netwons 1st law link
https://brainly.in/question/4132747#:~:text=Expert%2Dverified%20answer,question&text=According%20to%20Newton%27s%20first%20law,moving%20with%20the%20same%20velocity.
#SPJ1
Why are quarks important to physics and to the real, modern world?
Answers
Because Quarks are the fundamental building blocks of the universe.
Quark is any member of a group of elementary subatomic particles that interact by means of a strong force and are a fundamental constituent of matter.
The specific heat capacity of steel is 450 J/kg°C. If heat is added to 2 kg of steel for 30seconds, raising its temperature from 10°C to 32°C, what is the heat flow rate? Express your answer in W.
Answers
Given:
The specific heat capacity of steel is
[tex]c=\text{ 450 J/kg }^{\circ}C[/tex]The mass of the steel is m = 2 kg
The initial temperature of steel is
[tex]T_i=\text{ 10}^{\circ}\text{ C}[/tex]The final temperature of steel is
[tex]T_f\text{ = 32}^{\circ}C[/tex]
The time is t = 30 s
To find the heat flow rate.
Explanation:
The heat flow rate can be calculated by the formula
[tex]\frac{Q}{t}=\frac{mc(T_f-T_i)}{t}[/tex]On substituting the value, the heat flow rate will be
[tex]\begin{gathered} \frac{Q}{t}=\text{ }\frac{2\times450\times(32-10)}{30} \\ =\text{ 660 W} \end{gathered}[/tex]Thus, the heat flow rate is 660 W.
Why are metals the best conductors of electricity? Why are metals the best magnetisers ? What is the key characteristic of metals that allows us to answer these two apparently different questions?
Answers
Question: What is the key characteristic of metals that allows us to answer these two apparently different questions?
Answer:
Explanation:
The atoms of a metal contains at least one free electron. Conduction of electricity is due to the these free electrons.
Magnetism is also affected by the availability of free electrons.
Thus, metals are good conductors of electricity and also the best magnetisers because of the availability of free electrons in metals
Romeo and Juliet are sitting on a balcony 1.5 meters apart. If Romeo has a mass of 61.6 Kg and Juliet has a mass of 48.8 kg. What is the attractive force between them?
Answers
E. O 58.456 S48. A bullet is fired from the ground making an angle of 20 degwith the horizontal with a speed of 1500 m/s.Calculate the maximum height reached? (1 point)A. O20831.097 mB. 3642.875 mC. O8232.474 mD. O15803.894 mE. 13428.572 m9. A bullet is fired from the ground making an angle of 20 degwith the horizontal with a speed of 1500 m/s.How far (horizontally) will it fall? (1 point)
Answers
The maximum height of a projectile is given by:
[tex]h=\frac{v_0^2\sin^2\theta}{2g}[/tex]In this case, the initial velocity is 1500 m/s, the angle is 20°; then we have:
[tex]\begin{gathered} h=\frac{(1500^)^2(\sin20)^2}{(2)(9.8)} \\ h=13428.572 \end{gathered}[/tex]Therefore, the maximum height is 13428.572 m
Anyone available for teaching me simple pendulum in physics
Answers
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.
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
Answers
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
A block with a mass m1 is hit by a force of magnitude F which causes the block to have an acceleration of magnitude a. If a second block of mass m2 is hit by the same force of magnitude F which causes the block to have an acceleration of magnitude 2a, then which of these could be the two masses? A) m1= 200kg ; m2= 100kgB) m1= 50kg ; m2= 25 kgC) m1= 100kg ; m2= 50kgD)m1= 10kg ; m2= 50kgE)Any of these
Answers
We are given that a force "F" accelerates an object of mass "m1". According to Newton's second law, this can be represented by the following equation:
[tex]F=m_1a[/tex]Now, we are given that a second object of mass "m2" is accelerated by "2a" using the same force. Using Newton's second law we get:
[tex]F=m_2(2a)[/tex]Now, we will divide both equations, we get:
[tex]\frac{F}{F}=\frac{m_1a}{m_2(2a)}[/tex]Now, we simplify by canceling put the "F" and the "a":
[tex]1=\frac{m_1}{2m_2}[/tex]Now, we multiply both sides by "2m2", we get:
[tex]2m_2=m_1[/tex]Therefore, the first mass must be twice the second mass.
The options that meet this condition are:
[tex]m_1=200kg,m_2=100kg\text{ }[/tex][tex]m_1=50kg,m_2=25kg[/tex][tex]m_1=100kg,m_2=50kg[/tex]2. A parallel-plate capacitor has an area of 2.0 cm², and the plates are separated by 2.0 mm. a. What is the capacitance? b. How much charge does this capacitor store when connected to a 6.0 V battery?
Answers
Given data:
* The area of the parallel plate capacitor is,
[tex]\begin{gathered} A=2cm^2 \\ A=2\times10^{-4}m^2^{} \end{gathered}[/tex]* The distance between the plates is,
[tex]\begin{gathered} d=2\text{ mm} \\ d=2\times10^{-3}\text{ m} \end{gathered}[/tex]Solution:
(a). The capacitance of the capacitor in terms of area and distance between the plates is,
[tex]C=\frac{\epsilon_{\circ}A}{d}[/tex][tex]\text{where }\epsilon_{\circ}\text{ is the electrical permittivity of the fr}ee\text{ spaces}[/tex]Substituting the known values,
[tex]\begin{gathered} C=8.85\times10^{-12}\times\frac{2\times10^{-4}}{2\times10^{-3}} \\ C=8.85\times10^{-13}\text{ F} \end{gathered}[/tex]Thus, the value of the capacitanc is 8.85 times 10 power -13 Farad.
(b). The voltage across the battery is,
[tex]V=6\text{ Volts}[/tex]The charge stored in the capacitor in terms of the voltage and the capacitance is,
[tex]\begin{gathered} C=\frac{Q}{V} \\ Q=CV \end{gathered}[/tex]where Q is the charge stored in the capacitor
Substituting the known values,
[tex]\begin{gathered} Q=8.85\times10^{-13}\times6 \\ Q=53.1\times10^{-13}\text{ Coulomb} \end{gathered}[/tex]Thus, the charge stored in the parallel plate capacitor is 53.1 times 10 power -13 coulomb.
Calculate the torque experienced by the door due to this force using torque is equal to force times lever armLength of the lever arm = 1 mForce = 5 N
Answers
Given:
The applied force on the door is F = 5 N
The length of the lever arm is l = 1 m
Required: Torque experienced by the torque.
Explanation:
Torque is the product of force and the distance between the force applied and the rotational axis.
The force is applied on one side of the lever arm while the rotational axis will be at the other end of the lever arm.
So, the distance between the rotational axis and the force applied is the length of the arm.
Torque can be calculated by the formula
[tex]\tau=F\times l[/tex]On substituting the values, the torque will be
[tex]\begin{gathered} \tau=5\times1 \\ =5\text{ N m } \end{gathered}[/tex]Final Answer: The torque experienced by the door is 5 N m.
