We are given the following information
Mass of Cart 1 = m₁ = 1 kg
Mass of Cart 2 = m₂ = 3 kg
Initial velocity of Cart 1 = u₁ = 5 m/s
Initial velocity of Cart 2 = u₂ = -2 m/s
Final velocity of Cart 1 = v₁ = -4 m/s
Final velocity of Cart 2 = v₂ = ?
Please note that velocity is taken positive when moving right and negative when moving left.
Recall that the total momentum before the collision must be equal to the momentum after the collision.
Mathematically,
[tex]\begin{gathered} p_{\text{before}}=p_{\text{after}} \\ m_1u_1+m_2u_2=m_1v_1+m_2v_2 \end{gathered}[/tex]Substitute the given values and solve for v₂
[tex]\begin{gathered} m_1u_1+m_2u_2=m_1v_1+m_2v_2 \\ 1\cdot_{}5_{}+3\cdot_{}(-2)_{}=1\cdot_{}(-4)_{}+3\cdot_{}v_2 \\ 5_{}-6_{}=-4_{}+3\cdot_{}v_2 \\ 5_{}-6_{}+4=3\cdot_{}v_2 \\ 3=3\cdot_{}v_2 \\ \frac{3}{3}=v_2 \\ 1=v_2 \\ v_2=1\: \end{gathered}[/tex]Therefore, the final speed of the 3.0 kg cart is 1 m/s
Also, note that cart 2 will be moving right after the collision since the speed is positive.
A block B of mass 0.4 kg rests on a smooth horizontal table 1m away from a smooth pulley. Block B is connected to particle P of mass 0.3 kg by a light inextensible string which passes over the smooth pulley. Initially particle P is 0.8m above a horizontal floor. B is in contact with the table and the part of the string between B and the pulley is horizontal. P hangs freely below the pulley. Calculate the acceleration of block B and the tension in the string.
Given,
The mass of block B, M=0.4 kg
The mass of particle P, m=0.3 kg
The height at which the particle P is situated, h=0.8 m
As the pulley is smooth, the tension in the horizontal part and the vertical part of the string is the same and the acceleration of the particle and the block is the same.
The only force acting on block B is the tension in the string. Thus the net force on the block is given by,
[tex]Ma=T\text{ }\rightarrow\text{ (i)}[/tex]Where a is the acceleration of the block and the particle and T is the tension in the string.
The forces acting on the particle are gravity and the tension in the string,
The net force on the particle is given by,
[tex]ma=mg-T[/tex]Where g is the acceleration due to gravity.
On substituting for T in the above equation,
[tex]\begin{gathered} ma=mg-Ma \\ \Rightarrow a(m+M)=mg \\ \Rightarrow a=\frac{mg}{(m+M)} \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} a=\frac{0.3\times9.8}{0.3+0.4} \\ =4.2\text{ m/s}^2 \end{gathered}[/tex]Thus the acceleration of the block B is 4.2 m/s²
On substituting the known values in the equation (i),
[tex]\begin{gathered} T=0.4\times4.2 \\ =1.68\text{ N} \end{gathered}[/tex]Thus the tension in the string is 1.68 N
As you will see in a later chapter, forces are vector quantities, and the total force on an object is the vector sum of all forces acting on it.In the figure below, a force F1 of magnitude 5.90 units acts on a crate at the origin in a direction = 26.0° above the positive x-axis. A second force F2 of magnitude 5.00 units acts on the crate in the direction of the positive y-axis. Find graphically the magnitude and direction (in degrees counterclockwise from the +x-axis) of the resultant force F1 + F2.
First, let's draw the resultant vector. To add the vectors, let's draw the start point of one vector in the endpoint of the other vector. Then, the resultant vector will have the start point of the first vector and the endpoint of the second vector:
To find the magnitude and direction of the resultant vector, let's decompose F1 in horizontal and vertical directions:
[tex]\begin{gathered} F_{1x}=F_1\cos26°\\ \\ F_{1x}=5.9\cdot0.8988\\ \\ F_{1x}=5.3\\ \\ \\ \\ F_{1y}=F_1\sin26°\\ \\ F_{1y}=5.9\cdot0.438\\ \\ F_{1y}=2.58 \end{gathered}[/tex]Since F2 is in the vertical direction, we have F2x = 0 and F2y = F2 = 5.
Now, calculating the magnitude and direction of the resultant vector:
[tex]\begin{gathered} F_{rx}=F_{1x}+F_{2x}=5.3+0=5.3\\ \\ F_{ry}=F_{1y}+F_{2y}=2.58+5=7.58\\ \\ \\ \\ F_r=\sqrt{F_{rx}^2+F_{ry}^2}\\ \\ F_r=\sqrt{5.3^2+7.58^2}\\ \\ F_r=9.25\\ \\ \\ \\ \theta=\tan^{-1}(\frac{F_{ry}}{F_{rx}})\\ \\ \theta=\tan^{-1}(\frac{7.58}{5.3})\\ \\ \theta=55.03° \end{gathered}[/tex]Therefore the magnitude is 9.25 units and the direction is 55.03°.
Mario drives his Kart at an average speed of 42 miles per hour for 1.7 minutes. How far does he drive? Include units in your answer
We will have the following:
First, we transform the velocity unit of time, that is:
[tex]42\frac{mi}{h}\ast\frac{1h}{60min}=0.7\frac{mi}{min}[/tex]Then, we will have that:
[tex]x=\frac{0.7mi\ast1.7min}{1min}\Rightarrow x=1.19mi[/tex]So, it moved 1.19 miles.
how does the frequency of the two waves compare?
Given
frequency of two waves
Procedure
the rate at which something occurs over a particular period of time or in a given sample.
In this case, Wave A has more frequency than Wave B
And Wave B has half the Frequency of Wave A
The circular pistons in a hydraulic press are 8 and 40 cm in diameter. What is the force in the bigger piston when a 50 N force is applied to the small piston?
ANSWER
1250 N
EXPLANATION
We have a hydraulic press, which is something like this:
The pressure of the fluid inside the press is the same on both pistons. The pressure is related to the force and the area of the piston by,
[tex]P=\frac{F}{A}[/tex]As stated before, P1 = P2,
[tex]\frac{F_1}{A_1}=\frac{F_2}{A_2}[/tex]We know the diameter of the pistons d1 = 8cm and d2 = 40cm, and the force applied to the small piston is F1 = 50N. We have to solve the equation above for F2,
[tex]F_2=F_1\cdot\frac{A_2}{A_1}[/tex]The pistons are circular, thus the area is.
[tex]A=\pi\cdot r^2=\pi\cdot(\frac{d}{2})^2=\pi\cdot\frac{d^2}{4}_{}[/tex]Replace into the equation of the force,
[tex]F_2=F_1\cdot\frac{\pi\cdot\frac{d^2_2}{4}}{\pi\cdot\frac{d^2_1}{4}}[/tex]π/4 cancels out,
[tex]F_2=F_1\cdot\frac{d^2_2}{d^2_1}[/tex]Usually, we have to put all the units so that they are consistent - in this case, that would mean to convert the diameters to meters, but since the diameters are dividing their units will cancel out, so we don't need to convert those measures from centimeters to meters.
Replace F1, d1, and d2 for their values,
[tex]F_2=50N\cdot\frac{40^2cm^2}{8^2cm^2}[/tex][tex]F_2=50N\cdot25=1250N[/tex]The force in the bigger piston is 1250 N.
I need to find my weight in newtons using newtons law of gravitym=56 kg
Newton's law of gravity states that the gravitational force between two objects is given by:
[tex]F=G\frac{m_1m_2}{r^2}[/tex]where G is the gravitational constant, m1 and m2 are the masses of the objects and r is the distance between them.
In this case we would like to calculate your weight, that is, the gravitational force between earth and you. Then we need to use the mass of the earth, and its radius; their values are given as:
[tex]\begin{gathered} M=5.9722\times10^{24} \\ r=6.3781\times10^6 \end{gathered}[/tex]Plugging our values we have that:
[tex]\begin{gathered} F=6.67\times10^{-11}\cdot\frac{56\cdot5.9722\times10^{24}}{(6.3781\times10^6)} \\ F=548.36 \end{gathered}[/tex]Therefore your weight is 548.36 N
What is the ratio of thicknesses of polystyrene and flint glass that would contain the same number of wavelengths of light? (Use any necessary data found in this table.)tpolystyrene/flint glass =
Let's find the ratio of thickness of polystyrene and flint glass that would contain the same number of wavelengths of light.
The relation of wavelength for any given medium is:
[tex]\lambda_n=\frac{\lambda}{n}[/tex]Where:
λn is the wavelength of the wave medium.
λ is the wavelength in vacuum.
n is the refractive index of the medium.
Now, for the thickness, we have:
[tex]{\frac{d_p}{\lambda_p}}=\frac{d_f}{\lambda_f}[/tex]dp is the thickness of of polystyrene
df is the thickness of flint glass.
λp is the wavelength of light in polystyrene.
λf is the wavelength of light in flint glass.
Substitute λ/np for λp and λ/nf for λf:
[tex]\begin{gathered} \frac{d_p}{\frac{\lambda}{n_p}}=\frac{d_f}{\frac{\lambda}{n_f}} \\ \\ \frac{d_p}{d_f}=\frac{\frac{\lambda}{n_p}}{\frac{\lambda}{n_f}} \\ \\ \frac{d_p}{d_f}=\frac{n_f}{n_p} \end{gathered}[/tex]Where:
nf is the refractive index of flint glass = 1.66
n is the refractive index of polystyrene = 1.49
Thus, we have:
[tex]\begin{gathered} \frac{d_p}{d_f}=\frac{1.66}{1.49} \\ \\ \frac{d_p}{d_f}=1.114 \end{gathered}[/tex]Therefore the ratio of thicknesses of polystyrene and flint glass that would contain the same number of wavelengths of light is 1.114
ANSWER:
1.114
Determine the image distance and image height for a 5.00 cm tall object placed 10.0 cm from a double convex lens with a focal length of 15.0 cm.
Hello I really need help with this question please!! Explain the energy conversations that occur as a basketball falls, hits the ground, and bounces up ignore frictional forces. ???
We know that, if we ignore frictional forces, the mechanical energy of the system is conserved, this means that all point of time we have that:
[tex]K_i+U_i=K_f+U_f[/tex]This means that as the ball is falling the potential energy is being converted to kinetic energy until all the potential energy becomes zero (when the ball hits the ground); at this point all the energy is kinetic. After this point all the kinetic energy of the ball is being converted to potential energy while the ball rises until it reaches the maximum height (which has to be the same from where the ball started); at this point all the energy is potential energy since the ball stops for a fraction of time. Then the process begins again and the potential energy is being converted into kinetic energy as we explained earlier and the process goes on.
A light ray enters a square block of plastic at an angle 1 = 54° (measured from the surface normal) and emerges at an angle 2 = 67° (measured from its surface normal) as shown. The light ray enters the plastic a distance = 75 above the corner and emerges a distance from the corner.What is the index of refraction of the plastic? (Assume air is the surrounding medium, with an index of refraction of 1.00.)What is , in centimeters?
Let θ be the angle from the normal to the surface to the ray of light inside the block after the first refraction:
Let n_1 be the index of refraction that corresponds to the medium of the angle θ_1. According to Snell's Law:
[tex]\begin{gathered} n_1\sin (\theta_1)=n\sin (\theta) \\ n\sin (90-\theta)=n_2\sin (\theta_2) \end{gathered}[/tex]Since the angles θ_1 and θ_2 correspond to the air, then n_1=1 and n_2=1, Then:
[tex]\begin{gathered} \sin (\theta_1)=n\sin (\theta) \\ n\sin (90-\theta)=\sin (\theta_2) \end{gathered}[/tex]Since sin(90-θ)=cos(θ), then:
[tex]n\cos (\theta)=\sin (\theta_2)[/tex]Isolate sin(θ) and cos(θ) from their corresponding equations:
[tex]\begin{gathered} \sin (\theta)=\frac{\sin(\theta_1)_{}}{n} \\ \cos (\theta)=\frac{\sin(\theta_2)}{n} \end{gathered}[/tex]Recall the Pythagorean Identity:
[tex]\sin ^2(\theta)+\cos ^2(\theta)=1[/tex]Replace the expressions for sin(θ) and cos(θ):
[tex]\begin{gathered} (\frac{\sin (\theta_1)}{n})^2+(\frac{\sin (\theta_2)}{n})^2=1 \\ \Rightarrow\frac{\sin^2(\theta_1)}{n^2}+\frac{\sin ^2(\theta_2)}{n^2}=1 \\ \Rightarrow\sin ^2(\theta_1)+\sin ^2(\theta_2)=n^2 \\ \Rightarrow\sqrt[]{\sin ^2(\theta_1)+\sin ^2(\theta_2)}=n \end{gathered}[/tex]Therefore, the index of refraction is given by:
[tex]n=\sqrt[]{\sin^2(\theta_1)+\sin^2(\theta_2)}[/tex]Substitute θ_1=54° and θ_2=67° to find the index of refraction of the material:
[tex]\begin{gathered} n=\sqrt[]{\sin^2(54)+\sin^2(67)} \\ =1.22549\ldots \\ \approx1.23 \end{gathered}[/tex]From the diagram, notice that:
[tex]\begin{gathered} \tan \theta=\frac{L}{X} \\ \Rightarrow X=\frac{L}{\tan \theta} \end{gathered}[/tex]On the other hand:
[tex]\tan (\theta)=\frac{\sin (\theta)}{\cos (\theta)}[/tex]Substitute the expressions for sin(θ) and cos(θ):
[tex]\begin{gathered} \tan (\theta)=\frac{(\frac{\sin(\theta_1)_{}}{n})}{(\frac{\sin(\theta_2)_{}}{n})} \\ =\frac{\sin(\theta_1)}{\sin(\theta_2)} \end{gathered}[/tex]Substitute the expression for tan(θ) into the formula to find X:
[tex]\begin{gathered} X=\frac{L}{(\frac{\sin(\theta_1)}{\sin(\theta_2)})} \\ =\frac{\sin (\theta_2)}{\sin (\theta_1)}L \end{gathered}[/tex]Substitute L=75cm, θ_1=54° and θ_2=67° to find the value of X:
[tex]\begin{gathered} X=\frac{\sin(67)}{\sin(54)}\times75\operatorname{cm} \\ =85.3355\ldots cm \end{gathered}[/tex]Therefore, the answers are:
[tex]\begin{gathered} n=1.23 \\ X=85\operatorname{cm} \end{gathered}[/tex]Sound waves will travel faster on a dry day than on a humid day of the same temperature. Is this true or false?
The speed of a sound wave depends on the density of the medium on which it propagates. Then, it is affected by temperature and humidity.
The speed of sound in dry air is lower than in moist air.
Then, it is not true that sound waves will travel faster on a dry day.
Therefore, the answer is:
[tex]\text{False}[/tex]When two waves interfere completely destructively, what is true of the resultant wave?A. has a shorter wavelengthB. has a larger amplitude.c. has a smaller amplitude.D. has a longer wavelength.
When two waves interefere destructively, the two waves cancel each other out. This results in a lower amplitude.
Therefore, when two waves interfere destructively, the resultant wave will have smaller amplitude.
Which is equivalent to three (3) elementary charges?(a) 1.6 x 10-19 C (b) 2.0 x 10-19 C (c) 4.8 x 10-19 C (d) 5.4 x 10-19 C
The value of the elementary charge is
[tex]e=1.6\times10^{-19}\text{ C}[/tex]We have to find a charge equivalent to 3e.
The charge equivalent to 3e will be
[tex]\begin{gathered} 3e\text{ = 3}\times1.6\times10^{-19}\text{ C} \\ =4.8\times10^{-19}\text{ C} \end{gathered}[/tex]Thus, the correct option is c.
Two blocks are stacked on top of each other on the floor of an elevator as shown. The mass of Block 2 (bottom) is 3.0 kg and the mass of Block 1 (upper) is 2.0 kg. The elevator is accelerating upwards at a constant rate of 1.6 m/s2.Find the magnitude of the force that Block 2 exerts on the floor of the elevator.
The magnitude of the force exerted by Block 2 takes into account the mass of Block 1 because it's above it, so the force must include both masses.
Use Newton's Second Law.
[tex]\Sigma F_y=ma[/tex][tex]\begin{gathered} N-W=(m_1+m_2)a \\ N-(m_1+m_2)g=(m_1+m_2)a_{} \\ N=(m_1+m_2)(g+a) \end{gathered}[/tex]Use the given magnitudes to find the force N.
[tex]\begin{gathered} N=(2\operatorname{kg}+3\operatorname{kg})(9.8\cdot\frac{m}{s^2}+1.6\cdot\frac{m}{s^2}) \\ N=5\operatorname{kg}\cdot11.4\cdot\frac{m}{s^2} \\ N=57N \end{gathered}[/tex]Therefore, the force exerted by Block 2 is 57 Newtons.
The velocity as a function of time of a moving object is presented by the graph to the right. Use the graph below for the following questions.What is the magnitude of acceleration of the object between 6 s and 10 s in m/s per second?
ANSWER:
1 m/s per second
STEP-BY-STEP EXPLANATION:
We can calculate the acceleration by the following formula and with the help of the graph.
Just like this:
[tex]\begin{gathered} a=\frac{\Delta v}{\Delta t}=\frac{v_f-v_i}{t_f-t_i} \\ \\ \text{ We replacing} \\ \\ a=\frac{0-4}{10-6}=\frac{-4}{4}=-1\text{ m/s}^2 \end{gathered}[/tex]The acceleration is negative, this tells us that from the second 6 the object began to slow down, but its magnitude is 1 m/s²
The speed of transverse waves along a stretched spring is 6.18 meters per second. The spring is 4.02 meters long and is held in place at both ends. What are the first three natural frequencies of the spring? Include units in your answers.
The speed v of a transverse wave is related to its wavelength λ and its frequency f by the equation:
[tex]v=\lambda f[/tex]When a string or any string-like object (like the stretched spring) vibrates at its fundamental frequency f₁, the wavelength of the standing wave is equal to 2 times the length of the string:
[tex]\lambda=2L[/tex]Then:
[tex]v=2Lf_1[/tex]Isolate the fundamental frequency from the equation:
[tex]f_1=\frac{v}{2L}[/tex]The n-th natural frequency is given as a multiple of the fundamental frequency:
[tex]f_n=n\times f_1[/tex]Then, to find the first three natural frequencies, find the value of the fundamental frequency by replacing the speed of the transverse wave v=6.18m/s and the length of the stretched spring L=4.02m:
[tex]f_1=\frac{6.18\frac{m}{s}}{2\times4.02m}=0.7686567\ldots Hz[/tex]Then, the next two natural frequencies are:
[tex]\begin{gathered} f_2=2\times f_1=2\times(0.7686567\ldots Hz)=1.5373\ldots Hz \\ \\ f_3=3\times f_1=3\times(0.7686567\ldots Hz)=2.30597\ldots Hz \end{gathered}[/tex]Therefore, the first three natural frequencies of the spring are approximately 0.769Hz, 1.54Hz and 2.31Hz.
A 6.00 µF parallel plate capacitor has a charge of +40 μC. What is the potential energy stored in this capacitor?Answer Choices103 μJ113 μJ123 μJ 133 μJ
In order to calculate the energy stored in a capacitor, we can use the formula below:
[tex]E=\frac{Q^2}{2C}[/tex]Where Q is the charge and C is the capacitance.
So, for Q = 40 * 10^-6 C and C = 6 * 10^-6 F, we have:
[tex]\begin{gathered} E=\frac{(40\cdot10^{-6})^2}{2\cdot6\cdot10^{-6}} \\ E=\frac{1600\cdot10^{-12}}{12\cdot10^{-6}} \\ E=133\cdot10^{-6}\text{ J} \\ E=133\text{ }\mu J \end{gathered}[/tex]Therefore the correct option is the fourth one.
Choose the best answer from the choices below:The perpendicular bisector of a chord ____.A. bisects the chordB. is less than the radiusC. is perpendicular to any chordD. passes through the center of the circle
We have the next theorem.
The perpendicular bisector of any chord of a circle will pass through the center of the circle.
In our case
The perpendicular bisector of a chord passes through the center of the circle.
Therefore the correct choice
Which of the following is a theoretical particle thought to be the force carrier for gravity?Select one:a.bosonb.photonc.gravitond.photon
To find the theoretical particle thought to be the force carrier for gravity.
Explanation:
Graviton is the theoretical particle thought to be the force carrier for gravity.
A 1.20 × 10^3 -kilogram car is traveling east at 25 meters per second. The brakes are applied and the car is brought to rest in 5.00 seconds. Calculate the magnitude and the direction of the total impulse applied to the car to bring it to rest
Given data:
* The mass of the car is,
[tex]m=1.2\times10^3\text{ kg}[/tex]* The initial velocity of the car is,
[tex]u=25\text{ m/s}[/tex]* The time taken by the car is t = 5 seconds.
* The final velocity of the car is v = 0 m/s.
Solution:
The impulse applied to the car is,
[tex]J=mv-mu[/tex]Substituting the known values,
[tex]\begin{gathered} J=1.2\times10^3\times0-1.2\times10^3\times25 \\ J=0-30\times10^3 \\ J=-30\times10^3\text{ kgm/s} \end{gathered}[/tex]Thus, the magnitude of the impulse applied on the car is,
[tex]30\times10^3\text{ kgm/s}[/tex]As the car comes to rest after the impulse is applied on the car.
Thus, the direction of impulse is opposite to the direction of motion of the car.
A 1200 N cart rests on a 36 degree incline. Calculate the value of the component of the cart’s weight that holds it against the inclined plane.
Given,
The weight of the cat, W=1200 N
The angle of inclination, θ=36°
The component of the weight that holds the cat against the inclined plane is the vertical component.
The vertical component of the cat's weight is given by,
[tex]W_y=W\cos \theta[/tex]On substituting the known values,
[tex]\begin{gathered} W_y=1200\times\cos 36^{\circ} \\ =970.8\text{ N} \\ \approx970\text{ N} \end{gathered}[/tex]Thus the value of the component of the cat's weight that holds it against the inclined plane is 970 N
Therefore the correct answer is option 2.
Cleveland is performing an experiment, testing to see if increasing the acidity of the soil in which sunflower plants are growing makes them flower sooner. In this experiment, what are some values that Cleveland will want to hold constant.
Answer:
The temperature the plants are grown in
The amount of water the sunflowers get
The amount of light the sunflowers get
Explanation:
They want to know if increasing the acidity of the soil makes the plants flower sooner. So, they need to hold constant all the other variables in that area not the acidity of the soil or the day the plants begin to flower.
Therefore, the answers are
The temperature the plants are grown in
The amount of water the sunflowers get
The amount of light the sunflowers get
What is the mass of 61.5 mL of ethanol? The density of ethanol at room temperature = 0.789 g/mL.
We are given that the density of ethanol is 0.789 g/mL. Density is defined as:
[tex]D=\frac{m}{V}[/tex]Where:
[tex]\begin{gathered} D=\text{ density} \\ m=\text{ mass} \\ V=\text{ volume} \end{gathered}[/tex]Since we are asked about the mass we will multiply the formula by "V" on both sides:
[tex]DV=m[/tex]Substituting the values we get:
[tex](\frac{0.789g}{mL})(61.5mL)=m[/tex]Solving the operations we get:
[tex]48.5g=m[/tex]Therefore, the mass of ethanol is 48.5 grams.
Derivation of Gas Law?
The derivation of general gas law from Charles law, Boyles law, and pressure law gives P₁V₁/T₁ = P₂V₂/T₂.
What is general gas law?The general gas law is obtained from the combination of Charles law, Boyle's law, and Pressure law.
Charles Law states that the volume of a fixed mass of a gas is directly proportional to its absolute temperature, provided the pressure of the gas is constant.
V₁/T₂ = V₂/T₂
Boyle's law states that, the volume of a fixed mass of a gas is inversely proportional to its pressure, provided the temperature of the gas is constant.
P₁V₁ = P₂V₂
The pressure law states that, the pressure of a fixed mass of a gas is directly proportional to its absolute temperature provide the volume of the gas is constant.
P₁/T₁ = P₂/T₂
Combining the three laws, P₁V₁/T₁ = P₂V₂/T₂
Learn more about general gas law here: https://brainly.com/question/4147359
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has a mass of 81 grams. The sample was put into 150 mL of water. After dropping the mineral sample into the water, the volume on the graduated cylinder is 180 mL.what is the volume of the mineral sample? 60mL30mL150mL180mL
In order to calculate the volume of the mineral sample, we can subtract the volume after the addition of the minerals by the volume before dropping the mineral sample.
So we have:
[tex]\begin{gathered} \text{mineral volume}=\text{ after-before} \\ \text{mineral volume}=180-150 \\ \text{mineral volume}=30 \end{gathered}[/tex]So the volume is 30 mL, therefore the correct option is the second one.
When building an electronic toy, a buzzer requires 30 mA (.03 amps) of current. If the voltage from the batteries for the toy add up to 12 volts, what sized resister must be used to ensure the buzzer receives the necessary 30 mA of current. Enter the numerical value rounded to the nearest whole number. For example if you calculate 327.45 ohms, enter 327.
Ohm's law states that the resistance, current and voltage are related from the equation:
[tex]R=\frac{V}{I}[/tex]In this case we know the voltage is 12 V and the current is 0.3 A, plugging these values we have:
[tex]\begin{gathered} R=\frac{12}{0.3} \\ R=400 \end{gathered}[/tex]Therefore, the resitsance need is 400 Ohms.
A person who weighs 764 N supports himself on the ball of one foot. The normal force N = 764 N pushes up on the ball of the foot on one side of the ankle joint, while the Achilles tendon pulls up on the foot on the other side of the joint. The center of gravity of the person is located right above the tibia. A. What is the tension in the Achilles tendon? If the force acting is upward, enter a positive value and if the force acting is downward, enter a negative value. (N) B. What is the magnitude of the downward force exerted on the ankle joint by the tibia? (N)
ANSWER:
A. 2125.91 N
B. 2889.91 N
STEP-BY-STEP EXPLANATION:
A.
Balacing moment,
[tex]\begin{gathered} F_{\text{achiles}}\cdot d_1=N\cdot d_2 \\ F_{\text{achiles}}\cdot4.6=N\cdot12.8 \\ F_{\text{achiles}}=\frac{764\cdot12.8}{4.6} \\ F_{\text{achiles}}=2125.91\text{ N} \end{gathered}[/tex]B.
Balacing forces,
[tex]\begin{gathered} F_{\text{tibia}}=F_{\text{achiles}}+N_{} \\ \text{ replacing} \\ F_{\text{tibia}}=2125.91+764 \\ F_{\text{tibia}}=2889.91\text{ N} \end{gathered}[/tex]If the current is 10A (Ampere) ,when connected to 1,000 volt line, find the resistance.
Apply Ohm's Law
V = I x R
Where:
V= voltage = 1,000 volt
I = current = 10 A
R= resistance = ?
Isolate R
R = V / I = 1,000 / 10 = 100 ohms
In front of a door, there is a very heavy bookcase. A boy decides to shove it out of the way, so he stands next to it and begins to push. However, instead of moving it, he begins to move backward. Why does this happen? (im guessing because the bookcase is pushing back on him with a force equal to the one he exerted? not sure)
When the boy pushes the bookcase, he moves backward instead of moving the bookcase.
According to Newton's law,
[tex]F=\frac{dp}{dt}[/tex]Here, p is momentum which is equal to mass times velocity.
Also, According to Newton's third law, the bookcase also applies equal force on the boy.
As the mass of the bookcase is heavier than the mass of the boy.
Thus, the boy moves backward instead of pushing back the bookcase.
A conventional current is the flow of positive charges that move from higher to lower potential energy. Is this true or false?
A conventional current is the flow of positive charges that move from higher to lower potential energy.
The above-given statement is a false statement.
Conventional current is the current that flows out of the positive terminal, through the circuit, and into the negative terminal of the source.