Given:
The weight of John Glenn, w=640 N
To find:
a) The weight if the distance was twice that of the initial value.
b) Why is an astronaut never weightless.
Explanation:
a)
Let the distance between the spacecraft and the earth be r.
If it becomes twice, then the distance is 2r.
The initial gravitational force on John Glenn is,
[tex]F=w=\frac{GMm}{r^2}[/tex]Where G is the gravitational constant, M is the mass of the earth and m is the mass of John Glenn.
The force when the distance is twice,
[tex]\begin{gathered} w_n=\frac{GMm}{(2r)^2} \\ =\frac{GMm}{4r^2} \\ =\frac{w}{4} \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} w_n=\frac{640}{4} \\ =160\text{ N} \end{gathered}[/tex]b)
Even when the astronaut is in space they still have the mass and so does the earth. Thus there will always be a gravitational force of attraction between the earth and the astronaut. The astronaut does not feel the weight because there will be nothing in space that pushes them back. That is why an astronaut is never truly weightless.
Final answer:
a) Thus the weight of John Glenn will be 160 N
You turn a corner and are driving up asteep hill. Suddenly, a small boy runs out on the street chasing a ball. You slam on the brakes and skid to astop leaving a 50-foot-long skid mark on the street. The boy calmly walks away but a policemen watching fromthe sidewalk walks over and gives you a speeding ticket. He points out that the speed limit on this street is 25mph. After you recover your wits, you begin to examine the situation. You determine that the street makes anangle of 25◦with the horizontal and that the coefficient of static friction between your tires and the street is0.80. You also find that the coefficient of kinetic friction between your tires and the street is 0.60. Your car’sinformation book tells you that the mass of your car is 1600 kg. You weigh 140 lbs.
Given data:
Total displacement of the car;
[tex]s=50\text{ ft}[/tex]Speed limit;
[tex]v_m=25\text{ mph}[/tex]The angle of street from horizontal;
[tex]\theta=25\degree[/tex]Coefficient of static friction;
[tex]\mu_s=0.80[/tex]Coefficient of kinetic friction;
[tex]\mu_k=0.60[/tex]Mass of the car;
[tex]M=1600\text{ kg}[/tex]Weight of the man;
[tex]W=140\text{ lbs}[/tex]The kinetic friction force is given as,
[tex]F_k=\mu_k(M+m)g\cos \theta[/tex]Here, m is the mass of the man and g is the acceleration due to gravity.
The acceleration of the car driving up a steep hill is given as,
[tex]\begin{gathered} (M+m)g\sin \theta+F_k=(M+m)a \\ (M+m)g\sin \theta+\mu_k(M+m)g\cos \theta=(M+m)a \\ g\sin \theta+\mu_kg\cos \theta=a \end{gathered}[/tex]Substituting all known values,
[tex]\begin{gathered} (32\text{ ft/s}^2)\times\sin (25\degree)+0.6\times(32\text{ ft/s}^2)\times\cos (25\degree)=a \\ \approx30.92\text{ ft/s}^2 \end{gathered}[/tex]The velocity of the car is given as,
[tex]v^2=u^2-2as[/tex]Here, v is the final velocity (v=0, as the car stops), and u is the initial velocity.
The initial velocity of the car is given as,
[tex]u=\sqrt[]{v^2+2as}[/tex]Substituting all known values,
[tex]\begin{gathered} u=\sqrt[]{0^2+2\times(30.92\text{ ft/s}^2)\times(50\text{ ft})} \\ \approx55.61\text{ ft/s} \\ \approx37.91\text{ mph} \end{gathered}[/tex]Therefore, your speed is greater than the speed limit. Thus, you can not fight the ticket in the court.
Bart, mass 32.4 kilograms, and Milhouse, mass 27.6 kilograms, play on the schoolyard seesaw. If Bart and Milhouse want to sit 4.0 meters apart, how far from the center of the seesaw should Bart sit? Include units in your answer. Answer must be in 3 significant digits.
Given data:
* The mass of the Bart is m_1 = 32.4 kg.
* The mass of the Milhouse is m_2 = 27.6 kg.
* The distance between the Millhouse and Bart is d = 4 m.
Solution:
To balance the seesaw, the net moment about the center should be zero.
The diagrammatic representation of the given system is,
The distance between the Bart and Milhouse can be written as,
[tex]\begin{gathered} d=d_1+d_2 \\ 4=d_1+d_2 \\ d_2=4-d_1 \end{gathered}[/tex]where d_2 is the distance of Milhouse from the center and d_1 is the distance of Bart from the center,
Consider the moment as positive if it is in an anticlockwise direction and negative if it is a clockwise direction.
Thus, the net moment about the center is,
[tex]M=m_1d_1-m_2d_2[/tex]Substituting the known values,
[tex]\begin{gathered} 0=32.4\times d_1-27.6\times(4-d_1) \\ 0=32.4\times d_1-110.4+27.6d_1 \\ 0=60d_1-110.4 \end{gathered}[/tex]By simplifying,
[tex]\begin{gathered} 60d_1=110.4 \\ d_1=\frac{110.4}{60} \\ d_1=1.84\text{ m} \end{gathered}[/tex]Thus, the distance of the Bart from the center is 1.84 meters.
Starting from rest, billy goes down a slide (height of 2.5m above ground) and billy’s mass is 35kg. There is friction acting on billy, when he reaches the ground his speed is 3.5 m/s. How much mechanical energy was lost due to friction (give answer in joules)
In order to determune the amount of mechanicak energy lost, consider that the potential energy at the starting point must be equal to the sum of the energy lost due to the friction, plus the kinetic energy on the ground.
Then, you can write:
[tex]U=E_r+K[/tex]the amount of energy lost is energy associated to the friction.
Solve the previou equation for Er, replace the expressions for U and K, and replace the values of the given parameters, as follow:
[tex]\begin{gathered} E_r=U-K=(35kg)(9.8\frac{m}{s^2})(2.5m)-\frac{1}{2}(35kg)(3.5\frac{m}{s})^2 \\ E_r=643.12J \end{gathered}[/tex]Hence, the amoun of the energy lost was approximately 643.12J
A pottery wheel with rotational inertia 40 kgm^2 rotates at 10 rev/s. 4 kg of clay is dropped onto the wheel 1.2 m from the axis. What angular speed will the wheel have after this?1. 55 rad/s2. 8.7 rad/s3. 70 rad/s4. 0 rad/s
Given:
• Rotational inertia = 40 kg.m²
,• Initial angula speed = 10 rev/s
,• Mass, m = 4 kg
,• Diameter, d = 1.2 m
Let's find the angular speed of the wheel.
To find the angular speed, apply the formula:
[tex]L_i=(I+md^2)*w_f[/tex]Where wf is the final angular speed
I is the rotational inertia
m is the mass
d = 1.2
Li is the angular momentum.
To find the angular momentum, we have:
[tex]\begin{gathered} L_i=40*10*2\pi \\ L_i=2513.27\text{ kg.m}^2\text{ rad/s} \end{gathered}[/tex]Now, to find the final angular speed, wf, plug in values in the first equation and solve for wf:
[tex]\begin{gathered} Li=(I+md^2)w_f \\ \\ 2513.27=(40+4*1.2^2)w_f \\ \\ 2513.27=45.76w_f \\ \\ w_f=\frac{2513.27}{45.76} \\ \\ w_f=54.9\approx55\text{ rad/s} \end{gathered}[/tex]Therefore, the final angular speed is 55 rad/s.
ANSWER:
1.) 55 rad/s
Be the action of a force of 51N, a spring measures 39cm. Its length becomes 40.8 cm when subjected to another force of 61N. 1)Determine the empty length of the spring 2)Determine an elongation which will correspond to a force of 32N.3) So what is its length
Answer:
1) 29.82 cm
2) 5.76 cm
3) 35.58 cm
Explanation:
Part 1)
The force of a spring is equal to:
F = kΔx
Where k is the constant of the spring and Δx is the elongation. Δx = xf - xi, where xf is the length of the spring when the force is applied and xi is the empty length. Then
F = k(xf - xi)
Now, by the action of a force of 51N, a spring measures 39 cm, so
51 = k(39 - xi)
And by the action of a force of 61N, the spring length is 40.8 cm, so
61 = k(40.8 - xi)
To find the empty length, we need to solve the system of equations
51 = k(39 - xi)
61 = k(40.8 - xi)
First, solve the first equation for k
[tex]k=\frac{51}{39-x_i}[/tex]Then, replace this on the second equation and solve for xi
[tex]\begin{gathered} 61=k(40.8-x_i) \\ 61=\frac{51}{(39-x_i)}(40.8-x_i) \\ 61(39-x_i)=51(40.8-x_i) \\ 61(39)-61(x_i)=51(40.8)-51(x_i) \\ 2379-61x_i=2080.8-51x_i \\ 2379-2080.8=61x_i-51x_i \\ 298.2=10x_i \\ \frac{298.2}{10}=x_i \\ 29.82=x_i \end{gathered}[/tex]Therefore, the empty length of the spring is 29.82 cm
Part 2)
Now, we need to calculate the value of k, so replacing xi = 29.82, we get:
[tex]k=\frac{51}{39-29.82}=5.556[/tex]Therefore, the equation for the force is
F = 5.556Δx
Solving for Δx, we get:
Δx = F/5.556
Replacing the force by 32N, we can calculate the elongation as
Δx = 32/5.556 = 5.76 cm
Part 3)
Then, the length can be calculated by solving the following equation for xf
Δx = xf - xi
xf = Δx + xi
Replacing Δx = 5.76 cm and xi = 29.82 cm, we get:
xf = 5.76 cm + 29.82 cm
xf = 35.58 cm
So, its length is 35.58 cm
Therefore, the answers are
1) 29.82 cm
2) 5.76 cm
3) 35.58 cm
a car goes from 32 m/s to a complete stop in 4.8 seconds. calculate the average stopping force of the car if has a mass of 2500 kg
The average stopping force is 16,500 N
Initial velocity of car (v₁)= 32m/s
Final velocity (v₂) = 0m/s
Time to stop= 4.8 seconds
Mass of car= 2500 kg
we need to apply the concept of laws of motion
Acceleration of car (a)= Change in velocity/time
a= v₂-v₁/t
a= 0-32/4.8
a= -6.6 m/s² ( deceleration)
Force= mass x acceleration
Force= 2500x 6.6
Force= 16500 N
Therefore the average stopping force is 16,500 N.
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How to do Projectile Motion?
A thrown ball undergoes projectile motion so throwing a ball in the air is an example of projectile motion.
What is Projectile Motion?Projectile motion is the motion of an object pitched (projected) into the air. After the starting force that launches the object, the only occurrence of the force of gravity in the object is called a projectile motion, and its path is called its trajectory. Projectile motion is a form of motion in which an object or particle ( called a projectile, is thrown near the earth's surface and moves along a curved path under the action of gravity only. Throwing a ball or a cannonball. The motion of a billiard ball on the billiard table. t. The motion of the earth around the un-projectile motion is a special case of two-dimensional motion. A particle in motion at a vertical level with an initial velocity and experiencing a free-fall (downward) acceleration, displays projectile motion.
So we can conclude that Projectile motion is applicable in both throwing and hitting. A thrown ball undergoes projectile motion.
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An object is projected or flung into the air, and only gravity's acceleration affects the object's velocity. A projectile is what it is, and its trajectory is what it took to get there.
What is Projectile motion?An item or particle that is projected toward the surface of the Earth and moves along a curved path only under the influence of gravity is said to be experiencing projectile motion. Galileo demonstrated that this curving path was a parabola, however it can also be a straight line in the unique situation where it is hurled straight up.
Ballistics is the study of such motions, and this trajectory is a ballistic trajectory. Gravity, which works downward and gives the item a downward acceleration toward the Earth's center of mass, is the sole force of mathematical significance that is actively acting on it.
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What is the energy of a proton accelerated through a potential difference of 500,000 V?
ANSWER
[tex]8.01\cdot10^{-14}J[/tex]EXPLANATION
We want to find the energy of the proton accelerated through the given potential.
To do this, apply the relationship between energy and potential:
[tex]V=\frac{E}{q}[/tex]where q = charge
V = potential
The charge of a proton is:
[tex]1.602\cdot10^{-19}C[/tex]Therefore, we have that the energy of the proton is:
[tex]\begin{gathered} E=V\cdot q \\ E=500000\cdot1.602\cdot10^{-19} \\ E=8.01\cdot10^{-14}J \end{gathered}[/tex]That is the answer.
A spring with spring constant 40 N/m is compressed .1m past it natural length. A mass of .5kg is attached to the spring. A. What is the elastic potential energy stored in the spring?B. The spring is released. What is the speed of the masses as it reaches the natural length of the spring?
Given data
*The given spring constant is k = 40 N/m
*The given compressed length is x = 0.1 m
*The given mass is m = 0.5 kg
(a)
The formula for the elastic potential energy stored in the spring is given as
[tex]U_p=\frac{1}{2}kx^2[/tex]Substitute the values in the above expression as
[tex]\begin{gathered} U_p=\frac{1}{2}(40)(0.1)^2 \\ =0.2\text{ J} \end{gathered}[/tex]Hence, the elastic potential energy stored in the spring is 0.2 J
(b)
The formula for the speed of the masses is given by the conservation of energy as
[tex]\begin{gathered} U_p=U_k \\ \frac{1}{2}kx^2=\frac{1}{2}mv^2 \\ v=x\sqrt[]{\frac{k}{m}} \end{gathered}[/tex]Substitute the values in the above expression as
[tex]\begin{gathered} v=(0.1)\sqrt[]{\frac{40}{0.5}} \\ =0.89\text{ m/s} \end{gathered}[/tex]Hence, the speed of the masses as it reaches the length of the spring is v = 0.89 m/s
Please help me, it's asking me here how does static energy work?
Static energy is the energy due to motionless state of a particle. If a particle is at rest it possess static energy also known as potential energy. When some force is applied to the particle then the static energy gets converted into kinetic energy of the particle.
Please do this step-by-step how do you do it when it’s between
Given:
• Mass of block A = 6.0 kg
,• Mass of block B = 7.0 kg
,• Mass of block C = 13.0 kg
,• Force, F = 13.0 N
Let's find the magnitude of the tension in the rope between B and C.
Let's first find the acceleration.
We have:
[tex]13-T_B+T_B-T_A+T_A=6a+7a+13a[/tex]Thus, we have:
[tex]\begin{gathered} 13=26a \\ \\ a=\frac{13}{26} \\ \\ a=0.5\text{ m/s}^2 \end{gathered}[/tex]To find the tension between blocks B and C, we have the equation:
[tex]\begin{gathered} F-T_B=M_C*a \\ \\ T_B=F-M_c*a \end{gathered}[/tex]Where:
F = 13 N
Mc is the mass of block C = 13 kg
a is the acceleration = 0.5 m/s²
Thus, we have:
[tex]\begin{gathered} T_B=13-13*0.5 \\ \\ T_B=13-6.5 \\ \\ T_B=6.5\text{ N} \end{gathered}[/tex]Therefore, the magnitude of the tension in the rope between blocks B and C is 6.5 N
ANSWER:
6.5 N
At what point, if any, during a dive is a skydiver experiencing complete freefal? Explain. (1 point)•skydiver will experience complete freefall the moment right before they jump out of the plane because they are free to start fallingat any moment.•A skydiver will experience complete freefall when they first jump out of the plane because they only experience air resistance oncethey deploy their parachute.•A skydiver will never experience complete freefall until after they have deployed their parachute because they are now falling at asafe speed for their landingA skydiver will never experience complete freefall because as soon as they start their dive, they will experience air resistance.
To find:
Which of the given statements is true.
Explanation:
The free fall is defined as the motion of an object under the influence of only gravitational force. In free fall, there will be no forces, except gravitational force acting on the object.
When the skydiver jumps, gravity will pull the diver downwards. And the air resistance due to the air in the atmosphere will oppose this motion of the diver. Thus there will be two forces acting on the diver, gravity and the air resistance. Thus the skydiver will never experience free fall.
Final answer:
Thus the correct answer is option D.
A 10.0 cm tall object is placed 6.00 cm in front of a curved mirror and produces an image 2.00 cm behind the mirror. What is the focal length of the mirror?0.667 cm1.50 cm-3.00 cm-0.333 cm
Given data:
The height of object is h₀=10.0 cm.
The object distance is u=6 cm.
The image distance is v=-2.00 cm.(negative because the image is behind the mirror)
The focal length can be calculated by the mirror's formula as,
[tex]\begin{gathered} \frac{1}{f}=\frac{1}{u}+\frac{1}{v} \\ \frac{1}{f}=\frac{1}{6}+\frac{1}{-2} \\ f=-3.00\text{ cm} \end{gathered}[/tex]Thus, the focal length of the mirror is -3.00 cm.
PLEASE HELPPPPPPPPPPPPPPPPPPPPPPP
Answer:
right answer is death valley
Explanation:
because it is close to surface gravitational field
mexico
cus its the farthest
A spring of length 9.7 meters stretches to 9.8 meters when a 0.4 kg mass is hung vertically from one end. What is the spring constant?
Given,
The initial length of the spring, l=9.7 m
The length of the spring after stretching, L=9.8 m
The mass, m=0.4 kg
The magnitude of the restoring force of the spring due to the stretching from the mass will be equal to the force applied by the mass, which is nothing but the weight of the mass.
Thus,
[tex]\begin{gathered} mg=k\Delta x \\ =k(L-l) \end{gathered}[/tex]Where g is the acceleration due to gravity, k is the spring constant, and Δx is the stretch in the length of the spring.
On substituting the known values,
[tex]\begin{gathered} k=\frac{mg}{(L-l)_{}} \\ =\frac{0.4\times9.8}{9.8-9.7} \\ =\frac{3.92}{0.1} \\ =39.2\text{ N/m} \end{gathered}[/tex]Thus the spring constant is 39.2 m
what is electric power
Answer:
Definition- Electric power is the rate at which electrical energy is transferred by an electric circuit.
Answer:
the rate at which electrical energy is transferred by an electric circuit.
Explanation:
Find the magnitude of the sumof these two vectors:B63.5 m101 m57.0°
Vector diagram:
The resultant vector is given as,
[tex]R=\sqrt[]{A^2+B^2+2AB\cos \theta}[/tex]Here, θ is the angle between vector A and B.
Substituting all known values,
[tex]\begin{gathered} R=\sqrt[]{(63.5)^2+(101)^2+2\times101\times63.5\times\cos (33^{\circ})} \\ =158.08\text{ m} \end{gathered}[/tex]Therefore, the resultant magnitue of the sum of these two vectors are 158.08 m.
The x-component of the magnitude is given as,
[tex]\begin{gathered} R_x=101\cos (57^{\circ})+63.5\cos (90^{\circ}) \\ =55.0\text{ m} \end{gathered}[/tex]The y- component of the magnitude is given as,
[tex]\begin{gathered} R_y=63.5\sin (90^{\circ})+101\sin (57^{\circ}) \\ =148.2\text{ m} \end{gathered}[/tex]Therefore, the direction is given as,
[tex]\begin{gathered} \phi=\tan ^{-1}(\frac{R_y}{R_x}) \\ =\tan ^{-1}(\frac{148.2\text{ m}}{55.0\text{ m}}) \\ =69.63^{\circ} \end{gathered}[/tex]Therefore, the direction of the resultant vector is 69.63°.
part B:
Calculate the magnitude of the acceleration of the box if you push on the box with a constant force 170.0 N that is parallel to the ramp surface and directed up the ramp, moving the box up the ramp.
The magnitude of the acceleration of the box is 9.65 m/s².
What is the net force of the box?
The net force on the box is calculated as follows;
F(net) = F - Ff
where;
F is the applied forceFf is the force of frictionF(net) = F - μmgcosθ
where;
μ is the coefficient of friction given as 0.3θ is the angle of inclination of the plane = 55⁰m is the mass of the box = 15 kgF(net) = 170 - (0.3 x 15 x 9.8 x cos55)
F(net) = 144.71 N
The magnitude of the acceleration of the box is calculated as;
a = F(net) / m
a = (144.71) / (15)
a = 9.65 m/s²
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Question 24 of 25What disadvantage of analog signals is overcome by sending digital signals?A. The waves used to transmit analog signals carry more energy.B. The waves used to transmit analog signals are more dangerous.dC. Noise decreases the quality of analog signals.O0D. Noise decreases the loudness of analog signals.SUBMIT
The correct answer is option C, "Noise decreases the quality of the analog signals."
The anlog signals q
A 244 kg motorcycle is travelling with aspeed of 14.7 m-s-1A) Calculate the kinetic energy (in J) of themotorcycle.B) If the speed of the motorcycle is increasedby a factor of 1.6, by what factor does itskinetic energy change?C) Calculate the speed (in m-s-1) of themotorcycle if its kinetic energy is 1/3 of thevaluefound in (a).
Given data:
* The mass of the motorcycle is m = 244 kg.
* The speed of the motorcycle is u = 14.7 m/s.
Solution:
(A). The kinetic energy of the motorcycle is,
[tex]K_1=\frac{1}{2}mu^2[/tex]Substituting the known values,
[tex]\begin{gathered} K_1=\frac{1}{2}\times244\times(14.7)^2_{} \\ K_1=26362.98\text{ J} \end{gathered}[/tex]Thus, the value of kinetic energy is 26362.98 J.
(B). If the speed of the motorcycle is increased by a factor of 1.6,
[tex]\begin{gathered} v=14.7\times1.6 \\ v=23.52\text{ m/s} \end{gathered}[/tex]Thus, the kinetic energy of the motorcycle becomes,
[tex]\begin{gathered} K_2=\frac{1}{2}mv^2 \\ K_2=\frac{1}{2}\times244\times(23.52)^2 \\ K_2=67489.23\text{ m/s} \end{gathered}[/tex]Dividing K_2 by K_1,
[tex]\begin{gathered} \frac{K_2}{K_1}=\frac{67489.23}{26362.98} \\ \frac{K_2}{K_1}=2.56 \end{gathered}[/tex]Thus, the kinetic energy is increased by the factor of 2.56.
(C). The 1/3 of the kinetic energy in the first part is,
[tex]\begin{gathered} K=\frac{1}{3}\times K_1 \\ K=\frac{1}{3}\times26362.98 \\ K=8787.66\text{ J} \end{gathered}[/tex]Thus, the speed of the motorcycle with the kinetic energy K is,
[tex]\begin{gathered} K=\frac{1}{2}mv^2_{}_{} \\ 8787.66=\frac{1}{2}\times244\times v^2 \\ 8787.66=122\times v^2 \end{gathered}[/tex]By simplifying,
[tex]\begin{gathered} v^2=\frac{8787.66}{122} \\ v^2=72.03 \\ v\approx8.5\text{ m/s} \end{gathered}[/tex]Thus, the speed of the motorcycle is 8.5 m/s.
I’m confused about which electromagnetic waves have the lowest frequency
The eletromagnetic wave that has the highest frequency is the gamma rays. It also has the highest energy and shortest wavelengths.
On the other hand, the type of eletromagnetic wave that has the lowest frequency, lowest energy and longest wavelength is radio waves.
The wiring in a house must be thick enough so it does not become so hot as to start a fire.part aWhat diameter must a copper wire be if it is to carry a maximum current of 34 A and produce no more than 1.6 W of heat per meter of length?
Given:
The maximum current in the circuit is,
[tex]i=34\text{ A}[/tex]The power per length is,
[tex]\frac{P}{l}=1.6\text{ W/m}[/tex]To find:
The diameter of the copper wire
Explanation:
The power (P) produced by current i, through a copper wire of resistance R and length l is given by,
[tex]\begin{gathered} Pl=i^2R \\ \frac{R}{l}=\frac{P}{i^2} \\ \frac{R}{l}=\frac{1.6}{34\times34} \end{gathered}[/tex]Now,
[tex]\begin{gathered} R=\frac{\rho l}{A} \\ R=\frac{\rho l}{\pi r^2} \end{gathered}[/tex]The resistivity of copper is,
[tex]\rho=1.72\times10^{-8}\text{ ohm.m}[/tex]So, we can write,
[tex]\begin{gathered} \frac{R}{l}=\frac{\rho}{\pi r^2} \\ \frac{1.6}{34\times34}=\frac{1.72\times10^{-8}}{\pi r^2} \\ r^2=\frac{1.72\times10^{-8}\times34\times34}{1.6} \\ r=3.5\times10^{-3}\text{ m} \\ diamer\text{ is,} \\ 2r=7.0\times10^{-3}\text{ m} \end{gathered}[/tex]Hence, the diameter is,
[tex]7.0\times10^{-3}\text{ m}[/tex]The inductor in the RLC tuning circuit of an AM radio has a value of 9 mH. What should be the value of the variable capacitor, in picofarads, in the circuit to tune the radio to 94 kHz
We are asked to determine the capacitance of an RLC circuit given the frequency. To do that we will use the following formula:
[tex]C=\frac{1}{4\pi^2f^2L}[/tex]Where:
[tex]\begin{gathered} C=\text{ capacitance} \\ f=\text{ frequency} \\ L=\text{ inductance} \end{gathered}[/tex]Now, We plug in the values:
[tex]C=\frac{1}{4\pi^2(94\times10^3Hz)^2(9\times10^{-3}H)}[/tex]Now, we solve the operations:
[tex]C=3.19\times10^{-10}F[/tex]The capacitance is 3.19x10^-10 farads. In Picofarads this is equivalent to:
[tex]C=0.0319pF[/tex]A jet engine applies a force of 300 N horizontally to the right against a 50 kg rocket. The force of air friction 200 N. If the rocket is thrust horizontally 2.0 m, the kinetic energy gained by the rocket is
We will have the following:
[tex](300N)(2m)+(-200N)(2m)=200J[/tex]Then, we from the work-kinetic force theorem we will have that the total kinetic energy gained by the rocket was 200 Joules.
Two 4.587 cm by 4.587 cm plates that form a parallel-plate capacitor are charged to +/- 0.671 nC. What is the electric field strength inside the capacitor if the spacing between the plates is 1.257 mm?
ANSWER:
3.6 x 10^6 N/C
STEP-BY-STEP EXPLANATION:
Given:
Charge (q) = 0.671 nC = 0.671 x 10^-9 C
Side (s) = 4.587 cm = 4.587 x 10^-3 m
Vacuum permittivity (ε0) = 8.85 x 10^-12 F/m
We can calculate the electric field using the following formula:
[tex]\begin{gathered} E=\frac{q}{ε_0\cdot A} \\ \\ \text{ We replacing:} \\ \\ E=\frac{0.671\cdot10^{-9}}{(8.85\cdot10^{-12})(4.587\cdot10^{-3})(4.587\cdot10^{-3})} \\ \\ E=\:3603477.12=3.6\cdot10^6\text{ N/C} \end{gathered}[/tex]The electric field is equal to 3.6 x 10^6 N/C
a 2403 kg racecar has a total momentum of 9.912*10^4kgm/s at one point in the race. calculate the speed of the racecar at that point
In order to calculate the speed, we can use the formula for the momentum:
[tex]p=m\cdot v[/tex]Where p is the momentum (in kg m/s), m is the mass (in kg) and v is the speed (in m/s).
So, using p = 99120 kg m/s and m = 2403 kg, we have:
[tex]\begin{gathered} 99120=2403\cdot v\\ \\ v=\frac{99120}{2403}\\ \\ v=41.25\text{ m/s} \end{gathered}[/tex]Problem Try to answer the following questions:(a) What is the maximum height above ground reached by the ball?(b) What are the magnitude and the direction of the velocity of the ball just before it hits the ground? Show Your Problem Solving Steps: Show these below:1) Draw a Sketch2) Choose origin, coordinate direction3) Inventory List – What is known?4) Write the kinematics equation(s) and solution of Part (a):5) Write the kinematics equation(s) and solution of Part (b):Problem 3 A small ball is launched at an angle of 30.0 degrees above the horizontal. It reaches a maximum height of 2.5 m with respect to the launch position. Find (a) the initial velocity of the ball when it’s launched and (b) its range, defined as the horizontal distance traveled until it returns to his original height. As always you can ignore air resistance.(a) Initial velocity [Hints: How is v0 related to vx0 and vy0. How can you use the information given to calculate either or both of the components of the initial velocity?](b) Range [Hints: This problem is very similar to today’s Lab Challenge except that for the challenge the ball will land at a different height.]
3.
[tex]\begin{gathered} \theta=30^{\circ} \\ y_{\max }=2.5m \end{gathered}[/tex]a)
[tex]\begin{gathered} y_{\max }=\frac{v^2\sin ^2(\theta)}{g} \\ \end{gathered}[/tex]Solve for v:
[tex]\begin{gathered} v=\sqrt[]{\frac{y_{\max }\cdot g}{\sin ^2(\theta)}} \\ v=98\cdot\frac{m}{s} \end{gathered}[/tex]b)
[tex]\begin{gathered} r=\frac{v^2}{9}\sin (2\theta) \\ r=\frac{98^2}{9.8}\cdot\sin (2\cdot30) \\ r=\frac{98^2}{9.8}\sin (60) \\ r=848.7m \end{gathered}[/tex]Convert each quantity to the indicated units.
a. 3.01 g to cg
b. 6200 m to km
c. 0.13 cal/g to kcal/g
Answer:
Explanation:
a) 301 cg
b) 6.2 km
c) 0.00013
3. A rescuer jumped from an airship in the ocean 1.20 x 102 m above the water's surface. Whatwas her kinetic energy at the moment she was 30.0 m from the water's surface? What was herspeed at that moment assuming her mass is 60.0 kg?
Given data,
The initial velocity of the body is zero.
The distance travelled by the rescuer upto the height of 30 m from the water surface is,
[tex]\begin{gathered} S=102-30 \\ S=72\text{ m} \end{gathered}[/tex]The final velocity of the rescuer at the height 30 m is,
[tex]v^2-u^2=2gS[/tex]where g is the acceleration due to gravity.
Substituting the known values,
[tex]\begin{gathered} v^2=2\times9.8\times72 \\ v^2=1411.2 \\ v=37.6ms^2 \end{gathered}[/tex]Thus, the kinetic energy of the rescuer is,
[tex]K=\frac{1}{2}mv^2[/tex]Substituting the known values,
[tex]\begin{gathered} K=\frac{1}{2}\times60\times1411.2 \\ K=42336 \\ K=42.3\text{ KJ} \end{gathered}[/tex]Thus, the kinetic energy of the rescuer is 42.3 KJ and speed of the rescuer is 37.6 meter per second square.
A construction worker pushes a wheelbarrow 5.0 m with a horizontal force of 250.0 N. How much work is done by the worker on the wheelbarrow
Given data
*A construction worker pushes a wheelbarrow at a distance is d = 5.0 m
*The given horizontal force is F = 250.0 N
The formula for the work is done by the worker on the wheelbarrow is given as
[tex]W=F\times d[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} W=(250.0)(5.0) \\ =1250\text{ J} \end{gathered}[/tex]Hence, the work done by the worker on the wheelbarrow is W = 1250 J
