A baseball is rolling along a tabletop with avelocity of 3.9 m/s to the right. The tabletopis 1.1 m above the ground. The ball rolls offthe edge of the table and falls to theground.A.) What is the ball's final vertical Velocity?B.) How long does the ball take to fall?C.) how far from the table does the ball land?
Answers
To answer this question we need to notice that once the ball starts falling we have a projectile motion; which means that horizontally we have a rectilinear motion and vertically we have an uniformly accelerated motion.
Then we can use the following equations for each direction:
[tex]\begin{gathered} \text{ Horizontal motion:} \\ x=x_0+v_{0x}t \\ \text{ Vertical motion:} \\ a=\frac{v_f-v_0}{t} \\ y=y_0+v_0t+\frac{1}{2}at^2 \\ v_f^2-v_0^2=2a(y-y_0) \end{gathered}[/tex]Since the ball is moving down in the vertical direction we will think that down is the positive direction vertically.
a)
We know that the ball is rolling to the right when it rolls off the edge of the table, this means that vertically the initial velocity is zero; we also know that the ball will fall for 1.1 m and that the acceleration is the gravitational acceleration. Then we can use the third vertical motion equation to find the final velocity, plugging the values we know we have that:
[tex]\begin{gathered} v_f^2-0^2=2(9.8)(1.1) \\ v_f=\sqrt{2(9.8)(1.1)} \\ v_f=4.64 \end{gathered}[/tex]Therefore, the final vertical velocity is 4.64 m/s.
b)
To determine the time we can use the second vertical equation with the values we know:
[tex]\begin{gathered} 1.1=0+0t+\frac{1}{2}(9.8)t^2 \\ 4.9t^2=1.1 \\ t^2=\frac{1.1}{4.9} \\ t=\sqrt{\frac{1.1}{4.9}} \\ t=0.474 \end{gathered}[/tex]Therefore, it takes 0.474 s for the ball to fall.
c)
While the ball is falling it is also moving horizontally, in this direction we know the initial velocity is 3.9 m/s; using the horizontal equations we have:
[tex]\begin{gathered} x=0+(3.9)(0.474) \\ x=1.85 \end{gathered}[/tex]Therefore, the ball lads 1.85 m from the table.
Related Questions
critical mass depends on ___. Check all that apply.A. the polarityB. the purityC. the densityD. the shape
Answers
Related to the amount of a fissionable material's critical mass depends on different factors.
These factors are:
- the shape of the material
- the density
- the purity
all last factors are related to the critical mass, becasue of all of them change the efficiency at which neutrons continue the fission procedure.
Problem 42 (p.228) Different radioisotopes have different half-lives. For example, the half-life of carbon-14 is 5700 years, the half-life of uranium-235 is 704 million years , the half-life of potassium-40 is 1.3 billion years, and the half-life of rubidium-87 is 49 billion years.
Why would an isotope with a half life like that of carbon-14 be a poor choice to get the age of the Solar System?
b) The age of the universe is approximately 14 billion years. Does that mean that no
rubidium-87 has decayed yet?
Answers
a) The choice of carbon - 14 is not pat because it has a very small half life and would vanish quickly
b) The rubidium-87 is yet to decay since it is older than the universe.
What is the solar system?The solar system comprises of the sun and all the planets that move round the sun. Recall that the sun lies at the center of the solar system. This implies that we have the planets as they move round the sun in concentric circles.
Now we can see that the half lives of all the isotopes that were listed in the question are;
carbon-14 - 5700 yearsuranium-235 - 704 million yearspotassium-40 - 1.3 billion yearsrubidium-87 - 49 billion yearsGiven that the estimated age of the sun is 4.603 billion years, it is clear that the carbon-14 would be a poor choice for the dating of the sun since it has a half life of only a few thousand years.
b) Given the fact that the half life of the rubidium-87 isotope is 49 billion years, it follows that none of it has decayed as yet.
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8) If the volume of the liquid in graduated cylinder B is 90 mL, then whatis the volume of the rock?AYour answer8060B100180
Answers
Answer:
20 mL
Explanation:
The volume of the rock is equal to the difference of volume of A and B. So, it is equal to
90 mL - 70 mL = 20 mL
Because 90 mL is the volue in cylinder B and 70 mL is the volume in cylinder A.
Therefore, the volume of the rock is 20 mL
What does this image reveal about gravityon the moon compared to Earth?
Answers
ANSWER:
The Moon's gravity is less than Earth's.
STEP-BY-STEP EXPLANATION:
When you jump, you fall back to the ground. Apples or leaves also fall: we are all attracted to the Earth. It is the terrestrial attraction due to the force of gravity.
The force of gravity also exists on the Moon. But since the Moon is smaller than the Earth, the attraction felt on the Moon is smaller than the Earth's attraction.
As the gravity is less, you can do things such as the one shown in the image.
As the force of attraction is less, the weight is less on the Moon, which can cause things that would be impossible on Earth.
Need help with this question Short straight forward answers please :)
Answers
We will have the following:
a. The gravitational potential energy will be:
[tex]P_C=(15kg)(9.8m/s^2)(6m)\Rightarrow P_C=882J[/tex]So, the gravitational potential energy of C is 882 J.
b. The velocity of C right before it hits the ground will be:
[tex]\begin{gathered} 882J=\frac{1}{2}(15kg)v^2\Rightarrow\frac{1764J}{15kg}=v^2 \\ \\ \Rightarrow v=\frac{14\sqrt{15}}{5}m/s\Rightarrow v\approx10.84m/s \end{gathered}[/tex]So, the velocity will be approximately 10.84 m/s.
c.
1. We will have that Eg at the initial position will be: B < C
2. Vfinal upon impact with ground: B = C
3. Ek right before hitting he ground: B < C
f.
1. Eg: A > B
2. V final: A > B
3. Ek: A > B
4. V at 2 meters above the ground: A > B
5. Total energy at 2 m above the ground: A > B.
Carol Gillian theorized that when it comes to a perspective of Justice,males per socialized for a blank environment while females are socialized for a blank environment
Answers
Answer: men = work environment , women = home environment
Explanation: Gillian proposed that women come to prioritize as “ethics of care” and men as “ethics of justice”.
Imagine that you are on board a ship that was struck by a rogue wave.
Tell your story, from the calm before the wave hit to its aftermath.
Answers
The possibility of a wave toppling cruise liner is extremely low. They are made wide and have enough ballast on the lower decks to be heavy enough rogue waves. On the side, the crew's neglect also be necessary.
What are rogue waves?A wave that is double the region's major wave height is typically considered a rogue wave. The highest one-third of waves on average over a period of time make up the noteworthy wave height. Even the biggest ships and oil rigs can be rendered useless and sunk by rogue waves.
Have rogue waves ever struck a cruise ship?Rogue waves have occasionally hit cruise ships, although it is not frequently. Four cruise ships have collided with rogue waves since rogue wave records were first kept in 1995. All sustained damage, and some people reported injuries, but there have been no confirmed fatalities on cruise ships due to rogue waves.
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A particular cookie provides 54.0 kcal of energy. An athlete does an exercise that involves repeatedly lifting (without acceleration) a 103-kilogram weight 2.45-decimeters above the ground with an energy efficiency of 25%. How many repetitions of this exercise can the athlete do with the energy supplied from one of these cookies?
Answers
A maximum of about 229 repetitions of something like the exercise can be performed by that of the athlete utilizing the energy provided by each of the biscuits.
The proportion of input to produced energy can be used to define energy consumption.
A cookie, therefore, therefore has 54.0 kcal of calories. The 54.0 kcal throughout this croissant is used as power input by the athlete.
Efficiency = output energy / input energy
It can be written as:
Output energy = efficiency × input energy
Puting the values of efficiency and input energy.
Output energy = 0.25 × 54 kcal = 13.5 kcal.
The weightlifting exercise can be done n times for the output energy. This outgoing energy comes from mgh in the shape of potential energy. So,
Energy per repetition = [tex]mgh[/tex]
Put the values of m, g and h in above equation.
Energy per repetition = 103 kg × 9.8 m/ × (2.45 × 0.1m) = 247.303 J
Energy per repetition = 0.059 kcal.
So,
amount of repetitions = sum of output energy / energy per repetition
amount of repetitions = 13.5 kcal / 0.059 kcal = 229 repetitions.
Therefore, amount of repetition can be be 229.
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Three people are driving their cars in different directions, in an open field. At one point, while they are all driving, they each measure the other drivers’ velocities. When they compare measurements afterward, they notice that they all got different measurements from each other. Why do their measurements not match?
Answers
Answer:
because of their change in momentum
Multiple part question Here are the needed details:Five rotations took 5.15 seconds 1 rotation took 1.07s Distance from shoulder to elbow is 29 cm distance from shoulder to middle of the hand is 57cm.Questions:2. A how far in degrees did the hand travel during the five rotations?B. How far in radians did the hand travel during the five rotations?C. How far in meters did the hand travel during the five rotations?3. A. What was the average angular speed (degrees/s and rad/s) of the hand?B. What was the average linear speed (m/s) of the hand?C. Are the answers to a and b the same or different? Explain.4. A. What was the average angular acceleration (degrees/s squared and rad/s squared) of the hand. How do you know?B. What was the average centripetal acceleration (m/s squared) of the hand?C. Are the answers to a and b the same or different. Explain.5. A. How far (degrees and rad) did the elbow travel during the five rotations?B. How far (m) did the elbow travel during the five rotations?C. How do these compare to the hand? Why are they the same and or/different?6. A. What was the average angular speed (degrees/s and rad/s) of the elbow?B. What was the average linear speed (m/s) of the elbow?C. How do these compare to the hand? Why are they the same and or/different?7. A. What was the average angular acceleration (degrees/s squared and rad/s squared) of the elbow?B. What was the average centripetal acceleration (m/s squared) of the elbow?C. How do these compare to the hand? Why are they the same and or/ different?
Answers
Given:
Time taken for 5 rotations = 5.15 seconds
Time for 1 rotation = 1.07 seconds
Distance from shoulder to elbow = 29 cm
Distance from shoulder to the middle of the hand = 57 cm
Let's use the information above to answer the following questions.
Question 2:
Let's determine how far in degrees the hand travelled during the five rotations.
In one full rotation, we have 360 degrees.
Thus, 5 full rotations = 5 * 360 = 1800 degrees
Therefore, in 5 full rotations, the hand travelled 1800 degrees.
B. In radians, we have:
180 degrees = π rad
[tex]1800\degree=\frac{\pi}{180}\ast1800=10\pi\text{ radians}[/tex]C. To find the distance in meters, we have:
Distance from elbow to shoulder = 29 cm = 0.29 meters
[tex]2\pi\ast5\ast0.29=9.11\text{meters}[/tex]Therefore, the hand travelled 9.11 meters during the five rotations.
Question 3:
A. To find the average angular speed, apply the formula:
[tex]\begin{gathered} w=\frac{10\pi}{t}\text{ (rad/s)} \\ \\ w=\frac{1800}{t}\text{ (degre}es\text{/s)} \end{gathered}[/tex]Where t = 5.15 seconds
Thus, we have:
[tex]\begin{gathered} w=\frac{10\pi}{5.15}=6.1\text{ rad/s} \\ \\ w=\frac{1800}{5.15}=349.5\text{ degre}es\text{/s} \end{gathered}[/tex]B. Average linear speed of the hand.
To find the average linear speed of the hand, we have:
[tex]v=\frac{10\pi r}{t}=\frac{10\pi}{5.15}\ast\frac{1}{2}=3.05\text{ m/s}[/tex]C. The average angular speed and average linear speed are the same
An object is dropped from rest out of the window of a building, and the time to hit the ground is found to be 5 seconds. The same object is then dropped from rest out of a window twice as high above the ground as the original window. The time it takes the object to hit the ground is closest to:
Answers
ANSWER:
7 s
STEP-BY-STEP EXPLANATION:
Given:
u = 0m/s
t = 5 sec
g = 9.8 m/s^2
The first thing is to calculate the height of the building, using the following formula:
[tex]\begin{gathered} s=ut+\frac{1}{2}gt^2 \\ \text{ Replacing} \\ s=0\cdot5+\frac{1}{2}\cdot9.8\cdot5^2 \\ s=122.5\text{ m} \end{gathered}[/tex]Now, we apply the same formula, but we substitute the double value of the distance and solve for t, just like this:
[tex]\begin{gathered} 2\cdot122.5=\frac{1}{2}\cdot9.8\cdot\: t^2 \\ 9.8\cdot t^2=245\cdot2 \\ t^2=\frac{490}{9.8} \\ t=\sqrt[]{50} \\ t=7.07\text{ sec} \\ t\approx7\text{ sec} \end{gathered}[/tex]The time it takes for the object to fall is 7 seconds.
8.1 kg of copper sits at a temperature of 64 oF. How much heat is required to raise its temperature to 743 oF? The specific heat of copper is 385 J/kg-oC. Submit your answer in exponential form.
Answers
ANSWER:
STEP-BY-STEP EXPLANATION:
Given:
Mass (m) = 8.1 kg
T1 = 64 °F
T2 = 743 °F
Specific heat (C) = 385 J/kg*°C
[tex]undefined[/tex]If John Glenn weighed 640 N on Earth's surface, a) how much would he haveweighed if his Mercury spacecraft had (hypothetically) remained at twice thedistance from the center of Earth? b) Why is it said that an astronaut is nevertruly "weightless?"
Answers
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
A star of mass 3.0e30 kg is moving in a circle of radius 1.0e12 metres, with a period of 100 years. This is due to the gravity of a second unseen object. what is the mass of the unseen object in kg
Answers
The mass of the unseen object is 1.21 x 10²⁴ kg.
What is the centripetal acceleration of the star?
The centripetal acceleration of the star is calculated as follows;
a = v²/r
where;
v is the linear speed of the starr is the radiusv = 2πr/T
where;
T is period of the motion = 100 years = 3.154 x 10⁹ sv = (2π x 1 x 10¹²) / (3.154 x 10⁹)
v = 1,992.1 m/s
a = (1992.1)² / ( 1 x 10¹²)
a = 3.97 x 10⁻⁶ m/s²
The centripetal force of the star is calculated as follows;
F = ma
F = (3 x 10³⁰) x ( 3.97 x 10⁻⁶)
F = 1.19 x 10²⁵ N
The mass of the unseen object is calculated as follows;
F = mg
m = F/g
where;
g is the acceleration due to gravitym = (1.19 x 10²⁵) / (9.8)
m = 1.21 x 10²⁴ kg
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An object is dropped from a height of 65 m above ground level. A) determine the final speed in m/s, at which the object hits the ground c) determine the distance in meters, traveled during the last second of motion before hitting the ground.
Answers
Given:
height = 65 m
Given that the object is in free fall, let's solve for the following:
• (a). determine the final speed in m/s.
To find the final velocity, apply the kinematics equation:
[tex]v^2=u^2-2ax[/tex]Where:
v is the final velocity
u is the initial velocity = 0
a is the acceleration due to gravity = 9.8 m/s²
x is the displacement = 65 m
Thus, we have:
[tex]\begin{gathered} v^2=0^2-2(-9.8)(65) \\ \\ v^2=-(-1274) \\ \\ v^2=1274 \\ \\ \text{ Take the square root of both sides:} \\ \sqrt{v^2}=\sqrt{1274} \\ \\ v=35.69\text{ m/s} \end{gathered}[/tex]Therefore the final speed will be -35.69 m/s.
• (c). The distance traveled during the last second of motion before hitting the ground.
To find the distance, apply the formula:
[tex]H=ut+\frac{1}{2}at^2[/tex]Where:
H is the height.
u is the initial velocity = 0 m/s
t is the time
a is acceleration due to gravity.
Let's rewrite the formula to find the time traveled.
[tex]\begin{gathered} H=0t+\frac{1}{2}at^2 \\ \\ H=\frac{1}{2}at^2 \\ \\ t=\sqrt{\frac{2H}{a}} \end{gathered}[/tex]Thus, we have:
[tex]\begin{gathered} t=\sqrt{\frac{2*65}{9.8}} \\ \\ t=\sqrt{\frac{130}{9.8}} \\ \\ t=\sqrt{13.26} \\ \\ t=3.64\text{ s} \end{gathered}[/tex]Therefore, the time is 3.64 seconds.
Now, to find the distance traveled during the last second of motion, apply the formula:
[tex]s=\frac{1}{2}a(t_2^2-t_1^2)[/tex]Where:
t2 = 3.64 seconds
t1 = 3.64 seconds - 1 second = 2.64 seconds
Thus, we have:
[tex]\begin{gathered} s=\frac{1}{2}(9.8)((3.64)^2-(2.64)^2) \\ \\ s=4.9(13.2496-6.9696) \\ \\ s=4.9(6.28) \\ \\ s=30.77 \end{gathered}[/tex]Therefore, the distance in meters, traveled during the last second of motion before hitting the ground is 30.77 meters.
ANSWER:
(A). -35.69 m/s
(C). 30.77 m
Mr.D soars over a large group of zombies and is in the air for a total of 5s. How high did he go?
Answers
Use the momentum equation for photons found in this week's notes, the wavelength you found in #3, and Planck’s constant (6.63E-34) to calculate the momentum of this photon:
Answers
The wavelength is divided by Plank's constant to get the momentum equation for photons: p = h /λ.
What is photon?The electromagnetic force is carried by a photon, a basic particle that is a quantum of the electromagnetic field and includes electromagnetic radiation like light and radio waves. Since photons have no mass, they constantly move at the 299792458 m/s speed of light in a vacuum.
Assuming the wavelength determined in a prior issue, = 656 nm = 656 * 10 - 9 m, you get:
p = (6.63 * 10 ^-34) / (656 * 10 ^ -9) kg * m/s
P, which must be rounded to three significant numbers, is equal to 1.01067 * 10 - 27 kg*m/s.
Consequently, p = 1.01 * 10 -27 kg*m/s
Our number, rounded to two significant figures, is 1.0 * 10 - 27 kg*m/s because the answers are only rounded to two significant values.
Therefore, given that the wavelength is 656 nm, the first option—1.0*10-27 kg*m/s—is the correct response.
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Analyze the benefits and consequences of sleep create (continue) their own fitness programs incorporating sleep as a fundamental pillar of fitness
Answers
Explanation:
The Intimate Relationship Between Fitness and Sleep
*“If you don’t sleep, you undermine your body,” says W. Christopher Winter, MD, the president of Charlottesville Neurology and Sleep Medicine and the author of "The Sleep Solution: Why Your Sleep Is Broken and How to Fix It."
When it comes to working out, you know that what you do in the gym is important. But what you do outside the gym — what you eat, what you drink, and especially how you sleep, is just as crucial. In fact, you must sleep in order for exercise to actually work.
“We exercise for a purpose: for cardiovascular health, to increase lean muscle mass, to improve endurance, and more. All of these 'goals' require sleep,” says W. Christopher Winter, MD, the president of Charlottesville Neurology and Sleep Medicine and the author of The Sleep Solution: Why Your Sleep Is Broken and How to Fix It.
In other words, without sleep, exercise does not deliver those benefits, Dr. Winter explains. “If you don’t sleep, you undermine your body.”
A negative charge of .30 c and a positive charge of .50 are separated by .40What is the force between the charges?
Answers
Given:
The negative charge q1 = 0.3 C
The positive charge q2 = 0.5 C
The distance between the charges is 0.4 m
To find the magnitude of the force between them.
Explanation:
The formula to calculate the magnitude of the force is
[tex]F=\frac{kq1q2}{r^2}[/tex]Here, k is Coulomb's constant whose value is
[tex]k=\text{ 9}\times10^9Nm^2C^{-2}[/tex]On substituting the values, the magnitude of force will be
[tex]\begin{gathered} F=\frac{9\times10^9\times0.3\times0.5}{(0.4)^2} \\ =8.4375\text{ }\times10^9\text{ N} \end{gathered}[/tex]Final Answer: The magnitude of the force is 8.4375 x 10^(9) N.
A car is going at a speed of 25m/s when the driver puts her foot on the gas pedal. The carfeels a net force of 2000N for 50m. The car's mass is 1000kg.How much kinetic energy does the car have initially?
Answers
The initial kinetic energy of the car = 312.5 kJ
Explanation:The initial volume of the car, v = 25 m/s
The mass of the car, m = 1000 kg
The initial kinetic energy is given by the formula
[tex]\begin{gathered} KE=\frac{1}{2}mv^2 \\ \end{gathered}[/tex]Substitute m = 1000 kg, and v = 25 m/s into the formula
[tex]\begin{gathered} KE=\frac{1}{2}\times1000\times25^2 \\ KE=500\times625 \\ KE=312500J \\ KE=312.5kJ \end{gathered}[/tex]The initial kinetic energy of the car = 312.5 kJ
An x-ray with a wavelength of 3.5 × 10^-9 m travels with a speed of 3.0 × 10^8 m/s. What is the frequency of this electromagnetic wave? A.9.52 × 10^-1 Hz B.8.57 × 10^16 Hz C.1.17 × 10^-17 Hz D.1.05 Hz
Answers
Answer. B
A stone is thrown vertically upwards from a height of 1.5m and lands on the ground 6s later. What was the magnitude of the initial velocity?
Answers
Answer:
2.5m/s is the correct answer
A person is at the top of a tower. He takes a segment of a string which measures 30 cm long when at rest and hooks his 3 kg sword at the end of it. The spring extends to 35 cm long. He will use this spring to get to the ground. What is the spring constant of the spring, and how much of the spring (measured at equibilirum) does he need in order to have a net force of 0 upon himself when he touches the ground? Assume he hangs the spring from a hook located exactly 30 m above the ground. Be certain to draw a free body diagram of the forces on him the moment he hits the ground.
Answers
The given problem can be exemplified in the following diagram:
To determine the constant of the spring we can use Hook's law, which is the following:
[tex]F=k\Delta x[/tex]Where:
[tex]\begin{gathered} F=\text{ force on the string} \\ k=\text{ string constant} \\ \Delta x=\text{ difference in length} \end{gathered}[/tex]Now, we solve for "k" by dividing both sides by the difference in length:
[tex]\frac{F}{\Delta x}=k[/tex]The force on the string is equivalent to the weight attached to it. The weight is given by:
[tex]W=mg[/tex]Where:
[tex]\begin{gathered} W=\text{ weight} \\ m=\text{ mass} \\ g=\text{ acceleration of gravity} \end{gathered}[/tex]Substituting in the formula for the constant of the spring we get:
[tex]\frac{mg}{\Delta x}=k[/tex]Now, we substitute the values:
[tex]\frac{(3kg)(9.8\frac{m}{s^2})}{35\operatorname{cm}-30\operatorname{cm}}=k[/tex]Before solving we need to convert the centimeters into meters. To do that we use the following conversion factor:
[tex]100\operatorname{cm}=1m[/tex]Therefore, we get:
[tex]\begin{gathered} 35\operatorname{cm}\times\frac{1m}{100\operatorname{cm}}=0.35m \\ \\ 30\operatorname{cm}\times\frac{1m}{100\operatorname{cm}}=0.30m \end{gathered}[/tex]Substituting in the formula we get:
[tex]\frac{(3kg)(9.8\frac{m}{s^2})}{0.35m-0.30m}=k[/tex]Solving the operations:
[tex]588\frac{N}{m}=k[/tex]Therefore, the constant of the spring is 588 N/m.
A box is standing on a conveyor belt that is not in motion. At one point the belt starts moving with some acceleration. At that point the box starts moving too (without slipping). Which force is responsible for the acceleration of the box. a. The air resistance force. b. The force of the pull. c. The force of friction. d. The normal force.
Answers
Given that a box is standing on a conveyor belt that is not in motion.
When the belt starts moving with some acceleration, the box starts moving too without slipping.
Let's determine the force that is responsible for the acceleration of the box.
Here, since the box starts moving without slipping when the belt starts moving, there will be static friction between the box and the belt since the belt was fixed.
Now, the force which is responsible for the acceleration of the box will be the force of gravity and the normal force.
Applying the Newton's second law, if the there is only force of gravity and the normal force acting on the box, there will be zero horizontal acceleration.
In order for the box to accelerate without slipping, the force responsible will be the static frictional force.
ANSWER:
c. The force of friction.
A 6 kg object is being pulled by a horizontal force F=120 N on a friction-less horizontal surface. It moved a distance of 18 m. If its initial kinetic energy was 100 Joules, what is the final kinetic energy in Joules?
Answers
The final kinetic energy is 120m.
What is Work-Energy Theorem?
The Work-Energy Theorem states that the work done is equal to the change in the K.E. i.e Kinetic Energy of the object.
W = Δ(K.E.)
In the given question we had,
Mass = 6 kg,
Force = 120 N,
Distance = 18 m,
Initial Kinetic Energy ( KE1 ) = 100 Joules
According to Work-Energy Theorem,
W = Δ(K.E.)
W = KE2-KE1
F x S = KE2 - 100
120 x 18 = KE2 - 100
2160 + 100 = KE2
2260J = KE2
So, the final kinetic energy is 2260J.
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A particular cookie provides 54.0 kcal of energy. An athlete does an exercise that involves repeatedly lifting (without acceleration) a 103-kilogram weight 2.45-decimeters above the ground with an energy efficiency of 25%. How many repetitions of this exercise can the athlete do with the energy supplied from one of these cookies?
Answers
Energy effectiveness would be a term that refers to the proportion of input power over output. Power generation, as well as simply energy utilization, is the process of reducing the amount of energy used to produce goods and services.
Considering that,
A cookie contains 54.0 kcal of energy. An athlete utilizes the 54.0 kcal inside this cookie from input energy.
The following diagram illustrates the relationship among input as well as output energy:
Efficiency = output energy / input energy...(i)
Output energy = efficiency × input energy
By using equation (i)
⇒ output energy = 0.25 × 54 kcal = 13.5 kcal.
The lifting exercise has been performed n times for the output energy.
In terms of potential energy, such output energy could be written as follows:
Mass × gravity ×height.
So, energy per repetition = mgh = 103 kg × 9.8 m/ × (2.45 × 0.1m) = 247.303 J = 0.059 kcal.
So, Count of repetitions = sim of output energy / energy per repetition..(ii)
By using equation (ii)
Count of repetitions = 13.5 kcal / 0.059 kcal =229 repetitions.
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There is no _________ movement in a longitudinal wave.A. HorizontalB. Back and forthC. VerticalD. Parallel
Answers
Explanation
A longitudinalwave is in which the particles of the medium vibrate in the direction of the line of advance of the wave.Longitudinal waves cause the medium to move parallel to the direction of the wave.
A longitudinal wave can be set up for example in a streched spring by compressing the coils in a small region, and releasing the compressed region,
the back and forth motions of the coils of the spring is in the same direction that the wave travels
so, in a longitudinal wave there is not Vertical movement, so the answer is
C. Vertical
the spaceship is flying through space far from planets and stars with the engines firing.
The astronaut shuts the engines off.
The spaceship will….
a. stop moving immediately
b. slow down gradually and stop
c. continue with whatever speed it had when the engines were cut off
d. speed up for just a little while, then slow down
Answers
When a 6.0-F capacitor is connected to a generator whose rms output is 25 V, the current in the circuit is observed to be 0.40 A. What is the frequency of the source?
Answers
The frequency of the source is 1.66 Hz.
What is the impedance?Let us recall that the impedance of the circuit is the opposition that is offered to the flow of current by a circuit component that is not a resistor. Now let us find the impedance.
I = V/Z
I = current
V = voltage
Z = impedance
Z = V/I
Z = 25/0.4
Z = 62.5 ohm
Z^2 = R^2 + Xc^2
Z^2 = Xc^2
Xc= Z
Xc = 2πfC
f = frequency
C = capacitance
f= Xc/2πC
f = 62.5/2 * 3.142 * 6
f = 1.66 Hz
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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 66 kHz
Answers
Given:
The inductance is,
[tex]\begin{gathered} L=9\text{ mH} \\ =9\times10^{-3}\text{ H} \end{gathered}[/tex]The radio frequency is,
[tex]\begin{gathered} f=66\text{ kHz} \\ =66\times10^3\text{ Hz} \end{gathered}[/tex]To find:
value of the variable capacitor, in picofarads
Explanation:
The frequency of the AM is,
[tex]\begin{gathered} f=\frac{1}{2\pi\sqrt{LC}} \\ \end{gathered}[/tex]Substituting the values we get,
[tex]\begin{gathered} 66\times10^3=\frac{1}{2\pi\sqrt{9\times10^{-3}\times C}} \\ \sqrt{9\times10^{-3}\times C}=\frac{1}{2\pi\times66\times10^3} \\ \sqrt{9\times10^{-3}\times C}=2.41\times10^{-6} \\ 9\times10^{-3}\times C=5.81\times10^{-12} \\ C=6.45\times10^{-10} \\ C=645\times10^{-12}\text{ F} \\ C=645\text{ pF} \end{gathered}[/tex]Hence, the capacitance is 645 pF.
