if i climb a tree and drop a rock and it take 1.3 seconds to fall how high have i climbed
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Answer:
If we assume that the rock was dropped from rest, we can use the following equation of motion to determine the height you climbed:
h = (1/2) * g * t^2
where h is the height you climbed, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time it took for the rock to fall (1.3 seconds in this case).
Plugging in these values, we get:
h = (1/2) * 9.8 m/s^2 * (1.3 s)^2 = 10.16 meters
Therefore, you climbed approximately 10.16 meters (or about 33.3 feet).
Related Questions
____________ are forces in which two interacting objects are physically in contact with each other; examples include ___________ and ______________. a. Contact forces , friction, air resistance c. Action-at-a-distance fields , gravity, electric field b. Contact forces , gravity, electric field d. None of the above. Please select the best answer from the choices provided A B C D
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a. Contact forces, friction, air resistance
A battery-operated car utilizes a 12.0 V system. Find the charge the batteries must be able to move in order to accelerate the 750 kg car from rest to 25.0 m/s, make it climb a 2.00 x 10^2 m high hill, and then cause it to travel at a constant 25.0 m/s by exerting a 5.00 x 10^2 N force for an hour.
Answers
To find the charge the batteries must be able to move, we need to calculate the total work done by the car's motors, which is equal to the total energy required to perform the given tasks.
We can break down the problem into three parts: accelerating the car, lifting it to the top of the hill, and maintaining a constant speed against a resistive force.
Part 1: Accelerating the car
The work done in accelerating the car from rest to a speed of 25.0 m/s is given by:
\(W1 = (1/2) * m * v^2 = (1/2) * 750 kg * (25.0 m/s)^2 = 234,375 J\)
Part 2: Lifting the car to the top of the hill
The work done in lifting the car to a height of 2.00 x 10² m against gravity is given by:
\(W2 = m * g * h = 750 kg * 9.81 m/s^2 * 2.00 x 10^2 m = 1.47 x 10^6 J\)
Part 3: Maintaining constant speed against a resistive force
The work done in maintaining a constant speed of 25.0 m/s against a resistive force of 5.00 x 10² N for an hour (3600 seconds) is given by:
\(W3 = F * d = F * v * t = 5.00 x 10^2 N * 25.0 m/s * 3600 s = 4.50 x 10^7 J\)
The total work done by the car's motors is the sum of these three parts:
\(W = W1 + W2 + W3 = 4.65 x 10^7 J\)
The charge the batteries must be able to move is equal to the total energy required, divided by the voltage of the system:
\(Q = W / V = 4.65*10^7 J / 12.0 V=3.87*10^6 C\)
Therefore, the batteries must be able to move a charge of approximately 3.87 x 10⁶ coulombs to perform the given tasks.
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A baseball is thrown straight up with a velocity of 60 m/s. C) how many seconds later will the baseball reach its maximum height?
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The baseball will reach its maximum height after approximately 6.11 seconds using equations of motion where velocity is given
When a baseball is thrown straight up with an initial velocity of 60 m/s, it will eventually reach a maximum height before falling back down due to gravity. To determine how many seconds later the baseball will reach its maximum height, we need to use the equations of motion.
The equation we can use to find the time taken for the baseball to reach its maximum height is:
t = Vf / g
where t is the time taken, Vf is the final velocity (which is zero at the maximum height), and g is the acceleration due to gravity, which is approximately \(9.81 m/s^2\).
Substituting:
t = 60 / 9.81
t ≈ 6.11 seconds
Therefore, the baseball will reach its maximum height after approximately 6.11 seconds. After this point, the baseball will begin to fall back down to the ground due to the force of gravity, with its velocity increasing at a rate of \(9.81 m/s^2\)
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The force that opposes the motion of an object is called:.
Answers
Answer:
The Force of Friction!
Explanation:
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If Marissa has 52 molecules of Carbon Oxide, How many moles are there?
Answers
Answer:
8.6348020882604e-23
Explanation:
a car traveled 139.2km in 3.8 hours. what was the average speed?
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Answer:
36.63 km/hrExplanation:
The average speed of the car can be found by using the formula
\(v = \frac{d}{t } \\ \)
d is the distance
t is the time taken
From the question we have
\(v = \frac{139.2}{3.8} \\ = 36.63157...\)
We have the final answer as
36.63 km/hrHope this helps you
Greek engineers had the unenviable task of moving large columns from the quarries to the city. One engineer, Chersiphron, tried several different techniques to do this. One method was to cut pivot holes into the ends of the stone and then use oxen to pull the column. The 4-ft diameter column weighs 13200 lbs, and the team of oxen generates a constant pull force of 1500 lbs on the center of the cylinder G. Knowing that the column starts from rest and rolls without slipping, determine the velocity of its center G after it has moved 5 ft and the minimum static coefficient of friction that will keep it from slipping.
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The velocity is a vector quantity. The velocity of center G of the column after it has moved 5 ft is 5.18 ft.
From work Energy equation:\(U_{12} = T_2\)First, the potential energy of column,
\(U_{12} = \int\limits^5_0 {F} \, dx\)
Here, the column moves from zero and displaces 5 ft.
\(F\) - force = 1500 lbs
Thus,
\(U_{12 } = 7500\rm \ ft\ lbs\)
Now, total kinetic energy at position 2,
\(T_ 2 = \dfrac 12 mv_2^{-2} + \dfrac 12 \vec I \omega_2^2\)
Where,
m = 372.67 slugs
r = 2 ft
Since,
\(\omega_2 = \dfrac {\vec V}{r}\)
So,
\(7500 = \dfrac {12000}{2g}v_2^{-2}+\dfrac {12000}{4g}r^2 \times \dfrac {\vec V}{r}\\\\\vec v_2 = 5.18 \rm \ ft\)
Therefore, the velocity of center G of the column after it has moved 5 ft is 5.18 ft.
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How did harlow shapley conclude that the sun was not in the center of the milky way galaxy?
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We learned that We are in the disk of the Galaxy, about 5/8 of the way from the center.
What is the work of Harlow Shapley?Shapley, who was headquartered in Boulder, Colorado, used Cepheid variable stars to estimate the size of the Milky Way Galaxy and its position relative to the Sun. In 1953, he published his "liquid water belt" theory, today known as the concept of a livable zone.
There are many stars, grains of dust, and gas in the Milky Way. It is known as a spiral galaxy because, from the top or bottom, it would appear to be whirling like a pinwheel. About 25,000 light-years from the galaxy's nucleus, the Sun is situated on one of the spiral arms.
Approximately 5/8 of the way from the galaxy's nucleus, we are in the disc. William Herschel believed that the Sun and Earth were about in the middle of the vast cluster of stars known as the Milky Way.
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what kind of rock requires heat and preasure to form
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2. Using the same list of living things show how they could be classified into three groups.
Group 1 ________
Group 2 ________
Group 3 _________
Using the same list, show how they could be classified into four groups.
Group 1 ________
Group 2 ________
Group 3 _________
Group 4 __________
Please someone help as soon as possible.
and for this assignment group has have a different name while you are using these words in the picture.
oh almost forgot part 1 of this in the picture also.
Answers
Answer:
Also, you could do this for the groups:
Group 1- creatures with scales
Group 2- creatures with fur
Group 3- creatures with feathers
For 4 groups you could possible do:
Group 1- no legs
Group 2- 2 legs with wings
Group 3- 2 legs without wings
Group 4- 4 legs
what is a car doing as it rounds a curve
Answers
Answer:
its direction is changing
Convert 50,000 ML to mL
1. 5 x 10^5 mL
2. 5 x 10^-5 mL
3. 5 x 10^13 mL
4. 5 x 10^-13 mL
Answers
The answer is 1!
The answer is 1!
Squids propel themselves by expelling water. They do this by keeping water in a cavity and then suddenly contracting the cavity to force out the water through an opening. A 9 kg squid (including the water in the cavity) at rest suddenly sees a dangerous predator. If the squid expels 2 kg of water out of its body with a speed of 8 m/s, what would be its own escape speed
Answers
Answer:
v_squid = - 2,286 m / s
Explanation:
This exercise can be solved using conservation of the moment, the system is made up of the squid plus the water inside, therefore the force to expel the water is an internal force and the moment is conserved.
Initial moment. Before expelling the water
p₀ = 0
the squid is at rest
Final moment. After expelling the water
\(p_{f}\) = M V_squid + m v_water
p₀ = p_{f}
0 = M V_squid + m v_water
c_squid = -m v_water / M
The mass of the squid without water is
M = 9 -2 = 7 kg
let's calculate
v_squid = 2 8/7
v_squid = - 2,286 m / s
The negative sign indicates that the squid is moving in the opposite direction of the water
By solving the equation A) f(t)= = B) f(t): C) f(t) D) f(t)= = on [² f(u)du = t_ -L₁ €² 2 f(u)du is obtained: Jo 1+e²t 1 1+ e2t t = 1 1 2t 1-e²t
Answers
By solving the given equation on [² f(u)du = t_ -L₁ €² 2 f(u)du is obtained, we can find t.= J 1+e²t / 1 + e2t / 1-e²tdt. Now, we need to solve the integral,∫ 1+e²t / (1 + e2t)(1-e²t) dt.
For this integral, let u = 1+ e²tSo, du/dt = 2e²And, dt = du/2e²= 1/2e² ∫1+e²t / (u)(1-e²t) du= 1/2e² ∫ (1/u) - (e²/(1-e²t)) du= 1/2e² [ln|u| - ln|1-e²t|] + c.
Now, substituting back the value of u,= 1/2e² [ln|1+ e²t| - ln|1-e²t|] + c= 1/2e² ln|1+ e²t / 1-e²t| + c.
Now, putting the limits in the above expression and solving it, we get the value of t.= [1/2e² ln|1+ e²t / 1-e²t|] t = 1 2t / [1 + e²t] - L₁ 2t / [1-e²t].
Hence, the answer is D) f(t)= 2t / [1 + e²t] - L₁ 2t / [1-e²t].
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_True or False. A very massive object will free fall at the same rate of acceleration as a less massive object.
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Two flying fish have an inelastic collision while in mid-flight. One fish has a mass of 0. 650 kg and a velocity of 15. 0 m/s to the right; the other has a mass of 0. 950 kg and a velocity of 13. 5 m/s to the left. Find the change in their kinetic energy after the collision
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The change of kinetic energy is -156.7 joules.
\(V1_{i}\) =15
\(V2_{i}\)= -13.5(as towards left)
V= common velocity during collision
On applying momentum conservation during the collision
\(m_{1}V1_{i} + m_{2}V2_{i} = V ( m_{1} + m_{2})\)
\(0.650 \ 15 + 0.950 \ (-13.5) = (0.625 + 0.950) \ V\)
\(\frac{3.075}{1.6}\) =V
V= -1.92 m/s
To kind change in kinetic energy we will subtract final and initial kinetic energy
ΔK=\(\frac{1\ (m_{1} + m_{2} ) \ V}{2} ^{2}\) - \(\frac{1\ (m_{1} ) \ V1_{i} }{2} ^{2}\) - \(\frac{1\ (m_{2} ) \ V2_{i} }{2} ^{2}\)
ΔK= \(\frac{1\ (0.625 + 0.950 ) \ (-13.5)}{2} ^{2}\) - \(\frac{1\ (0.625 ) \ 15 }{2} ^{2}\)- \(\frac{1\ (0.950 ) \ 13.5 }{2} ^{2}\)
ΔK= = - 156.7 J
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A plane has an eastward heading with an airspeed of 156 m/s. A 20.0 m/s wind is blowing southward at the same time as the plane is flying. The velocity of the plane relative to the ground is
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To find the velocity of the plane relative to the ground, we need to use vector addition. The eastward airspeed of the plane is one vector, while the southward wind speed is another vector. The resulting vector is the velocity of the plane relative to the ground.
Using the Pythagorean theorem, we can find the magnitude of the resulting vector:
Velocity^2 = (156 m/s)^2 + (20.0 m/s)^2
Velocity = sqrt[(156 m/s)^2 + (20.0 m/s)^2]
Velocity = 158.1 m/s
The direction of the resulting vector can be found using trigonometry. We can use the tangent function to find the angle between the eastward direction and the direction of the resulting vector:
tan(theta) = opposite/adjacent
tan(theta) = (20.0 m/s)/(156 m/s)
theta = 7.3 degrees south of east
Therefore, the velocity of the plane relative to the ground is 158.1 m/s at an angle of 7.3 degrees south of east.
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A quality control engineer wants to determine if the diameters of ball bearings produced by a machine are normally distributed. From a random sample of 300 bearings, he determines that the sample mean is 10.00 mm with a sample standard deviation of ±0.10 mm. Moreover, he obtains the following frequency distribution for the diameters. Are the bearing diameters normally distributed at the 5% significance level?
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To determine if the bearing diameters are normally distributed at the 5% significance level, we can use a Chi-square goodness-of-fit test. Here's a step-by-step explanation:
1. Calculate the expected frequencies under the assumption of a normal distribution with a mean of 10.00 mm and a standard deviation of ±0.10 mm. You can use a standard normal distribution table or software to find the probabilities for each interval and then multiply these probabilities by the sample size (300) to obtain the expected frequencies.
2. Compare the observed frequencies (from the given frequency distribution) with the expected frequencies calculated in step 1.
3. Calculate the Chi-square statistic using the formula: χ² = Σ [(Observed frequency - Expected frequency)² / Expected frequency] for each interval.
4. Determine the degrees of freedom for the test. This is equal to the number of intervals minus one.
5. Find the critical value for the Chi-square distribution with the determined degrees of freedom and a significance level of 5%.
6. Compare the calculated Chi-square statistic with the critical value. If the calculated value is greater than the critical value, reject the null hypothesis and conclude that the bearing diameters are not normally distributed.
If the calculated value is less than the critical value, fail to reject the null hypothesis and conclude that there is not enough evidence to say that the bearing diameters are not normally distributed.
Following these steps will help you determine if the bearing diameters are normally distributed at the 5% significance level.
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2. Think about the relationships you observed and then answer these questions:
a. What the difference between the quantities that have vector drawings and the ones that don’t?
b. In your own words, what does "elastic collision" mean?
c. List quantities that have the same value (and direction if a vector) before and after the collision.
If a quantity has the same value (and direction if a vector), it is said to be "conserved"
d. What quantities are not "conserved"?
e. Run one more experiment to check your answer to 2c and 2d. Describe your experiment and
explain how it supports your answers.
3. Try some of the experiments again, varying the elasticity. Record your results in a similar data table, but
add a column for "elasticity".
4. Describe:
a. Any changes you need to make to your definition of "elastic collision" from 2b.
b. Adaptations to your ideas about quantities that are conserved when the elasticity is varied.
When you are finished compile all your images, your table, and your answers to the questions into one
document and submit it in Canvas.
This is a flash si
Answers
A. The difference between the quantities that have vector drawings and the ones that don't is that the vector drawings represent physical quantities that have a direction, such as velocity, momentum, and force.
What is vector drawings?Vector drawings are digital images created using mathematical equations that define geometric shapes, lines, curves, and other objects. Vector drawings are composed of paths, which are the lines, curves, and shapes that make up the artwork.
b. In an elastic collision, the total energy and momentum of the colliding objects remain the same before and after the collision.
c. Quantities that have the same value (and direction if a vector) before and after the collision are momentum, kinetic energy, and velocity.
d. Quantities that are not conserved are total energy, momentum, and angular momentum.
e. To check my answers, I could run another experiment using objects of different masses to see if the momentum and kinetic energy remain the same after the collision.
a. No changes need to be made to my definition of elastic collision from 2b.
b. When the elasticity is varied, the momentum and kinetic energy remain the same before and after the collision, but the total energy may not.
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In a skateboard race, the winner covered 26 meters in 3.88 seconds . What was his speed?
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An air-filled cavity for producing electromagnetic standing waves has two parallel, highly conducting walls separated by a distance L. One standing-wave pattern in the cavity produces nodal planes of the electric field with a spacing of 1.50 cm. The next-higher-frequency standing wave in the cavity produces nodal planes with a spacing of 1.25 cm. What is the distance L between the walls of the cavity?
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If higher-frequency standing wave in the cavity produces nodal planes with a spacing of 1.25 cm, the distance L between the walls of the cavity is 7.5cm.
The general formula for both ends closed is (n+1) λ/2
Let's say for n, \(\frac{\pi }{2} n = 1.5 = \pi = 3cm\)
for (n+1), λ(n+1)/2 = 1.25 = λ(n+1) = 2.5cm
∴ (n+1) (1.5) = (n+2)(1.25) = n = 4
∴ L = (4+1)(3/2) = L = 7.5cm.
Electromagnetic waves, or EM waves, are waves produced by oscillations between electric and magnetic fields. In other words, EM waves consist of oscillating magnetic and electric fields. Gamma rays have the highest frequencies in the electromagnetic spectrum.
Most sunrise alarm clocks are yellow to simulate the brightness of the sun. This energizing color is the color your body naturally associates with the morning. Colors like purple and red are often used for simulating the sunset and may be more conducive to falling asleep.
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A person in a kayak paddles down river at an average speed of 10 m/s. After 2 hours, how
far has she traveled? Hint: Change the speed to km/h first.
Answers
Answer:
72 km
Explanation:
The formula for Distance = Speed × Time
Speed = 10 m/s
1 metre per second = 3.6 kilometres per hour
10 metre per second =
10 metre per second = × 3.6 kilometres per hour/1 metre per second
= 36km/hr
How far he has travelled = Distance =
Speed × Time
= 36km/hr × 2 hrs
= 72 km
A horizontal line on a phase-change graph indicates
Answers
Answer:
A horizontal line on a phase change graph means there has been
no change. Often longer periods of research are needed to see
significant change.
If a fisherman applies a horizontal force with magnitude 51.5 n to the box and produces an acceleration of magnitude 3.00 m/s2, what is the mass of the box?
Answers
The mass of the box which the fisherman applies a horizontal force is 17.16 Kg
To solve this exercise, the formula and procedure to be applied are:
F = m * a
Where:
m = massF = Forcea = accelerationInformation about the problem:
a= 3.00 m/s2F= 51.5 N1 N = kg * m/s²m = ?Applying the force formula and isolating the mass, we get:
F = m * a
m = F/a
m = 51.5 Kg * m/s² / 3.00 m/s2
m = 17.16 Kg
What is force?It is the physical magnitude that expresses the effort necessary to move the mass of one kilogram with an acceleration of one meter per second squared, this is expressed in units of the international system in Newton.
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Advanced Wireless Services (AWS) is a wireless telecommunications spectrum band used for mobile voice and data services, video, and messaging. The AWS band uses frequencies in several segments: from 1695 to 2200 MHz. a) Determine the corresponding range of wavelengths used by the AWS mobile devices. b) To what region/band of electromagnetic spectrum does the range belong
Answers
Given data
*The given frequency is
\(f_1_{}=1695\text{ MHz}=1695\times10^6\text{ Hz}\)*The another given frequency is
\(f_2=2200\text{ MHz=}2200\times10^6\text{ Hz}\)*The given speed of light is c = 3.0 × 10^8 m/s
(a)
The formula for the wavelength is given as
\(\lambda_1=\frac{c}{f_1}\)Substitute the known values in the above expression as
\(\begin{gathered} \lambda_1=\frac{(3\times10^8)}{(1695\times10^6)^{}} \\ =0.176\text{ m} \end{gathered}\)The another wavelength for the another frequency is calculated as
\(\begin{gathered} \lambda_2=\frac{c}{f_2} \\ =\frac{(3.0\times10^8)}{(2200\times10^6)} \\ =0.136\text{ m} \end{gathered}\)(b)
Radio spectrum is an electromagnetic spectrum where the frequency range belongs of wireless telecommunication.
Starting from rest, a marble first rolls down a steeper hill, then down a less steep hill of the same height. For which is it going faster at the bottom?.
Answers
The first and second marble have the same speed at the bottom.
We need to know about the conservation of energy to solve this problem. The energy can not be created or destroyed, the only way is to change to another form. It can be written as
Ei = Ef
where Ei is initial energy and Ef is the final energy
The potential energy of the first marble is the same as the second marble because they have the same height. According to conservation of energy, the potential energy will be converted to kinetic energy and the speed of the marbles will same at the bottom.
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A box of books weighing 325 N is shoved across the floor by a force of 475 N exerted downward at an angle of 35° below the horizontal. (a) If ????k between the box and the floor is 0. 55, how long does it take to move the box 5. 80 m starting from rest? s (b) What is the maximum coefficient of friction between the box and the floor that allows the box to move from this applied force.
Answers
The net forces on the box acting perpendicular and parallel to the floor are
∑ F[perp] = F[normal] - 325 N + (475 N) sin(-35°) = 0
∑ F[perp] = (475 N) cos(-35°) - F[friction] = ma
where m is the mass of the box and a is its acceleration.
Solve for F[normal] :
F[normal] = 325 N + (475 N) sin(35°) ≈ 597 N
Then the frictional force has magnitude
F[friction] = 0.55 F[normal] ≈ 329 N
and so
60.5 N ≈ (325 N) a/g
(note that sin(-35°) = -sin(35°), cos(-35°) = cos(35°), and mg = 325 N so m = (325 N)/g)
Solve for a :
a = (60.5 N) / (325 N) g ≈ 1.82 m/s²
(a) Assuming this acceleration is constant, starting from rest, the box achieves a final velocity v such that
v² = 2a∆x
v² = 2 (1.82 m/s²) (5.80 m)
⇒ v ≈ 4.60 m/s
which happens in time t such that
v = at
4.60 m/s = (1.82 m/s²) t
⇒ t ≈ 0.177 s
(b) Let µ be the coefficient of static friction. The box just begins to slide if the magnitude of the parallel component of the applied force matches the magnitude of friction, i.e.
∑ F[para] = (475 N) cos(-35°) - F[friction] = 0
We have
F[friction] = µ F[normal] = (597 N) µ
so that
(597 N) µ = (475 N) cos(35°)
⇒ µ ≈ 0.651
Mr. Seifert needs to push a cardboard box down the hallway for Ms. Wang. The box has a mass of 40 kg and he is pushing it with an acceleration of 2 m/s/s. Because the cardboard does not slide easily, there is a friction force of 25 Newtons acting on the box to the LEFT. How much force is Mr. Seifert applying to the box to move it forward to the RIGHT?
Answers
Answer:
105N
Explanation:
force = mass x acceleration
force without the 25N = 40 x 2
= 80N
On top of this, he has to counteract the 25N so the actual force is 80N + 25N which is 105N
A ball is kicked off of a roof at 23 m/s [R 25° U]. What is the height of
the building if it lands 2.8 s later?
Answers
Answer:
Explanation:
Considering the fact that we ave been given an angle of inclination here, we best use it! That means that the velocity of 23 m/s is actually NOT the velocity we need; I tell my students that it is a "blanket" velocity but is not accurate in either the x or the y dimension of parabolic motion. In order to find the actual velocity in the dimension in which we are working, which is the y-dimension, we use the formula:
\(v_{0y}=v_0sin\theta\) and filling in:
\(v_{0y}=23sin(25)\) which gives us an upwards velocity of 9.7 m/s. So here's what we have to work with in its entirety:
\(v_{0y}=9.7m/s\)
a = -9.8 m/s/s
t = 2.8 seconds
Δx = ?? m
The one-dimensional motion equation that utilizes all of these variables is
Δx = \(v_0t+\frac{1}{2}at^2\) and filling in:
Δx = \(9.7(2.8)+\frac{1}{2}(-9.8)(2.8)^2\) I am going to do the math according to the correct rules of significant digits, so to the left of the + sign and to 2 sig fig, we have
Δx = 27 + \(\frac{1}{2}(-9.8)(2.8)^2\) and then to the right of the + sign and to 2 significant digits we have
Δx = 27 - 38 so
Δx = -11 meters. Now, we all know that distance is not a negative value, but what this negative number tells us is that the ball fell 11 meters BELOW the point from which it was kicked, which is the same thing as being kicked from a building that is 11 meters high.
How would doubling the mass of Earth affect the gravity we experience?