Answer:
B
Explanation:
Because the force has 2 components (horizontal and vertical), the horizontal component must be smaller than the total force. The Pythagorean theorem only adds positive values (because they're squared) so it makes sense. Using trigonometry, 100*cos(-20) yields a horizontal force of around 939.7N, which is less than 1000N.
two blocks with masses 2 kg and 4 kg are pushed from rest by the same amount of fore for a distance of 100 m on a frictionless floor. the final kinetic energy of the 2 kg block after the 100 m distance is
Answer:
the kinetic energy of the 2 kg mass after the 100 m is equal to 1962 J
Explanation:
mass of block A = 2 kg
mass of block B = 4 kg
distance the blocks were pushed = 100 m
NB: Blocks were pushed the same distance at the same equal time period. And the ground is without friction.
Work done in moving the 2 kg mass along the 100 m distance is,
work = force x distance moved
force exerted by the 2 kg mass = 2 x 9.81 m/s^2(acceleration due to gravity)
force = 19.62 N
therefore,
work done = 19.62 x 100 = 1962 Joules of work.
According to energy conservation principles, the kinetic energy impacted of the 2 kg mass through this distance will be equal to the work done in moving the 2 kg mass through this distance.
Therefore, the kinetic energy of the 2 kg mass after the 100 m is equal to 1962 J
In a Venn diagram, the separate circles contain characteristics unique to each compared and the intersection contains characteristics that are common to both items being compared. This Venn diagram compares the inner and outer planets. What belongs in the center section?
a. -Revolve around the Sun
-Rotate on an axis
-Generally have rings
b. -Revolve around the Sun
-Rotate on axis
-Generally have moons
c. -Rotate around the Sun
-Revolve on an axis
-Generally have moons
d. -Rotate around the Sun
-Revolve on an axis
-Generally have rings
Answer:
B. revolve around the sun
rotate on an axis
generally have moons
Explanation:
edge 2021
a research submarine what is the maximum depth it can go
Answer: 36, 200 feet deep according to information on google
Explanation:
A small submarine, the bathyscape Trieste, made it to 10,916 meters (35,813 feet) below sea level in the deepest point in the ocean, the Challenger Deep in the Marianas Trench, a few hundred miles east of the Philippines. This part of the ocean is 11,034 m (36,200 ft) deep, so it seems that a submarine can make it as deep as it's theoretically possible to go
A person sits on a freely spinning lab stool that has no friction in its axle. When this person extends her arms, A. her moment of inertia decreases and her angular speed decreases B. her moment of inertia decreases and her angular speed increases C. her moment of inertia increases and her angular speed decreases D. her moment of inertia increases and her angular speed decreases E. her moment of inertia increases and her angular speed remains the same.
Answer:
C. her moment of inertia increases and her angular speed decreases
D. her moment of inertia increases and her angular speed decreases
Explanation:
The moment of inertia of a body is the sum of the products of an increment of mass and the square of its distance from the center of rotation. When a spinning person extends her arms, part of her mass increases its distance from the center of rotation, so increases the moment of inertia.
The kinetic energy of a spinning body is jointly proportional to the moment of inertia and the square of the angular speed. Hence an increase in moment of inertia will result in a decrease in angular speed unless there is a change in the rotational kinetic energy.
This effect is used by figure skaters to increase their spin rate by drawing their arms and legs closer to the axis of rotation. Similarly, they can slow the spin by extending arms and legs.
When the person extends her arms, her moment of inertia increases and her angular speed decreases.
_____
Note to those looking for a letter answer
Both choices C and D have identical (correct) wording the way the problem is presented here. You will need to check carefully the wording in any problem you may think is similar.
PLEASE HELP !
Complete the following sentence. Choose the right answer from the given ones. The internal energy of the body can be changed A / B / C. A. only when the body is warmed or cooled B. when work is done on the body or heat flow C. only when the body does work
B
HOPE IT HELPS LET ME KNOW IF U NEED EXPLANATION
A lawn mower has a flat, rod-shaped steel blade that rotates about its center. The mass of the blade is 0.65 kg and its length is 0.55 m. You may want to review (Pages 314 - 318) . Part A What is the rotational energy of the blade at its operating angular speed of 3510 rpm
Complete Question
A lawn mower has a flat, rod-shaped steel blade that rotates about its center. The mass of the blade is 0.65 kg and its length is 0.55 m. You may want to review (Pages 314 - 318) .
Part A What is the rotational energy of the blade at its operating angular speed of 3510 rpm
Part B
If all of the rotational kinetic energy of the blade could be converted to gravitational potential energy, to what height would the blade rise?
Answer:
Part A
[tex]R = 1081 \ J[/tex]
Part B
[tex]h = 169.7 \ m[/tex]
Explanation:
From the question we are told that
The mass of the blade is [tex]m_b = 0.65 \ kg[/tex]
The length is [tex]l = 0.55 \ m[/tex]
The angular speed is [tex]w = 3510 rpm = 3510 * \frac{2 \pi }{60} = 367.6 \ rad/sec[/tex]
Generally the moment of inertia of the of this mower is mathematically evaluated as
[tex]I = \frac{m_b * l^2 }{12}[/tex]
substituting values
[tex]I = \frac{0.65 * 0.55^2 }{12}[/tex]
[tex]I = 0.016 \ kg m^2[/tex]
Generally the rotational kinetic energy of the bland is
[tex]R = \frac{1}{2} * I * w^2[/tex]
substituting values
[tex]R = \frac{1}{2} * 0.016 * 367.6^2[/tex]
[tex]R = 1081 \ J[/tex]
At point where the gravitational potential energy is equal to the rotational kinetic energy we have that
[tex]P = R = m_b * h * g[/tex]
Where P is the gravitational potential energy
substituting values
[tex]1081 = 0.65 * 9.8 * h[/tex]
=> [tex]h = 169.7 \ m[/tex]
The index of refraction of Sophia's cornea is 1.387 and that of the aqueous fluid behind the cornea is 1.36. Light is incident from air onto her cornea at an angle of 17.5° from the normal to the surface. At what angle to the normal is the light traveling in the aqueous fluid?
Answer:
17.85°
Explanation:
To find the angle to the normal in which the light travels in the aqueous fluid you use the Snell's law:
[tex]n_1sin\theta_1=n_2sin\theta_2[/tex]
n1: index of refraction of Sophia's cornea = 1.387
n2: index of refraction of aqueous fluid = 1.36
θ1: angle to normal in the first medium = 17.5°
θ2: angle to normal in the second medium
You solve the equation (1) for θ2, next, you replace the values of the rest of the variables:
[tex]\theta_2=sin^{-1}(\frac{n_1sin\theta_1}{n_2})\\\\\theta_2=sin^{-1}(\frac{(1.387)(sin17.5\°)}{1.36})=17.85\°[/tex]
hence, the angle to normal in the aqueous medium is 17.85°
Use the Bohr model to address this question. When a hydrogen atom makes a transition from the 5 th energy level to the 2nd, counting the ground level as the first,
A. What is the energy E of the emitted photon in electron volts?、
B. What is the wavelength in nanometers of the emitted photon?
C. What is the radius of the hydrogen atom in nanometers in its initial 5th energy level?
Answer:
A. 2.82 eV
B. 439nm
C. 59.5 angstroms
Explanation:
A. To calculate the energy of the photon emitted you use the following formula:
[tex]E_{n1,n2}=-13.4(\frac{1}{n_2^2}-\frac{1}{n_1^2})[/tex] (1)
n1: final state = 5
n2: initial state = 2
Where the energy is electron volts. You replace the values of n1 and n2 in the equation (1):
[tex]E_{5,2}=-13.6(\frac{1}{5^2}-\frac{1}{2^2})=2.82eV[/tex]
B. The energy of the emitted photon is given by the following formula:
[tex]E=h\frac{c}{\lambda}[/tex] (2)
h: Planck's constant = 6.62*10^{-34} kgm^2/s
c: speed of light = 3*10^8 m/s
λ: wavelength of the photon
You first convert the energy from eV to J:
[tex]2.82eV*\frac{1J}{6.242*10^{18}eV}=4.517*10^{-19}J[/tex]
Next, you use the equation (2) and solve for λ:
[tex]\lambda=\frac{hc}{E}=\frac{(6.62*10^{-34} kg m^2/s)(3*10^8m/s)}{4.517*10^{-19}J}=4.39*10^{-7}m=439*10^{-9}m=439nm[/tex]
C. The radius of the orbit is given by:
[tex]r_n=n^2a_o[/tex] (3)
where ao is the Bohr's radius = 2.380 Angstroms
You use the equation (3) with n=5:
[tex]r_5=5^2(2.380)=59.5[/tex]
hence, the radius of the atom in its 5-th state is 59.5 anstrongs
A) The energy E of the emitted photon in electron volts is; E = 2.856 eV
B) The wavelength in nanometers of the emitted photon is; λ = 434.4nm
C) The radius of the hydrogen atom in nanometers in its initial 5th energy level is; rₙ = 1.323 nm
A) Formula for the energy E of the emitted photons is;
E = -13.6([tex]\frac{1}{n_{2}^2} - \frac{1}{n_{1}^2}[/tex])
We are given;
n₂ = 5
n₁ = 2
Thus;
E = -13.6([tex]\frac{1}{5^2} - \frac{1}{2^2}[/tex])
E = 2.856 eV
B) The formula for the wavelength is;
λ = hc/E
where;
h is Planck's constant = 6.626 × 10⁻³⁴ m².kg/s
c is speed of light = 3 × 10⁸ m/s
E is energy of photon
λ is wavelength of the photon
Earlier we saw that E = 2.856 eV. Converting to Joules gives;
E = 4.5758 × 10⁻¹⁹ J
Thus;
λ = (6.626 × 10⁻³⁴ × 3 × 10⁸)/(4.5758 × 10⁻¹⁹)
λ = 4.344 × 10⁻⁷ m
Converting to nm gives;
λ = 434.4nm
C) Formula for the radius of the hydrogen atom is;
rₙ = n²a₀
where;
a₀ is bohr's radius = 5.292 × 10⁻¹¹ m
n = 5
Thus;
rₙ = 5² × 5.292 × 10⁻¹¹
rₙ = 1.323 × 10⁻⁹
rₙ = 1.323 nm
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Two cars start moving from the same point. One travels south at 60 miyh and the other travels west at 25 miyh. At what rate is the distance between the cars increasing two hours later?
Answer:
65 m/h
Explanation:
Let the distance of the car moving south be y.
Let the distance of the car moving west be x.
Let the distance between the two cars be a.
These three distances can be represented as a right angled triangle. So we can say:
[tex]a^2 = x^2 + y ^2[/tex]
Let us differentiate with respect to time, since the distances are changing with respect to time:
[tex]2a\frac{da}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt} \\\\=>a\frac{da}{dt} = x\frac{dx}{dt} + y\frac{dy}{dt}[/tex]__________(1)
da/dt = rate of change of distance between two cars
The speed of the car moving south (dy/dt) is 60 m/h and the speed of the car moving west (dx/dt) is 25 m/h.
Therefore:
dy/dt = 60 m/h and dx/dt = 25 m/h
After two hours, the distance of the two cars will be:
y = 2 * 60 = 120 miles
x = 2 * 25 = 50 miles
Therefore:
[tex]a^2 = 50^2 + 120^2\\\\a^2 = 2500 + 14400 = 16900\\\\a = \sqrt{16900}\\ \\a = 130 miles[/tex]
From (1):
130(da/dt) = 50(25) + 120(60)
130(da/dt) = 1250 + 7200 = 8450
da/dt = 8450/130 = 65 m/h
Therefore, after two hours, the distance between the two cars is changing at a rate of 65 m/h.
Sr-90 has a half-life of T1/2 = 2.85 a (years). How much Sr-90 will remain in a 5.00 g sample after 5.00 a? Show all of your work. (2 marks)
Answer:
1.48 g
Explanation:
A = A₀ (½)^(t / T)
where A is the final amount,
A₀ is the initial amount,
t is time,
and T is the half life.
A = (5.00 g) (½)^(5.00 a / 2.85 a)
A = 1.48 g
A water slide is constructed so that swimmers, starting from rest at the top of the slide, leave the end of the slide traveling horizontally. One person hits the water 5.00 m from the end of the slide in a time of 0.504 s after leaving the slide. Ignore friction and air resistance. Find the height H.
Answer:
4.93 m
Explanation:
According to the question, the computation of the height is shown below:
But before that first we need to find out the speed which is shown below:
As we know that
[tex]Speed = \frac{Distance}{Time}[/tex]
[tex]Speed = \frac{5}{0.504}[/tex]
= 9.92 m/s
Now
[tex]v^2 - u^2 = 2\times g\times h[/tex]
[tex]9.92^2 = 2\times 9.98 \times h[/tex]
98.4064 = 19.96 × height
So, the height is 4.93 m
We simply applied the above formulas so that the height i.e H could arrive
A person is swimming 1.1 m beneath the surface of the water in a swimming pool. A child standing on the diving board drops a ball into the pool directly above the swimmer. The swimmer sees the ball dropped from a height of 4.2 m above the water. From what height was the ball actually dropped?
Answer:
The actual height is [tex]A =3.158 \ m[/tex]
Explanation:
From the question we are told that
The depth of the person is [tex]d = 1.1 \ m[/tex]
The apparent height is [tex]D = 4.2 \ m[/tex]
Generally
The refractive index of water is [tex]n_w = 1.33[/tex]
The refractive index of the air is [tex]n_a = 1[/tex]
The apparent depth is mathematically represented as
[tex]D = A [\frac{n_w}{n_a} ][/tex]
substituting values
[tex]4.2 = A [\frac{1.33}{1} ][/tex]
=> [tex]A = \frac{4.2 }{1.33}[/tex]
[tex]A =3.158 \ m[/tex]
The ball was dropped at the height of "3.158 m". To understand the calculation, check below.
Refractive IndexAccording to the question,
Water's refractive index, [tex]n_w[/tex] = 1.33
Air's refractive index, [tex]n_a[/tex] = 1
Apparent height, D = 4.2 m
Person's depth, d = 1.1 m
We know the relation,
→ D = A[[tex]\frac{n_w}{n_a}[/tex]]
By substituting the values, we get
4.2 = A[[tex]\frac{1.33}{1}[/tex]]
By applying cross-multiplication,
A = [tex]\frac{4.2}{1.33}[/tex]
= 3.158 m
Thus the approach above is correct.
Find out more information about refractive index here:
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A student at another university repeats the experiment you did in lab. Her target ball is 0.860 m above the floor when it is in the target holder and the steel ball she uses has a mass of 0.0120 kg. She finds that the target ball travels a distance of 1.40 m after it is struck. Assume g = 9.80 m/s2. What is the kinetic energy (in joules) of the target ball just after it is struck?
Answer:
11.8 Joules
Explanation:
Given:-
- The height of the target ball, si = 0.860 m
- The mass of target and steel ball, m = 0.012 kg
- The target ball travels a distance ( x ) after being struck = 1.40 m
Find:-
What is the kinetic energy (in joules) of the target ball just after it is struck?
Solution:-
- We are given the initial distance of the target ball as 0.86 m above the floor which travels a distance ( x ) after being struck.
- We will employ the one dimensional kinematic equation of motion to determine the initial velocity ( vi ) of the target ball as follows:
[tex]vf^2 = vi^2 - 2*g*x[/tex]
Where,
vf: The final velocity of target ball at maximum height = 0
g: The gravitational acceleration constant = 9.8 m/s^2
- Plug in the required parameters and evaluate the ( vi ) as follows:
[tex]0^2 = vi^2 - 2*( 9.80 )*( 1.40 )\\\\vi^2 = 27.44\\\\vi = \sqrt{27.44} = 5.24 m/s[/tex]
- The kinetic energy ( Ek ) of an object with mass ( m ) and initial velocity ( vi ) is expressed as:
[tex]E_k = 0.5*m*(vi)^2\\\\E_k = 0.5*0.86*27.44\\\\E_k = 11.8 J[/tex]
Answer: The kinetic energy of the target ball just after it is struck is 11.8 Joules.
I need someone that knows physics. I have a test in 10 hrs and Im not good at it. Can anyone help me?
Answer:
I can help! What level of physics is it and what are your main topics?
510 g squirrel with a surface area of 935 cm2 falls from a 4.8-m tree to the ground. Estimate its terminal velocity. (Use the drag coefficient for a horizontal skydiver. Assume that the squirrel can be approximated as a rectanglar prism with cross-sectional area of width 11.6 cm and length 23.2 cm. Note, the squirrel may not reach terminal velocity by the time it hits the gr
Answer:
The terminal velocity is [tex]v_t =17.5 \ m/s[/tex]
Explanation:
From the question we are told that
The mass of the squirrel is [tex]m_s = 50\ g = \frac{50}{1000} = 0.05 \ kg[/tex]
The surface area is [tex]A_s = 935 cm^2 = \frac{935}{10000} = 0.0935 \ m^2[/tex]
The height of fall is h =4.8 m
The length of the prism is [tex]l = 23.2 = 0.232 \ m[/tex]
The width of the prism is [tex]w = 11.6 = 0.116 \ m[/tex]
The terminal velocity is mathematically represented as
[tex]v_t = \sqrt{\frac{2 * m_s * g }{\dho_s * C * A } }[/tex]
Where [tex]\rho[/tex] is the density of a rectangular prism with a constant values of [tex]\rho = 1.21 \ kg/m^3[/tex]
[tex]C[/tex] is the drag coefficient for a horizontal skydiver with a value = 1
A is the area of the prism the squirrel is assumed to be which is mathematically represented as
[tex]A = 0.116 * 0.232[/tex]
[tex]A = 0.026912 \ m^2[/tex]
substituting values
[tex]v_t = \sqrt{\frac{2 * 0.510 * 9.8 }{1.21 * 1 * 0.026912 } }[/tex]
[tex]v_t =17.5 \ m/s[/tex]
Some types of spiders build webs that consist of threads made of dry silk coated with a solution of a variety of compounds. This coating leaves the threads, which are used to capture prey, hygroscopic - that is, they attract water from the atmosphere. It has been hypothesized that this aqueous coating makes the threads good electrical conductors. To test the electrical properties of coated thread, researchers placed a 5-mm length of thread between two electrical contacts. The researchers stretched the thread in 1-mm increments to more than twice its original length, and then allowed it to return to its original length, again in 1-mm increments. Some of the resistance measurements are shown.If the conductivity of the thread results from the aqueous coating only, how does the cross-sectional area A of the coating compare when the thread is 13 mm long versus the starting length of 5 mm? Assume that the resistivity of the coating remains constant and the coating is uniform along the thread.If the conductivity of the thread results from the aqueous coating only, how does the cross-sectional area of the coating compare when the thread is 13 long versus the starting length of 5 ? Assume that the resistivity of the coating remains constant and the coating is uniform along the thread.A13mm is about 1/10 A5mm.A13mm is about 1/4 A5mm. === correct answer... I figured it out. R = pL/A. L is 2.5 times. Therefore, A must be 1/4 times.A13mm is about 2/5 A5mm.A13mm is the same as A5mm.
Answer:
A13 mm is about 1/4 A5 mm
Explanation:
Find the attachment
When you "crack" a knuckle, you suddenly widen the knuckle cavity, allowing more volume for the synovial fluid inside it and causing a gas bubble suddenly to appear in the fluid. The sudden production of the bubble, called "cavitation", produces a sound pulse---the cracking sound. Assume that the sound is transmitted uniformly in all directions and that it fully passes from the knuckle interior to the outside, at a distance of 0.29 m from your ear. If the pulse has a sound level of 61 dB at your ear, what is the rate at which energy is produced by the cavitation
Answer:If a wave y(x, t) = (6.0 mm) sin(kx + (600 rad/s)t + Φ) travels along a string, how much time does any given point on the string take to move between displacements y = +2.0 mm and y = -2.0 mm?
Explanation:
A light spring having a force constant of 115 N/m is used to pull a 9.00 kg sled on a horizontal frictionless ice rink. The sled has an acceleration of 2.10 m/s2. Part A By how much does the spring stretch if it pulls on the sled horizontally
Answer:
Stretch in the spring = 0.1643 (Approx)
Explanation:
Given:
Mass of the sled (m) = 9 kg
Acceleration of the sled (a) = 2.10 m/s ²
Spring constant (k) = 115 N/m
Computation:
Tension force in the spring (T) = ma
Tension force in the spring (T) = 9 × 2.10
Tension force in the spring (T) = 18.9 N
Tension force in the spring = Spring constant (k) × Stretch in the spring
18.9 N = 115 N × Stretch in the spring
Stretch in the spring = 18.9 / 115
Stretch in the spring = 0.1643 (Approx)
A 18-kg sled is being pulled along the horizontal snow-covered ground by a horizontal force of 30 N. Starting from rest, the sled attains a speed of 2.0 m/s in 8.5 m. Find the coefficient of kinetic friction between the runners of the sled and the snow. Num
Answer:
Coefficient of kinetic friction = 0.146
Explanation:
Given:
Mass of sled (m) = 18 kg
Horizontal force (F) = 30 N
FInal speed (v) = 2 m/s
Distance (s) = 8.5 m
Find:
Coefficient of kinetic friction.
Computation:
Initial speed (u) = 0 m/s
v² - u² = 2as
2(8.5)a = 2² - 0²
a = 0.2352 m/s²
Nweton's law of :
F (net) = ma
30N - μf = 18 (0.2352)
30 - 4.2336 = μ(mg)
25.7664 = μ(18)(9.8)
μ = 0.146
Coefficient of kinetic friction = 0.146
Q) A particle in simple harmonic motion starts its motion from its mean position. If T be the time period, calculate the ratio of kinetic energy and potential energy of the particle at the instant when t = T/12.
t\12 and the parties are spreading ever
Explanation:
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Penny is adjusting the position of a stand up piano of mass mp = 150 kg in her living room. The piano is lp = 1.35 m in length. The piano is currently at an angle of θp = 36 degrees to the wall. Penny wants to rotate the piano across the carpeted floor so that it is flat up against the wall. To move the piano, Penny pushes on it at the point furthest from the wall. This piano does not have wheels, so you can assume that the friction between the piano and the rug acts at the center of mass of the piano.
Required:
a. Write an expression for the minimum magnitude of the force FS in N Penny needs to exert on the piano to get it moving. Assume the corner of the piano on the wall doesn't slide and the static friction between the rug and the piano is µs.
b. The coefficient of kinetic friction between the carpet and the piano is uk = 0.27. Once the piano starts moving, calculate the torque τ in N·m that Penny needs to apply to keep moving the piano at a constant angular velocity.
c. Calculate the amount of work Wp, in J Penny does on the piano as she rotates it.
Answer:
a) Fs = (μs*mp*g)/2 | b) τ = Fs*lp | c) Wτ,constant = τΘ
Explanation:
a) Fs = (μs*mp*g)/2
b) τ = Fs*lp
c) Wτ,constant = τΘ
A jeep starts from rest with a constant acceleration of 4m/s2.At the same time a car travels with a constant speed of 36km/h overtake and passes the jeep how far beyond the starting point will the jeep overtakes the car?
Answer:
25m
Explanation:
Let's assume the Jeep attains a velocity of 36km/h ; a constant speed same with that of the car.
While the Jeep is accelerating to that speed, the car with that speed passes it.
Now we can calculate the time taken for the Jeep to attain the velocity of 36km/h on her constant acceleration.
This time is t = v/a; from Newton's Law of Motion:
a = V-U / t ; a-acceleration
V is final velocity = 36km/h
U is initial velocity 0 since the body starts from rest.
Hence t = 36000/3600 ÷ 4 = 2.5s
Note conversting from km/h to m/s we multiply by 1000/3600.
But the distance covered by the car while the Jeep just accelerates is
S = U × t = 10× 2.5 = 25m.
Note From Newton's law of Motion, distance for constant speed is defined as: U × t
Hence the Car would be 25m off the starting point just as the Jeep accelerates. It would overtake the Jeep when it just covers 25m from the Jeep starting point.
During last year’s diving competition, the divers always pull their limbs in and curl up their bodies when they do flips. Just before entering the water, they fully extend their limbs to enter straight down as shown. Explain the effect of both actions on their angular velocities and kinetic energy (support your answer with working). Also explain the effect on their angular momentum.
Answer:
the angular speed of the person increases, being able to make more turns and faster.
K₂ = K₁ I₁ / I₂
Explanation:
When the divers are turning the system is isolated, so all the forces are internal and therefore also the torque, therefore the angular momentum is conserved
initial, joint when starting to turn
L₀ = I₁ w₁
final. When you shrink your arms and legs
Lf = I₂ w₂
L₀ = Lf
I₁ w₁ = I₂ w₂
when you shrink your arms and legs the distance to the turning point decreases and since the moment of inertia depends on the distance squared, the moment of inertia also decreases
I₂ <I₁
w₂ = I₁ / I₂ w₁
therefore the angular speed of the person increases, being able to make more turns and faster.
When it goes into the water it straightens the arm and leg, so the moment of inertia increases
I₁> I₂
w₁ = I₂ / I₁ w₂
therefore we see that the angular velocity decreases, therefore the person trains the water like a stone and can go deeper faster.
In both cases the kinetic energy is
K = ½ I w²
the initial kinetic energy is
K₁ = ½ I₁ w₁²
the final kinetic energy is
K₂ = ½ I₂ w₂²
we substitute
K₂ = ½ I₂ (I₁ / I₂ w1² 2
K₂ = ½ I₁² / I₂ w₁² = (½ I₁ w₁²) I₁ / I₂
K₂ = K₁ I₁ / I₂
therefore we see that the kinetic energy increases by factor I₁/I₂
The starships of the Solar Federation are marked with the symbol of the Federation, a circle, whereas starships of the Denebian Empire are marked with the Empire's symbol, an ellipse whose major axis is n times its minor axis (a=nb in the figure ).
How fast, relative to an observer, does an Empire ship have to travel for its markings to be confused with those of a Federation ship? Use c for the speed of light in a vacuum.
Express your answer in terms of n and c.
Complete question
The complete question is shown on the first uploaded image
Answer:
The velocity is [tex]v = c* \sqrt{1 - \frac{1}{n^2} }[/tex]
Explanation:
From the question we are told that
a = nb
The length of the minor axis of the symbol of the Federation, a circle, seen by the observer at velocity v must be equal to the minor axis(b) of the Empire's symbol, (an ellipse)
Now this length seen by the observer can be mathematically represented as
[tex]h = t \sqrt{1 - \frac{v^2}{c^2} }[/tex]
Here t is the actual length of the major axis of of the Empire's symbol, (an ellipse)
So t = a = nb
and b is the length of the minor axis of the symbol of the Federation, (a circle) when seen by an observer at velocity v which from the question must be the length of the minor axis of the of the Empire's symbol, (an ellipse)
i.e h = b
So
[tex]b = nb [\sqrt{1 - \frac{v^2}{c^2} } ][/tex]
[tex][\frac{1}{n} ]^2 = 1 - \frac{v^2}{c^2}[/tex]
[tex]v^2 =c^2 [1- \frac{1}{n^2} ][/tex]
[tex]v^2 =c^2 [\frac{n^2 -1}{n^2} ][/tex]
[tex]v = c* \sqrt{1 - \frac{1}{n^2} }[/tex]
A 110-kg football player running at 8.00 m/s catches a 0.410-kg football that is traveling at 25.0 m/s. Assuming the football player catches the ball with his feet off the ground with both of them moving horizontally, calculate: the final velocity if the ball and player are going in the same directio
Answer:[tex]8.062\ m/s[/tex]
Explanation:
Given
masss of football player [tex]M=110\ kg[/tex]
Velocity of football player [tex]u_1=8\ m/s[/tex]
mass of football [tex]m=0.41\ kg[/tex]
velocity of football [tex]u_2=25\ m/s[/tex]
Final velocity will be given by applying conservation of linear momentum
After catching the ball Player and ball moves with same velocity
[tex]\Rightarrow Mu_1+mu_2=(M+m)v[/tex]
[tex]\Rightarrow 110\times 8+0.41\times 25=(110+0.41)v[/tex]
[tex]\Rightarrow 880+10.25=110.41\times v[/tex]
[tex]\Rightarrow v=\frac{890.25}{110.41}=8.063\ m/s[/tex]
So, final velocity will be [tex]8.062\ m/s[/tex]
Consider a system of a cliff diver and the earth. The gravitational potential energy of the system decreases by 24,500 J as the diver drops to the water from a height of 44.0 m. Determine her weight in newtons.
Before she jumped from the cliff, her gravitational potential energy was
GPE = (her weight) x (height of the cliff) .
That's exactly the GPE that she lost on the way down to the water. So we can write
24,500 J = (her weight) x (44.0 m)
Divide each side by 44.0 m:
Her weight = 24,500 J / 44 m
Her weight = 556.8 Newtons
(about 125 pounds)
A string is stretched between fixed supports separated by 72.0 cm. It is observed to have resonant frequencies of 370 and 555 Hz, and no other resonant frequencies between these two.(a) What is the lowest resonant frequency for this string?(b) What is the wave speed for this string?
Answer:
(a) f = 185 Hz
(b) v = 266.4 m/s
Explanation:
(a) The lowest frequency can be calculated by using the following formula for the calculation of the modes (resonant frequencies) in a string:
[tex]f_n=\frac{nv}{2L}[/tex]
[tex]f_n=nf[/tex]
n: order of the mode
v: velocity of the waves in the string
L: length of the string = 72.0cm = 0.72m
fn: frequency of the n-th mode
With the information about two consecutive modes you can find the lowest resonant frequency. First you find the resonant mode n:
[tex]f_n=nf\\\\f_{n-1}=(n-1)f\\\\\frac{f_n}{f_{n-1}}=\frac{n}{n-1}[/tex]
you solve the previous equation for n:
[tex](n-1)f_n=nf_{n-1}\\\\555n-555=370n\\\\n=3[/tex]
With this information you can calculate the lowest resonant frequency:
[tex]f_n=nf\\\\f=\frac{f_n}{n}=\frac{555}{3}=185Hz[/tex]
b) You have information about two consecutive modes fn, fn-1. Then, you can calculate the velocity of the waves:
[tex]f_{n}-f_{n-1}=n\frac{v}{2L}-(n-1)\frac{v}{2L}\\\\f_n-f_{n-1}=\frac{v}{2L}\\\\v=2L(f_n-f_{n-1})[/tex]
fn = 555 Hz
fn-1: 370 Hz
[tex]v=2(0.72m)(555-370)Hz=266.4\frac{m}{s}[/tex]´
hence, the velocityof the waves in the string is 266.4 m/s
A flock of ducks is trying to migrate south for the winter, but they keep being blown off course by a wind blowing from the west at 5.0 m/s . A wise elder duck finally realizes that the solution is to fly at an angle to the wind.If the ducks can fly at 7.0 m/s relative to the air, what direction should they head in order to move directly south?
The ducks' flight path as observed by someone standing on the ground is the sum of the wind velocity and the ducks' velocity relative to the wind:
ducks (relative to wind) + wind (relative to Earth) = ducks (relative to Earth)
or equivalently,
[tex]\vec v_{D/W}+\vec v_{W/E}=\vec v_{D/E}[/tex]
(see the attached graphic)
We have
ducks (relative to wind) = 7.0 m/s in some direction θ relative to the positive horizontal direction, or[tex]\vec v_{D/W}=\left(7.0\dfrac{\rm m}{\rm s}\right)(\cos\theta\,\vec\imath+\sin\theta\,\vec\jmath)[/tex]
wind (relative to Earth) = 5.0 m/s due East, or[tex]\vec v_{W/E}=\left(5.0\dfrac{\rm m}{\rm s}\right)(\cos0^\circ\,\vec\imath+\sin0^\circ\,\vec\jmath)[/tex]
ducks (relative to earth) = some speed v due South, or[tex]\vec v_{D/E}=v(\cos270^\circ\,\vec\imath+\sin270^\circ\,\vec\jmath)[/tex]
Then by setting components equal, we have
[tex]\left(7.0\dfrac{\rm m}{\rm s}\right)\cos\theta+5.0\dfrac{\rm m}{\rm s}=0[/tex]
[tex]\left(7.0\dfrac{\rm m}{\rm s}\right)\sin\theta=-v[/tex]
We only care about the direction for this question, which we get from the first equation:
[tex]\left(7.0\dfrac{\rm m}{\rm s}\right)\cos\theta=-5.0\dfrac{\rm m}{\rm s}[/tex]
[tex]\cos\theta=-\dfrac57[/tex]
[tex]\theta=\cos^{-1}\left(-\dfrac57\right)\text{ OR }\theta=360^\circ-\cos^{-1}\left(-\dfrac57\right)[/tex]
or approximately 136º or 224º.
Only one of these directions must be correct. Choosing between them is a matter of picking the one that satisfies both equations. We want
[tex]\left(7.0\dfrac{\rm m}{\rm s}\right)\sin\theta=-v[/tex]
which means θ must be between 180º and 360º (since angles in this range have negative sine).
So the ducks must fly (relative to the air) in a direction 224º relative to the positive horizontal direction, or about 44º South of West.
An airplane flies 2500 miles east in 245 seconds what is the velocity of the plane?
Speed = (distance) / (time)
Speed = (
Velocity = speed, and its direction
The velocity of the plane is 10.2 miles per second East.
(about 48 times the speed of sound)
Escaping from a tomb raid gone wrong, Lara Croft (m = 62.0 kg) swings across an alligator-infested river from an 8.80-m-long vine. If her speed at the bottom of the swing is 6.30 m/s and she makes it safely across the river, what is the minimum breaking strength of the vine?
Answer:
887.2 N
Explanation:
Given that
Mass of Lara, m = 62 kg
Length of the vine, l = 8.8 m
Speed of the swing, v = 6.3 m/s Then,
We start by calculating her weight.
w = mass * acceleration
w = mg
w = 62 * 9.8
w = 607.6 N
F(c) = mv²/l, on substituting
F(c) = (62 * 6.3²) / 8.8
F(c) = 2460.78 / 8.8
F(c) = 279.6 N
T - 607.6 N = 279.6 N
T = 607.6 N + 279.6 N
T = 887.2 N
Thus, the minimum breaking strength of the vine is 887.2 N
The minimum breaking strength of the vine is about 900N.
TensionTension is a force developed by a rope, string, or cable when stretched under an applied force.
Given:
Mass (m) = 62 kg, length (l) = 8.8 m, velocity (v) = 6.3 m/s, g = 10 m/s².
Weight(W) = m * g = 62 * 10 = 620N
centripetal force = F(c) = mv²/l = 62 * 6.3² / 8.8 = 279.6N
T - W = F(c)
T - 620 = 279.6
T = 900N
The minimum breaking strength of the vine is about 900N.
Find out more on Tension at: https://brainly.com/question/24994188
