Answer:
The work done by the friction force is - 139927.5 J.
Explanation:
mass of diver, m = 45 kg
distance falls, h = 450 m
initial speed, u = 0 m/s
final speed = 51 m/s
According to the work energy theorem,
Work done by the gravity + work done by the friction force = change in kinetic energy
[tex]m g h + W' = 0.5 m ()v^2 - u^2)\\\\45\times 9.8\times 450 + W' = 0.5\times 45\times (51^2 - 0)\\\\198450 + W' = 58522.5\\\\W' = - 139927.5 J[/tex]
A wave has a frequency of 87.00 Hz and has a wavelength of 74.62 m. What is its
velocity?
Answer:
v = 6491.94 m/s
Explanation:
We are given;
Frequency; f = 87 Hz
Wavelength;λ = 74.62 m
Formula for velocity(v) of waves from the wave equation is;
v = fλ
Thus;
v = 87 × 74.62
v = 6491.94 m/s
Need an answer in hurry u can make the pic big
Give the number of protons and the number of neutrons in the nucleus of each of the following isotopes Aluminum 25 :13 protons and 12 neutrons
Answer:
No of proton is 13 and nucleus is 13
Bola bermassa 200 gram dilempar
ke bawah dari ketinggian 20 m
dengan kecepatan 2 m/s. Jika
percepatan gravitasi bumi 10
m/s2 energi kinetik pada
ketinggian 8 m adalah ......
Answer:
0.4
Explanation:
[tex] \frac{1}{2} mv ^{2} [/tex]
kinetic energy formula , potential energy is not considered
0.5×0.2×2×2
The peak value of the electric field component of an electromagnetic wave is E. At a particular instant, the intensity of the wave is of 0.020 W/m2. If the electric field were increased to 5E, what would be the intensity of the wave?
Answer:
[tex]I_2=0.50 w/m^2[/tex]
Explanation:
From the question we are told that:
initial Intensity [tex]I_1=0.020 w/m^2[/tex]
Final Electric field [tex]E_2=5E[/tex]
Generally the equation for Relation ship between intensity and Electric field is mathematically given by
[tex]\frac{I_1}{I_2}= \frac{E_1^2}{E_2^2}[/tex]
Therefore
[tex]I_2=\frac{I_1}{ \frac{E_1^2}{E_2^2}}[/tex]
[tex]I_2=\frac{0.020}{ \frac{E^2}{5E^2}}[/tex]
[tex]I_2=0.50 w/m^2[/tex]
A 7.5-kg rock and a 8.9 × 10-4-kg pebble are held near the surface of the earth. (a) Determine the magnitude of the gravitational force exerted on each by the earth. (b) Calculate the magnitude of the acceleration of each object when released.
Answer:
F' = 73.7 N
F = 8.749×10⁻³ N
a' = a = 9.83 m/s²
Explanation:
(a)
For the rock
Applying
F' = Gm'm/r²................... Equation 1
Where F = magnitude of the gravitational force on the rock, G = Gravitational constant, m' = mass of the rock, m = mass of the earth, r = radius of the earth.
From the question,
Given: m' = 7.5 kg
Constant: m = 5.98×10²⁴ kg, G = 6.67×10⁻¹¹ Nm²/kg², r = 6.37×10⁶ m
Substitute these values into equation 1
F' = 6.67×10⁻¹¹ (7.5)(5.98×10²⁴)/(6.37×10⁶)²
F' = 7.37×10¹ N
F' = 73.7 N
Also, For the pebble,
F = GMm/r².............. Equation 2
Where M = mass of the pebble, F = Gravitational force exerted on the pebble by the earth
Given: M = 8.9×10⁻⁴ kg,
Substitute into equation 2
F = 6.67×10⁻¹¹(8.9×10⁻⁴)(5.98×10²⁴)/(6.37×10⁶)²
F = 8.749×10⁻³ N
(b)
For the rock,
a' = F'/m'
Where a' = magnitude of the acceleration of the rock
a' = 73.7/7.5
a' = 9.83 m/s²
For the pebble,
a = F/M
Where a = acceleration of the pebble
a = (8.749×10⁻³)/(8.9×10⁻⁴)
a = 9.83 m/s²
A small object with mass 0.200 kg moves with constant speed in a vertical circle of radius 0.500 m. It takes the object 0.500 s to complete one revolution. (a) What is the translational speed of the object
Answer:
6.28 m/s.
Explanation:
Given that,
The mass of the object, m = 0.2 kg
The radius of the circle, r = 0.5 m
It takes the object 0.500 s to complete one revolution.
We need to find the translational speed of the object. Let it is v. We know that,
[tex]v=\dfrac{2\pi r}{t}\\\\v=\dfrac{2\pi \times 0.5}{0.5}\\\\v=6.28\ m/s[/tex]
So, the transalational speed of the object is 6.28 m/s.
The laboratory exercise for this chapter addresses kinematic motion. You have experienced these motions in everyday life. Instead of a discussion board requiring posts in a forum, this assignment has been modified to accept your response to the following questions in this assignment. Be sure to clearly address each of the points below and show all of your work • What is the difference between a vector and a scalar quantity? • List two examples each of vector and scalar quantities • Write the due date of this assignment: Month and Day (For example, July 15 would be Month - 7. Day = 15) • For a building having a height equal to the quantity you have recorded for Day in meters in our example 15 meters), compute the time required for the ball to fall to the ground while experiencing acceleration due to gravity (g=9.8m/s) • How fast was the ball traveling when it hit the ground? Submit the kinetics Assignment by 11:59 p.m. (ET) on Monday,
Answer:
A) vectors: veloicty, force
scalar: speed, work
B) t = 1.75 s, C) v = - 17 2 m / s
Explanation:
We answer each part separately
A) A vector magnitude has magnitude and direction instead a scalar magnitude has only magnitude
vector quantities: the speed of a car number is the magnitude and direction is where it goes
Force, the number is the magnitude and above that applies gives direction
Scalar magnitude: how quickly the number of the speedometer of the car
Temperature, work
B) I = 15 m height to the soil and get to calculate time = 0
y = y₀ + v₀ t - ½ g t²
as the ball is loose its initial velocity is zero
0 = 0 +0 - ½ g t²
t = [tex]\sqrt{2y_o/g}[/tex]
t = [tex]\sqrt{2 \ 15/ 9.8}[/tex]
t = 1.75 s
C) the velocity to the reach the floor
v = vo - g t
v = 0 - g t
v = - 9.8 1.74
v = - 17 2 m / s
The negative signt iindicates that the speed goes down
According to ____________ , the randomness of the universe is constantly increasing.
a. The first law of thermodynamics
b. The zeroth law of thermodynamics
c. The second law of thermodynamics
Answer:
According to " The second law of thermodynamics", the randomness of the universe is constantly increasing?
Explanation:
So answer option C. Have a great summer.
Which statements describe using genetic factors to influence the growth of organisms? Select the three (3) that apply.
-increasing use of hybrid crops
-altering genes in DNA to create new plants
-increasing human population
-increasing climate change
-developing disease or pest resistant crops
Answer:
- increasing use of hybrid crops
- altering genes in DNA to create new plants
- developing disease or pest resistant crops
Explanation:
The use of genetic factors to influence the growth of a plant encompasses manipulating the genetic constituent (gene) of such plant.
For example,
- Increasing use of hybrid crops entails mating two pure bred plants based on a gene of interest responsible for a particular trait, to form a hybrid.
- Altering genes in DNA to create new plants is also a genetic factor as it has to with gene modification.
- developing disease or pest resistant crops means that the genetic make up of such plant has been modified to be resistant to pest/disease.
The two most prominent wavelengths in the light emitted by a hydrogen discharge lamp are 656 nm (red) and 486 nm (blue). Light from a hydrogen lamp illuminates a diffraction grating with 550 lines/mm , and the light is observed on a screen 1.7 m behind the grating.
What is the distance between the first-order red and blue fringes?
Express your answer to two significant figures and include the appropriate units.
Answer:
Δd = 7.22 10⁻² m
Explanation:
For this exercise we must use the dispersion relationship of a diffraction grating
d sin θ = m λ
let's use trigonometry
tan θ = y / L
how the angles are small
tant θ = sinθ /cos θ = sin θ
we substitute
sin θ = y / L
d y / L = m λ
y = m λ L / d
let's use direct ruler rule to find the distance between two slits
If there are 500 lines in 1 me, what distance is there between two lines
d = 2/500
d = 0.004 me = 4 10⁻⁶ m
diffraction gratings are built so that most of the energy is in the first order of diffraction m = 1
let's calculate for each wavelength
λ = 656 nm = 656 10⁻⁹ m
d₁ = 1 656 10⁻⁹ 1.7 / 4 10⁻⁶
d₁ = 2.788 10⁻¹ m
λ = 486 nm = 486 10⁻⁹ m
d₂ = 1 486 10⁻⁹ 1.7 / 4 10⁻⁶
d₂ = 2.066 10⁻¹ m
the distance between the two lines is
Δd = d1 -d2
Δd = (2,788 - 2,066) 10⁻¹
Δd = 7.22 10⁻² m
In the Biomedical and Physical Sciences building at MSU there are 135 steps from the ground floor to the sixth floor. Each step is 16.6 cm tall. It takes 5 minutes and 30 seconds for a person with a mass of 73.5 kg to walk all the way up. How much work did the person do?
Answer:
W = 16.4 kJ
Explanation:
Given that,
There are 135 steps from the ground floor to the sixth floor.
Each step is 16.6 cm tall.
The mass of a person, m = 73.5 kg
We need to find the work done by the person. We know that,
Work done = Fd
Where
d is the displacement, d = 135 × 0.166 = 22.41
So,
W = 73.5 × 10 × 22.41
= 16471.35 J
or
W = 16.4 kJ
So, 16.4 kJ is the work done by the person.
Which of the statements below are TRUE! Group of answer choices The carbon rod in batteries react to form a carbon cation. A good car battery gives you a little bit of power for a long period of time. A good car battery gives you a lot of power in a short period of time. The carbon rod in batteries is used as an inert electrode.
Answer:
The carbon rod in batteries is used as an inert electrode
Explanation:
A battery is considered as a power source that consists of one or more electrochemical cells having an external connections to provide power to the electrical devices such as the lights, bulbs, fans, mobile phones, etc.
It contains a positive terminal and a negative terminal.
The carbon rod in the battery does not help in the electrochemical reactions. It acts as an inert electrode and helps to flow the electrons only.
Thus the true statement is :
The carbon rod in batteries is used as an inert electrode.
How does gravity affect your ability to live on a planet?
find the resistance of wire of
0.65m Radius 0.25
and
resistivity 3x10-6 OHM
Complete Question:
Find the resistance of a wire of length 0.65 m, radius 0.25 mm and resistivity 3 * 10^{-6} ohm-metre.
Answer:
Resistance = 9.95 Ohms
Explanation:
Given the following data;
Length = 0.65 m
Radius = 0.25 mm to meters = 0.00025 m
Resistivity = 3 * 10^{-6} ohm-metre.
To find the resistance of the wire;
Mathematically, resistance is given by the formula;
[tex] Resistance = P \frac {L}{A} [/tex]
Where;
P is the resistivity of the material. L is the length of the material.A is the cross-sectional area of the material.First of all, we would find the cross-sectional area of the wire.
Area of circle = πr²
Substituting into the equation, we have;
Area = 3.142 * (0.00025)²
Area = 3.142 * 6.25 * 10^{-8}
Area = 1.96 * 10^{-7} m²
Now, to find the resistance of the wire;
[tex] Resistance = 3 * 10^{-6} * \frac {0.65}{1.96 * 10^{-7}} [/tex]
[tex] Resistance = 3 * 10^{-6} * 3316326.531 [/tex]
Resistance = 9.95 Ohms
Who Knows?
Thank You
Answer:
Use below table to read the color codes.
The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have studied the change in length of the knee extensor tendon in sprinters and nonathletes. They find (on average) that the sprinters' tendons stretch 40 mm, while nonathletes' stretch only 32 mm. The spring constant for the tendon is the same for both groups, 32 N/mm. What is the difference in maximum stored energy between the sprinters and the nonathlethes?
Answer:
Explanation:
From the given information:
The difference in the maximum energy stored is can be determined by finding the difference in the maximum stored energy in the sprinters and that of the non-athlete:
[tex]\Delta U = \dfrac{1}{2}k(x_2^2 - x_1^2)[/tex]
[tex]\Delta U = \dfrac{1}{2} (32 \ N/mm) (\dfrac{1 \ mm}{10^{-3} \ m}) ((40\times 10^{-3})^2 - (32\times 10^{-3})^2)[/tex]
[tex]\Delta U =16000 \times (5.76\times 10^{-4})[/tex]
[tex]\mathbf{\Delta U =9.216\ J}[/tex]
An electric drill starts from rest and rotates with a constant angular acceleration. After the drill has rotated through a certain angle, the magnitude of the centripetal acceleration of a point on the drill is 8.2 times the magnitude of the tangential acceleration. What is the angle?
Answer:
The angle is 4.1 rad.
Explanation:
The centripetal acceleration (α) is given by:
[tex] \alpha = \omega^{2} r [/tex] (1)
Where:
ω: is the angular velocity
r: is the radius
And the tangential acceleration (a) is:
[tex] a = \alpha r [/tex] (2)
Since the magnitude of "α" is 8.2 times the magnitude of "a" (equating (2) and (1)) we have:
[tex] \omega^{2} r = 8.2\alpha r [/tex]
[tex] \omega^{2} = 8.2\alpha [/tex] (3)
Now, we can find the angle with the following equation:
[tex] \omega_{f}^{2} = \omega_{0}^{2} + 2\alpha \Delta \theta [/tex]
Where:
[tex] \omega_{f}[/tex]: is the final angular velocity [tex] \omega_{0}[/tex]: is the initial angular velocity = 0 (it starts from rest)
[tex]\Delta \theta[/tex]: is the angle
[tex] \omega^{2} = 2\alpha \Delta \theta [/tex] (4)
By entering equation (3) into (4) we can calculate the angle:
[tex] 8.2\alpha = 2\alpha \Delta \theta [/tex]
[tex] \Delta \theta = 4.1 rad [/tex]
Therefore, the angle is 4.1 rad.
I hope it helps you!
In air an object weighs 15N, when immersed in water it weighs 12N, when immersed in another liquid, it weighs 13N, Calculate the density of the object and that of the other liquid?
M1 = 15/g = 15/9.8 = 1.53 kg = mass of object in air. M2 = 12/9.8 = 1.22 kg = mass of object immersed. M1-M2 = 1.53-1.22 = 0.31 kg lost by object = mass of water displaced. ... Do = 4.94 g/cm^3 = density of object.
Which phenomenon occurs when one wave is superimposed on another?
A. Interference
B. Refraction
C. Diffraction
D. Polarization
Answer:Alternativa A. Damos o nome de interferência a superposição de efeitos que ocorre ao ser produzido dois pulsos de onda, que serão propagados e acabarão inevitavelmente por se encontrar. No instante em que os pulsos se cruzarem, há então, uma superposição de efeitos individuais de cada um deles. Se durante o cruzamento, houver um reforço das ondas, estará ocorrendo a este fenômeno.
Identify each statement as an example of melting or sublimation,
lodide produces fumes when heated.
Melting
Sublimation
An iceberg turns to ocean water.
Candle wax turns to liquid when hot.
"Fog" is created from dry ice.
1st is sublimation
2and is melting
3red is melting
4th is sublimation
sublimation is just "skipping" the liquid phase / state
You are using a constant force to speed up a toy car from an initial speed of 6.5 m/s
to a final speed of 22.9 m/s. If the toy car has a mass of 340 g, what is the work
needed to speed this car up?
By the work-energy theorem, the total work done on the car is equal to the change in its kinetic energy:
W = ∆K
W = 1/2 (0.34 kg) (22.9 m/s)² - 1/2 (0.34 kg) (6.5 m/s)²
W ≈ 82 J
what is the suitable way of using social media
Answer:
not using it too much and getting addicted
Explanation:
a 230 kg roller coaster reaches the top of the steepest hill with a speed of 6.2 km/h. It then descends the hill, which is at an angle of 45 and is 51.0 m long/ What will its kinetic energy be wehn it reaches the bottom
Answer: 81.619 kJ
Explanation:
Given
Mass of roller coaster is [tex]m=230\ kg[/tex]
It reaches the steepest hill with speed of [tex]u=6.2\ km/h\ or \ 1.72\ m/s[/tex]
Hill to bottom is 51 m long with inclination of [tex]45^{\circ}[/tex]
Height of the hill is [tex]h=51\sin 45^{\circ}=36.06\ m[/tex]
Conserving energy to get kinetic energy at bottom
Energy at top=Energy at bottom
[tex]\Rightarrow K_t+U_t=K_b+U_b\\\Rightarrow \dfrac{1}{2}mu^2+mgh=K_b+0\\\\\Rightarrow K_b=0.5\times 230\times 1.72^2+230\times 9.8\times 36.06\\\Rightarrow K_b=340.216+81,279.24\\\Rightarrow K_b=81,619.456\ J\\\Rightarrow K_b=81.619\ kJ[/tex]
(c) Two argon atoms form the molecule Ar2 as a result of a van der Waals interaction with U0= 1.68×10-21 J and R0= 3.82×10-10 m. Find the frequency of small oscillations of one Ar atom about its equilibrium position.
Answer:
Explanation:
Answer:
[tex]\mathbf{f_o =1.87 \times 10^{11} \ Hz}[/tex]
Explanation:
The formula for calculating the elastic potential energy is:
[tex]U_o = \dfrac{1}{2}kR_o^2[/tex]
By rearrangement and using (K) as the subject;
[tex]K = \dfrac{2 U_o}{R_o^2}[/tex]
[tex]k = \dfrac{2\times 1.68 \times 10^{-21}}{(3.82\times 10^{-10})^2}[/tex]
k = 2.3 × 10⁻² N/m
Now; the formula used to calculate the frequency of the small oscillation is:
[tex]f_o = \dfrac{1}{2 \pi}\sqrt{\dfrac{k}{m}}[/tex]
where;
m = mass of each atom
assuming
m = 1.66 × 10⁻²⁶ kg
Then:
[tex]f_o = \dfrac{1}{2 \pi}\sqrt{\dfrac{2.3 \times 10^{-2} N/m}{1.66 \times 10^{-26} \ kg}}[/tex]
[tex]\mathbf{f_o =1.87 \times 10^{11} \ Hz}[/tex]
Study the position-time graph for a bicycle. Which statement is supported by the graph? Position vs Time O The bicycle has speed but not velocity. O The bicycle is moving at a constant velocity. O The bicycle has a displacement of 3 m. O The bicycle is not in motion. 3 Position (m) 0 1 2 3 4 5 Time (s) Next Submit Save and Exit Mark this and return tViewers/AssessmentViewer/Activit. 0 M M
Answer:
D) The bicycle is not in motion.
Explanation:
Study the position-time graph for a bicycle.
Which statement is supported by the graph?
A) The bicycle has speed but not velocity.
B) The bicycle is moving at a constant velocity.
C) The bicycle has a displacement of 3 m.
D) The bicycle is not in motion.
Solution:
Velocity is the time rate of change of displacement. It is the ratio of displacement to time taken.
Speed is the time rate of change of distance. It is the ratio of distance to time taken.
From the position-time graph, we can see that the bicycle has a constant positon of 3 m for the whole of the time. That is the position remains 3 m even as the time changes. Therefore, we can conclude that the bicycle is not in motion.
From the position-time data provided, it can concluded that the bicycle is not in motion.
MotionMotion of a body involves a change in the position of that body with time.
A body in motion is constantly changing position or orientation as time passes.
The body may move with constant velocity/speed or changes in its velocity.
A position-time graph provides information about the motion of a body.
From the data provided:
At time 0, the bicycle is at position 3At time 1, the bicycle is at position 3At time 2, the bicycle is at position 3At time 3, the bicycle is at position 3At time 4, the bicycle is at position 3At time 5, the bicycle is at position 3The position of the bicycle remains the same for all time intervals.
Therefore, from the position-time data provided, it can concluded that the bicycle is not in motion.
Learn more about motion and position-time graph at: https://brainly.com/question/2356782
A ball is thrown horizontally from the top of a building 59 m high. The ball strikes the ground at a point 65 m horizontally away from and below the point of release. What is the speed of the ball just before it strikes the ground
Answer:
Explanation:
We are looking for final velocity. Since the ball is thrown horizontally, there is no upwards velocity, so the y dimension here is only useful to us for finding how long the ball was in the air. In the y dimension, here's what we know:
a = -9.8 m/s/s
Δx = -59 m
[tex]v_0=0[/tex] (again, initial upwards velocity is 0 because the ball was thrown horizontally)
We can put all that together in the equation:
Δx = [tex]v_0t+\frac{1}{2}at^2[/tex] and filling in:
[tex]-59=0t+\frac{1}{2}(-9.8)t^2[/tex] which simplifies to
[tex]-59=\frac{1}{2}(-9.8)t^2[/tex] and solving for t:
[tex]t=\sqrt{\frac{2(-59)}{-9.8} }[/tex] and
t = 3.5 sec
Now we can use that time in the d = rt equation, which is all we need for the horizontal dimension (I'll show you why in just a second). In the horizontal dimension, here's what we know:
a = 0 m/s/s
Δx = 65 m
t = 3.5 sec
Putting that all together in our one-dimensional equation for displacement:
Δx = [tex]v_0t+\frac{1}{2}at^2[/tex] and acceleration is 0, we can simplify that down to
Δx = [tex]v_0t[/tex] which is the exact same thing as d = rt where r is the velocity we are looking for. Filling in:
65 = v(3.5) so
v = 18.6 m/s
That's the velocity with which the ball strikes the ground.
A Michelson interferometer operating at a 600nm wavelength has a 2.02-cm-long glass cell in one arm. To begin, the air is pumped out of the cell and mirror M2 is adjusted to produce a bright spot at the center of the interference pattern. Then a valve is opened and air is slowly admitted into the cell. The index of refraction of air at 1.00 atm pressure is 1.00028.
How many bright-dark-bright fringe shifts are observed as the cell fills with air?
Answer:
19
Explanation:
Given that:
wavelength = 600 nm
Distance (d) = 2.02 cm = 2.02 × 10⁻² m
refraction index of air (n) = 1.00028
Pressure = 1.00 atm
∴
The number of bright-dark-bright fringe shifts can be determined by using the formula:
[tex]\Delta m = \dfrac{2d}{\lambda} (n -1 ) \\ \\ \Delta m = \dfrac{2\times2.02 \times 10^{-2}}{600\times 10^{-9}} (1.00028 -1 ) \\ \\ \Delta m = 67333.33 \times 10^{-5}(1.00028 -1) \\ \\ \Delta m = 67333.33 \times 10^{-5}(2.8\times 10^{-4}) \\ \\ \Delta m = 18.853 \\ \\ \mathbf{\Delta m = 19}[/tex]
Which of the following best describes the upper respiratory tract?
O It takes air in from outside the body.
O It is where oxygen and carbon dioxide are exchanged.
O It is located inside the thorax.
O It is not directly involved in respiration.
A 0.25 kg mass is placed on a vertically oriented spring that is stretched 0.56 meters from its equilibrium position. If the spring constant is 105 N/m, how fast will the mass be moving when it reaches the equilibrium position? Hint: You cannot ignore the change in gravitational potential energy in this problem. Please give your answer in units of m/s, however, do not explicitly include units when typing your answer into the answer box.
Answer:
The speed of the ball when it reaches equilibrium position is 3.31 m/s
Explanation:
Given;
mass of the object, m = 0.25 kg
initial displacement of the object, h₁ = 0.56 m
spring constant, k = 105 N/m
displacement at equilibrium position, h₂ = 0
initial velocity of the object, v₁ = 0
velocity of the object at equilibrium position = v₂
The change in gravitational potential energy at the equilibrium position is given as;
ΔP.E = mg(h₂ - h₁)
The change in kinetic energy of the object at the equilibrium position is given as;
ΔK.E = ¹/₂m(v₂² - v₁²)
Apply the principle of conservation of mechanical energy;
ΔK.E + ΔP.E = 0
¹/₂m(v₂² - v₁²) + mg(h₂ - h₁) = 0
¹/₂m(v₂² - 0) + mg(0 - h₁) = 0
¹/₂mv₂² - mgh₁ = 0
¹/₂mv₂² = mgh
¹/₂v₂² = gh
v₂² = 2gh
v₂ = √2gh
v₂ = √(2 x 9.8 x 0.56)
v₂ = 3.31 m/s
Therefore, the speed of the ball when it reaches equilibrium position is 3.31 m/s
