Answer:
0.035 J
Explanation:
Applying,
W = ke²/2.............. Equation 1
Where W = workdone by the stretching the spring, k = spring constant, e = extension.
make k the subject of the equation
k = 2W/e²............... Equation 2
From the question
Given: W = 2 J, e = (43-28) = 15 cm = 0.15 m
Substitute these values into equation 2
k = (2×2)/(0.15²)
k = 177.78 N/m
Hence, work need to stretch the spring from 33 cm to 35 cm
therefore,
e = 35-33 = 2 cm = 0.02 m
Substitute into equation 1
W = 177.78(0.02²)/2
W = 0.035 J
The launching catapult of the aircraft carrier gives the jet fighter a constant acceleration of 59 m/s2 from rest relative to the flight deck and launches the aircraft in a distance of 97 m measured along the angled takeoff ramp. If the carrier is moving at a steady 26 knots (1 knot = 1.852 km/h), determine the magnitude v of the actual velocity of the fighter when it is launched.
Answer:
Explanation:
We shall find the final velocity of aircraft with respect to aircraft carrier using the following relation.
v² = u² + 2as
v² = 0² + 2 x 59 x 97
v² = 11446
v = 107 m /s
velocity of aircraft carrier = 1.852 x 26 = 48.152 km/h
= 48.152 x 1000 / (60 x 60) m/s
= 13.37 m /s
This velocity of aircraft carrier will be added to the velocity of aircraft .
So absolute velocity of aircraft = 107 m /s + 13.37 m/s
= 120.37 m/s
all
What is Velocity
Answer:
noun
the speed of something in a given direction.
"the velocities of the emitted particles"
(in general use) speed.
"the tank shot backwards at an incredible velocity"
Similar:
speed
pace
rate
tempo
momentum
impetus
swiftness
swift/fast pace
fastness
quickness
speediness
rapidity
briskness
expeditiousness
expedition
dispatch
acceleration
clip
fair old rate
fair lick
steam
nippiness
fleetness
celerity
ECONOMICS
the rate at which money changes hands within an economy.
noun: velocity of circulation; plural noun: velocities of circulation
Answer:
i hope this helps you
Explanation:
hii
Radiation exerts pressure on surfaces on which it lalls (radintion pressure). Will this pressure be greater on a shiny surface or a dark surface
Answer:
Shiny surface.
Explanation:
We know that radiation pressure is the pressure over a surface exposed to electromagnetic radiation.
Where if the radiation is absorbed by the material (like in the case of a dark surface), the pressure is the energy density flux divided by the speed of light, while if the radiation is totally reflected (idealized case, but we can suppose that this happens for a shiny surface) the pressure is twice pressure for the absorbed case.
This is a simplification for the radiation pressure but is enough to conclude that the radiation pressure is always greater on reflective surfaces, then for this case, the pressure will be greater on a shiny surface than in a dark surface,
What are 3 artificial and 2 natural sources of electromagnetic radiation?
Answer: its b bro
Explanation:
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quick answer please
.
.
.Five identical capacitor plates each of area A are arranged such that adjacent plates are at a distance d apart . The plates are connected to a potential difference of EMF V as shown in the figure. The capacitance :
Answer:
tq1=EoA/D
-2q1=-2EoAV/d
Option C.
Answer:
tq1=EoA/D
-2q1=-2EoAV/D
Explanation:
option C is correct
You throw a glob of putty straight up toward the ceiling, which is 3.50 m above the point where the putty leaves your hand. The initial speed of the putty as it leaves your hand is 9.10 m/s.
1. What is the speed of the putty just before it strikes the ceiling? Express your answer with the appropriate units.
2. How much time from when it leaves your hand does it take the putty to reach the ceiling? Express your answer with the appropriate units.
Answer:
Explanation:
Given that:
the putty initial speed (u) = 9.10 m/s
distance (s) between hand and the ceiling = 3.50 m
the speed of the putty prior to the time it hits the ceiling can be determined by considering the second equation of motion.
v² - u² = 2as
Since the putty is moving in a vertical motion(i.e. in an upward direction)
v² - u² = -2gs
v² = u² - 2gs
[tex]v = \sqrt{u^2 - 2gs}[/tex]
[tex]v = \sqrt{(9.10)^2 -( 2* 9.8) (3.50 -0)}[/tex]
[tex]v = \sqrt{82.81 -19.6 (3.50)}[/tex]
[tex]v = \sqrt{82.81 -68.6}[/tex]
[tex]v = \sqrt{14.21}[/tex]
v = 3.77 m/s
2.
The time it takes to reach the ceiling from the moment it leaves your hand can be calculated by using the first equation of motion:
v = u + at
In an upward direction
v = u - gt
making time t the subject;
v - u = -gt
[tex]t = \dfrac{v-u}{-g}[/tex]
[tex]t = \dfrac{3.77 - 9.10}{-9.8}[/tex]
[tex]t = \dfrac{-5.33}{-9.8}[/tex]
t = 0.54s
sometimes balance point may not be obtained on the potentiometer wire why
2 Lights slows down when it enters water from air.
a What happens to its speed?
b What happens to its wavelength?
c What happens to its frequency?
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
With neat circuit diagrams where possible show any two (2) ways of direct current motor
excitations,
what are the symptoms of hepatitis 'b'
the velocity of a ship in the unit of m/s moving initially along west is given by V(t) = 5-2t. It’s acceleration at t=1 s is given by:
1. 0 m/s^2
2. 2m/s^2 along west
3. 2m/s^2 along east
4. None
Whis one is correct?
Answer:
4. None
Explanation:
Applying,
a(t) = dV(t)/dt
Where a(t) = Acceleration of the ship at a given time.
From the question,
Given: V(t) = 5-2t
Therefore,
a(t) = dV(t)/dt = 5 m/s²
Hence it's acceleration is 5 m/s²
From the question,
The right option is 4. None
You send a traveling wave along a particular string by oscillating one end. If you increase the frequency of oscillations, does the speed of the wave increase, decrease, or remain the same?
Answer:
The speed of the wave remains the same
Explanation:
Since the speed of the wave v = √(T/μ) where T is the tension in the string and μ is the linear density of the string.
We observed that the speed, v is independent of the frequency of the wave in the string. So, increasing the frequency of the wave has no effect on the speed of the wave in the string, since the speed of the wave in the string is only dependent on the properties of the string.
So, If you increase the frequency of oscillations, the speed of the wave remains the same.
The tires of a car make 60 revolutions as the car reduces its speed uniformly from 95.0 km/h to 60.0 km/h. The tires have a diameter of 0.88 m. If the car continues to decelerate at this rate, how far does it go
Answer:
-2.869 rad/s2
Explanation:
Data given:
speed, vi at 95.0 km/h = 95 X (1 hour /3600 seconds) X (1000m / 1km)
Note that, for every 1 hour, there will be 60sec X 60sec = 3600 seconds
And for every 1km, there will be 1000m.
So, speed of 95.0 km/h = 26.389 m/s
speed, vi = r ω (radius X angular velocity)
angular velocity, ωi = v/r
ωi = 26.389 m/s ÷ half of 0.88 m diameter
= 59.975 rad/s
decelerating to speed, vf at 60.0 km/h = 60 X X (1 hour /3600 seconds) X (1000m / 1km)
= 16.667m/s
The angular velocity for this speed = 16.667m/s ÷ half of 0.88 m diameter
= 37.879rad/s
How far the car goes is equivalent to the angular acceleration which equals to (ωf^2 - ωi^2) ÷ 2θ
= (37.879rad/s)^2 - (59.975 rad/s)^2 ÷ 2 (60 rev X 2π rad/rev)
= -2.869 rad/s2
Find the weight of a man whose mass is 40 kg on earth.
(also
write complete data plus proper formula).
I am sure it help you with that much ☺️
Explanation:
pleasae give me some thanks please good morning sister
Does the same battery always deliver the same amount of flow to any circuit? Mention two observations of any circuits in this lab that support your answer. Explain.
Answer:
Yes
Explanation:
Given that the battery is the same the PD ( potential difference ) in the circuit will also be the same likewise the flow of charge in the circuit,
Hence the same amount of charge flow is delivered to any circuit.
attached below are examples
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.
A photon with a frequency of 5.02 × 1014 hertz is absorbed by an excited hydrogen atom. This causes the electron to be ejected from the atom, forming an ion. Calculate the energy of this photon in joules. [Show all work, including the equation and substitution with units.] Determine the energy of this photon in electron-volts. What is the number of the lowest energy level (closest to the ground state) of a hydrogen atom that contains an electron that would be ejected by the absorption of this photon?
Answer:
Explanation:
An atom emits a photon (particle of light) when transitioning from a ground state to its excited state. To obey conservation of energy, the energy gained by the atom when an electron moves to a lower energy level is equal to the energy it loses in emitting the photon. (The energy of a photon is E = hf, where E is the energy, h is Planck's constant, and f is the frequency of the photon.) Conversely, when an atom absorbs a photon (as is the case in absorption spectra), the electron absorbing the photon moves to a higher energy level.
A circle has a radius of 20 cm and a central angle that measures 216° find the length of the arc defined by the central angle
Answer:
4320cm
Explanation:
(length of the arc)/(length of the radius)=central angle
lengthof the arc/20=216
lengthof the arc=216×20
length of the arc=4320cm
hope this helps, let me know if I'm right!
The length measured along the arc is called length of the arc. The length of the arc defined by the central angle is 4320cm.
What is central angle?Central angle is the angle subtended by the arc at the center of circle.
central angle = length of the arc / length of the radius
216 = length of the arc/20
length of the arc = 216×20
length of the arc = 4320cm
Thus, the length of the arc defined by the central angle is 4320cm.
Learn more about central angle.
https://brainly.com/question/15698342
#SPJ2
A ball is thrown vertically upward at 24.0 ms can reach a height of 28.8m ( neglecting air resistance).The speed,in m/s,when it is halfway to its highest point is (using g= 10 ms ^2)
Answer:
The answer is "[tex]16.79\ \frac{m}{s}[/tex]"
Explanation:
In this question, the halfway indicates the height that is [tex]\frac{28.8}{2}=14.4 \ m[/tex]
Using formula:
[tex]v^2=u^2+2as\\\\v^2=24^2+2(-10)(14.4)\\\\[/tex]
[tex]v^2=576-288\\\\v^2=288\\\\v=\sqrt{288}\\\\v=16.97 \ \frac{m}{s}[/tex]
How long does take for a freely falling object to reach 4.0 m/s
Answer :
Considering initial velocity is 0,
It takes about 0.4 seconds.
Use the equation v-u =at
4-0 = 9.8×t
t = 4/9.8 = 0.4 seconds approximately.
what is the suitable way of using social media
Answer:
not using it too much and getting addicted
Explanation:
A body of mass 2.0 kg makes an elastic collision with another body at rest and continues to move in the original direction but with one-fourth of its original speed.
(a) What is the mass of the other body?
(b) What is the speed of the two-body center of mass if the initial speed of the 2.0 kg body was 4.0 m/s?
Answer:
(a) the mass of the second body is 1.2 kg
(b) the speed of the two-body center of mass 2.5 m/s
Explanation:
Given;
mass of the body, m₁ = 2 kg
let the mass of the second body = m₂
let the initial speed of the first body, = u₁ = 4 m/s
then, the final speed of the first body, v₁= ¹/₄u₁ = 0.25u₁
initial speed of the second body, u₂ = 0
let the final speed of the second body = v₂
Apply principle of conservation of linear momentum to determine the mass of the second body;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
2u₁ + 0(m₂) = 2(0.25u₁) + m₂v₂
2u₁ = 0.5u₁ + m₂v₂
1.5(4) = m₂v₂
6 = m₂v₂
Apply one-directional velocity
u₁ + v₁ = u₂ + v₂
u₁ + (0.25u₁) = 0 + v₂
1.25u₁ = v₂
1.25(4) = v₂
5 = v₂
Then, the mass of the second body is calculated as;
m₂v₂ = 6
5m₂ = 6
m₂ = 6/5
m₂ = 1.2 kg
(b) the speed of the two-body center of mass after collision;
[tex]V_c_m = \frac{m_1v_1 + m_2v_2}{m_1 + m_2} \\\\V_c_m = \frac{2(0.25\times 4) \ + \ 1.2(5)}{2\ + \ 1.2} \\\\V_c_m = 2. 5 \ m/s[/tex]
an object is 70 um long and 47.66um wide. how long and wide is the object in km?
Answer:
length = 7*10^(-8)km
width = 4.666*10^(-8) km
Explanation:
We know that:
1 μm = 1*10^(-6) m
and
1km = 1*10^3 m
or
1m = 1*10^(-3) km
if we replace the meter in the first equation, we get:
1 μm = 1*10^(-6)*1*10^(-3) km
1 μm = 1*10^(-6 - 3)km
1 μm = 1*10^(-9)km
Now with this relationship we can transform our measures:
Length: 70 μm is 70 times 1*10^(-9)km, or:
L = 70*1*10^(-9)km = 7*10^(-8)km
And for width, we have 47.66um, this is 46.66 times 1*10^(-9)km, or:
W = 46.66*1*10^(-9)km = 4.666*10^(-8) km
An air-filled pipe is found to have successive harmonics at 700 Hz , 900 Hz , and 1100 Hz . It is unknown whether harmonics below 700 Hz and above 1100 Hz exist in the pipe. What is the length of the pipe
Answer:
the length of the pipe is 0.85 m or 85 cm
Explanation:
Given the data in the question;
The successive harmonics are; 700 Hz , 900 Hz , and 1100 H
Now, for a closed pipe,
length of pipe (L) = λ/4
Harmonics; 1x, 3x, 5x, 7x, 9x, 11x
1100Hz - 900Hz = 200Hz
⇒ 2x = 200Hz
x = 100Hz ( fundamental frequency )
λ = V/f = 340 /100 = 3.4 m
Now
Length L = λ / 4
L = 3.4 / 4
L = 0.85 m or 85 cm
Therefore, the length of the pipe is 0.85 m or 85 cm
79. If the takeoff velocity of an airplane on a runway is 300 km/hr with an acceleration of 2 m/s. What is the takeoff time
of the airplane.
Answer:
The take off time is equal to 41.66s.
Explanation:
Given that,
The initial speed of the airplane, u = 0
Acceleration, a = 2 m/s²
Let v is the take off time. Using first equation of motion,
v = u +at
Put v = 300 km/hr = 83.3 m/s
So,
[tex]t=\dfrac{v}{a}\\\\t=\dfrac{83.33}{2}\\\\t=41.66\ s[/tex]
So, the take off time is equal to 41.66s.
Jacie made a model to show the water cycle. The model she made is shown
below.
Which process in the model represents condensation?
A. As water vapor transfers heat to ice cubes, it forms clear droplets outside the
plastic wrap.
B. As water vapor gains heat from ice cubes, it forms clear droplets outside the
plastic wrap.
C. As water vapor transfers heat to ice cubes, it forms colored droplets inside the
plastic wrap.
D. As water vapor gains heat from ice cubes, it forms colored droplets inside the
plastic wrap
Answer:
option C
Explanation:
as water vapor transfer heat, colored drops are seen inside the wrap.
An air-filled parallel-plate capacitor has plates of area 2.90 cm2 separated by 2.50 mm. The capacitor is connected to a(n) 18.0 V battery.
Complete question:
An air-filled parallel-plate capacitor has plates of area 2.90 cm2 separated by 2.50 mm. The capacitor is connected to a(n) 18.0 V battery. Find the value of its capacitance.
Answer:
The value of its capacitance is 1.027 x 10⁻¹² F
Explanation:
Given;
area of the plate, A = 2.9 cm² = 2.9 x 10⁻⁴ m²
separation distance of the plates, d = 2.5 mm = 2.5 x 10⁻³ m
voltage of the battery, V = 18 V
The value of its capacitance is calculated as;
[tex]C = \frac{k\epsilon_0A}{d} \\\\C = \frac{(1)(8.85\times 10^{-12})(2.9 \times 10^{-4})}{2.5 \times 10^{-3}} \\\\C = 1.027 \times 10^{-12} \ F[/tex]
Therefore, the value of its capacitance is 1.027 x 10⁻¹² F
15. Muous produced by snails help them move along the ground How does the mucus help maits
move?
A Mucus is cold
C. Mucus reduces friction
B. Mucus leaves a trail D. Mucus makes the snail lighter
16. Which of the following statements does NOT tell about the effects of pravitational force
on objects?
A. The more distant the body from the earth, the lesser is the gravitational force of attraction
B. Gravitational force pushes the objects upward
C. Gravitational force is greater when the objects are closer together
D. The farther you are from earth, the lesser your weight
17. Why is it difficult for us to go up in a mountain than go down?
A We are moving against gravity
B. We are pulled by the wind
C. We are moving towards gravity
D. None of the above
18. Which of these will have the strongest gravitational pull on Earth'?
A a baby
C. a drinking glass
B. a half sack of nice
D. an elephant
19. Without air resistance, all objects fall at the same rate.
D. Neither True or False
C Maybe
B. False
20. The gravity between two objects increases as the distance between them
D stabilizes
C neutralizes
B. increases
A. decreases
A True
Answer:
15.C Mucus reduces friction
A pressure sensor was used to measure the unsteady pressure in a cylinder. The sensor output was acquired for 15 seconds at a rate of 202 Hz and spectral analysis was performed using FFT. What is the maximum frequency that will be displayed on the power spectrum plot
Answer:
Maximum frequency on power spectrum plot = 101 Hz
Explanation:
Given:
Time taken for output = 15 seconds
Frequency rate = 202 Hz
Find:
Maximum frequency on power spectrum plot
Computation:
Maximum frequency = Given frequency rate / 2
Maximum frequency on power spectrum plot = Frequency rate / 2
Maximum frequency on power spectrum plot = 202 / 2
Maximum frequency on power spectrum plot = 101 Hz