Answer:
the tension in the tightrope is 233.68 N
Explanation:
Given the data in the question;
Time taken to reach the opposite tower t = 0.9 s
Distance between the two towers S = 26 m
mass per one meter length = 0.28 kg
First we calculate the velocity;
Velocity V = Distance / time
we substitute
Velocity V = 26 m / 0.9 s
Velocity V = 28.889 m/s
We know that Velocity V can also be expressed as;
V = √( T / m )
we make T the subject of formula
V² = T / m
T = mV²
we substitute
T = 0.28 × ( 28.889 )²
T = 233.68 N
Therefore, the tension in the tightrope is 233.68 N
what are 2 ways that we can express to show our connection to our culture
Answer:
Food, clothes, language, and belief
___________________ uses the heat energy from the Earth. This type of alternative energy can be installed basically anywhere in North Carolina.
A) wave energy
B) solar energy
C) nuclear energy
D) geothermal energy
Answer:
Geothermal!!
Explanation:
Geo means earth! Just like Hydro means water, and Pyro means fire :)
Answer:
The Answer is D). Geothermal energy
Explanation:
Hope this helps!!!!!
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
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.
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
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
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]
A small block, with a mass of 0.05 kg compresses a spring with spring constant 350 N/m a distance of 4 cm. It is released from rest, then slides around the loop and up the incline before momentarily comes to rest at point A. The radius of the loop is 0.1 m.
Required:
Find the elastic potential energy.
Answer:
The elastic potential energy of the spring is 0.28 J
Explanation:
Given;
mass of the block, m = 0.05 kg
spring constant, k = 350 N/m
extension of the spring, x = 4 cm = 0.04 m
The elastic potential energy of the spring is calculated as;
[tex]U_x = \frac{1}{2}kx^2\\\\U_x = \frac{1}{2} \times 350 \times (0.04)^2\\\\U_x = 0.28 \ J[/tex]
Therefore, the elastic potential energy of the spring is 0.28 J
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
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
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
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
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
what is the suitable way of using social media
Answer:
not using it too much and getting addicted
Explanation:
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.
I NEED HELP WITH THIS PLEASE
Answer:
D
Explanation:
One of the factors of increasing the rate of a reaction is increasing concentration. Therefore adding more people increases the number of people on the dance floor , therefore increasing the concentration increases the rate of reaction.
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?
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
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
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]
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,
a man is standing near the edge of a cliff 85 meters high. he throws a stone upward vertically with an intial velocity of 10 m/s. the stone clears the cliff edge on the way down and falls all the way to the ground. what is the maximum height of the stone above the ground
Answer:
h = 90.10 m
Explanation:
Given that,
A man is standing near the edge of a cliff 85 meters high, h₀ = 85 m
The initial speed of the stone, u = 10 m/s
The path followed by the projectile is given by :
[tex]h(t)=-4.9t^2+10t+85[/tex] ....(1)
For maximum height,
Put dh/dt = 0
So,
[tex]\dfrac{dh}{dt}=-9.8t+10=0\\\\t=\dfrac{10}{9.8}\\\\=1.02\ s[/tex]
Put the value of t in equation (1).
[tex]h(t)=-4.9(1.02)^2+10(1.02)+85\\\\=90.10\ m[/tex]
So, the maximum height of the stone is equal to 90.10 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.
With neat circuit diagrams where possible show any two (2) ways of direct current motor
excitations,
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
calculate the correct fuse that should be used for a 230V,1KW electric hair dryer.
Answer: 4 A
Explanation:
Given
Voltage [tex]V=230\ V[/tex]
Power [tex]P=1\ kW[/tex]
Power is given by [tex]P=VI\\[/tex]
Insert the values
[tex]\Rightarrow 1000=230\times I\\\\\Rightarrow I=\dfrac{1000}{230}\\\\\Rightarrow I=3.84\ A[/tex]
The rating of fuse is slightly higher than the normal operating conditions. Therefore, a 4 A fuse should be used here.
A mass of 4 kg is traveling over a quarter circular ramp with a radius of 10 meters. At the bottom of the incline the mass is moving at 21.3 m/s and at the top of the incline the mass is moving at 2.8 m/s. What is the work done by all non-conservative force in Joules?
Answer:
499.7 J
Explanation:
Since total mechanical energy is conserved,
U₁ + K₁ + W₁ = U₂ + K₂ + W₂ where U₁ = potential energy at bottom of incline = mgh₁, K₁ = kinetic energy at bottom of incline = 1/2mv₁² and W₁ = work done by friction at bottom of incline, and U₂ = potential energy at top of incline = mgh₂, K₁ = kinetic energy at top of incline = 1/2mv₂² and W₂ = work done by friction at top of incline. m = mass = 4 kg, h₁ = 0 m, v₁ = 21.3 m/s, W₁ = 0 J, h₂ = radius of circular ramp = 10 m, v₂ = 2.8 m/s, W₂ = unknown.
So, U₁ + K₁ + W₁ = U₂ + K₂ + W₂
mgh₁ + 1/2mv₁² + W₁ = mgh₂ + 1/2mv₂² + W₂
Substituting the values of the variables into the equation, we have
mgh₁ + 1/2mv₁² + W₁ = mgh₂ + 1/2mv₂² + W₂
4 kg × 9.8 m/s²(0) + 1/2 × 4 kg × (21.3 m/s)² + 0 = 4 kg × 9.8 m/s² × 10 m + 1/2 × 4 kg × (2.8 m/s)² + W₂
0 + 2 kg × 453.69 m²/s² = 392 kgm²/s² + 2 kg × 7.84 m²/s² + W₂
907.38 kgm²/s² = 392 kgm²/s² + 15.68 kgm²/s² + W₂
907.38 kgm²/s² = 407.68 kgm²/s² + W₂
W₂ = 907.38 kgm²/s² - 407.68 kgm²/s²
W₂ = 499.7 kgm²/s²
W₂ = 499.7 J
Since friction is a non-conservative force, the work done by all the non-conservative forces is thus W₂ = 499.7 J
PLEASE HELP ME WITH THIS ONE QUESTION
Answer:
[tex]^{214} _{83} Bi[/tex] → [tex]^{210}_{81}Tl[/tex] + [tex]^4_2He[/tex]
Explanation:
[tex]^{214} _{83} Bi[/tex] → [tex]x[/tex] + [tex]^4_2He[/tex]
Subtract the [tex]^4_2He[/tex] from the [tex]^{214} _{83} Bi[/tex]
214 - 4 = 210
83 -2 = 81
Therefore, [tex]x[/tex] = [tex]^{210}_{81}Tl[/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
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