Answer:
Your fart only has so much force, not nearly enough to launch you into oblivion. Your fart and you still exert a force onto each other, so I guess, hypothetically, you could fly if you really, really try hard enough. Just make sure you don't try too hard and prolapse as a result :)
How is centripetal force affected if an object increases its speed?
A. Decreases
B. Increases
C. Cut in half
D. No effect
Answer:
B. Increases.
Explanation:
[tex]{ \bf{F = \frac{m {v}^{2} }{r} }} \\ { \bf{F \: \alpha \: {v}^{2} }}[/tex]
Keeping mass, and radius constant, speed or velocity is directly proportioanal to centripetal force.
,In order to increase the speed of an object on a circular path, YOU have to increase the centripetal force acting on it.
A 4kg block is attached to a vertical spring with a spring constant of 800N/m. How much elastic potential energy is stored in the system?
E= [tex]\frac{1}{2}[/tex]×k×x² = 1J
given k=800n/m x=0.05m
A block of mass 2 kg starts from rest at the top of a friction quarter of a circle of radius R. The block then slides over frictionless curved surface in the shape of a eventually comes to rest 8 m from the beginning s a horizontal rough surface where e of the horizontal surface. The coefficient kinetic friction between the rough surface and the block is 0.4 . determine the acceleration of the block over the rough surface length 8m
The acceleration of the block over the rough surface is 1.22625 m/s²
The process through which the acceleration is obtained is presented as follows of approach to
The given parameters are;
Mass of block, m = 2 kg
Nature of the surface of the quarter circle = Frictionless
The length of the horizontal, d = 8 m
The coefficient of friction of the horizontal surface, μ = 0.4
The unknown parameter;
The acceleration of the block over the rough surface
Method;
Find the work done by friction to stop the block and divide the result by the mass of the block
The work done by friction, [tex]W_f[/tex] = (Force of friction) × (Distance the block moves on the rough surface before coming to rest)
[tex]\mathbf{W_f}[/tex] = [tex]\mathbf{F_f}[/tex] × d
[tex]F_f[/tex] = Normal reaction of surface on block, [tex]N_r[/tex] × μ
Normal reaction on block, [tex]\mathbf{N_r}[/tex] = Weight of block
[tex]\mathbf{N_r}[/tex] ≈ 2 kg × 9.81 m/s² = 19.62 N
Therefore;
The work done by friction [tex]\mathbf{W_f}[/tex] = [tex]\mathbf{F_f}[/tex] × d = [tex]\mathbf{N_r}[/tex] × μ × d
[tex]\mathbf{W_f}[/tex] = 19.62 N × 0.4 × 8 m = 62.784 J
The work done by the block, W = Force, F × d
Force, F = m × a
Where;
a = The acceleration of the block
According to the principle of conservation of energy, we have;
[tex]\mathbf{W_f}[/tex] = W
∴ 19.62 J = 2 kg × a × 8 m
a = 19.62/(2 kg × 8 m) = 1.22625 m/s²
The acceleration of the block over the rough surface, a = 1.22625 m/s²
Learn more about work done due and friction here;
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An object of mass 80 kg is released from rest from a boat into the water and allowed to sink. While gravity is pulling the object down, a buoyancy force of 1/50 times the weight of the object is pushing the object up (weight=mg). If we assume that water resistance exerts a force on the abject that is proportional to the velocity of the object, with proportionality constant 10 N-sec/m, find the equation of motion of the object. After how many seconds will the velocity of the object be 40 m/s? Assume that the acceleration due to gravity is 9.81 m/sec^2.
Answer:
a) Fnet = mg - Fb - Fr
b) 8.67 secs
Explanation:
mass of object = 80 kg
Buoyancy force = 1/50 * weight ( 80 * 9.81 ) = 15.696
Proportionality constant = 10 N-sec/m
a) Calculate equation of motion of the object
Force of resistance on object due to water = Fr ∝ V
= Fr = Kv = 10 V
Given that : Fb( due to buoyancy ) , Fr ( Force of resistance ) acts in the positive y-direction on the object while mg ( weight ) acts in the negative y - direction on the object.
Fnet = mg - Fb - Fr
∴ Equation of motion of the object ( Ma = mg - Fb - Fr )
b) Calculate how long before velocity of the object hits 40 m/s
Ma = mg - Fb - Fr
a = 9.81 - 0.1962 - 0.125 V = 9.6138 - 0.125 V
V = u + at ---- ( 1 )
u = 0
V = 40 m/s
a = 9.6138 - 0.125 V
back to equation 1
40 = 0 + ( 9.6138 - 0.125 (40) ) t
40 = 4.6138 t
∴ t = 40 / 4.6138 = 8.67 secs
Vặt nhỏ được ném lên từ điểm A trên mặt đất với vận tốc đầu 20m/s theo phương thẳng đứng. Xác định độ cao của điểm O mà vật đạt được. Bỏ qua ma sát
Explanation:
mặt đất với vận tốc ban đầu 20m/s. Bỏ qua mọi ma sát, lấy g = 10 m/s2. Độ cao cực đại mà vật đạt được là.
the mass of an object is 10 kg and the velocity is 4 m/s, what is the momentum?
The answer is 40 kg. m/s.
Formula for momentum:
p=mv
p=(10 kg.)(4 m/s)
So, therefore, the final answer is p=40 kg. m/s.
I hope this helped answer your question. Enjoy your day, and take care!
Answer: its 400 n/s
Explanation:
cus my thing said it was right
Grog (born in 600 BC) Euriados (born in 50 AD) Nicholas (born in 1600 AD) Describe how each one of these Men would perform their duties as an astronomer, and what information would be important to them?
All these people perform similar duties due to their similar profession.
All these People are the astronomers so they used various instruments that are present at their time in order to study the heavenly bodies such as stars and planets present in the sky. They also analyze their findings with the help of researches and experiments. They also develop theories that are based on personal observations and tested the theories of other astronomers in order to verify their theories.
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A sound wave made up of large number of unrelated frequencies superposted on each other is
Since the frequencies are unrelated, and there are a large number of them, I'll say this represents an example of noise.
why do atom absorb photon since it makes it unstable??
[tex]\textsf{When an electron is hit by a }[/tex] [tex]\textsf{photon of light, it absorbs the quanta}[/tex] [tex]\textsf{of energy the photon was carrying}[/tex] [tex]\textsf{and moves to a higher energya}[/tex] [tex]\textsf{ state. Electrons therefore have to }[/tex] [tex]\textsf{jump around within the atom as }[/tex] [tex]\textsf{they either gain or lose energy. }[/tex]
When an electron is hit by a photon of light, it absorbs the quanta of energy the photon was carrying and moves to a higher energy state. Electrons therefore have to jump around within the atom as they either gain or lose energy.
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Calculate the self-inductance (in mH) of a 45.0 cm long, 10.0 cm diameter solenoid having 1000 loops. mH (b) How much energy (in J) is stored in this inductor when 21.0 A of current flows through it? J (c) How fast (in s) can it be turned off if the induced emf cannot exceed 3.00 V? s
Answer:
(a) The self inductance, L = 21.95 mH
(b) The energy stored, E = 4.84 J
(c) the time, t = 0.154 s
Explanation:
(a) Self inductance is calculated as;
[tex]L = \frac{N^2 \mu_0 A}{l}[/tex]
where;
N is the number of turns = 1000 loops
μ is the permeability of free space = 4π x 10⁻⁷ H/m
l is the length of the inductor, = 45 cm = 0.45 m
A is the area of the inductor (given diameter = 10 cm = 0.1 m)
[tex]A = \pi r^2 = \frac{\pi d^2}{4} = \frac{\pi \times (0.1)^2}{4} = 0.00786 \ m^2[/tex]
[tex]L = \frac{(1000)^2 \times (4\pi \times 10^{-7}) \times (0.00786)}{0.45} \\\\L = 0.02195 \ H\\\\L = 21.95 \ mH[/tex]
(b) The energy stored in the inductor when 21 A current ;
[tex]E = \frac{1}{2}LI^2\\\\E = \frac{1}{2} \times (0.02195) \times (21) ^2\\\\E = 4.84 \ J[/tex]
(c) time it can be turned off if the induced emf cannot exceed 3.0 V;
[tex]emf = L \frac{\Delta I}{\Delta t} \\\\t = \frac{LI}{emf} \\\\t = \frac{0.02195 \times 21}{3} \\\\t = 0.154 \ s[/tex]
You have 150 W/m^2 hitting your roof each day. You can convert 13% of it into
usable energy, and you need 3.5 kW to run your house for a day. Show the MATH,
answer and units, to determine the size solar panel you will need to succeed.
Answer:
Energy = .13 W / m^2 energy of incident energy
N = 3500 Watts / day power needed
N = 3500 Watts (3600 * 24 sec) = .0405 Watts/sec
The problem must mean that one needs 3.5 Kw-days
3.5 Kw-days = 3500 watts * 86400 sec = 3.02E8 joules
150 J/sec-m^2 * .13 = 19.5 J / sec-m^2 usable energy
In one day 19.5 J/sec-m^2 = 1.68E6 J/m^2 usable energy received
Area = 3.028E8 J / 1.68E6 J/m2 = 180 m^2
One would need 180 m^2 of solar panels
That's quite a lot of energy
A 1100 watt microwave oven uses 1.1 kW while running so 3.5 kW for 24 hours seems to be quite a lot.
Fast please!!. Serious answers only.
Considering that Susan is 38 years old, what is the probability that she lives to the age of 85?
Answer:
The probability that Susan turns 85 years old is 45,436/97,825= 0.4644.
Explanation:
Edmentum
Answer:
45,436/97,825= 0.4644.
A/An is a type of blood cell that's also called a red blood cell. a) Jeukocyte O b) thrombocyte c) plasma d) erythrocyte
Answer:
red blood cell, also called erythrocyte
Explanation:
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Meaning of power in physics
Answer:
The rate of doing work is called power.
Answer:
The amount of energy transported or transformed per unit time is referred to as power in physics. The watt, which is equal to one joule per second in the International System of Units, is the unit of power.
OAmalOHopeO
A 1050 kg car accelerates from 11.3 m/s to 26.2 m/s . What impulse does the engine give?
Answer:
I = 15,645. kg*m/s or 15,645 N*s
Explanation:
I = m(^v)
I = 1050kg((26.2m/s-11.3m/s)
I = 15,645. kg*m/s
An organ pipe of length 3.0 m has one end closed. The longest and next-longest possible wavelengths for standing waves inside the pipe are
Answer:
The longest wavelength for closed at one end and open at the other is
y / 4 where y is the wavelength - that is node - antinode
The next possible wavelength is 3 y / 4 - node - antinode - node -antinode
y / 4 = 3 m y = 12 meters the longest wavelength
3 y / 4 = 3 m y = 4 meters 1 / 3 times as long
What is tensor quantity?
Is Inertia a tensor? give reason
Answer:
A tensor is a quantity, for example a stress or a strain, which has magnitude, direction, and a plane in which it acts. Stress and strain are both tensor quantities. ... A tensor is a quantity, for example a stress or a strain, which has magnitude, direction, and a plane in which it acts.
Inertia Tensor. where I = the inertia tensor. The angular momentum of a rigid body rotating about an axis passing through the origin of the local reference frame is in fact the product of the inertia tensor of the object and the angular velocity. ... As shown in [7], the inertia tensor is symmetric.
Explanation:
Hope dis help
2. What is the average speed of an athlete who runs 1500 m in 4 minutes?
Answer:
375 is the answer.
Explanation:
Speed : Distance / Time taken
S: m/ s
s: 1500/4
375 m / s answer
Answer:
375m per minute
Explanation:
if you are looking for a diffrent unit just multiply your answer by however many minutes are in that time frame
A projectile is fired into the air from the top of a 200-m cliff above a valley as shown below. Its initial velocity is 60 m/s at 60° above the horizontal. Calculate (a) the maximum height, (b) the time required to reach its highest point, (c) the total time of flight, (d) the components of its velocity just before striking the ground, and (e) the horizontal distance traveled from the base of the cliff.
a) y(max) = 337.76 m
b) t₁ = 5.30 s the time for y maximum
c)t₂ = 13.60 s time for y = 0 time when the fly finish
d) vₓ = 30 m/s vy = - 81.32 m/s
e)x = 408 m
Equations for projectile motion:
v₀ₓ = v₀ * cosα v₀ₓ = 60*(1/2) v₀ₓ = 30 m/s ( constant )
v₀y = v₀ * sinα v₀y = 60*(√3/2) v₀y = 30*√3 m/s
a) Maximum height:
The following equation describes the motion in y coordinates
y = y₀ + v₀y*t - (1/2)*g*t² (1)
To find h(max), we need to calculate t₁ ( time for h maximum)
we take derivative on both sides of the equation
dy/dt = v₀y - g*t
dy/dt = 0 v₀y - g*t₁ = 0 t₁ = v₀y/g
v₀y = 60*sin60° = 60*√3/2 = 30*√3
g = 9.8 m/s²
t₁ = 5.30 s the time for y maximum
And y maximum is obtained from the substitution of t₁ in equation (1)
y (max) = 200 + 30*√3 * (5.30) - (1/2)*9.8*(5.3)²
y (max) = 200 + 275.40 - 137.64
y(max) = 337.76 m
Total time of flying (t₂) is when coordinate y = 0
y = 0 = y₀ + v₀y*t₂ - (1/2)* g*t₂²
0 = 200 + 30*√3*t₂ - 4.9*t₂² 4.9 t₂² - 51.96*t₂ - 200 = 0
The above equation is a second-degree equation, solving for t₂
t = [51.96 ±√ (51.96)² + 4*4.9*200]/9.8
t = [51.96 ±√2700 + 3920]/9.8
t = [51.96 ± 81.36]/9.8
t = 51.96 - 81.36)/9.8 we dismiss this solution ( negative time)
t₂ = 13.60 s time for y = 0 time when the fly finish
The components of the velocity just before striking the ground are:
vₓ = v₀ *cos60° vₓ = 30 m/s as we said before v₀ₓ is constant
vy = v₀y - g *t vy = 30*√3 - 9.8 * (13.60)
vy = 51.96 - 133.28 vy = - 81.32 m/s
The sign minus means that vy change direction
Finally the horizontal distance is:
x = vₓ * t
x = 30 * 13.60 m
x = 408 m
You are outdoors when you hear the constant chirp of a still cricket. You start walking toward the cricket and at some point you are able to detect that the intensity of the chirp of the cricket has increased by a factor of 4. What of the following statements is true at your new position with respect to the cricket?
a. The power delivered by the sound wave you hear has doubled.
b. The speed of the sound wave emitted by the cricket has decreased by a factor of 4.
c. The distance between you and the cricket has decreased by a factor of 2
Answer:
C
Explanation:
intensity = Power delivered by the sound (Watt)/ Surounding Area (m²)
I = P/A
A = πr²
r = is the distance between you and the cricket.
so in other form we can get
I = P/πr²
let take I(1) as first intensitilynyou heard and I(2) as the increased intensity.
I(1) / I(2) = r(2)² / r(1)²
1/4 = r(2)²/r(1)²
1/2 = r(2) / r(1)
r(2) = ½ r(1)
or r(2) is decreaases by a factor of 2.
Simple Pendulum: A 34-kg child on an 18-kg swing set swings back and forth through small angles. If the length of the very light supporting cables for the swing is 4.9 m, how long does it take for each complete back-and-forth swing
Answer:
The correct answer is "4.443 sec".
Explanation:
Given:
Mass of child,
= 34 kg
Mass of swing,
= 18 kg
Length,
= 4.9 m
The time period of pendulum will be:
T = [tex]2 \pi \sqrt{4g}[/tex]
= [tex]2 \pi \sqrt{\frac{4.9}{9.8} }[/tex]
= [tex]4.443 \ sec[/tex]
Answer:
The time taken to back and forth is 4.4 s .
Explanation:
Length, L = 4.9 m
let the time period is T.
Acceleration due to gravity, g = 9.8 m/s^2
Use the formula of time period
[tex]T = 2 \pi\sqrt{L}{g}\\\\T = 2 \times 3.14\sqrt{4.9}{9.8}\\\\T = 4.4 s[/tex]
a stone is thrown vertically upwards with a velocity of 20 m per second determine the total time of flight of stone in air
Answer:
Explanation:
The best way to do this is to remember the rule about the halfway mark in a parabolic path. At a trajectory's half way point in its travels, it will be at its max height. To get the total time in the air, we take that time at half way and double it. Here's what we know that we are told:
initial velocity is 20 m/s
Here's what we know that we are NOT told:
a = -9.8 m/s/s and
final velocity is 0 at an object's max height in parabolic motion.
We will use the equation:
[tex]v=v_0+at[/tex] where v is final velocity and v0 is initial velocity. Filling in:
0 = 20 + (-9.8)t and
-20 = -9.8t so
t = 2 seconds. The stone reaches its max height 2 seconds after it is thrown; that means that after another 2 seconds it will be on the ground. Total air time is 4 seconds.
what is the major difference between the natural frequency and the damped frequency of oscillation.
Answer:
This causes the amplitude of the oscillation to decay over time. The damped oscillation frequency does not equal the natural frequency. Damping causes the frequency of the damped oscillation to be slightly less than the natural frequency
When a charged particle moves at an angle of 26.1 with respect to a magnetic field, it experiences a magnetic force of magnitude F. At what angle (less than 90o) with respect to this field will this particle, moving at the same speed?
Answer:
The angle is 153.9 degree.
Explanation:
Let the magnetic field is B and the charge is q. Angle = 26.1 degree
The force is F.
Let the angle is A'.
Now equate the magnetic forces
[tex]q v B sin 26.1 = q v B sin A'\\\\A' = 180 - 26.1 = 153.9[/tex]
A 1-cm long wire carrying 15 A is inside a solenoid 4 cm in radius with 800 turns/m carrying a current of 40 mA. The wire segment is oriented perpendicularly to the axis of the solenoid. What is the magnitude of the magnetic force on this wire segment in ???? N?
Answer:
the magnitude of the magnetic force on the wire segment is 6.03 x 10⁻⁶ N
Explanation:
Given;
length of the conductor, L = 1 cm = 0.01 m
current carried by the solenoid, I₁ = 15 A
radius of the solenoid, r = 4 cm
number of turns per length of the solenoid, n = 800 turns/m
current carried by the solenoid, I₂ = 40 mA = 0.04 A
The magnetic field of the solenoid is calculated as;
B = μnI₂
where;
μ is the permeability of free space = 4π x 10⁻⁷ Tm/A
B = ( 4π x 10⁻⁷) x (800) x (0.04)
B = 4.022 x 10⁻⁵ T
The magnitude of the magnetic force on the wire segment is calculated as;
F = BI₁L sinθ
where
θ is the angle made by the wire segment against the solenoid = 90⁰
F = (4.022 x 10⁻⁵) x (15) x (0.01) x sin(90)
F = 6.03 x 10⁻⁶ N
Therefore, the magnitude of the magnetic force on the wire segment is 6.03 x 10⁻⁶ N
: A fan is placed on a horizontal track and given a slight push toward an end stop 1.80 meters away. Immediately after the push, the fan of the cart engages and slows the cart with an acceleration of -0.45 m/s2. What is the maximum possible velocity (magnitude) the cart can have after the push so that the cart turns around just before it hits the end-stop
Answer:
The initial velocity is 1.27 m/s.
Explanation:
distance, s = 1.8 m
acceleration, a = - 0.45 m/s^2
final velocity, v = 0
let the initial velocity is u.
Use third equation of motion
[tex]v^2 = u^2 + 2 a s \\\\0 = u^2 - 2 \times 0.45\times 1.8\\\\u = 1.27 m/s[/tex]
We have that the Initial velocity is mathematically given as
u=1.27m/s
Maximum possible velocity
Question Parameters:
a slight push toward an end stop 1.80 meters away
he fan of the cart engages and slows the cart with an acceleration of -0.45 m/s2
Generally the equation for the third equation of motion is mathematically given as
Vf^2 = Vi^2 + 2ad
Therefore
0=u^2+0.45*1.8
u=1.27m/s
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May someone help...please. Pretty please...
If a person is 18 % shorter than average, what is the ratio of his walking pace (that is, the frequency 'f' of his motion) to the walking pace of a person of average height? Assume that a person's leg swings like a pendulum and that the angular amplitude of everybody's stride is about the same.
f(short)/f(avg)=?
We have that the ratio of his walking pace to the walking pace of a person of average height is
[tex]\frac{V_2}{V_1}=1.10[/tex]
given the assumption and the calculation given below
From the question we are told that:
Consider a person 18\% shorter than average
Let average height of a person be [tex]10m[/tex]
Therefore
The height of an [tex]18\%[/tex] shorter man is mathematically given as
H=10*0.18
H=8.2m
Generally, the equation for velocity is mathematically given by
[tex]v=\frac{1}{2\pi} \sqrt{{g}{l}}[/tex]
Where we have the Assumption that a person's leg swings like a pendulum and that the angular amplitude of everybody's stride is about the same
Therefore
[tex]\frac{V_1}{V_2}=\frac{l_1}{l_2}[/tex]
[tex]\frac{V_1}{V_2}={82}{100}[/tex]
[tex]\frac{V_2}{V_1}=1.10[/tex]
In conclusion
The ratio of his walking pace (that is, the frequency 'f' of his motion) to the walking pace of a person of average height is
[tex]\frac{V_2}{V_1}=1.10[/tex]
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A parallel plate air capacitor has a circular disc of diameter 0.1 m, 2 mm apart and potential difference of 300 V is connected between the plates Calculate: (i) Energy of the capacitor and (ii) Electric intensity between the plates.
Your Answer Is in this attachment
trong cùng một nhiệt độ, lượng năng lượng trên mỗi mol của chất khí nào lớn nhất
a) Khí đơn nguyên tử
b) Khí có từ ba nguyên tử
c) Khí lưỡng nguyên tử
Susan is quite nearsighted; without her glasses, her far point is 34 cm and her near point is 17 cm . Her glasses allow her to view distant objects with her eye relaxed. With her glasses on, what is the closest object on which she can focus?
Answer:
[tex]u=34cm[/tex]
Explanation:
From the question we are told that:
Far point is [tex]V=34 cm[/tex]
Near point is [tex]u=17 cm[/tex]
Therefore
Focal Length
[tex]f=-34cm[/tex]
Generally the equation for the Lens is mathematically given by
[tex]\frac{1}{u}=\frac{1}{f}-\frac{1}{v}[/tex]
[tex]\frac{1}{u}=\frac{1}{-34}-\frac{1}{-17}[/tex]
[tex]u=34cm[/tex]
