Solution-1:-
[tex]\boxed{\sf \dfrac{10\times 1000}{60\times 60}}[/tex]
Solution:-2
[tex]\boxed{\sf Sodium\:and\:Potassium}[/tex]
Solution:-3
[tex]\boxed{\sf 320m}[/tex]
Solution:-4
[tex]\boxed{\sf Rough\:tiles\:are\:used\:in\:bathroom}[/tex]
Solution:-5
[tex]\boxed{\sf Mg_3N_2}[/tex]
Solution:-6
[tex]\boxed{\sf Grapes\:and\:Rambutan}[/tex]
Solution:-7
[tex]\boxed{\sf {}^{}_{}N}[/tex]
Solution:-8
[tex]\boxed{\sf Galactuse}[/tex]
Solution:-9
[tex]\boxed{\sf Y-X}[/tex]
Solution:-10
[tex]\boxed{\sf Cell\:wall}[/tex]
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
A diffraction grating has 6000 lines per centimeter ruled on it. What is the angular separation (in degrees) between the second and the third orders on the same side of the central bright fringe when the grating is illuminated with a beam of light of wavelength 500 nm
Explanation:
Hope it Will help he hsuejwoamxgehanwpalasmbwfwfqoqlmdbehendalmZbgevzuxwllw. yeh we pabdvddxhspapalw. X
The angular separation (in degrees) between the second and the third orders on the same side of the central bright fringe if the wavelength is 500 nm and A diffraction grating has 6000 lines per centimeter ruled on it, is 27.29°.
What is diffraction?Waves spreading outward around obstructions are known as diffraction. Sound, electromagnetic radiation like light, X-rays, and gamma rays, as well as very small moving particles like atoms, neutrons, and electrons that exhibit wavelike qualities all exhibit diffraction.
Given:
The number of lines = 6000 per cm,
The Wavelength, λ = 500 nm = 500 × 10 ⁻⁹ m
Calculate the diffraction grating,
[tex]d = 1 / no\ of\ lines[/tex]
d = 10⁻² / 6000 m,
Calculate the second-order maxima angle and third-order maxima angle by the formula given below,
[tex]dsin\theta_1 = n_1 \lambda[/tex]
[tex]sin\theta_1 = n_1\lambda / d[/tex]
[tex]\theta _1 = sin^{-1}[2\times 500\times 10 ^{-9}/10^{-2}\times 6000][/tex]
θ₁ = sin⁻¹(0.6)
θ₁ = 36.87°
Similarly, for θ₂,
θ₂ = sin⁻¹(3 × 500 × 10 ⁻⁹ / 10⁻² × 6000)
θ₂ = sin⁻¹(0.9)
θ₂ = 64.16°
Calculate the separation as follows,
θ₂ - θ₁ = 64.16° - 36.87°
θ₂ - θ₁ = 27.29°
Therefore, the angular separation (in degrees) between the second and the third orders on the same side of the central bright fringe if the wavelength is 500 nm and A diffraction grating has 6000 lines per centimeter ruled on it, is 27.29°.
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Air is compressed polytropically from 150 kPa, 5 meter cube to 800 kPa. The polytropic exponent for the process is 1.28. Determine the work per unit mass of air required for the process in kilojoules
a) 1184
b) -1184
c) 678
d) -678
Answer:
wegkwe fhkrbhefdb
Explanation:B
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 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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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
what is simple machine?
Explanation:
Those tools that helps to make our work easier ,faster and more convenient in our daily life it is called simple Machine.
What is cubical expansivity of liquid while freezing
Answer:
"the ratio of increase in the volume of a solid per degree rise of temperature to its initial volume" -web
Explanation:
tbh up above ✅
Answer:
cubic meter
Explanation:
Increase in volume of a body on heating is referred to as volumetric expansion or cubical expansion
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.
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
You're carrying a 3.0-m-long, 24 kg pole to a construction site when you decide to stop for a rest. You place one end of the pole on a fence post and hold the other end of the pole 35 cm from its tip. How much force must you exert to keep the pole motionless in a horizontal position?
Answer:
[tex]F=133N[/tex]
Explanation:
From the question we are told that:
Length [tex]l=3.0m[/tex]
Mass [tex]m=24kg[/tex]
Distance from Tip [tex]d=35cm[/tex]
Generally, the equation for Torque Balance is mathematically given by
[tex]mg(l/2)=F(l-d)[/tex]
[tex]2*9.81(3/2)=F(3-35*10^-2)[/tex]
Therefore
[tex]F=133N[/tex]
A mass is tired to spring and begins vibration periodically the distance between it's lowest position is 48cm what is the Amplitude of the vibration
Answer:
The amplitude of vibration of the spring is "24 cm"
The periodic vibrating body's motion follows a sinusoidal path. This sinusoidal path is illustrated in the attached picture.
From the picture, it can be clearly seen that the amplitude of the periodic vibration motion is the distance from its mean position to the highest point.
Since the distance of both the highest and the lowest points from the mean position is the same. Therefore, the distance between the lowest and the highest point must be equal to two times the amplitude of the wave.
Amplitude = 24 cm
A train starts from rest (at position zero) and moves with constant acceleration. On the first observation, its velocity is 20m/s and 80seconds later the velocity became 60m/s. At 80s calculate the position, average velocity, and the constant acceleration over the interval.(7-points)
The value of the acceleration is a = 0.5 m/s². The position at 80 s is x = 3200 m and finally the average velocity is v = 40 m/s.
Acceleration:
We can use the fallowing kinematic equation to get the acceleration at 80 s.
[tex]a=\frac{v_{f}-v_{i}}{t}[/tex]
Where:
v(i) is the initial velocity (20 m/s)v(f) is the final velocity (60 m/s)t is the interval (80 s)The, we have:
[tex]\vec{a}=\frac{60-20}{80}[/tex]
[tex]\vec{a}=0.5\: m/s^{2}[/tex]
Position:
Knowing the acceleration we can find the position using the falling equation.
[tex]\vec{x}=v_{i}t+0.5at^{2}[/tex]
[tex]\vec{x}=20*80+0.5*0.5*80^{2}[/tex]
[tex]\vec{x}=3200 m[/tex]
Average velocity:
The definition of the average velocity is:
[tex]\vec{v}=\frac{\Delta x}{t}[/tex]
[tex]\vec{v}=\frac{x_{f}-x_{i}}{t}[/tex]
[tex]\vec{v}=\frac{3200-0}{80}[/tex]
[tex]\vec{v}=40\: m/s[/tex]
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what effect does the force of gravity have on a stone thrown vertically upwards
Answer:
rock go down
Explanation:
what comes up must come down.
A car is traveling at 118 km/h when the driver sees an accident 85 m ahead and slams on the brakes. What minimum constant deceleration is required to stop the car in time to avoid a pileup
Answer:
The constant minimum deceleration required to stop the car in time to avoid pileup is 6.32 m/s²
Explanation:
From the question, the car is traveling at 118 km/h, that is the initial velocity, u = 118km/h
The distance between the car and the accident at the moment when the driver sees the accident is 85 m, that is s = 85 ,
Since the driver slams on the brakes and the car will come to a stop, then the final velocity, v = 0 km/h = 0 m/s
First, convert 118 km/h to m/s
118 km/h = (118 × 1000) /3600 = 32.7778 m/s
∴ u = 32.7778 m/s
Now, to determine the deceleration, a, required to stop,
From one of the equations of motion for linear motion,
v² = u² + 2as
Then
0² = (32.7778)² + 2×a×85
0 = 1074.3841 + 170a
∴ 170a = - 1074.3841
a = - 1074.3841 / 170
a = - 6.3199
a ≅ - 6.32 m/s²
Hence, the constant minimum deceleration required to stop the car in time to avoid pileup is 6.32 m/s²
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There is more than 1 answer,
The picture is down
Answer:
test her prototype and collect data about its flight
a bullet is dropped from the same height when another bullet is fired horizontally they will hit the ground
Answer:
simultaneously
Time taken to reach the ground depends on the vertical component of velocity, not horizontal component of velocity.
As a skydiver accelerates downward, what force increases? A. Gravity B. Thrust C. Air resistance D. Centripetal
Answer:
(A) Gravity is you're answer.
Explanation:
When an object or human is falling at an increased rate, The force of gravity is taking place.
14. What's one of the two requirements electric current?
A. There must be an electric potential between two bodies
B. There must be no valence electrons that make their element unstable
C. There must be a carbon element present in the electric current
D. There must be a magnetic force between two bodies
Marko
One of the two requirements of electric current is there must be an electric potential between two bodies
For electric current to flow, there must be an electric potential between two bodies.
This is because electric charge flows from a higher electric potential to a lower electric potential just as, water flows from a higher gravitational potential to a lower gravitational potential.
The difference between the electric potential between the two bodies causes the electric charge to flow between the two bodies.
This flow of electric charge constitutes electric current and electric current will only flow when there is an electric potential between two bodies.
So, one of the two requirements of electric current is there must be an electric potential between two bodies.
So, the answer is A
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Two resistances, R1 and R2, are connected in series across a 12-V battery. The current increases by 0.500 A when R2 is removed, leaving R1 connected across the battery. However, the current increases by just 0.250 A when R1 is removed, leaving R2 connected across the battery.
(a) Find R1.
Ω
(b) Find R2.
Ω
Answer:
a) R₁ = 14.1 Ω, b) R₂ = 19.9 Ω
Explanation:
For this exercise we must use ohm's law remembering that in a series circuit the equivalent resistance is the sum of the resistances
all resistors connected
V = i (R₁ + R₂)
with R₁ connected
V = (i + 0.5) R₁
with R₂ connected
V = (i + 0.25) R₂
We have a system of three equations with three unknowns for which we can solve it
We substitute the last two equations in the first
V = i ( [tex]\frac{V}{ i+0.5} + \frac{V}{i+0.25}[/tex] )
1 = i ( [tex]\frac{1}{i+0.5} + \frac{1}{i+0.25}[/tex] )
1 = i ( [tex]\frac{i+0.5+i+0.25}{(i+0.5) \ ( i+0.25) }[/tex] ) = [tex]\frac{i^2 + 0.75i}{i^2 + 0.75 i + 0.125}[/tex]
i² + 0.75 i + 0.125 = 2i² + 0.75 i
i² - 0.125 = 0
i = √0.125
i = 0.35355 A
with the second equation we look for R1
R₁ = [tex]\frac{V}{i+0.5}[/tex]
R₁ = 12 /( 0.35355 +0.5)
R₁ = 14.1 Ω
with the third equation we look for R2
R₂ = [tex]\frac{V}{i+0.25}[/tex]
R₂ =[tex]\frac{12}{0.35355+0.25}[/tex]
R₂ = 19.9 Ω
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.
Your little sister (mass 25 kg) is sitting in her little red wagon (mass
8.5 kg) at rest. You begin pulling her forward, accelerating her with a
constant force for 2.35 s to a speed of 1.8 m/s. Calculate the impulse
you imparted to the wagon and its passenger.
Answer:
p = 60.6N*s
Explanation:
v_f = v_0+a*t
a = (v_f-v_0)/t
a = (1.8m/s)/2.35s
a = 0.77m/s²
F = m*a
F = (25kg+8.5kg)*0.77m/s²
F = 25.8N
^p = F*t
p = 25.8N*2.35s
p = 60.6N*s
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]
The temperature of a body falls from 30°C to 20°C in 5 minutes. The air
temperature is 13°C. Find the temperature after a further 5 minutes.
Answer:
15.88
is the correct answer
If a conducting loop of radius 10 cm is onboard an instrument on Jupiter at 45 degree latitude, and is rotating with a frequency 2 rev/s; What is the maximum emf induced in this loop? If its resistance is 0.00336 ohms, how much current is induced in this loop? And what is the maximum power dissipated in the loop due to its rotation in Jupiter's magnetic field?
Answer:
a) fem = - 2.1514 10⁻⁴ V, b) I = - 64.0 10⁻³ A, c) P = 1.38 10⁻⁶ W
Explanation:
This exercise is about Faraday's law
fem = [tex]- \frac{ d \Phi_B}{dt}[/tex]
where the magnetic flux is
Ф = B x A
the bold are vectors
A = π r²
we assume that the angle between the magnetic field and the normal to the area is zero
fem = - B π 2r dr/dt = - 2π B r v
linear and angular velocity are related
v = w r
w = 2π f
v = 2π f r
we substitute
fem = - 2π B r (2π f r)
fem = -4π² B f r²
For the magnetic field of Jupiter we use the equatorial field B = 428 10⁻⁶T
we reduce the magnitudes to the SI system
f = 2 rev / s (2π rad / 1 rev) = 4π Hz
we calculate
fem = - 4π² 428 10⁻⁶ 4π 0.10²
fem = - 16π³ 428 10⁻⁶ 0.010
fem = - 2.1514 10⁻⁴ V
for the current let's use Ohm's law
V = I R
I = V / R
I = -2.1514 10⁻⁴ / 0.00336
I = - 64.0 10⁻³ A
Electric power is
P = V I
P = 2.1514 10⁻⁴ 64.0 10⁻³
P = 1.38 10⁻⁶ W
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]
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
Light takes 1.2 sec to get from the moon to the Earth. Assume you are looking at the moon with noticeable earth shine. If the Sun burned out, you would eventually see the crescent of the moon disappear. The earth shine part of the moon would disappear Answer 2.4 s after the crescent disappeared.
Answer:
1.2 seconds
Explanation:
Answer to the following question is 1.2 seconds
Because light from the moon takes 1.2 seconds to reach Earth, the light released from the crescent immediately before it vanishes will also take 1.2 seconds to reach Earth. As a result, the earth-shine portion of the moon will vanish 1.2 seconds after the crescent has vanished.
A mass is hanging from the end of a horizontal bar that pivots around an axis through its center, but it is being held stationary. The bar is then released and begins to rotate. As the bar rotates from horizontal to vertical, the magnitude of the torque on the bar: ________
Answer:
The torque decreases because as the hanging mass goes down, the moment arm about the pivot point decreases. Since the torque is directly proportional to the length of the moment arm, torque decreases.
The torque decreases because as the hanging mass goes down, the moment arm about the pivot point decreases. Since the torque is directly proportional to the length of the moment arm, torque decreases.
What is the difference between atomic number and atomic mass ?Atomic number of an element is defined as total number of protons present in the nucleus, neutrons carry no net electrical charge, so it is the charge number of the nucleus.
atomic mass of an element can be defined as the atomic weight is measured total mass of an element’s atom, the total number of neutrons and protons in the nucleus of an atom.
Both Atomic mass and an atomic number of elements are closely related if atomic number is high, then the atomic mass is also said to be high.
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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