Answer:
h = 31.46 m
Explanation:
We have,
Initial speed of a projectile is 37.7 m/s
It was projected at an angle of 41.2° above the horizontal on a long flat firing range.
It is required to find the maximum height reached by the projectile. The formula used to find it is given by :
[tex]h=\dfrac{u^2\sin^2\theta}{2g}[/tex]
Plugging all the known values,
[tex]h=\dfrac{(37.7)^2\times \sin^2(41.2)}{2\times 9.8}\\\\h=31.46\ m[/tex]
So, the maximum height reached by the projectile is 31.46 m.
What effect does the velocity of a rotating object have on the centripetal acceleration?
Answer:
as centripatal force acts upon an object moving at a circle in constant speed de force acts always inwards as de velocity of object is directed tangent to de circle de force can accelerate de object by changing its direction bt not actually de speed
The two quantities are closely related, but the cause/effect is the other way around.
-- The centripetal force is caused by something outside this discussion, not by the object.
-- The centripetal force acting on the object determines the object's centripetal acceleration.
-- The centripetal acceleration is the cause of whatever the object's velocity (speed and direction) turns out to be.
-- It's the centripetal acceleration that has the effect on the object's velocity.
As an example, you wouldn't say that the orbiting of a TV satellite is what causes the Earth's centripetal force that acts on it.
Light bulb 1 operates with a filament temperature of 2800 K, whereas light bulb 2 has a filament temperature of 1700 K. Both filaments have the same emissivity, and both bulbs radiate the same power. Find the ratio A1/A2 of the filament areas of the bulbs. A1/A2
Answer:
A₁/A₂ = 0.136
Explanation:
The power radiated by a filament bulb is given by the following formula:
E = σεAT⁴
where,
E = Emissive Power
σ = Stephen Boltzman Constant
ε = emissivity
A = Area
T = Absolute Temperature
Therefore, for bulb 1:
E₁ = σε₁A₁T₁⁴
And for bulb 2:
E₂ = σε₂A₂T₂⁴
Dividing both the equations:
E₁/E₂ = σε₁A₁T₁⁴/σε₂A₂T₂⁴
According to given condition, the emissive power and the emissivity is same for both the bulbs. Therefore,
E/E = σεA₁T₁⁴/σεA₂T₂⁴
1 = A₁T₁⁴/A₂T₂⁴
A₁/A₂ = (T₂/T₁)⁴
where,
T₁ = 2800 K
T₂ = 1700 K
Therefore,
A₁/A₂ = (1700 K/2800 K)⁴
A₁/A₂ = 0.136
Which of the following are true?
a) the total momentum of an isolated system is constant.
b) the total momentum of any number of particles is equal to the algebraic sum of the momenta of individual particles.
c) the total momentum of any number of particles is equal to the vector sum of the momenta of individual particles.
d) the vector sum of forces acting on a particle equals the rate of change of momentum of the particle with respect to time.
e) the total momentum of any system is constant.
f) the vector sum of forces acting on a particle equals the rate of change of velocity of the particle with respect to time.
A vertical spring-mass system undergoes damped oscillations due to air resistance. The spring constant is 2.65 ✕ 104 N/m and the mass at the end of the spring is 11.7 kg. (a) If the damping coefficient is b = 4.50 N · s/m, what is the frequency of the oscillator? Hz
Answer:
f = 7.57 Hz
Explanation:
To find the frequency of the damping oscillator, you first use the following formula for the angular frequency:
[tex]\omega=\sqrt{\omega_o-(\frac{b}{2m})^2}=\sqrt{\frac{k}{m}-(\frac{b}{2m})^2}\\\\[/tex] (1)
k: spring constant = 2.65*10^4 N/m
m: mass = 11.7 kg
b: damping coefficient = 4.50 Ns/m
You replace the values of k, m and b in the equation (1):
[tex]\omega=\sqrt{\frac{2.65*10^4N/m}{11.7kg}-(\frac{4.50Ns/m}{2(11.7kg)})^2}\\\\\omega=47.59\frac{rad}{s}[/tex]
Finally, you calculate the frequency:
[tex]f=\frac{\omega}{2\pi}=\frac{47.59}{2\pi}Hz=7.57\ Hz[/tex]
hence, the frequency of the oscillator is 7.57 Hz
1.How do we define electric potential at any point in space ?
2.How does the electric potential change along the direction of the net electric field?Explain them in details.
Answer:
1. The electric potential at any point in space, is the work done in moving a unit charge from infinity to that point.
2.The electric potential varies radially by decreasing as the distance from the electric field increases.
Explanation:
1. The electric potential at any point in space, is the work done in moving a unit charge from infinity to that point. It is defined mathematically as, V = W/q where V = electric potential, W = work done in moving charge q into electric field and q = charge. For the work done of moving a charge q into an electric field of charge q' at a distance r between them, W = kqq'/r and work done per unit charge in moving q in electric field is V = W/q = kqq'/r ÷q = kq'/r, Also, for an electric field, E, V = ∫E.dr where r is the direction moved by the charge in the electric field.
2. Since the electric potential V = kq'/r, V ∝ 1/r since k and q' are constant. So, the electric potential varies radially by decreasing as the distance from the electric field increases.
Do wave properties affect wave speed
Answer:
Nope!
Explanation:
The amplitude of a wave does not affect the speed at which the wave travels. Both Wave A and Wave B travel at the same speed. The speed of a wave is only altered by alterations in the properties of the medium through which it travels.
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An 80-kg quarterback jumps straight up in the air right before throwing a 0.43-kg football horizontally at 15 m/s . How fast will he be moving backward just after releasing the ball?PLEASE SORT QUESTION BELOWSort the following quantities as known or unknown. Take the horizontal direction to be along the x axis.mQ: the mass of the quarterbackmB: the mass of the football(vQx)i: the horizontal velocity of quarterback before throwing the ball(vBx)i: the horizontal velocity of football before being thrown(vQx)f: the horizontal velocity of quarterback after throwing the ball(vBx)f: the horizontal velocity of football after being thrown
Answer:
a
The speed of the quarterback backward is [tex]v_q = 0.08 \ m/s[/tex]
b
Known are
[tex]m_Q , m_B , (v_{Bx})_i (v_{Qx})_f, (v_{Bx})_f[/tex]
Unknown
[tex](v_{Qx})_f[/tex]
Explanation:
From the question we are told that
The mass of the quarterback is [tex]m_Q = 80 \ kg[/tex]
The mass of the ball is [tex]m_B = 0.43 \ kg[/tex]
The speed of the ball is [tex]v_{B x}= 15 \ m/s[/tex]
The law of momentum conservation can be mathematically represented as
[tex]m_Q u_{Qx} + m_Bu_{Bx} = - m_{Q} v_{Qx} + m_B v_{Bx}[/tex]
Now at initial both ball and quarterback are at rest and the negative sign signify that the quarterback moved backwards after throwing the ball
So
[tex]m_Q v_{Qx} = m_B v_ {Bx}[/tex]
=> [tex]v_{Qx} = \frac{m_Bv_{Bx}}{m_Q}[/tex]
substituting values
[tex]v_q = \frac{0.43 * 15}{80}[/tex]
[tex]v_q = 0.08 \ m/s[/tex]
A 15 g toy car moving to the right at 24 cm/s has a head-on nearly elastic collision with a 21 g toy car moving in the opposite direction at 31 cm/s. After colliding, the 15 g car moves with a velocity of 41 cm/s to the left. Find the speed of the second car after the collision.
Answer:
The speed of the second toy car after collision is [tex]v_2 = 0.155 \ m/s[/tex]
Explanation:
Let movement to the right be positive and the opposite negative
From the question we are told that
The mass of the car is [tex]m_1 = 15 \ g = \frac{15}{1000} = 0.015 \ kg[/tex]
The initial velocity of the car is [tex]u_1 = 24 \ cm /s = 0.24 m/s[/tex]
The mass of the second toy car [tex]m_2 = 21 g = 0.021 \ kg[/tex]
The initial velocity of the car is [tex]u_2 = 31 \ cm/s =- 0.31 m/s[/tex]
The final velocity of the first car is [tex]v = 41cm/s = - 0.41 m/s[/tex]
From law of momentum conservation we have that
[tex]m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2[/tex]
substituting values
[tex](0.015* 0.24) +( 0.021 * -0.31) = (0.015 * -0.41 ) + 0.021 v_2[/tex]
[tex]-0.00291 = -0.0615 + 0.021 v_2[/tex]
[tex]v_2 = 0.155 \ m/s[/tex]
A coat rack weighs 65.0 lbs when it is filled with winter coats and 40.0 lbs when it is empty. The base of the coat rack has an area of 452.4 in2. How much more pressure, in psi (pounds per square inch), is exerted by the coat rack on the floor when it is filled with winter coats than when it is empty
Answer:
0.056 psi more pressure is exerted by filled coat rack than an empty coat rack.
Explanation:
First we find the pressure exerted by the rack without coat. So, for that purpose, we use formula:
P₁ = F/A
where,
P₁ = Pressure exerted by empty rack = ?
F = Force exerted by empty rack = Weight of Empty Rack = 40 lb
A = Base Area = 452.4 in²
Therefore,
P₁ = 40 lb/452.4 in²
P₁ = 0.088 psi
Now, we calculate the pressure exerted by the rack along with the coat.
P₂ = F/A
where,
P₂ = Pressure exerted by rack filled with coats= ?
F = Force exerted by filled rack = Weight of Filled Rack = 65 lb
A = Base Area = 452.4 in²
Therefore,
P₂ = 65 lb/452.4 in²
P₂ = 0.144 psi
Now, the difference between both pressures is:
ΔP = P₂ - P₁
ΔP = 0.144 psi - 0.088 psi
ΔP = 0.056 psi
What is the effect on the period of a pendulum if you decrease its length by 6.35%? (Answer this question in terms of the initial period T.) T' = 0.87703 Incorrect: Your answer is incorrect. T
Answer:
T' = 0.9677T
Explanation:
The period of a pendulum is given by the following formula:
[tex]T=2\pi \sqrt{\frac{l}{g}}[/tex]
l: length of the pendulum
g: gravitational acceleration
If the length of the pendulum is decreased in 6.35% the length of the pendulum becomes:
[tex]l'=l-0.0635l=0.9365l[/tex]
The new period for a length of l' is:
[tex]T'=2\pi \sqrt{\frac{l'}{g}}=2\pi \sqrt{\frac{0.9365l}{g}}=\sqrt{0.9365}(2\pi \sqrt{\frac{l}{g}})=0.9677(2\pi \sqrt{\frac{l}{g}})\\\\T'=0.9677T[/tex]
hence, the new period is 0.9677T
The amplitudes and phase differences for four pairs of waves of equal wavelengths are (a) 2 mm, 6 mm, and π rad; (b) 3 mm, 5 mm, and π rad; (c) 7 mm, 9 mm, and π rad; (d) 2 mm, 2 mm, and 0 rad. Each pair travels in the same direction along the same string. Without written calculation, rank the four pairs according to the amplitude of their resultant wave, greatest first. (Hint: Construct phasor diagrams.)
Answer:
a = d > b = c
Explanation:
The information about amplitudes and phase differences for four pairs of waves of equal wavelengths are given below:
(a) 2 mm, 6 mm, and π rad
(b) 3 mm, 5 mm, and π rad
(c) 7 mm, 9 mm, and π rad
(d) 2 mm, 2 mm, and 0 rad
Whenever a wave has zero phase difference, its amplitude of the resultant wave will be twice the amplitude of any of the two waves. Nevertheless, let assume that the amplitude is a vector having angle Ø between them. The resultant vector will help us rank the four pairs according to the amplitude of their resultant wave by using phasor diagrams.
a.) 6 - 2 = 4mm
b.) 5 - 3 = 2mm
c.) 9 - 7 = 2mm
d.) 2 + 2 = 4mm
Therefore,
a = d and b = c
a = d > b = c
Please find the attached file for the phasor diagrams
Need help on this question I’d really appreciate it thanks!
Answer:radiation
Explanation:
radiation is the only one that makes sense
A group of college students eager to get to Florida on a spring break drove the 710-mi trip with only minimum stops. They computed their average speed for the trip to be 55.7 mi/h.How many hours did the trip take?
Answer:
Time taken for trip = 12.74 hour (Approx)
Explanation:
Given:
Distance of trip = 710-mi
Average speed for the trip = 55.7 mi/h
Find:
Time taken for trip = ?
Computation:
⇒ Time = Distance / Speed
⇒ Time taken for trip = Distance of trip / Average speed for the trip
⇒ Time taken for trip = 710-mi / 55.7 mi/h
⇒ Time taken for trip = 12.74 hour (Approx)
What does it mean that an exoplanet exists in the "habitable zone"? Question 6 options: (a) It is located the same distance from its star that Earth is from the Sun. (b) It is located the right distance from its star to enable liquid water to exist on its surface. (c) It means that the exoplanet has an oxygen rich atmosphere. (d) It is located a distance from its star that makes extinction level asteroid impacts unlikely.
Answer:
(b) It is located the right distance from its star to enable liquid water to exist on its surface.
Explanation:
The habitable zone according to astronomy and astrobiology is a region around a star that the planets around the star can hold and support liquid water. For our solar system, the habitable zone coincides approximately with the distance from the sun to Earth.
Um corpo de massa m= 2,0Kg é lançado horizontalmente, de uma altura h= 125m, com velocidade de módulo Vo =10m/s, como mostra a figura. Desprezando a resistência do ar e adotando g= 10m/s2 , determine: a) A energia mecânica total do corpo; b) A energia cinética do corpo a meia altura em relação ao solo; c) O tempo gasto até que o corpo atinja o solo; d) O alcance do movimento.
Answer:
A) E = 2550 J
B) K = 1325 J
C) t = 5,05 s
Explanation:
A) The total mechanical energy is given by the sum of the gravitational potential energy and the kinetic energy of the body:
[tex]E=U+K=mgh+\frac{1}{2}mv^2[/tex] (1)
m: mass of the body = 2,0 kg
g: gravitational acceleration = 9,8 m/s^2
h: height = 125 m
v: initial velocity of the body = 10 m/s
You replace the values of all variables h, m, g and v in the equation (1):
[tex]E=(2,0kg)(9,8m/s^2)(125m)+\frac{1}{2}(2,0kg)(10m/s)^2=2550\ J[/tex]
the total mechanical energy is 2550 J
B) The kinetic energy of the corp, when it is at a height of h/2 is given by:
[tex]K=\frac{1}{2}mv^2[/tex]
where
[tex]v=\sqrt{(v_x)^2+(v_y)^2}[/tex]
The x component of the velocity is constant in the complete trajectory, which is the initial velocity, that is, vo = vx
The y component is given by:
[tex]v_y^2=v_{oy}^2+2gy[/tex]
voy: vertical initial velocity = 0m/s
y: height = h/2 = 125/2 = 62.5 m
[tex]v_y=\sqrt{2g\frac{h}{2}}=\sqrt{2(9.8m/s^2)(62.5m)}=35m/s[/tex]
Then, you can calculate the velocity of the body and next, you can calculate the kinetic energy:
[tex]v=\sqrt{(10m/s)^2+(35m/s)^2}=36,40\frac{m}{s}\\\\K=\frac{1}{2}(2,0kg)(36,40m/s)^2=1325\ J[/tex]
C) The time that body takes in all its trajectory is:
[tex]t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(125m)}{9,8m/s^2}}=5,05s[/tex]
What is most often given a value of zero to describe an object's position on a straight line?
O displacement
O reference point
O distance
O ending location
Answer:
O reference point
Explanation:
A reference point is often given the value of zero to describe an object position on a straight line, or when it didn't move. If the object doesn't move, that means that there is no displacement, and it is a reference point. The answer to the question is reference point.
The electric potential in a region that is within 2.00 mm of the origin of a rectangular coordinate system is given by V=Axl+Bym+Czn+DV=Axl+Bym+Czn+Dwhere AA, BB, CC, DD, ll, mm, and nn are constants. The units of AA, BB, CC, and DD are such that if xx, yy, and zz are in meters, then VV is in volts. You measure VV and each component of the electric field at four points and obtain these results:Point (x,y,z)(m) V(V) Ex(V/m) Ey(V/m) Ez(V/m) 1 (0, 0, 0) 10.0 0 0 0 2 (1.00, 0, 0) 4.0 16.0 0 0 3 (0, 1.00, 0) 6.0 0 16.0 0 4 (0, 0, 1.00) 8.0 0 0 16.01. Use the data to calculate A.2. Use the data to calculate B3. Use the data to calculate C4. Use the data to calculate D5. Use the data to calculate E6. Use the data to calculate l7. Use the data to calculate m8. Use the data to calculate n
Answer:
Given the potential, [tex] V = Ax^l+By^m+Cz^n+D [/tex]
The components of the electric field are:
[tex]E_x = \frac{-dV}{dx} = -Alx^l^-^1[/tex]
[tex]E_y = \frac{-dV}{dy} = - Bmy^m^-^1[/tex]
[tex]E_z = \frac{-dV}{dz} = - nCzn^n^-^1[/tex]
Let's calculate the potential difference for all given points.
[tex] V(0, 0, 0) = 10V => Ax^l+By^m+Cz^n+D = 10 [/tex]
[tex]=> D = 10[/tex]
[tex] V(1, 0, 0) = 4V => A + 10 = 4 [/tex]
Solving for A, we have:
[tex] A = 4 - 10 [/tex]
[tex] A = -6 [/tex]
[tex] V(0, 1, 0) = 6V => B + 10 = 6 [/tex]
Solving for B, we have:
[tex] B = 6 - 10[/tex]
[tex] B = -4 [/tex]
[tex] V(0, 0, 1) = 8V => C + 10 = 4 [/tex]
Solving for C, we have:
[tex] C = 8 - 10 [/tex]
[tex] C = -2 [/tex]
For all given points, let's calculate the magnitude of electric field as follow:
[tex]E_x (1, 0, 0) = 16 => - Alx^l^-^1 = 16[/tex]
[tex]Al = -16[/tex]
Solving for l, we have:
[tex]l = \frac{-16}{A}[/tex]
From above, A = -6
[tex]l = \frac{-16}{-6}[/tex]
[tex]l = \frac{8}{3}[/tex]
[tex] E_y (0, 1, 0) = 16=> Bmy^m^-^1 = 16 [/tex]
[tex]Bm = -16[/tex]
Solving for m, we have:
[tex]m = \frac{-16}{A}[/tex]
From above, B = -4
[tex]m = \frac{-16}{-4}[/tex]
[tex]m = 4[/tex]
[tex] E_y (0, 0, 1) = 16=> nCz^n^-^1 = 16 [/tex]
[tex]nC = - 16[/tex]
Solving for n, we have:
[tex]n = \frac{-16}{C}[/tex]
From above, C = -2
[tex]n = \frac{-16}{-2}[/tex]
[tex]n = 8[/tex]
Two vectors A and B are such that A =1,B=2,A.B=1 find angle
Answer:[tex]60^{\circ}[/tex]
Explanation:
Given
[tex]\mid\Vec{A}\mid=1[/tex]
[tex]\mid\Vec{B}\mid=2[/tex]
And [tex]A\cdot B=1[/tex]
We know [tex]\vec{A}\cdot \vec{B}=\mid\Vec{A}\mid\mid\Vec{B}\mid\cos \theta[/tex]
Where [tex]\theta[/tex] is the angle between them
Substituting the values
[tex]1=1\times 2\cos \theta[/tex]
[tex]\cos \theta =\dfrac{1}{2}[/tex]
[tex]\theta =60^{\circ}[/tex]
Thus the angle between [tex]A[/tex] and [tex]B[/tex] is [tex]60^{\circ}[/tex]
The manufacturer of a 9 [V] flashlight battery says that the battery will deliver 20 [mA] for 80 continuous hours. However, during that time the voltage will drop from 9 [V] to 6 [V]. Assume the voltage drop is linear in time, but that the current is constant. How much energy does the batteiy deliver in 80 [h]? Battery "capacity" can be stated in terms of the total charge the battery will deliver, with units of milliamp-hours [mA-h]. If the battery can be considered "dead" when the voltage reaches 6 V, what is the capacity of the flashlight battery in mA-h?
Answer:
a. Energy ≅ 43.2 kJ
b. Capacity of battery = 1600-mAh
Explanation:
average voltage value = (9 + 6)/2 = 15/2 = 7.5
current = 20 mA = 20 x [tex]10^{-3}[/tex]
time duration = 80 hrs = 80 x 60 x 60 = 288000 sec
If the current is assumed to be constant, then the energy delivered is the product of voltage across, current delivered and the time duration during which it is delivered.
Energy delivered by battery = 7.5 x 20 x [tex]10^{-3}[/tex] x 288000 = 43200 J
Energy ≅ 43.2 kJ
If the battery is considered dead when it reaches 6-v, then that means that at 6-v, there is no potential difference between them.
capacity of flashlight battery is the product of current delivered and the time duration of delivery.
capacity = 20-mA x 80-hr = 1600-mAh
How does the centripetal force depend on speed, radius and the revolving mass?
Answer:
Mass velocity and radius are all related to centripetal force
Explanation:
By frequency of its rotation and the radius of the circular path along which objects moves
Which of these is not a factor that influences animal growth? *
2 points
A. inherited diseases
B. gravitational force
C. genetic information
D. availability of food
How much work is done (by a battery, generator, or some other source of potential difference) in moving Avogadro's number of electrons from an initial point where the electric potential is 6.70 V to a point where the electric potential is -8.90 V? (The potential in each case is measured relative to a common reference point.)
Answer:
W = 1.5 x 10⁶ J = 1.5 MJ
Explanation:
First, we calculate the potential difference between the given 2 points. So, we have:
V₁ = Electric Potential at Initial Position = 6.7 V
V₂ = Electric Potential at Final Position = - 8.9 V
Therefore,
ΔV = Potential Difference = V₂ - V₁ = -8.9 V - 6.7 V = - 15.6 V
Since, we use magnitude in calculation only. Therefore,
ΔV = 15.6 V
Now, we calculate total charge:
Total Charge = q = (No. of Electrons)(Charge on 1 Electron)
where,
No. of Electrons = Avagadro's No. = 6.022 x 10²³
Charge on 1 electron = 1.6 x 10⁻¹⁹ C
Therefore,
q = (6.022 x 10²³)(1.6 x 10⁻¹⁹ C)
q = 96352 C
Now, from the definition of potential difference, we know that it is equal to the worked done on a unit charge moving it between the two points of different potentials:
ΔV = W/q
W = (ΔV )(q)
where,
W = work done = ?
W = (15.6 V)(96352 C)
W = 1.5 x 10⁶ J = 1.5 MJ
A radar antenna is rotating and makes one revolution every 24 s, as measured on earth. However, instruments on a spaceship moving with respect to the earth at a speed v measure that the antenna makes one revolution every 44 s. What is the ratio v/c of the speed v to the speed c of light in a vacuum
Answer:
0.838
Explanation:
The ratio v/c of the speed v to the speed c of light in a vacuum is shown below:
Given that
[tex]\triangle t_0 = 24\ seconds[/tex] = time interval for one revolution
[tex]\triangle t = 44\ seconds[/tex] = time interval measured with speed v
based on the given information, the ratio v/c of the speed v to the speed c of light in a vacuum is
[tex]\triangle t = \frac{\triangle t_0}{\sqrt{1 - \frac{v^2}{c^2}}}[/tex]
[tex]{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{\triangle t_0}{\triangle t}[/tex]
Now squaring both the sides
[tex]\frac{v^2}{c^2} = 1 - \frac{(\triangle t_0)^2}{(\triangle t)^2}[/tex]
Now remove the squaring root from both the sides and putting the values
[tex]\frac{v}{c} = {\sqrt{1 - \frac{(\triangle t_0)^2}{(\triangle t)^2}[/tex]
[tex]= {\sqrt{1 - \frac{(24)^2}{(44)^2}[/tex]
= 0.838
A carousel has a radius of 1.70 m and a moment of inertia of 110 kg · m2. A girl of mass 44.0 kg is standing at the edge of the carousel, which is rotating with an angular speed of 3.40 rad/s. Now the girl walks toward the center of the carousel and stops at a certain distance from the center d. The angular speed of the carousel is now 5.4 rad/s. How far from the center, in meters, did the girl stop?
Answer:
Explanation:
Initial moment of inertia of the carousel + girl
I₁ = 110 + 44 x 1.7²
= 110 + 127.16
= 237.16 kgm².
final moment of inertia of carousel + girl
I₂ = 110 + 44 x d²
applying law of conservation of angular momentum
I₁ ω₁ = I₂ω₂
ω₁ and ω₂ are angular velocities of the carousel before and after .
237.16 x 3.4 = (110 + 44 x d²)x 5.4
806.34 = 594 + 237.6 d²
237.6 d² = 212.34
d²= .8936
d = .9453 m
An ideal gas in a balloon is kept in thermal equilibrium with its constant-temperature surroundings. How much work is done if the outside pressure is slowly reduced, allowing the balloon to expand to 50 times its original size
Answer:
w = 252.32 N
Explanation:
given data
balloon expand = 50 times its original size
we consider here initially pressure and volume
pressure = 645 pa
volume = 0.10 m³
solution
as in isothermala process ideal gas
PV = mRT
P = [tex]\frac{mRT}{v}[/tex]
P = [tex]\frac{c}{v}[/tex]
here c is constant
so work done is express as
[tex]w = c \int\limits^{V2}_{V1} {\frac{dv}{v}}[/tex]
w = [tex]c \times ln( \frac{v2}{v1})[/tex]
and we know c = p1 × v1
so
w = p1 × v1 × [tex]ln (\frac{50v1}{v1} )[/tex]
w = 645 × 0.1 × ln(50)
w = 252.32 N
A peak with a retention time of 407 s has a width at half-height (w1/2) of 7.6 s. A neighboring peak is eluted 17 s later with a w1/2 of 9.4 s. A compound that is known not to be retained was eluted in 2.5 s. The peaks are not baseline resolved. How many theoretical plates would be needed to achieve a resolution of 1.5?
Answer:
2.46 x 104
Explanation:
Solution
Recall that:
The retention time of a peak = 407 s
with a width at half-height of = 7.6 s
A compound is retained in 2.5 s.
resolution to be achieved = 1.5
Thus,
The number of plates (theoretical)= 16(tr2 / w2)
The R Resolution R= 0.589 Δtr / w1/2av = 0.589(17s) / 1/2(7.6s + 9.4s) = 1.18
Supposed that applied column contains 10,000 theoretical plates and the resolution of two peaks is 1.18
So if the column is replaced to obtain 1.5 resolution, the number of theoretical plates is needed is stated below;
width at the base = 9.4 - 7.6 = 1.8; tr = 0.786
N = 5.55tr2 / w21/2 = 5.55 (0.7862/ 1.182) x 104
= 2.46 x 104
Therefore, required theoretical plates to achieve a resolution of 1.5 is 2.46 x 104
Beats are the result of
Answer:
The phenomenon of beats is the result of sound interference and sound diffraction in periodic vibrations. In periodic vibration, the beats of sounds produced by the interference and diffraction of the waves of different frequencies.
Explanation:
Beats. When two sound waves of different frequency approach your ear, the alternating constructive and destructive interference causes the sound to be alternatively soft and loud - a phenomenon which is called "beating" or producing beats.
What is the acceleration of an object that takes 20 sec to change from a speed of 200 m/s to 300 m/s ?
Given:
initial velocity, u = 200 m/sFinal velocity, v = 300 m/s Time taken, t = 20 secTo be calculated:
Calculate the acceleration of given object ?
Formula used:
Acceleration = v - u / t
Solution:
We know that,
Acceleration = v - u / t
☆ Substituting the values in the above formula,we get
Acceleration ⇒ 300 - 200 / 20
⇒ 100/20
⇒ 5 m/s²
Imagine an isolated positive point charge Q (many times larger than the charge on a single proton). There is a charged particle A (whose charge is much smaller than charge Q) at a distance from the point charge Q. On which of the following quantities does the magnitude of the electric force on this charged particle A depend:_____.
1. the size and shape of the point charge Q
2. the specific location of the point charge Q (while the distance between Q and A is fixed)
3. the mass of the charged particle A
4. the size and shape of the charged particle A
5. the distance between the point charge Q and the charged particle A
6. the type of charge on the charged particle A
7. the mass of the point charge Q
8. the amount of the charge on the point charge Q
9. the magnitude of charge on the charged particle A
10. the specific location of the charged particle A (while the distance between Q and A is fixed)
11. the relative orientation between Q and A (while the distance between Q and A is fixed)
Answer:
Explanation:
The magnitude of the electric force on this charged particle A depends upon the following
5. the distance between the point charge Q and the charged particle A
8. the amount of the charge on the point charge Q
9. the magnitude of charge on the charged particle A
Suppose multiple stations are connected in wired network where CSMA/CD implemented. Among them, station A wants to communicate with B and station C wants to communicate with D. Previously, both of these sender stations have faced collisions at different timings and are still unsuccessful in their transmission. The value of k for station A has reached to 6, whereas the value of k for station B has reached to 4. The maximum attempts allowed to any station is set to 10 and transmission time for a frame is 0.5ms.
This time, Both of the sender stations (A & C) sensed the medium, both got the channel idle and started transmission. Sooner, the station A detected the collision.
i.What will happen now?
ii.How long station A has wait before re-trying
iii.Meanwhile what will happen at station C
Answer:
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Explanation:
