Answer:
frequency wren hear when approaches is 470.29 Hz
frequency wren hear after hawk pass is 415.75 Hz
Explanation:
given data
hawk velocity Vs= 38 m/s
frequency f = 440 Hz
wren velocity Vo= 17 m/s
speed of sound s = 343 m/s
to find out
What frequency wren hear and What frequency wren hear after hawk pass
solution
we apply here frequency formula that is
[tex]f1=f ( \frac{s+V_o}{s+V_s} )[/tex] .........1
here f1 is frequency hear by observer
put here all value as Vo and Vs negative because it approaches
[tex]f1=f ( \frac{s-V_o}{s-V_s} )[/tex]
[tex]f1=440 ( \frac{343-17}{343-38} )[/tex]
f1 = 470.29 Hz
so frequency wren hear when approaches is 470.29 Hz
and
after passing from equation 1 we take both Vo and Vs as positive
[tex]f1=f ( \frac{s+V_o}{s+V_s} )[/tex]
[tex]f1=440 ( \frac{343+17}{343+38} )[/tex]
f1 = 415.75 Hz
so frequency wren hear after hawk pass is 415.75 Hz
A farm hand does 972 J of work pulling an empty hay wagon along level ground with a force of 310 N [23° below the horizontal]. The distance the wagon travels is: a) 32 m b) 1.7m c) 0.39 m d) 3.4 m
Answer:
Option d)
Solution:
As per the question:
Work done by farm hand, [tex]W_{FH} = 972J[/tex]
Force exerted, F' = 310 N
Angle, [tex]\theta = 23^{\circ}[/tex]
Now,
The component of force acting horizontally is F'cos[tex]\theta[/tex]
Also, we know that the work done is the dot or scalar product of force and the displacement in the direction of the force acting on an object.
Thus
[tex]W_{FH} = \vec{F'}.\vec{d}[/tex]
[tex]972 = 310\times dcos23^{\circ}[/tex]
d = 3.406 m = 3.4 m
The distance the wagon travels is 3.4 m. So the correct option is d.
Explanation:To find the distance the wagon travels, we need to use the work-energy principle. Since the force is applied at an angle, we can find the horizontal component of the force by multiplying the force by the cosine of the angle. The work done is equal to the force multiplied by the distance, so we can rearrange the equation to solve for distance. Therefore, the distance the wagon travels is equal to the work done divided by the force component, or 972 J / (310 N * cos(23°)) = 3.4 m.
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An electrostatic paint sprayer has a potential of 25.0 kV that repels paint droplets onto a grounded object. What charge must a 0.63 mg drop of paint have to arrive at the object with a speed of 9 m/s? Give answer in terms of 10^-10 C.
Answer:
10⁻⁶ C
Explanation:
Let the charge be Q
Gain of energy in electric field
= potential x charge
= 25 x 10³ x Q
Kinetic energy of droplet
= .5 x .63 x 10⁻³ x 9 x 9
= 25.515 x 10⁻³ J
So , equating the energy gained
25 x 10³ x Q = 25.515 x 10⁻³
Q = 10⁻⁶ C
Why does an astronaut in a spacecraft orbiting Earthexperience
a feeling of weightlessness?
Answer:
Astronaut in spacecraft while orbiting earth experience weightlessness because there is no gravity of earth or moon is acting on the body of an astronaut.
while on earth, we experience weight because the gravity of earth is acting on our body which is pulling us downward.
Both spacecraft and the astronauts both are in a free-fall condition.
A 12.0-cm long cylindrical rod has a uniform cross-sectional area A = 5.00 cm2. However, its density increases linearly from 2.60 g/cm3 at one end to 18.5 g/cm3 at the other end. This linearly increasing density can be described using the equation ???? = B + Cx. (a) Find the constants B and C required for this rod, assuming the low-density end is placed at x = 0 cm and the high-density end is at x = 12 cm. (b) The mass of the rod can be found using:m=dV=Adx=(B+Cx)Adx
Answer:
(a) The constants required describing the rod's density are B=2.6 and C=1.325.
(b) The mass of the road can be found using [tex]A\int_0^{12}\left(B+Cx)dx[/tex]
Explanation:
(a) Since the density variation is linear and the coordinate x begins at the low-density end of the rod, we have a density given by
[tex]2.6\frac{g}{cm^3}+\frac{18.5\frac{g}{cm^3}-2.6\frac{g}{cm^3}}{12 cm}x = 2.6\frac{g}{cm^3}+1.325x\frac{g}{cm^2}[/tex]
recalling that the coordinate x is measured in centimeters.
(b) The mass of the rod can be found by having into account the density, which is x-dependent, and the volume differential for the rod:
[tex]m=\int\rho dv=\int\left(B+Cx\right)Adx=5\int_0^{12}\left(2.6+1.325x\right)dx=126.6[/tex],
hence, the mass of the rod is 126.6 g.
A baseball catcher is performing a stunt for a television commercial. He will catch a baseball (mass 145 g) dropped from a height of 66.0 m above his glove. His glove stops the ball in 0.0118 s. What is the force (in N) exerted by his glove on the ball? (Indicate the direction with the sign of your answer. Assume the baseball is traveling in the negative direction.)
Answer:
Explanation:
The ball is going down with velocity. It must have momentum . It is stopped by
catcher so that its momentum becomes zero . There is change in momentum . So force is applied on the ball by the gloves .
The rate of change of momentum gives the magnitude of force. This force must be in upward direction to stop the ball. So force is in positive direction .
Let us measure the force applied on the ball .
Final velocity after the fall by 66 m
V = √ 2gh
= √ 2x9.8 x 66
35.97 m /s
Momentum = m v
0.145 x 35.97
= 5.2156 kgms⁻¹
Change in momentum
= 5.2156 - 0
= 5.2156
Rate of change of momentum
= Change of momentum / time = 5.2156 / .0118
= 442 N
if R be the force exerted by gloves to stop the ball
R - mg represents the net force which stops the ball so
R - mg = 442
R = 442 + mg
= 442 + .145 x 908
443.43 N
According to Newton's second law of motion rate of change of momentum is equal to the applied force. The force exerted by the glove of the player on the ball will be 497.4 N.
What is Newton's second law of motion?According to Newton's second law of motion rate of change of momentum is equal to the applied force.
Momentum is given by the product of mass and velocity. it is denoted by P.it leads to the impulsive force.
P = mv
ΔP = mΔv
[tex] \rm{ F = \frac{\delP}{\delt} } [/tex]
[tex]V = \sqrt{2gh} [/tex]
[tex]\rm{V = \sqrt{2\times9.81\times60} [/tex]
v = 34.31 m/sec
ΔV = v -u
ΔV = 34.31-0
ΔV = 34.31 m /sec.
Δt = 0.0118 sec
[tex]\rm{\frac{\delv}{\delt} } [/tex]= 34.31
F = [tex]m\rm{\frac{\delv}{\delt} }[/tex]
F = [tex] 0.145\times3430 [/tex]
F = 497.4 N
Thus the force exerted by the glove of the player on the ball will be 497.4 N.
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The coefficient of static friction between a block and a horizontal floor is 0.4500, while the coefficient of kinetic friction is 0.1500. The mass of the block is 4.400 kg. If a horizontal force is slowly increased until it is barely enough to make the block start moving, what is the net force on the block the instant that it starts to slide?
Answer:
The net force on the block is 19.40 N.
Explanation:
Given that,
Coefficient of static friction = 0.4500
Coefficient of kinetic friction = 0.1500
Mass of block = 4.400 kg
We need to calculate the net force required to make the block slide
[tex]F=\mu_{s} mg[/tex]
Put the value into the formula
[tex]F=0.4500\times4.400\times9.8[/tex]
[tex]F=19.40\ N[/tex]
Hence, The net force on the block is 19.40 N.
Final answer:
The net force on the block the instant it starts to slide is 19.548 N.
Explanation:
The net force on the block the instant it starts to slide can be found by multiplying the coefficient of static friction (μmk) by the normal force (N).
Given that the coefficient of static friction is 0.4500 and the mass of the block is 4.400 kg, we can calculate the normal force using the formula N = mg, where g is the acceleration due to gravity (9.8 m/s²).
Therefore, the net force on the block the instant it starts to slide is N × μmk = 4.400 kg × 9.8 m/s × 0.4500 = 19.548 N.
What is the electric field in N/C) at a point where the force on a -2.1 x 10-5 C charge is (4.31 -6.89) x 10-6 N?
Answer: 0.123 N/C
Explanation: In order to solve this question we have to use the electric Force on a particle produced by an electric field which is given by:
F=q*E
so E=F/q= -2,58* 10^-6/-2.1*10^-5= 0.123 N/C
Three balls are kicked from the ground level at some angles above horizontal with different initial speeds. All three balls reached in their paths the samemaximum height. Ball A covered the total horizontal range D; ball B covered the total horizontal range 1.5D and ball C covered total horizontal range 2D when they hit the ground. Which of these three balls spent the greatest time in flight? (section 4.3) They all have the same time of flight ОА OB oc Band A and B
Answer:
Time of flight A is greatest
Explanation:
Let u₁ , u₂, u₃ be their initial velocity and θ₁ , θ₂ and θ₃ be their angle of projection. They all achieve a common highest height of H.
So
H = u₁² sin²θ₁ /2g
H = u₂² sin²θ₂ /2g
H = u₃² sin²θ₃ /2g
On the basis of these equation we can write
u₁ sinθ₁ =u₂ sinθ₂=u₃ sinθ₃
For maximum range we can write
D = u₁² sin2θ₁ /g
1.5 D = u₂² sin2θ₂ / g
2 D =u₃² sin2θ₃ / g
1.5 D / D = u₂² sin2θ₂ /u₁² sin2θ₁
1.5 = u₂ cosθ₂ /u₁ cosθ₁ ( since , u₁ sinθ₁ =u₂ sinθ₂ )
u₂ cosθ₂ >u₁ cosθ₁
u₂ sinθ₂ < u₁ sinθ₁
2u₂ sinθ₂ / g < 2u₁ sinθ₁ /g
Time of flight B < Time of flight A
Similarly we can prove
Time of flight C < Time of flight B
Hence Time of flight A is greatest .
Suppose that we choose inches as our basic unit of distance and days as our basic unit of time. a. What would the units of velocity and acceleration be in this system? Explain. b. Would this be a good choice of units for measuring the acceleration of an automobile? Explain.
Answer:
a)
velocity unit = [tex]\frac{inches}{days}[/tex]
acceleration unit = [tex]\frac{inches}{days^{2} }[/tex]
b) No
Explanation:
a)
Velocity is a vector expression of the displacement respect to time of something, with magnitude and a defined direction. Then, to know the magnitude of the vector (v) we have to divide the distance (Δx) by the time (Δt) it takes to travel that distance:
v = Δx / Δt
Besides, the unit of velocity is a compound unit between distance units and time units. If we choose inches as our basic unit of distance and days as our basic unit of time, then the unit of velocity would be:
[tex]\frac{inches}{days}[/tex]
On the other hand, acceleration is a vector defined as the rate at which an object changes its velocity. The math expression divides the change o velocity (Δv) by the time (Δt) it takes to make this change to obtain the acceleration magnitude (a):
a = Δv / Δt
So, the unit of acceleration is a compound unit between velocity units and time units. Again, if we choose inches as our basic unit of distance and days as our basic unit of time, then the unit of velocity in this system would be [tex]\frac{inches}{days}[/tex] , as shown. Finally, the unit of acceleration would be:
[tex]\frac{inches}{days^{2} }[/tex]
In resume:
velocity unit = [tex]\frac{inches}{days}[/tex]
acceleration unit = [tex]\frac{inches}{days^{2} }[/tex]
b)
To answer whether this units´ system is a good choice or not for measuring the acceleration of an automobile, let´s think about its normal values.
A car normally takes between 4 to 8 seconds to change its velocity from 0 to 100 km/h, then normal acceleration values for the media of 6 seconds (or its equivalent in hours 0.0016 h) are:
a = Δv / Δt = [tex]\frac{100\frac{km}{h}-0\frac{km}{h} }{0.0016 h}[/tex] = 62,500 [tex]\frac{km}{h^{2} }[/tex]
If we want to use the units system of this exercise, then we have to use the equivalences between inches and kilometers (1 km = 39,370.1 inches) and the other between days and hours (24 h = 1 day). Then,
a = 62,500 [tex]\frac{km}{h^{2} }[/tex] * [tex]\frac{39370.1 inches}{1 km}[/tex] * [tex](\frac{24 h}{1 days}) ^{2}[/tex]
a = 62500 [tex]\frac{km}{h^{2} }[/tex] * [tex]\frac{39,370.1 inches}{1 km}[/tex] * [tex]\frac{576 h^{2} }{days^{2} }[/tex]
a = 1.42 ˣ10¹² [tex]\frac{inches}{days^{2} }[/tex]
If we choose this units´ system to express the acceleration of an automobile, it results in a very high number that introduces a difficulty just to quantify the acceleration. Then, this system is not a good choice for that purpose.
In the return stroke of a typical lightning bolt, a current of 7.01×10^4 A (70.1 kA). exists for 20 μs. How much charge is transferred in this event ?
Answer:
Charge, q = 1.402 C
Explanation:
Given that,
Current from lightning bolt, [tex]I=7.01\times 10^4\ A[/tex]
Time, t = [tex]t=20\ \mu s=2\times 10^{-5}\ s[/tex]
Let q is the charge transferred in this event. We know that the total charge divided by time is called current. Mathematically, it is given by :
[tex]I=\dfrac{q}{t}[/tex]
[tex]q=I\times t[/tex]
[tex]q=7.01\times 10^4\times 2\times 10^{-5}[/tex]
q = 1.402 C
So, 1.402 coulomb of charge is transferred in this event. Hence, this is the required solution.
The radii of curvature of a biconvex lens are 4 cm and 15 cm. The lens is in air, its index of refraction is 1.5. An object is at 1 m before the front surface of the lens. Calculate the distance of the image from the back surface of the lens.
Answer:
-1.19 m
Explanation:
R1 = + 4 cm
R2 = - 15 cm
n = 1.5
distance of object, u = - 1 m
let the focal length of the lens is f and the distance of image is v.
use lens makers formula to find the focal length of the lens
[tex]\frac{1}{f}=\left ( n-1 \right )\left ( \frac{1}{R_{1}}-\frac{1}{R_{2}} \right )[/tex]
By substituting the values, we get
[tex]\frac{1}{f}=\left ( 1.5-1 \right )\left ( \frac{1}{4}+\frac{1}{15} \right )[/tex]
[tex]\frac{1}{f}=\frac{19}{120}[/tex] .... (1)
By using the lens equation
[tex]\frac{1}{f}=\frac{1}{v}-\frac{1}{u}[/tex]
[tex]\frac{19}{120}=\frac{1}{v}+\frac{1}{1}[/tex] from equation (1)
[tex]\frac{1}{v}=\frac{19-120}{120}[/tex]
v = -1.19 m
Find the wavelength in meters for a transverse mechanical wave with an amplitude of 10 cm and a radian frequency of 20π rad/s if the medium through which it travels has a bulk modulus of 40 MPa and a density of 1000 kg/m^3.
Answer:
The wavelength of the wave is 20 m.
Explanation:
Given that,
Amplitude = 10 cm
Radial frequency [tex]\omega = 20\pi\ rad/s[/tex]
Bulk modulus = 40 MPa
Density = 1000 kg/m³
We need to calculate the velocity of the wave in the medium
Using formula of velocity
[tex]v=\sqrt{\dfrac{k}{\rho}}[/tex]
Put the value into the formula
[tex]v=\sqrt{\dfrac{40\times10^{6}}{10^3}}[/tex]
[tex]v=200\ m/s[/tex]
We need to calculate the wavelength
Using formula of wavelength
[tex]\lambda =\dfrac{v}{f}[/tex]
[tex]\lambda=\dfrac{v\times2\pi}{\omega}[/tex]
Put the value into the formula
[tex]\lambda=\dfrac{200\times2\pi}{20\pi}[/tex]
[tex]\lambda=20\ m[/tex]
Hence, The wavelength of the wave is 20 m.
Some homes that use baseboard heating use copper tubing. Hot water runs through and heats the copper tubing, which in turn heats aluminum fins. It is actually the aluminum fins that heat the air rising through the fins. How much energy would it take to heat a section of the copper tubing that weighs about 715.0 g , from 13.65 ∘C to 26.60 ∘C ? Copper has a specific heat of 0.3850 J/(g⋅∘C).
Answer:
19 degrees is the answer
Explanation:
A freight train has a mass of 1.5 X 10^7 kg. If the
locomotivecan exert a constant pull of 7.5 X1^5 N, how long does it
take toincrease the speed of the train from rest to 80 km/h?
Answer:
t=444.4s
Explanation:
m=1.5*10^7 kg
F=7.5*10^5 N
v=80km/h*(1h/3600s)*(1000m/1km)=22.22m/s
Second Newton's Law:
F=ma
a=F/m=7.5*10^5/(1.5*10^7)=0.05m/s^2
Kinematics equation:
vf=vo+at=at
vo: initial velocity equal zero
t=vf/a=22.22/0.05=444.4s
To calculate the time it takes to increase the speed of a train from rest to 80 km/h, you can use Newton's second law of motion.
Explanation:To calculate the time it takes to increase the speed of the train from rest to 80 km/h, we can use Newton's second law of motion. First, we need to calculate the force required to accelerate the train. Given the mass of the train, 1.5 x 10^7 kg, and the acceleration, we can use the formula: force = mass x acceleration
force = (1.5 x 10^7 kg) x (80 km/h to m/s conversion)
Next, we can use the formula: force = mass x acceleration to find the time it takes to accelerate the train:
time = force / (7.5 x 10^15 N)
Plugging in the values, we can calculate the time it takes to increase the speed of the train from rest to 80 km/h.
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A person standing on a cliff extends their arm past the cliff's edge and throws a stone with velocity 16.0 m/s downward. The stone is 80.0 m above ground when it leaves the person's hand. a) Draw a diagram of the problem and label the known and unknown quantities. b) When will the stone land? c) What is the stone's velocity when it lands? Show your work
Answer:
b) 6.03 seconds
c) 43.164 m/s
Explanation:
t = Time taken
u = Initial velocity = 16 m/s
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s²
[tex]v=u+at\\\Rightarrow 0=16-9.81\times t\\\Rightarrow \frac{-16}{-9.81}=t\\\Rightarrow t=1.63 \s[/tex]
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=16\times 1.63+\frac{1}{2}\times -9.81\times 1.63^2\\\Rightarrow s=13.05\ m[/tex]
So, the stone would travel 13.05 m up
So, total height of the stone would fall is 13.05+80 = 93.05 m
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 93.05=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{93.05\times 2}{9.81}}\\\Rightarrow t=4.4\ s[/tex]
b) The stone will land 1.63+4.4 = 6.03 seconds later
[tex]v=u+at\\\Rightarrow v=0+9.81\times 4.4\\\Rightarrow v=43.164\ m/s[/tex]
c) The stone's velocity when it lands is 43.164 m/s
A hot-air balloon is descending at a rate of 1.9 m/s when a passenger drops a camera. If the camera is 47 m above the ground when it is dropped, how much time does it take for the camera to reach the ground? If the camera is 47 m above the ground when it is dropped, what is its velocity just before it lands? Let upward be the positive direction for this problem.
Answer:
The camera lands in t=2.91s with a velocity:
[tex]v=-30.45m/s[/tex]
Explanation:
The initial velocity of the camera is the same as the hot-air ballon:
[tex]v_{o}=-1.9m/s[/tex]
[tex]y_{o}=47m[/tex]
Kinematics equation:
[tex]v(t)=v_{o}-g*t[/tex]
[tex]y(t)=y_{o}+v_{o}t-1/2*g*t^{2}[/tex]
when the camera lands, y=0:
[tex]0=47-1.9t-4.91*t^{2}[/tex]
We solve this equation to find t:
t1=-3.29s This solution have not sense in our physical point of view
t2=2.91s
So, the camera lands in t=2.91s
We replace this value in v(t):
[tex]v=v_{o}-g*t=-1.9-9.81*2.91=-30.45m/s[/tex]
A 24 cm candle is placed 6 m in front of a thin diverging lens with a focal length magnitude of 3 m. Find the image's height and classify it as real or virtual, upright or inverted, enlarged or reduced.
Answer:
Image is virtual and formed on the same side as the object. 2 m from the lens.
The size of the image is 7.97 cm
Image is upright as the magnification is positive and smaller than the object.
Explanation:
u = Object distance = 6 m
v = Image distance
f = Focal length = -3 m (concave lens)
[tex]h_u[/tex]= Object height = 24 cm
Lens Equation
[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{-3}-\frac{1}{6}\\\Rightarrow \frac{1}{v}=\frac{-1}{2} \\\Rightarrow v=\frac{-2}{1}=-2\ m[/tex]
Image is virtual and formed on the same side as the object. 2 m from the lens.
Magnification
[tex]m=-\frac{v}{u}\\\Rightarrow m=-\frac{-2}{6}\\\Rightarrow m=0.33[/tex]
[tex]m=\frac{h_v}{h_u}\\\Rightarrow 0.33=\frac{h_v}{0.24}\\\Rightarrow h_v=0.33\times 0.24=0.0797\ m[/tex]
The size of the image is 7.97 cm
Image is upright as the magnification is positive and smaller than the object.
A hanging magnesium wire with diameter 2.1 mm (2.1 × 10-3 m) is initially 1.15 m long. When a 104 kg mass is hung from it, the wire stretches an amount 8.46 mm. A mole of magnesium has a mass of 24 grams, and its density is 1.74 g/cm3. Find the approximate value of the effective spring stiffness of the interatomic force.
The effective spring stiffness of the interatomic force within a magnesium wire, when subject to stretching by a 104 kg mass, is approximately 120451.067 N/m. This value is calculated using Hooke's Law, which relates the force applied to the extension of the spring (wire) and the spring's stiffness.
Explanation:To find the effective spring stiffness of the interatomic force in a magnesium wire, we can use Hooke's Law, which states that the force F needed to extend or compress a spring by some distance x is proportional to that distance. This is mathematically represented as F = kx, where k is the spring stiffness (or force constant), and x is the extension or compression of the spring (in this case, the wire).
First, we calculate the force applied by the mass hung from the magnesium wire. The force due to gravity is F = mg, where m is the mass (104 kg), and g is the acceleration due to gravity (approximately 9.8 m/s2).
F = 104 kg * 9.8 m/s2 = 1019.2 N.
The elongation of the wire (x) is given as 8.46 mm, which is 8.46 * 10-3 m.
Using Hooke's Law, we can solve for the spring stiffness, k, by rearranging the equation: k = F / x.
k = 1019.2 N / (8.46 * 10-3 m) = 120451.067 N/m (approximately).
This calculation gives us the effective spring stiffness of the interatomic force within the magnesium wire when a 104 kg mass is hung from it, causing it to stretch by 8.46 mm.
The density of an alloy is 5000 kg/m3. What is the density in g/cm 0.005 @ 0.500 5.000 50.000
Answer:
5 g/cm³
Explanation:
Density of an alloy = 5000 kg/m³
The Density of a material is the ratio of the mass by the volume.
The units here are of the metric system
Mass is the resistance a body has which opposes motion when force is applied.
Volume is the amount of material in an object
Convert kg/m³ to g/cm³
1 kg = 1000 g
1 m³ = 10⁻⁶ cm
[tex]5000\ kg/m^3=5000\times 1000\times 10^{-6}=5\ g/cm^3[/tex]
∴ 5000 kg/m³ = 5 g/cm³
Starting from the front door of your ranch house, you walk 50.0 m due east to your windmill, and then you turn around and slowly walk 40.0 m west to a bench where you sit and watch the sunrise. It takes you 28.0 s to walk from your house to the windmill and then 42.0 s to walk from the windmill to the bench. For the entire trip from your front door to the bench, what is your average velocity?
For the entire trip from your front door to the bench, what is your average speed?
Final answer:
The average velocity for the trip from the front door to the bench is 0.143 m/s due east, and the average speed for the same trip is 1.286 m/s.
Explanation:
To calculate the average velocity for the entire trip, we need to consider the displacement (final position relative to the starting position) and the total time taken for the trip. The displacement is 10.0 m east (50.0 m due east - 40.0 m back west), and the total time is 28.0 s + 42.0 s, which equals 70.0 s. Therefore, the average velocity is displacement divided by time, calculated as follows:
Average Velocity = Displacement/Total Time = 10.0 m / 70.0 s = 0.143 m/s due east.
To calculate the average speed, we consider the total distance traveled and the total time. The distance is 50.0 m east + 40.0 m west = 90.0 m, and the time is the same 70.0 s. Thus, the average speed is:
Average Speed = Total Distance/Total Time = 90.0 m / 70.0 s = 1.286 m/s.
Find the values of the root mean square translational speed v, molecules in gaseous diatomic oxygen (O2), gaseous carbon dioxide (CO2), and gaseous diatomic hydrogen (H2), at temperature 100° C.
Answer:
For diatomic oxygen:V=539.06 m/s
For carbon dia oxide:V=459.71 m/s
For dia atomic hydrogen:V=2156.25 m/s
Explanation:
As we know that
Root mean square velocity V
[tex]V=\sqrt{\dfrac{3RT}{M}}[/tex]
Where
R is the gas constant
[tex]R=8.31\ \frac{kg.m^2}{s^2.mol.K}[/tex]
T is the temperature (K).
M is the molecular weight.
For diatomic oxygen:
M=32 g/mol
T=273+100 = 373 K
[tex]R=8.31\ \frac{kg.m^2}{s^2.mol.K}[/tex]
[tex]V=\sqrt{\dfrac{3RT}{M}}[/tex]
[tex]V=\sqrt{\dfrac{3\times 8.31\times 373}{32\times 10^{-3}}}[/tex]
V=539.06 m/s
For carbon dia oxide
M=44 g/mol
T=273+100 = 373 K
[tex]R=8.31\ \frac{kg.m^2}{s^2.mol.K}[/tex]
[tex]V=\sqrt{\dfrac{3RT}{M}}[/tex]
[tex]V=\sqrt{\dfrac{3\times 8.31\times 373}{44\times 10^{-3}}}[/tex]
V=459.71 m/s
For dia atomic hydrogen:
M= 2 g/mol
T=273+100 = 373 K
[tex]R=8.31\ \frac{kg.m^2}{s^2.mol.K}[/tex]
[tex]V=\sqrt{\dfrac{3RT}{M}}[/tex]
[tex]V=\sqrt{\dfrac{3\times 8.31\times 373}{2\times 10^{-3}}}[/tex]
V=2156.25 m/s
If a marathon runner averages 8.83 mi/h, how long does it take him or her to run a 26.22-mi marathon? Express your answers in h, min and s.
Answer:
t=2.97h
t= 10692s
t= 178.2 min
Explanation:
We propose the following ratio:
[tex]\frac{t}{26.22mi} =\frac{1h}{8.83mi}[/tex]
[tex]t=\frac{1h*26.22mi}{8.883mi}[/tex]
t=2.97h
Equivalences
1h=3600s
1h=60 min
Calculation of t in minutes (min) and seconds(s):
t=2.97h*3600s/h= 10692s
t=2.97h*60min/h= 178.2 min
A parallel-plate capacitor is constructed of two square plates, size L×L, separated by distance d. The plates are given charge ±Q. Let's consider how the electric field changes if one of these variables is changed while the others are held constant. Part A What is the ratio E(final)/E(initial) of the final to initial electric field strengths if Q is doubled?
Answer:
E(final)/E(initial)=2
Explanation:
Applying the law of gauss to two parallel plates with charge density equal σ:
[tex]E=\sigma/\epsilon_{o}=Q/(L^{2}*\epsilon_{o})\\[/tex]
So, if the charge is doubled the Electric field is doubled too
E(final)/E(initial)=2
An Airbus A380 airliner can takeoff when its speed reaches 80 m/s. Suppose its engines together can produce an acceleration of 3 m/s. (a) How long does it take the plane to accelerate from 0 m/s to 80 m/s? (b) What is the distance traveled during this period?
Answer:
(a). The time is 26.67 sec.
(b). The distance traveled during this period is 1066.9 m.
Explanation:
Given that,
Speed = 80 m/s
Acceleration = 3 m/s
Initial velocity = 0
(a). We need to calculate the time
Using equation of motion
[tex]v = u+at[/tex]
[tex]t = \dfrac{v-u}{a}[/tex]
Put the value into the formula
[tex]t = \dfrac{80-0}{3}[/tex]
[tex]t =26.67\ sec[/tex]
The time is 26.67 sec.
(b). We need to calculate the distance traveled during this period
Using equation of motion
[tex]s = ut+\dfrac{1}{2}at^2[/tex]
[tex]s = 0+\dfrac{1}{2}\times3\times(26.67)^2[/tex]
[tex]s =1066.9\ m[/tex]
The distance traveled during this period is 1066.9 m.
Hence, This is the required solution.
An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is = 2.2 × 10^-7 C/m^2, and the plates are separated by a distance of 1.3 × 10^-2 m. How fast is the electron moving just before it reaches the positive plate?
Answer:[tex]1.066\times 10^7 m/s[/tex]
Explanation:
Given
Charge per unit area on each plate([tex]\sigma [/tex])=[tex]2.2\times 10^{-7}[/tex]
Plate separation(y)=0.013 m
and velocity is given by
[tex]v^2-u^2=2ay[/tex]
where a=acceleration is given by
[tex]a=\frac{F}{m}=\frac{eE}{m}[/tex]
e=charge on electron
E=electric field
m=mass of electron
[tex]E=\frac{\sigma }{\epsilon _0}[/tex]
[tex]a=\frac{e\sigma }{m\epsilon _0}[/tex]
substituting values
[tex]v=sqrt{\frac{2e\sigma y}{m\epsilon _0}}[/tex]
[tex]v=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 2.2\times 10^{-7}\times 0.013}{9.1\times 10^{-31}\times 8.85\times 10^{-12}}}[/tex]
[tex]v=1.066\times 10^7 m/s[/tex]
The velocity of an electron just before it reaches the positive plate of a parallel plate capacitor can be calculated by equating the converted potential energy to kinetic energy, and then solving for velocity using the electric field, the charge and mass of the electron.
Explanation:Calculating the Final Velocity of an Electron in a Parallel Plate Capacitor
To find out how fast an electron is moving just before it reaches the positive plate of a parallel plate capacitor, we can use the concepts of electric fields and potential energy. The electric field E between the plates can be found using the charge per unit area σ (sigma) and the vacuum permittivity ε₀ (epsilon nought), given by E = σ / ε₀. Once the electric field is known, we can determine the force on the electron as F = qE, where q is the charge of the electron.
Since the electron starts from rest, the potential energy at the negative plate is converted entirely into kinetic energy just before it hits the positive plate. We can use the electron's charge and the potential difference (V) between the plates to find this energy: qV. We know that the kinetic energy is ½mv², where m is the mass of the electron and v is the final velocity.
Setting the potential energy equal to the kinetic energy, we get qV = ½mv². Solving for the final velocity, v, we find that v = √(2qV/m). Since the potential difference V can be calculated as the product of electric field E and the distance d between the plates, we can substitute V with Ed. The final equation for velocity is therefore v = √(2qEd/m). This will give us the velocity of the electron just before it reaches the positive plate.
A boy pulls with a 20 N force, at a 20 degree incline. What part of the force moves the wagon?
Answer:
horizontal direction force move wagon at 18.79 N
Explanation:
given data
force F = 20 N
angle = 20 degree
to find out
What part of the force moves the wagon
solution
we know here as per attach figure
boy pull a wagon at force 20 N at angle 20 degree
so there are 2 component
x in horizontal direction i.e F cos20
and y in vertical direction i.e F sin20
so we can say
horizontal direction force is move the wagon that is
horizontal direction force = F cos 20
horizontal direction force = 20× cos20
horizontal direction force = 18.79 N
so horizontal direction force move wagon at 18.79 N
The speed of light is 2.998*10^8 . How far does light travel in 1.0µs?Set the math up. But don't do any of it. Just leave your answer as a math expression. Also, be sure your answer includes all the correct unit symbols.
The distance light travels in 1.0 µs is approximately 3.00 × 10^2 m.
Explanation:The speed of light is approximately 3.00 × 108 m/s. To find out how far light travels in 1.0 µs, we need to multiply the speed of light by the time taken.
Distance = Speed × Time
Distance = (3.00 × 108 m/s) × (1.0 × 10-6 s)
Calculating this value gives us:
Distance = 3.00 × 102 m
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To find how far light travels in 1.0μs, multiply the speed of light (2.998×108 m/s) by the time interval (1.0×10-6 s), yielding an expression for distance with units in meters.
Explanation:To calculate the distance light travels in 1.0μs (microsecond), you can use the formula for distance where distance traveled is speed multiplied by time. The speed of light (c) in a vacuum is a fundamental constant, and we will use the value 2.998×108 m/s for our calculations.
The time given is 1.0μs, which needs to be converted into seconds to match the units of the speed of light. Remember that 1μs is equal to 1×10-6 seconds. Therefore, the distance light travels in 1.0μs is a matter of multiplying the speed of light by this time.
The math expression without doing any calculation will be:
Distance = c × time
Distance = (2.998×108 m/s) × (1.0×10-6 s)
The units for the final answer will be in meters (m).
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Abel is not sure why, from h = vi2 sin2 θi 2g , the height the tennis ball reaches is maximum when θi = 90°, and asks Kato to explain. Which of Kato's responses is correct? "When θi = 90°, sin2 θi = sin(2θi) = 1, which is its maximum value, so h is maximum." "When θi = 90°, sin2 θi is maximum, so h is maximum." "When θi = 90°, sin θi = 1, sin2 θi = 2 · 1 = 2, which is its maximum value, so that means h is maximum." "When θi = 90°, sin2 θi is minimum, so h is maximum."
The correct response from Kato is: 'When θi = 90°, sin^2 θi = sin(2θi) = 1, which is its maximum value, so h is maximum.
Explanation:
When a tennis ball is thrown into the air at an angle, its height can be calculated using the equation:
h = (vi^2 * sin^2(θi)) / (2g)
where:
h is the heightvi is the initial velocityθi is the launch angleg is the acceleration due to gravityThe term sin^2(θi) represents the square of the sine of the launch angle.
When the launch angle is 90°, sin^2(θi) equals 1, which is its maximum value. This means that the height the tennis ball reaches is maximum when the launch angle is 90°.
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Kato's second response, "When θi = 90°, sin^2(θi) is maximum, so h is maximum," is the correct explanation.
Explanation:
To determine the maximum height of a tennis ball, we can use the equation:
h = (vi^2 * sin^2(θi)) / (2g)
where h is the maximum height, vi is the initial velocity, θi is the launch angle, and g is the acceleration due to gravity.
Let's analyze Kato's responses:
"When θi = 90°, sin^2(θi) = sin(2θi) = 1, which is its maximum value, so h is maximum."Therefore, Kato's second response, "When θi = 90°, sin^2(θi) is maximum, so h is maximum," is the correct explanation.
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A 2450-Hz sound source and an observer are moving away from each other. The source is moving at 15.0m/s, and the observer is moving at 25.0m/s. Assuming the speed of sound to be 343 m/s, what is the frequency heard by the observer?
Answer:
[tex]f'=2176.256983Hz[/tex]
Explanation:
The relationship between observed frequency f' and the emitted frequency f is given by the doppler effect equation. In this case the observer and the source are moving in opposite direction away from each other, so:
[tex]f'=\frac{c-v_0}{c+v_s} f[/tex]
Where:
[tex]c=Speed-of-the-sound-waves=343m/s[/tex]
[tex]v_0=Velocity -of-observer=25m/s[/tex]
[tex]v_s=Velocity -of-source=15m/s[/tex]
[tex]f=Emitted -frequency=2450Hz[/tex]
[tex]f'=Observed-frequency[/tex]
Evaluating the data in the equation:
[tex]f'=\frac{343-25}{343+15}*2450=2176.256983Hz[/tex]
A uniform continuous line charge with net positive charge Q and length L lies on the x-axis from −L2 to +L2. This problem asks about the electric field at a point on the +y-axis: (0,a). continuous line charge
(1) What is the direction of the x-component of the electric field at (0,a)?
(A) +x(to the right)
(B) -x (to the left)
(C) zero (there is no net horizontal component of the E-field)
Answer:
(C) zero (there is no net horizontal component of the E-field)
Explanation:
If we subdivide the bar into small pieces, each piece (dx) contains a charge (dq), the electric field of each piece is equivalent to the field of a punctual electric charge, and has a direction as shown in the attached figure. For each piece (dx) in the negative axis there is another symmetric piece (dx) in the positive axis, and as we see in the figure for symmetry the sum of their electric fields gives a resultant in the Y axis (because its components in X are cancelled by symmetry).
Then the resultant of the electric field will be only in Y.
(C) zero (there is no net horizontal component of the E-field)
The x-component of the electric field at point (0,a) due to a uniform continuous line charge on the x-axis is zero, due to the symmetrical distribution of charge and corresponding cancellation of horizontal electric field components.
Explanation:The student is asking about the direction of the electric field at a point on the positive y-axis due to a uniform continuous line charge distributed along the x-axis. To find the direction of the x-component of the electric field at the point (0,a), we can consider the symmetry of the charge distribution. For any small element of charge on the positive side of the x-axis, there is an identical element of charge on the negative side at the same distance from the origin. The electric fields produced by these two elements at point (0,a) on the y-axis will have the same magnitude but opposite x-components. These x-components will cancel each other out, resulting in a net x-component of the electric field being zero. Therefore, the correct answer to the student's question is (C) zero (there is no net horizontal component of the electric field).