Answer:
a)
f = 1.59*10⁸ Hz.
λ = 0.42 m
vp = 0.67*10⁸ m/s = 0.22 c
b)
γ = 0.31 1/m
Explanation:
a)
In a transmission line, voltage varies sinusoidally with the distance and time.We can express mathematically this relationship in a cosine form as follows:V (x,t) = A cos (kx-ωt + φ₀) (1)where A = amplitude of the wave (Volts)k = number of wave = 2*π/λ (being λ the wavelength of the wave)ω = angular frequency = 2*π*f (being f the frequency of the wave)φ₀ = phase of the wave.In our case, the expression for the voltage is as follows:v (x,t) = 10 V Exp [-γx] cos (15x-10⁹ t) (2)From (1) and (2) we find the following equalities:k = 15 = 2*π/λ (3)Solving for λ:[tex]\lambda = \frac{2*\pi }{k} =\frac{2*\pi }{(15)1/m} = 0.42 m (4)[/tex]
ω = 2*π*f = 10⁹ rad/sec (5)Solving for f:[tex]f = \frac{\omega }{2*\pi } =\frac{10e9rad/sec }{(2*\pi rad} = 1.59*e8 (6)[/tex]
The phase velocity of the wave is just the product of the wavelength times the frequency, as follows:[tex]v_{phase} =\lambda * f = 0.42m * 1.59e8 1/sec = 0.67e8 m/s (7)[/tex]
b)
We can express the amplitude at any time as follows:[tex]V = 10 V *e^{-\gamma*x} (8)[/tex]
If we know that V= 4 V at x=3m, (8) becomes:[tex]10 V *e^{-\gamma*3m} = 4 V (9)[/tex]
Taking ln on both sides, we can solve for γ as follows:[tex]-\gamma*3m* ln e = -\gamma*3m = ln (\frac{4V}{10V}) = ln 0.4\\\gamma = -\frac{ln 0.4}{3m} = 0.31 (1/m) (10)[/tex]
γ = 0.31 1/m
1. An object with a mass of 5 kg is pushed by a force of 10 N. What is the object's acceleration?
Answer:2m/s^2
Explanation:
a=f/m
7. Reid runs away from Joharri after an intense political debate. His initial acceleration is
Okm/h and his final acceleration is 25 km/hour in 10 seconds. What is Reid's average
acceleration?
Answer:
0.6944 m/sec^2
Explanation:
The computation of the average acceleration is given below:
a = v - u ÷ t
where
a denotes average acceleration
v denotes final velocity
u denotes initial velocity
t denotes time
So, the average acceleration is
= (25 - 0) ÷ 10
= 0.6944 m/sec^2
What do Ice core samples with lower ratios of O-18 to O-16 isotopes tell scientist about past climates
Answer:
Ocean-floor sediments can also be used to determine past climate. ... Ice cores contain more 16O than ocean water, so ice cores have a lower 18O/ 16O ratio than ocean water or ocean-floor sediments. Water containing the lighter isotope 16O evaporates more readily than 18O in the warmer subtropical regions
Explanation:
In the legend of William Tell, Tell is forced to shoot an apple from his son's head for failing to show respect to a high official. In our case, let's say Tell stands 8.7 meters from his son while shooting. The speed of the 144-g arrow just before it strikes the apple is 20.4 m/s, and at the time of impact it is traveling horizontally. If the arrow sticks in the apple and the arrow/apple combination strikes the ground 8 m behind the son's feet, how massive was the apple
Answer:
M = 0.31 kg
Explanation:
This exercise must be done in parts, let's start by finding the speed of the set arrow plus apple, for this we define a system formed by the arrow and the apple, therefore the forces during the collision are internal and the moment is conserved
let's use m for the mass of the arrow with velocity v₁ = 20.4 m / s and M for the mass of the apple
initial instant. Just before the crash
p₀ = m v₁ + M 0
instant fianl. Right after the crash
p_f = (m + M) v
p₀ = p_f
m v₁ = (m + M) v
v =[tex]\frac{m}{m+M} \ v_1[/tex] (1)
now we can work the arrow plus apple set when it leaves the child's head with horizontal speed and reaches the floor at x = 8 m. We can use kinematics to find the velocity of the set
x = v t
y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²
when it reaches the ground, its height is y = 0 and as it comes out horizontally, [tex]v_{oy} = 0[/tex]
0 = h - ½ g t²
t² = 2h / g
For the solution of the exercise, the height of the child must be known, suppose that h = 1 m
t = [tex]\sqrt{ \frac{ 2 \ 1}{9.8} }[/tex]
t = 0.452 s
let's find the initial velocity
v = v / t
v = 8 / 0.452
v = 17.7 m / s
From equation 1
v = m / (m + M) v₁
m + M = [tex]m \ \frac{v_1}{v}[/tex]
M = m + m \ \frac{v_1}{v}
we calculate
M = 0.144 + 0.144 [tex]\frac{20.4}{17.7}[/tex]
M = 0.31 kg
Are we within earths Roche limit
Answer: Closer to the Roche limit, the body is deformed by tidal forces. Within the Roche limit, the mass' own gravity can no longer withstand the tidal forces, and the body disintegrates. hope this helps can u give me brainliest
Explanation:
Which of these would have the highest temperature?
ice
· Water
water vapor
Answer:
water vapor
Explanation:
did assignment on edge
A Sonometer wire of length
l vibrates at a
frequency of
350 Hz. when the length of the
wire is increased by 15cm, the
Wire Vibrates at 280 H₂ under
constant tension. Determine l
Answer:
-75 cm
Explanation:
At l ; F = 350 Hz
At l + 15 cm ; F = 280 Hz
I = 350
I + 15 = 280
280I = 350(I + 15)
280I = 350I + 5250
280I - 350I = 5250
-70I = 5250
I = - 75cm
The length is - 75 cm
There is a 247–m–high cliff at Half Dome in Yosemite National Park in California. Suppose a boulder breaks loose from the top of this cliff. What is the velocity of the boulder just before it strikes the ground?
Answer:
Vf = 69.61 m/s
Explanation:
We will use the third equation of motion to solve this problem:
[tex]2gh = V_{f}^2 - V_{i}^2\\[/tex]
where,
g = acceleration due to gravity = 9.81 m/s²
h = height of cliff = 247 m
Vf = final velocity = ?
Vi = initial velocity = 0 m/s (boulder breaks loose from rest)
Therefore,
[tex](2)(9.81\ m/s^2)(247\ m) = V_{f}^2 - (0\ m/s)^2\\V_{f} = \sqrt{4846.14\ m^2/s^2}\\[/tex]
Vf = 69.61 m/s
When an
object is placed 15 cm from "
a concave mirror, a real image magnified
3 times is formed - Find
a) the focal length of the mirror
b) where the object must be placed to
give a virtual image 3 times the height
of the object
Answer:
Focal length(f)= -15 cm, magnification= -3 (image is real). So, -v/u=-3 ; or v=3u {v=image distance and u=object distance}. Using mirror formula,
1/f= 1/v + 1/u
-1/15 = 1/3u + 1/u
4/3u = -1/15
u= -20 cm
v =3u= 3×(-20)= -60 cm
So object is 20 cm and image is 60 cm in front of the mirror.
Explanation:
A 45.0 kilogram boy is riding a 15.0-kilogram bicycle with a speed of 8.00 meters per second. What is the combined kinetic energy of the boy and the bicycle? A)480.J B)240.0J C)1920J D)1440J
Answer:
1920Joules
Explanation:
The formula for calculating the kinetic energy of a body is expressed as;
KE = 1/2 mv²
m isthe mass
V is the speed
For the two masses, the combined KE is expressed as;
KE = 1/2(m1+m2)v²
KE = 1/2(45+15)(8)²
KE = 1/2 * 60 * 64
KE = 30 * 64
KE = 1920J
Hence the combined kinetic energy of the boy and the bicycle is 1920Joules
The combined kinetic energy of the boy and the bicycle is of 1920 J.
Given data:
The mass of boy is, m = 45.0 kg.
The mass of bicycle is, M = 15.0 kg.
The speed of bicycle is, v = 8.00 m/s.
The kinetic energy of an object is defined as the energy possessed by an object by virtue of motion of object. The combined kinetic energy of the boy-bicycle system is given as,
[tex]KE = \dfrac{1}{2}(m+M)v^{2}[/tex]
Solve by substituting the values as,
[tex]KE = \dfrac{1}{2}(45+15) \times 8^{2}\\\\KE = 1920 \;\rm J[/tex]
Thus, we can conclude that the combined kinetic energy of the boy and the bicycle is of 1920 J.
Learn more about the concept of kinetic energy here:
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can someone please help me I am so behind I neee to catch up but I need it to be correct both of them
Answer:
1.B, 2.A
Explanation:
HELP DUE 3 MINUTESSSSD
Answer:
Break down small pebbles and sediments, like sand
Break down large rocks like mountains
Explanation:
6. As distance increases, gravitational force *
(10 Points)
increases
decreases
6. A garden hose attached to a nozzle is used to fill a 15-gal bucket. The inner diameter of the hose is 1.5 cm, and it reduces to 0.8 cm at the nozzle exit. If it takes 50 s to fill the bucket with water (density = 1 kg/L), determine (a) the volume and mass flow rates of water through the hose, and (b) the average velocity of water at the nozzle exit.
Answer: 1.135 L/s; 1.35 kg/s, 22.57 m/s
Explanation:
Given
Volume of bucket [tex]V=15\ gal\approx 56.78\ L[/tex]
time to fill it [tex]t=50\ s[/tex]
Volume flow rate
[tex]\dot{V}=\dfrac{56.78}{50}=1.135\ L/s\approx 1.135\times 10^{-3}\ m^3/s[/tex]
The inner diameter of the hose [tex]D=1.5\ cm[/tex]
diameter of the nozzle exit [tex]d=0.8\ cm[/tex]
we can volume flow rate as
[tex]\Rightarrow \dot{V}=Av\quad \quad \text{v=average velocity through nozzle exit}\\\\\Rightarrow 1.135\times 10^{-3}=\frac{\pi }{4}d^2\times v\\\\\Rightarrow 1.135\times 10^{-3}=\frac{\pi }{4}(0.8\times 10^{-2})^2\times v\\\\\Rightarrow v=\dfrac{4\times 1.135\times 10^{-3}}{\pi \times 64\times 10^{-6}}=22.57\ m/s[/tex]
Mass flow rate
[tex]\Rightarrow \dot{m}=\rho \times \dot{V}\\\Rightarrow \dot{m}=1\ kg/L\times 1.135\ L/s=1.35\ kg/s[/tex]
Someone help me I will mark brainless it’s 2 questions .
Answer:
1.C,2.A
Explanation:
given what you think employers value,how do you think should behave at work?
Explanation:
Employers value professionalism. They are looking for someone worth every penny they would pay. Someone who can delivery before deadlines, meet daily targets and has minimum complaints. One must be a team person and can coordinate with other members of the team well.
Therefore, employee must act accordingly. He must exhibit professionalism and delivery the task assigned on time. At the same employee must have his/her own opinions and vision for the organization.
If the net force acting on an object is 0 N, you can be sure that the forces acting on the object are
A. balanced B.Unbalanced C. acting at the same direction
I think the answer would be A.
After all, it is 0 which is technically a dead center number meaning that the net should be balanced and still.
Hope this helps and have a nice day.
-R3TR0 Z3R0
can someone help me with this
Answer:
8/3 m/s² or 2.67m/s²
Explanation:
According to the second Law of Newton, the acceleration of an object is directly related to the net force and inversely related to its mass. This can be translated in a mathematical equation:
a= F/m
a=4000N/1500kg
a=8/3m/s²
or in decimal form
2.67m/s²
which data set has the largest range?
A. 5,6,75
B. 102,105,103,110
C. 14,19,18,35,32,38,40
D. 1,9,2,5,4,19,2,4,3
Answer: A is correct because
A.70 as a range
B. 8 as an range
C.26 as an range
D.18 as an range
a force of 17 acts on a block of mass 4kg the forces of friction opposing the motion is 5n . the acceleration of the block is
The force applied on the block of mass 4 kg is 17 N and a frictional force of 5 N is opposing this force. Hence, the net force is 8N and the acceleration is 2 m/s².
What is net force?Force is a vector quantity thus, characterized by a magnitude and direction. If two same or different forces are acting on a body from the same direction, the net force will be the sum of the magnitude of the two forces.
If two forces are acting from opposite directions, they will cancel each other and the net force is the substracted value of their magnitudes.
Given that, mass of the block = 4 kg
force applied on the block = 17 N
frictional force = 5 N
Net force = 17 - 5 = 8 N.
Force = mass × acceleration. (by Newton's second law of motion)
acceleration = force/ mass
= 8 N / 4 Kg = 2 m/s²
Therefore, the acceleration of the block is 2 m/s².
To find more on net force, refer here:
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The velocity of an object is +47 m/s at 3.0 seconds and is +65 m/s at 12.0 seconds. Calculate the acceleration of the object
Answer:
[tex]\boxed{\text{\sf \Large 2.0 m/s^2 $}}[/tex]
Explanation:
Use acceleration formula
[tex]\displaystyle \text{$ \sf acceleration=\frac{change \ in\ velocity}{change \ in \ time} $}[/tex]
[tex]\displaystyle a=\frac{65-47}{12.0-3.0} =2.0[/tex]
A dock worker applies a constant horizontal force of 81.0 N to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves a distance 13.0 m in a time of 5.20 s .
Required:
a. What is the mass of the block of ice?
b. If the worker stops pushing at the end of 5.00 s, how far does the block move in the next 5.00 s?
Answer:
The correct answer is:
(a) 84.240 kg
(b) 24.038 m
Explanation:
The given values are:
Force,
F = 81.0 N
Distance,
S = 13.0 m
Time,
t = 5.20 s
As we know,
The acceleration of mass will be:
⇒ [tex]a=\frac{2S}{t^2}[/tex]
On substituting the given values, we get
⇒ [tex]=\frac{2\times 13.0}{(5.20)^2}[/tex]
⇒ [tex]=\frac{26}{27.04}[/tex]
⇒ [tex]=0.961538 \ m/s^2[/tex]
(a)
The mass of the block will be:
⇒ [tex]m=\frac{F}{a}[/tex]
On substituting the given values, we get
⇒ [tex]=\frac{81.0}{0.961538}[/tex]
⇒ [tex]=84.240 \ kg[/tex]
(b)
The final velocity after a given time i.e.,
t = 5.00 s
⇒ [tex]v=at[/tex]
On substituting the values, we get
⇒ [tex]=0.961538\times 5.00[/tex]
⇒ [tex]=4.8076 \ m/s[/tex]
In time, t = 5.00 s
The distance moved by the block will be:
⇒ [tex]d=vt[/tex]
On putting the values, we get
⇒ [tex]=4.8076\times 5.00[/tex]
⇒ [tex]=24.038 \ m[/tex]
5) In the last part of step 7 of the procedure, you measured the resistance of the flashlight when it had no current passing through it. The resistance of the flashlight is different, however, when current is passing through it. Explain how your measurement of the resistance of the variable resistor obtained in part 7 is a valid approximation of the resistance of the flashlight when it had current passing through it. Is the resistance higher when the flashlight is on or off
Answer:
Following are the responses to this question:
Explanation:
The small current passes thru the capacitor of the strain gauge and the current is generated throughout the resistor. For the very first time, in contrast to what we calculate, its resistance of the multimeter is quite high and therefore the small stream flowing through the bulb would have very little impact on the measure. Thus, as the current flows through the flashbulb, this same calculation is of excellent price, its material is heated and resistance varies with increase. Therefore, when the bulb will be on, sensitivity is greater.
g A rigid air cylinder with a volume of 100 cm3 is punctured with a hole having a crosssectional area of 0.3 mm2 . The original pressure and temperature of the air inside the cylinder are 800 kPa and 35 C. As the air leaves the hole in the cylinder, it reaches a pressure of 100 kPa and a temperature of 5 C. The velocity of the air as it escapes through the hole is 100 m/s. Calculate the original mass of air inside the tank and the mass in the tank 5 seconds after it is punctured, assuming the exit conditions of the air remain independent of time g
Solution :
The volume of the rigid cylinder = [tex]$100 \ cm^3 = 100 \times 10^{-6} \ m^3$[/tex]
Initial pressure inside the cylinder, [tex]$P_i = 800 \ kPa$[/tex]
Initial temperature inside the cylinder, [tex]$T_i = 35^\circ C= 308 \ K$[/tex]
Final temperature inside the cylinder, [tex]$T_f = 5^\circ C= 278 \ K$[/tex]
Final pressure inside the cylinder, [tex]$P_f = 100 \ kPa$[/tex]
Area of the hole, A = [tex]$0.3 \ mm^2 = 3 \times 10^{-7} \ m^2$[/tex]
Velocity of the air through the hole, V = 100 m/s
The final pressure and the temperature inside the cylinder will be the condition same as the ambient conditions.
At initial state, from the equation of state,
PV = mRT, where R = 287 J/kg-K for air
[tex]$800 \times 10^3 \times100 \times10^{-6} = m_1 \times 287 \times 308$[/tex]
∴ [tex]$m_1=9.1 \times 10^{-4} \ kg$[/tex]
Since the exit condition does not change with time, we have ,
At ambient condition, [tex]$P_f = 100 \kPa$[/tex] and [tex]$T_f= 278 \ K$[/tex].
Therefore, we can find the density of the air
[tex]$P=\rho R T$[/tex]
[tex]$\rho = \frac{P}{RT}$[/tex]
[tex]$=\frac{100 \times 10^3}{287 \times 278}$[/tex]
[tex]$= 1.25 \ kg/m^3$[/tex]
Mass flow rate of air from the cylinder = [tex]$\dot m$[/tex]
[tex]$\dot m$[/tex] can be written as [tex]$\dot m$[/tex] [tex]$=\rho_{f} \times A \times v$[/tex]
[tex]$\dot m$[/tex] = [tex]$1.25 \times 3 \times 10^{-7} \times 100$[/tex]
[tex]$\dot m$[/tex] = [tex]$3.75 \times 10^{-5}$[/tex] kg/s
Mass escaped from the cylinder in 5 seconds
[tex]$m=3.75 \times 10^{-5} \times 5$[/tex]
[tex]$= 1.875 \times 10^{-4} \ kg$[/tex]
Mass of air remaining in the cylinder after 5 seconds :
[tex]$m_2 = m_1 - m$[/tex]
[tex]$m_2 = 9.1 \times 10^{-4} - 1.875 \times 10^{-4}$[/tex]
[tex]$m_2 = 7.225 \times 10^{-4} \ kg$[/tex]
= 0.7225 grams
Mitch holds a pumpkin at waist level. How can he add potential energy to the pumpkin?
Answer:
raise it as high as he can
Explanation:
What happens to solar radiation when it is absorbed
Answer:
Absorbed sunlight is balanced by heat radiated from Earth's surface and atmosphere. ... The atmosphere radiates heat equivalent to 59 percent of incoming sunlight; the surface radiates only 12 percent. In other words, most solar heating happens at the surface, while most radiative cooling happens in the atmosphere
1. A basket coffee filter (see below) is very light and has a large drag coefficient. It is possible to stack several filters together so they have the same drag coefficient as a single filter. Suppose you tried dropping one filter from a ladder, then tried dropping two stacked filters from the same height, then 3 and so on. For each try you measure the time taken for the filters to fall to the floor. How would you expect the time for the filters to fall to compare to the number of filters
Answer:
he times are getting closer as we use each filter, in the expression n would be the number of filters
t = [tex]\sqrt{ \frac{2(y_o - (n-1) h)}{g} }[/tex]
Explanation:
For this exercise let's use the kinematics relations
y = y₀ + v₀ t - ½ g t²
When the first filter reaches the ground, its height is y = 0, as they release its initial velocity is zero
for the 1st filter
0 = y₀ - ½ g t²
t² = 2y₀ / g
t = [tex]\sqrt{ \frac{2y_o}{g} }[/tex]
when we release the second filter upon arrival it has a height y = h where h is the height of each filter
h = y₀ - ½ g t²
t = [tex]\sqrt{ \frac{2(y_o- h)}{g} }[/tex]
when we release the third filter it reaches y = 2h
2h = y₀ - ½ g t²
t = [tex]\sqrt{ \frac{2(y_o -2h)}{g} }[/tex]
we can write the terms of this succession
(n-1) = y₀ - [tex]\frac{1}{2}[/tex] g t²
t = [tex]\sqrt{ \frac{2(y_o - (n-1) h)}{g} }[/tex]
therefore we see that the times are getting closer as we use each filter, in the expression n would be the number of filters
An object swings in a horizontal circle, supported by a 1.8-m string. It swings at a speed of 3 m/s. What is the mass of the object given that the tension in the string is 90 N?
Answer:
Mass = 18 kg
Explanation:
Formula for force in centripetal motion is;
F = mv²/r
We have;
Mass; m.
Speed; v = 3 m/s
radius; r = 1.8 m
Force; F = 90 N
Thus;
Making m the subject;
m = Fr/v²
m = 90 × 1.8/3²
m = 18 kg
If the object on the Moon were raised to a height of 30.0 m, what would be the potential energy? PE=mgh (g on the Moon is 1.62m/s)
The tensile strength (the maximum tensile stress it can support without breaking) for a certain steel wire is 3000 MN/m2. What is the maximum load that can be applied to a wire with a diameter of 3.0 mm made of this steel without breaking the wire?
Answer:
The correct answer is "21195 N".
Explanation:
The given values are:
Tensile strength,
= 3000 MN/m²
Diameter,
= 3.0 mm
i.e.,
= 3×10⁻³ m
Now,
The maximum load will be:
= [tex]Tensile \ strength\times Area[/tex]
On substituting the values, we get
= [tex](3000\times 10^6)(\frac{\pi}{4} (3\times 10^{-3})^2)[/tex]
= [tex](3000\times 10^6)(\frac{3.14}{4} (3\times 10^{-3})^2)[/tex]
= [tex]21195 \ N[/tex]