Answer:
D
Explanation:
Continuous data is a characteristic of analog signals though been smooth is a subject of question. The smooth signal is after it's been processed. However a wave can still stay continuous defined by what we call duty circle.
Smooth and continuous data is one of the major features of an analog signal.
Analog signal isn't as accurate as the digital signal and it contains minimum
and maximum values which could be positive or negative.
Analog signal comprises of smooth and continuous data. This is explained
by one time-varying quantity representing another time-based variable.In
this type of signal , a variable is an analog of the other.
Analog signals examples include the following : Human voice, Thermometers, analog clocks etc
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You are designing a hydraulic lift for a machine shop. The average mass of a car it needs to lift is about 1500 kg. You wish to exert a force on a smaller piston of not more than 550N .a) What should be the specifications on the dimensions of the pistons?Asmall piston/Alarge piston = ???b) How far down will you need to push the piston in order to lift the car 50cm ?h = ???
Answer:
(a) Area(small piston)/Area(large piston) = 0.037
(b) h = 1336.36 cm = 13.36 m
Explanation:
(a)
The stress on the smaller piston is equally transmitted to the larger piston, in a hydraulic lift. Therefore,
Stress (small piston) = Stress (large piston)
Force (small piston)/Area (small piston) = Force (Large Piston)/Area (Large Piston)
Area(small piston)/Area(large piston) = Force (small piston)/Force(Large piston)
Area(small piston)/Area(large piston) = 550 N/(1500 kg)(9.8 m/s²)
Area(small piston)/Area(large piston) = 0.037
(b)
The work is also transmitted equally to the large piston. So,
Work(small piston) = Work(Large Piston)
Force(small piston).Displacement(small piston) = Force(large piston).Displacement(small piston)
(550 N)(h) = (1500 kg)(9.8 m/s²)(50 cm)
h = 735000 N.cm/550 N
h = 1336.36 cm = 13.36 m
(a) The ratio of area smaller piston to area of larger piston is 0.037.
(b) The distance the smaller piston will be pushed down to lift the car at the given height is 13.36 m.
The given parameters;
mass of the car, m = 1500 kgforce on the smaller piston, F₁ = 550 Nlet the area of the small piston = A₁
let the area of the large piston = A₂
Apply constant pressure principle as shown below;
[tex]P = \frac{F_1}{A_1} = \frac{F_2}{A_2} \\\\\frac{A_1}{A_2} = \frac{A_{small}}{A_{large}} = \frac{F_1}{F_2} = \frac{550}{mg} \\\\ \frac{A_{small}}{A_{large}} = \frac{550}{1500 \times 9.8} \\\\ \frac{A_{small}}{A_{large}} = 0.037[/tex]
The height the car was raised = 50 cm = 0.5 mThe distance the effort will be applied is calculated as follows;
[tex]550 d = mgh\\\\550d = (1500 \times 9.8 \times 0.5)\\\\550 d = 7350\\\\d = \frac{7350}{550} \\\\d = 13.36 \ m[/tex]
Thus, the distance the smaller piston will be pushed down to lift the car at the given height is 13.36 m.
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A long glass rod A is initially at 22.0°C. A second rod B is identical to rod A and has the same mass and initial temperature as A. The same amount of heat is supplied to both rods and the two rods A and B reach final temperatures of 86.3°C and 190.0°C respectively. If the specific heat of glass is 0.2007 kcal/(kg· °C), what is the specific heat of the material from which rod B is made?
Answer:
[tex]c_B=0.0768\frac{kcal}{kg\cdot\°C}[/tex]
Explanation:
In order to calculate the specific heat of the material B, you use the following formula for the change in the temperature of a substance, where an amount of heat Q is given to the substance:
[tex]Q=mc(T_2-T_1)[/tex]
Q: amount oh heat
m: mass of the substance
T2: final temperature
T1: initial temperature
c: specific heat of the substance.
If QA and QB are the heat of material A and B, you have:
[tex]Q_A=m_Ac_A(T_{2A}-T_{1A})\\\\Q_B=m_Bc_B(T_{2B}-T_{1B})[/tex]
both materials have the same mass, mA = mB
cA: specific heat of A = 0.2007 kcal/(kg.°C)
cB: specific heat of B = ?
T2A: final temperature of A = 86.3°C
T1A: initial temperature of A = 22.0°C
T2B: final temperature of B = 190.0°C
T1B: initial temperature of B = 22.0°C
In this case you have that both material A and B receive the same amount of heat Q. Then, you equal QA with QB and solve for cB:
[tex]m_Ac_A(T_{2A}-T_{1A})=m_Bc_B(T_{2B}-T_{1B})\\\\c_B=\frac{c_A(T_{2A}-T_{1A})}{(T_{2B}-T_{1B})}\\\\c_B=\frac{(0.2007kcal/(kg.\°C))(86.3\°C-22.0\°C)}{190.0\°C-22.0\°C}\\\\c_B=0.0768\frac{kcal}{kg\cdot\°C}[/tex]
hence, the specific heat of the second rod B is 0.0768kcal/(kg°C)
PLEASE HELP ME, A person has a 340 g can of organic frozen apple juice concentrate. In order to make the most dilute solution, which amount of water should he add? A. 709 mL B. 1064 mL C. 1419 mL D. 1774 mL
A distant galaxy emits light that has a wavelength of 434.1 nm. On earth, the wavelength of this light is measured to be 438.6 nm. A) Decide whether this galaxy is approaching or receding from the earth. Give your reasoning. B) Find the speed of the galaxy relative to the earth.
Answer:
A) receding from the earth
B) [tex]3.078x10^6m/s[/tex]
Explanation:
A) receding from the earthThe wavelength went from 434.1nm to 438.6nm, there was an increase in wavelength (also knowecn as redshift due to the doppler efft), this increase is due to the fact that the source that emits the radiation (the distant galaxy) is moving away and therefore the light waves it emits are "stretched", causing us to see a wavelength greater than the original.
B) [tex]3.078x10^6m/s[/tex]to calculate the relative speed we use the following formula:
[tex]v_{rel}=c(1-\frac{\lambda_{1}}{\lambda_{2}} )[/tex]
where [tex]c[/tex] is the speed of light: [tex]c=3x10^8m/s[/tex]
[tex]\lambda_{1}[/tex] is the wavelength emited by the source, and
[tex]\lambda_{2}[/tex] is the wavelength measured on earth.
we substitute all the values and do the calculations:
[tex]v_{rel}=(3x10^8m/s)(1-\frac{434.1nm}{438.6nm} )\\\\v_{rel}=(3x10^8m/s)(1-0.98974)\\\\v_{rel}=(3x10^8m/s)(0.01026)\\\\v_{rel}=3.078x10^6m/s[/tex]
the relative speed is: [tex]3.078x10^6m/s[/tex]
The energy delivered to the resistive coil is dissipated as heat at a rate equal to the power input of the circuit. However, not all of the energy in the circuit is dissipated by the coil. Because the emf source has internal resistance, energy is also dissipated by the battery as heat. Calculate the rate of dissipation of energy PbatPbatP_bat in the battery.
Answer:
P = I²r
Explanation:
ε= IR + Ir
where r is the internal resistance
A car 4m long moving at a velocity of 25m/s was beside a lorry 20m long with a velocity 19mls. At t =
0,
the distance between them was 10m. How long will it take the car to overtake the lorry (a) 9s (b) 5s (c) 3s
(d) 2s
Answer:
???????
Explanation:
????????
Tired of being chased by a jaguar, you set a trap. Hoping to drop it on the jaguar, you try to push a
44.0 kg stone boulder off of the edge of a cliff that slopes down at an angle of 15.0°. Being weak with
hunger, the best you can do is push the boulder with a force of 222 N. The coefficient of kinetic friction
between the boulder and the ground is is 0.700. (Ignore static friction.)
What is the acceleration of the boulder while you push it down the incline?
Answer: acceleration = 3.27m/s^2
Explanation:
Given that the
Mass M = 44kg
Angle Ø = 15 degree
Coefficient of friction ų = 0.7
Force F = 222N
F - Fr = ma ...... (1)
Where Fr = frictional force
Fr = ųN
N = normal reaction = mg
Fr = ųmgsinØ
Fr = 0.7 × 44 × 9.81 × sin 15
Fr = 78.2N
Substitutes Fr, F and M into equation one.
222 - 78.2 = 44a
143.79 = 44a
Make a the subject of formula
a = 143.79/44
Acceleration a = 3.27 m/s^2
An ideal diatomic gas undergoes a cyclic process. In the first step, the gas undergoes an isothermal expansion from V1 to 3.00 V1. In the second step of the process the gas undergoes an isovolumetric decrease in pressure. In the third step the gas undergoes an adiabatic compression from 3.00 V1 back to V1 completing the cycle.
Required:
a. Sketch the cycle.
b. In terms of P., V. and T., determine P2, P3, T3.
c. In terms of P., V. To determine W, Q and ΔE int for each step. Take T, to be between 100K and 1000K
Answer:
Step 1
Work done = -9134.4 J
ΔQ = -9134.4 J
Step 2
ΔQ = -3570.32 J = ΔU
W = 0
Step 3
The pdV work done = 3570.32 J
The Vdp work done = 11053.37 J
Heat transferred, ΔE = 0.
Explanation:
For diatomic gases γ = 1.4
Step 1
Where:
v₂ = 3.00·v₁
On isothermal expansion of an ideal gas by Boyle's law, we have;
p₁·v₁ = p₂·v₂ which gives;
p₁·v₁ = p₂×3·v₁
Dividing both sides by v₁, we have;
p₁= 3·p₂
[tex]p_2 = \dfrac{p_1}{3}[/tex]
Hence, the pressure is reduced by a factor of 3
Work done =
[tex]n\cdot R\cdot T\cdot ln\dfrac{v_{f}}{v_{i}}[/tex]
Where:
n = 1 mole
R = 8.3145 J/(mole·K)
T = 1000 K we have
[tex]1 \times 8.3145 \times 1000 \times ln\left (\dfrac{1}{3} \right ) = -9134.4 J[/tex]
Step 2
The gas undergoes a constant volume decrease in pressure given by Charles law as follows;
[tex]\dfrac{p_2}{p3} = \dfrac{T_1}{T_3}[/tex]
Whereby p₂ > p₃, T₁ will be larger than T₃
W = 0 for constant volume process
ΔQ = m×cv×ΔT = 1 × 3.97 × -900 = -3570.32 J = ΔU
Step 3
For adiabatic compression, we have;
[tex]\dfrac{p_3}{p_1} = \left (\dfrac{V_1}{V_3} \right )^{\gamma } = \left (\dfrac{T_3}{T_1} \right )^{\frac{\gamma }{\gamma -1}}[/tex]
Where:
T₁ = 1000 K
T₃ = 100 K
We have;
[tex]\left (\dfrac{V_1}{3\cdot V_1} \right )^{\gamma } = \left (\dfrac{100}{1000} \right )^{\dfrac{\gamma}{\gamma -1}}[/tex]
[tex]\left (\dfrac{1}{3} \right ) = \left (\dfrac{1}{10} \right )^{\dfrac{1}{\gamma -1}}[/tex]
[tex]log\left (\dfrac{1}{3} \right ) = {\dfrac{1}{\gamma -1}} \times log \left (\dfrac{1}{10} \right )^[/tex]
[tex]\gamma -1 =\dfrac{log \left (\dfrac{1}{10} \right )}{ log\left (\dfrac{1}{3} \right ) } {[/tex]
∴ γ-1 = 2.096
γ = 3.096
The pdV work done =
[tex]m \times c_v \times (T_1 - T_3)[/tex]
m×R/(γ - 1)×(T₁ - T₃) =
3.97×(1000 - 100) = 3570.32 J
The Vdp work done =
[tex]m \times c_p \times (T_1 - T_3)[/tex]
[tex]c_p = k \times c_v = 3.096 \times 3.97 = 12.3 \, J/(mol\cdot K)[/tex]
12.3×(1000 - 100) = 11053.37 J
Heat transferred, ΔE = 0.
On average, both arms and hands together account for 13% of a person's mass, while the head is 7.0% and the trunk and legs account for 80%. We can model a spinning skater with her arms outstretched as a vertical cylinder (head, trunk, and legs) with two solid uniform rods (arms and hands) extended horizontally. Suppose a 74.0 kg skater is 1.80 m tall, has arms that are each 70.0 cm long (including the hands), and a trunk that can be modeled as being 35.0 cm in diameter. If the skater is initially spinning at 68.0 rpm with her arms outstretched, what will her angular velocity 2 be (in rpm ) after she pulls in her arms and they are at her sides parallel to her trunk? Assume that friction between the skater and the ice is negligble.
Answer:
176.38 rpm
Explanation:
mass percentage of arms and legs = 13%
mass percentage of legs and trunk = 80%
mass percentage of head = 7%
Total mass of the skater = 74.0 kg
length of arms = 70 cm = 0.7 m
height of skater = 1.8 m
diameter of trunk = 35 cm = 0.35 m
Initial angular momentum = 68 rpm
We assume:
The spinning skater with her arms outstretched as a vertical cylinder (head, trunk, and legs) with two solid uniform rods (arms and hands) extended horizontally.friction between the skater and the ice is negligible.We split her body into two systems, the spinning hands as spinning rods
1. Each rod has moment of inertia = [tex]\frac{1}{3} mL^{2}[/tex]
mass m of the arms is 13% of 74 kg = 0.13 x 74 = 9.62 kg
mass of each side will be assumed to be 9.62/2 = 4.81 kg
L = length of each arm
therefore,
I = [tex]\frac{1}{3}[/tex] x 4.81 x [tex]0.7^{2}[/tex] = 0.79 kg-m for each arm
2. Her body as a cylinder has moment of inertia = [tex]\frac{1}{2} mr^{2}[/tex]
r = radius of her body = diameter/2 = 0.35/2 = 0.175 m
mass of body trunk = (80% + 7%) of 74 kg = 0.87 x 74 = 64.38 kg
I = [tex]\frac{1}{2}[/tex] x 64.38 x [tex]0.175^{2}[/tex] = 0.99 kg-m
We consider each case
case 1: Body spinning with arm outstretched
Total moment of inertia = sum of moments of inertia of both arms and moment of inertia of body trunk
I = (0.79 x 2) + 0.99 = 2.57 kg-m
angular momentum = Iω
where ω = angular speed = 68.0 rpm = [tex]\frac{2\pi }{60}[/tex] x 68 = 7.12 rad/s
angular momentum = 2.57 x 7.12 = 18.29 kg-rad/m-s
case 2: Arms pulled down parallel to trunk
The momentum of inertia will be due to her body trunk alone which is 0.91 kg-m
angular momentum = Iω
= 0.99 x ω = 0.91ω
according to conservation of angular momentum, both angular momentum must be equal, therefore,
18.29 = 0.99ω
ω = 18.29/0.99 = 18.47 rad/s
18.47 ÷ [tex]\frac{2\pi }{60}[/tex] = 176.38 rpm
which statement about the image Formed by a plane mirror is correct?
1. the image is larger then the object
2. the image is smaller then the object
3. the image is twice as far from the mirror as the object
4. the image is virtual.
Answer:
The image is virtual
number-4
Two fans are watching a baseball game from different positions. One fan is located directly behind home plate, 18.3 mfrom the batter. The other fan is located in the centerfield bleachers, 127 m from the batter. Both fans observe the batterstrike the ball at the same time(because the speed of light is about a million times faster than that of sound), but the fan behind home plate hears the sound first. What is the time difference between hearing the sound at the two locations? Use 345 m/s as the speed of sound.
Answer:
Δt = 0.315s
Explanation:
To calculate the time difference, in which both fans hear the batterstrike, you first calculate the time which takes the sound to travel the distances to both fans:
[tex]t_1=\frac{d_1}{v_s}[/tex]
[tex]t_2=\frac{d_2}{v_s}[/tex]
d1: distance to the first fan = 18.3 m
d2: distance to the second fan = 127 m
vs: speed of sound = 345 m/s
You replace the values of the parameters to calculate t1 and t2:
[tex]t_1=\frac{18.3m}{345m/s}=0.053s\\\\t_2=\frac{127m}{345m/s}=0.368s[/tex]
The difference in time will be:
[tex]\Delta t =t_2-t_2=0.368s-0.053s=0.315s[/tex]
Hence, the time difference between hearing the sound at the location s of both fans is 0.315s
Which of the following statements best describes the visible spectrum of light as seen by the human eye? The lowest frequency appears , and the highest frequency appears violet. B. The lowest frequency appears red, and the highest frequency appears yellow. C. The lowest frequency appears green, and the highest frequency appears violet. D. The lowest frequency appears green, and the highest frequency appears yellow.
Answer:
The Answer is red is the lowest and violet is the highest frequency
Explanation:
I think that means A, because the red isn't in the question. But I'm sure red is the lowest frequency and violet is the highest in the visible light spectrum
The visible spectrum as it appears to the human eye is that A. the lowest frequency appears red, and the highest frequency appears violet.
Humans can only view a portion of the electromagnetic spectrum and this portion is known as visible light.
The colors in this visible light have different frequencies which include:
Violet with a frequency range of 700 - 790 THzBlue with a frequency range of 600 - 700 THzGreen with a frequency range of 530 - 580 THz Yellow with a frequency range of 510–530 THzOrange with a frequency range of 480–510 THz and, Red with a frequency range of 400–480 THzNotice how red is the lowest frequency and violet is the highest so we can conclusively say that the lowest frequency appears red, and the highest frequency appears violet.
Find out more at https://brainly.com/question/15091042.
Consider a blackbody that radiates with an intensity I1I1I_1 at a room temperature of 300K300K. At what intensity I2I2I_2 will this blackbody radiate when it is at a temperature of 400K400K
Answer:
Explanation:
We shall apply Stefan's formula
E = AσT⁴
When T = 300
I₁ = Aσ x 300⁴
When T = 400K
I₂ = Aσ x 400⁴
I₂ / I₁ = 400⁴ / 300⁴
= 256 / 81
= 3.16
I₂ = 3.16 I₁ .
An iron railroad rail is 800 ft long when the temperature is 31°C. What is its length (in ft) when the temperature is −17°C?
Answer:
799.54 ft
Explanation:
Linear thermal expansion is:
ΔL = α L₀ ΔT
where ΔL is the change in length,
α is the linear thermal expansion coefficient,
L₀ is the original length,
and ΔT is the change in temperature.
Given:
α = 1.2×10⁻⁵ / °C
L₀ = 800 ft
ΔT = -17°C − 31°C = -48°C
Find: ΔL
ΔL = (1.2×10⁻⁵ / °C) (800 ft) (-48°C)
ΔL = -0.4608
Rounded to two significant figures, the change in length is -0.46 ft.
Therefore, the final length is approximately 800 ft − 0.46 ft = 799.54 ft.
help me i cant solve it
⛔⛔⛔⛔⛔⛔⛔⛔⛔⛔
mark it as brainliest.... ✌✌What is the mass of the heaviest book this person can hold onto vertically before it slips out of his or her fingers? The coefficient of static friction of the surface between the fingers and the book cover is 0.65.
Answer:
0.795 kg
Explanation:
Assuming the complete question:
A person with compromised pinch strength in their fingers can only exert a normal force of 6.0 N to either side of a pinch-held object. What is the mass of the heaviest book this person can hold onto vertically before it slips out of his or her fingers? The coefficient of static friction of the surface between the fingers and the book cover is 0.60
SOLUTION:
The maximum weight of the book will equal the maximal friction force that can be produced:
m g = 2 f [tex]F{normal}[/tex]
Note that there are two sides of the book, so the friction force equals 2 times the friction force on one side (hence the factor 2).
So the maximum mass of the book is
m = 2 f [tex]F_n[/tex]/ g
m = 2[tex]\times[/tex]0.65[tex]\times[/tex] 6.0N / (9.81N/kg)
m = 0.795 kg
what is the application of physics
Answer:
1) We can estimate the age of the earth
2) we can calculate the speed of anything
3) we can also calculate gravity, e.t.c.
Explanation:
I could give you more just ask
A student uses the right-hand rule as shown.
What is the direction of the magnetic field in front of the wire
closest to the student?
up
right
down
left
Answer:
right is the correct answer to the given question .
Explanation:
In this question figure is missing
The main objective right-hand rule to decide the position of the magnetic force on the positive force acting, either the position of the thumb of a right hand with in position of v, the fingers throughout the position of B1, and a right angles throughout the position of F1 to the hand positions.
So [tex]F1 \ =\ q\ v \ B1\ Sin\alpha[/tex]
So from the magnetic right hand rule the direction of the magnetic field in front of a wire is right .All the others options are incorrect because they do not give the direction of the magnetic field in front of a wire is right .Answer:
Right
Explanation:
Just did this on edgen, in the diagram, the fingers are pointing to the right, which indicates that the magnetic field is acting to the right.
Question 1 [7]
Hydrogen gas is used in a Carnot cycle having an efficiency of 60% with a low temperature of 300K. During the heat rejection the pressure changes from 90 kPa to 120 kPa. Find the high and low temperature heat transfer and the net cycle work per unit mass of hydrogen.
Question 2 [8]
A rigid insulated container has two rooms separated by a membrane. Room A contains 1 kg of air at 200°C and Room B contains 1.5 kg of air at 20°C, both rooms are at 100 kPa. Consider two different cases
A. The Heat transfer between A and B creates a final uniform T
B. The membrane breaks and air comes to a uniform state.
For both cases find the final temperature. Are the two-process reversible and different? Explain.
Daniel Levinson found that men and women
A. Go through the same stages of development
B. Differ in terms of their social roles and identities
C. Deal with the development tasks in each stage differently
D. All of the above
Answer:
Option D. is correct
Explanation:
Daniel Levinson, a psychologist developed an adult development theory. This theory was referred to as the Seasons of Life theory. This theory describes the stage in which a person leaves adolescence and start making choices about adult life.
Daniel Levinson found that men and women go through the same stages of development, differ in terms of their social roles and identities, and deal with the development tasks in each stage differently.
Option D. is correct
Answer:
Daniel Levinson found that men and women
d. all of the above
Explanation:
cause i got a 100 on edg;)
A stunt cyclist needs to make a calculation for an upcoming cycle jump. The cyclist is traveling 100 ft/sec toward an inclined ramp which ends 10 feet above a level landing zone. Assume the cyclist maintains a constant speed up the ramp and the ramp is inclined Ao (degrees) above horizontal. With the pictured imposed coordinate system, the parametric equations of the cyclist will be: x(t) = 100t cos(A) y(t) = –16t2 + 100t sin(A) + 10.
Calculate the horizontal velocity of the cyclist at time t; this is the function x'(t) = _______
What is the horizontal velocity if A = 20 degrees?
What is the horizontal velocity if A = 45 degrees? (Four decimal places.)
Calculate the vertical velocity of the cyclist at time t; this is the function y(t) =________.
What is the vertical velocity if A = 20 degrees? (Four decimal places.)
What is the vertical velocity if A = 45 degrees? (Four decimal places.)
The vertical velocity of the cyclist is zero at time ________ seconds.
If the cyclist wants to have a maximum height of 35 feet above the landing zone, then the required launch angle is A = _______ degrees. (Accurate to four decimal places.)
Answer:
Explanation:
The parametric equations of the cyclist are:
[tex]x(t)=100tcos(A)\\\\y(t)=-16t^2+100tsin(A)+10[/tex] (1)
A) The horizontal velocity is the derivative of x(t), in time:
[tex]x'(t)=100cos(A)[/tex]
B) For A=20° the horizontal velocity is:
[tex]v_x=x'(t)=100cos(20\°)=93.9692ft/s[/tex]
For A=45°:
[tex]v_x=100cos(45\°)=70.7106ft/s[/tex]
C) To find the time in which the vertical velocity is zero you first obtain the derivative of, in time:
[tex]v_y=y'(t)=-32t+100sin(A)+10[/tex]
Next, you equal the vertical velocity to zero and solve for time t:
[tex]-32t+100sin(A)+10=0\\\\t=\frac{100sin(A)+10}{32}[/tex]
D) The maximum height is reached when the derivative of y (height) is zero. You use the previous value of t in the equation (1), equals y to 35. Next, you solve for t:
[tex]y=35\\\\-16(\frac{100sin(A)+10}{32})^2+100(\frac{10sin(A)+10}{32})sinA+10=35\\\\-\frac{16}{1024}(10000sin^2A+2000sinA+100)+31.25sin^2A+31.25sinA+10=35\\\\-156.25sin^2A-31.25sinA-1.5625+31.25sin^2A+31.25sinA+10=35\\\\-125sin^2A-26.5625=0[/tex]
A red ball is thrown down with an initial speed of 1.6 m/s from a height of 25 meters above the ground. Then, 0.4 seconds after the red ball is thrown, a blue ball is thrown upward with an initial speed of 24 m/s, from a height of 1.2 meters above the ground. The force of gravity due to the earth results in the balls each having a constant downward acceleration of 9.81 m/s^2. 1) What is the speed of the red ball right before it hits the ground?
2) How long does it take the red ball to reach the ground?
3) What is the height of the blue ball 2 seconds after the red ball is thrown?
4) How long after the red ball is thrown are the two balls in the air at the same height?
Answer:
1. v = 22.2 m/s
2. t = 2.25 seconds
3. h = 27.05 m
4. t = 1.16 seconds
Explanation:
The questions involve motion under the influence of gravity
1. Using the formula v² = u² + 2gh
where u = 1.6 m/s; g= 9.81 m/s²; h = 25 m; v = ?
v² = (1.6)² + 2 * 9.81 * 25
√v² = √493.06
v = 22.2 m/s
2. Using h = ut + 1/2 gt²
where h = 25 m; u = 0 (since velocity on reaching the ground is zero); g = 9.81 m/s²; t = ?
therefore, h = 1/2 gt²
making t subject of the formula, t = √ (2*h /g)
t = √ (2 * 25 / 9.81)
t = 2.25 seconds
3. Time of travel for the blue ball, t = 2 - 0.4 = 1.6s
using h = ut - gt²
u = 24 m/s; t = 1.6 s; g = 9.81 m/s²
note: since the ball is travelling against gravity, g is negative
h = 24 * 1.6 - 11/2 * 9.81 * 1.6²
h = 38.4 - 12.55 = 25.85 m
since height above the ground is 1.2 m,
total height h = 25.85 m + 1.2 m
h = 27.05 m
4. Let the time of travel of the red ball be t seconds.
So the time of travel of the blue ball = (t - 0.4) seconds.
Both the balls are at the same height :
25 - s = 1.2 + h where s & h are the displacements of the red & the blue ball respectively.
25 - (ut + 1/2 gt2) = 1.2 + (ut - 1/2 gt2)
25 - (1.6 t + 0.5 * 9.8 t²) = 1.2 + (24(t-0.4) - 0.5*9.8*(t-0.4)²)
solving the equation above for the time after which both the balls are at the same height.
25 - 1.6t - 4.9t² = 1.2 + 24t - 9.6 - 4.9t² + 3.92t - 0.784
collecting like terms
(25 - 1.2 + 9.6 + 0.784) = (24 + 3.92 + 1.6) * t
t = 34.184 / 33.44
t = 1.16 seconds
You are standing on a large sheet of frictionless ice and holding a large rock. In order to get off the ice, you throw the rock so it has velocity 12.0 m/s relative to the earth at an angle of 35.0∘ above the horizontal. Your mass is 72.0 kg and the rock’s mass is 3.50 kg . What is your speed after you throw the rock?
Answer:
0.4778 m/s
Explanation:
To solve this question, we will make use of law of conservation of momentum.
We are given that the rock's velocity is 12 m/s at 35°. Thus, the horizontal component of this velocity is;
V_x = (12 m/s)(cos(35°)) = 9.83 m/s.
Thus, the horizontal component of the rock's momentum is;
(3.5 kg)(9.83 m/s) = 34.405 kg·m/s.
Since the person is not pushed up off the ice or down into it, his momentum will have no vertical component and so his momentum will have the same magnitude as the horizontal component of the rock's momentum.
Thus, to get the person's speed, we know that; momentum = mass x velocity
Mass of person = 72 kg and we have momentum as 34.405 kg·m/s
Thus;
34.405 = 72 x velocity
Velocity = 34.405/72
Velocity = 0.4778 m/s
4–72 A person puts a few apples into the freezer at 215°C to cool them quickly for guests who are about to arrive. Initially, the apples are at a uniform temperature of 20°C, and the heat transfer coefficient on the surfaces is 8 W/m2·K. Treating the apples as 9-cm-diameter spheres and taking their properties to be r 5 840 kg/m3, cp 5 3.81 kJ/kg·K, k 5 0.418 W/m·K, and a 5 1.3 3 1027 m2/s, determine the center and surface temperatures of the apples in 1 h. Also, determine the amount of heat transfer from each apple. Solve this problem using analytical one-term approximation method (not the Heisler charts).
Complete and Clear Question:
A person puts a few apples into the freezer at -15°C to cool them quickly for guests who are about to arrive. Initially, the apples are at a uniform temperature of 20°C, and the heat transfer coefficient on the surfaces is 8 W/m2·K. Treating the apples as 9-cm-diameter spheres and taking their properties to be [tex]\rho =[/tex] 840 kg/m3, [tex]c_{p} =[/tex] 3.81 kJ/kg·K, k = 0.418 W/m·K, and [tex]\alpha = 1.3 * 10^{-7} m^{2} /s[/tex], determine the center and surface temperatures of the apples in 1 h. Also, determine the amount of heat transfer from each apple. Solve this problem using analytical one-term approximation method (not the Heisler charts).
Answer:
Temperature at the center of the apple, T(t) = 11.215°C
Temperature at the surface of the apple, T(r,t) = 2.68°C
Amount of heat transfer from each apple, Q = 21.47 kJ
Explanation:
For clarity and easiness of expression, the calculations are handwritten and attached as a file. Check the attached files for the complete calculation.
A child bounces a 48 g superball on the sidewalk. The velocity change of the superball is from 28 m/s downward to 17 m/s upward. If the contact time with the sidewalk is 1 800 s, what is the magnitude of the average force exerted on the superball by the sidewalk
Answer:
F = 1.2×10⁻³ N
Explanation:
From the question,
Applying newton's second law of motion,
F = m(v-u)/t................... Equation 1
Given: F = magnitude of the average force exerted on the ball, m = mass of the ball, v = final velocity, u = initial velocity, t = time of contact.
Note: let downward be negative and upward be positive.
Given: m = 48 g = 48/1000 = 0.048 kg, v = 17 m/s, u = -28 m/s (downward),
t = 1800 s
Substitute into equation 1
F = 0.048(17-[28])/1800
F = 1.2×10⁻³ N
A small block is released from rest at the top of a frictionless incline. The block travels a distance 0.633 m in the first second after it is released. How far does it travel in the next second
Answer:1.89 m
Explanation:
Given
Block travels [tex]0.63\ m[/tex] in first second
It is released from rest i.e. initial speed is zero (u=0)
using
[tex]s=ut+\frac{1}{2}at^2[/tex]
where a=acceleration
here acceleration is the component of gravity on incline plane (say [tex]\theta [/tex])
so
[tex]s_1=\frac{1}{2}\times g\sin \theta (1)^2[/tex]
[tex]0.633\times 2=9.8\sin \theta \times 1^2[/tex]
[tex]\sin\theta =0.1291[/tex]
[tex]\theta =7.41^{\circ}[/tex]
So distance traveled in [tex]2\ sec[/tex]
[tex]s=\frac{1}{2}\times g\sin \theta (2)^2[/tex]
[tex]s=0.5\times 9.8\times \sin (7.41)\times 4[/tex]
[tex]s=2.52\ m[/tex]
So distance traveled in [tex]2^{nd}\ sec[/tex] is
[tex]s-s_1=2.52-0.633=1.89\ m[/tex]
A plane is flying to a city 756 km directly north of its initial location. The plane maintains a speed of 203 km/h relative to the air during its flight. (a) If the plane flies through a constant headwind blowing south at 53.5 km/h, how much time (in h) will it take to reach the city
Answer:
The answer is 5.05 hours.
Explanation:
If the plane has an airspeed of 203 km/h which only applies for air and not the ground speed, we can subtract the speed of the wind since it is a headwind in the directly opposite direction.
So the speed of the plane becomes 203 - 53.5 = 149.5 km/h which will give us the true airspeed of the plane and the ground speed as well.
From here we can calculate the time it will take to reach the city as
756 km / 149.5 km/h = 5.05 hours.
I hope this answer helps.
Use Hooke's Law to determine the variable force in the spring problem. A force of 450 newtons stretches a spring 30 centimeters. How much work is done in stretching the spring from 40 centimeters to 70 centimeters?
Answer:
Work Done = 67.5 J
Explanation:
First we find the value of spring constant (k) using Hooke's Law. Hooke's is formulated as:
F = kx
where,
F = Force Applied = 450 N
k = Spring Constant = ?
x = Stretched Length = 30 cm = 0.3 m
Therefore,
450 N = k(0.3 m)
k = 450 N/0.3 m
k = 1500 N/m
Now, the formula for the work done in stretching the spring is given as:
W = (1/2)kx²
Where,
W = Work done = ?
k = 1500 N/m
x = 70 cm - 40 cm = 0.3 m
Therefore,
W = (1/2)(1500 N/m)(0.3 m)²
W = 67.5 J
An electron moving in a wire collides again and again with atoms and travels an average distance between collisions that is called the mean free path. If the mean free path is less in some metals, what can you say about the resistance of these metals? For a given conductor, what can you do to lengthen the mean free path?
Explanation:
A substance with a short , medium, free path has improved electron flow resistance and a higher electrical resistance . Heat applications impose more molecular chaos on all materials and shorten the track further, increasing resistance of most materials. So, just refresh the material to expand the course. In certain materials, when cooled to the minimum temperature, the conductivity is substantially increased.
Two blocks with masses 1 and 2 are connected by a massless string that passes over a massless pulley as shown. 1 has a mass of 2.25 kg and is on an incline of 1=43.5∘ with coefficient of kinetic friction 1=0.205 . 2 has a mass of 5.45 kg and is on an incline of 2=32.5∘ with coefficient of kinetic friction 2=0.105 . The two‑block system is in
Answer:
The acceleration of [tex]M_2[/tex] is [tex]a = 0.7156 m/s^2[/tex]
Explanation:
From the question we are told that
The mass of first block is [tex]M_1 = 2.25 \ kg[/tex]
The angle of inclination of first block is [tex]\theta _1 = 43.5^o[/tex]
The coefficient of kinetic friction of the first block is [tex]\mu_1 = 0.205[/tex]
The mass of the second block is [tex]M_2 = 5.45 \ kg[/tex]
The angle of inclination of the second block is [tex]\theta _2 = 32.5^o[/tex]
The coefficient of kinetic friction of the second block is [tex]\mu _2 = 0.105[/tex]
The acceleration of [tex]M_1 \ and\ M_2[/tex] are same
The force acting on the mass [tex]M_1[/tex] is mathematically represented as
[tex]F_1 = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1[/tex]
=> [tex]M_1 a = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1[/tex]
Where T is the tension on the rope
The force acting on the mass [tex]M_2[/tex] is mathematically represented as
[tex]F_2 = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2[/tex]
[tex]M_2 a = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2[/tex]
At equilibrium
[tex]F_1 = F_2[/tex]
So
[tex]T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1 =M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2[/tex]
making a the subject of the formula
[tex]a = \frac{M_2 g sin \theta_2 - M_1 g sin \theta_1 - \mu_1 M_1g cos \theta - \mu_2 M_2 g cos \theta_2 }{M_1 +M_2}[/tex]
substituting values [tex]a = \frac{(5.45) (9.8) sin (32.5) - (2.25) (9.8) sin (43.5) - (0.205)*(2.25) *9.8cos (43.5) - (0.105)*(5.45) *(9.8) cos(32.5) }{2.25 +5.45}[/tex]
=> [tex]a = 0.7156 m/s^2[/tex]
The acceleration of the second block to the right is 2.21 m/s².
The normal force on block1 is calculated as follows;
[tex]F_n_1 = m_1g cos(\theta_1)[/tex]
The parallel force on block 1 is calculated as;
[tex]F_x_1 = m_1gsin(\theta)[/tex]
The frictional force on block 1 is calculated as;
[tex]F_k_1 = \mu_k F_n = \mu_k m_1gcos\theta_1[/tex]
The net force on block 1 is calculated as;
[tex]\Sigma F_x_1 = m_1gsin(\theta_1) - \mu_k_1m_1gcos(\theta_1)[/tex]
The normal force on block 2 is calculated as follows;
[tex]F_n_2 = m_2gcos\theta _2[/tex]
The frictional force on block 2 is calculated as;
[tex]F_k_2 = \mu k_2 m_2g cos\theta _2[/tex]
The net force on block 2 is calculated as follows;
[tex]\Sigma F_x_2 = m_2a_2\\\\ m_2g_2 sin(\theta _2) - F_k_2 - \Sigma F_x_1 = m_2a_2 \\\\m_2gsin(\theta) - F_k_2 - (m_1gsin(\theta) - \mu_k _1 m_1g cos(\theta)) = m_2a_2\\\\m_2gsin(\theta) -\mu_k_2 m_2gcos(\theta) + \mu_k _1 m_1g cos(\theta) - m_1gsin(\theta) = m_2a_2\\\\5.45( 9.8) sin(32.5) -(0.105)(5.45)(9.8)cos(32.5) + \\\\0.205( 2.25) ( 9.8)cos(43.5) - 2.25( 9.8) sin (43.5) = 5.45a_2\\\\ 12.07 = 5.45a_2\\\\a_2= \frac{12.07}{5.45} \\\\a_2 = 2.21 \ m/s^2[/tex]
Thus, the acceleration of the second block to the right is 2.21 m/s².
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