Final answer:
To find how far the suitcase lands from the dog, one must calculate the time it takes for the suitcase to hit the ground using its vertical motion and multiply that time by the horizontal component of its initial velocity, with no need to consider air resistance.
Explanation:
To determine how far from the dog the suitcase will land, we need to break down the initial velocity of the suitcase (v0) into its horizontal (vx) and vertical components (vy). The flight time of the suitcase is primarily determined by its vertical motion, governed by the equation y = vyt + 0.5gt2, where y is the vertical displacement (in this case, equal to -h since the suitcase is falling down), t is the time, and g is the acceleration due to gravity. Since we ignore air resistance, the horizontal velocity remains constant throughout the flight. Therefore, the horizontal distance d from the dog can be found using d = vxt.
To extract the horizontal (vx) and vertical (vy) components of the initial velocity v0, we use the equations: vx = v0cos(α) and vy = v0sin(α). Finally, solving for t using the vertical motion equation and substituting it into the horizontal distance equation gives us the required distance d.
Speed of a rocket At t sec after liftof, the height of a rocket is 3t 2 ft. How fast is the rocket climbing 10 sec after liftof?
Final answer:
The speed of the rocket 10 seconds after liftoff is 60 ft/s.
Explanation:
To find the speed of the rocket 10 seconds after liftoff, we can differentiate the equation for the height of the rocket with respect to time. In this case, the equation for the height of the rocket is given as h = 3t^2. Taking the derivative of this equation with respect to time will give us the rate of change of height with respect to time, which is the speed of the rocket. The derivative of 3t^2 with respect to t is 6t. Plugging in t = 10 into the derivative, we get 6(10) = 60.
A baseball player friend of yours wants to determine his pitching speed. You have him stand on a ledge and throw the ball horizontally from an elevation 6.0 m above the ground. The ball lands 25 m away.
What is his pitching speed?
To solve this problem we will apply the concepts related to the kinematic equations of linear motion. Punctually we will verify the vertical displacement and the horizontal displacement from their respective components. We will start by calculating the time it took to reach the objective and later with that time, we will find the horizontal velocity launch component. The position can be written as,
[tex]h= v_{0y}t+\frac{1}{2}a_yt^2[/tex]
Here,
h = Height
[tex]v_{0y}[/tex]= Initial velocity in vertical direction
[tex]a_{y}[/tex] = Vertical acceleration (At this case, due to gravity)
[tex]t[/tex] = Time
There is not vertical velocity because the ball was thrown horizontally), then we have that
[tex]6= (0)t+\frac{1}{2}(9.8)t^2[/tex]
[tex]t = 1.1065s[/tex]
Now using the equation of horizontal motion we have with this time that the initial velocity was,
[tex]x = v_{ox}t+\frac{1}{2}a_xt^2[/tex]
Here,
[tex]v_{0x}[/tex]= Horizontal initial velocity
[tex]t[/tex] = Time
[tex]a_x[/tex] = Acceleration in horizontal plane
There is not acceleration in horizontal plane, only in vertical plane, then we have
[tex]25= v_{0x}(1.1065)+\frac{1}{2}(0)(1.1065)^2[/tex]
[tex]v_{0x} = 22.5938m/s[/tex]
Therefore the pitching speed is 22.5938m/s
An object is dropped from rest at a height of 128 m. Find the distance it falls during its final second in the air.
Answer:
In the last second, the object traveled 45.6 m.
Explanation:
Hi there!
The equation of the height of the object at a time t is the following:
h = h0 + v0 · t + 1/2 · g · t²
Where:
h =height of the object at time t.
h0 = initial height.
v0 = initial velocity.
t = time.
g = acceleration due to gravity (-9.8 m/s²).
First, let's find how much time it takes the object to reach the ground. For that, we have to find the value of "t" for which h = 0.
h = h0 + v0 · t + 1/2 · g · t²
Since the object is dropped and not thrown (v0 = 0).
h = h0 + 1/2 · g · t²
0 m = 128 m - 1/2 · 9.8 m/s² · t²
-128 m / -4.9 m/s² = t²
t = 5.1 s
Now, let's find the height of the object 1 s before reaching the ground (at t = 4.1 s):
h = h0 + v0 · t + 1/2 · g · t²
h = 128 m - 1/2 · 9.8 m/s² · (4.1 s)²
h = 45.6 m
Then, in the last second (from 4.1 s to 5.1 s) the object traveled 45.6 m
A boy 12.0 m above the ground in a tree throws a ball for his dog, who is standing right below the tree and starts running the instant the ball is thrown. If the boy throws the ball horizontally at 8.50 m/s, (a) how fast must the dog run to catch the ball just as it reaches the ground, and (b) how far from the tree will the dog catch the ball?
Answer:
Explanation:
Height covered = 12m
time to fall by 12 m
s = 1/2 gt²
12 = 1/2 g t²
t = 1.565 s
Horizontal distance of throw
= 8.5 x 1.565
= 13.3 m
This distance is to be covered by dog during the time ball falls ie 1.565 s
Speed of dog required = 13.3 / 1.565
= 8.5 m /s
b ) dog will catch the ball at a distance of 13.3 m .
Can the sum of the magnitudes of two vectors ever be equal to the magnitude of the sum of the same two vectors? If no, why not? If yes, when?
Answer:
They cannot be equal
Explanation:
Let 2 vector equal in magnitude and opposite direction. One of them is 1 and the other is -1
The sum of their magnitudes is 1 + 1 = 2
The magnitude of their sum of 2 vector is 0 since the 2 are in opposite directions, they cancel out each other and their final magnitude is thus 0.
So the sum of the magnitudes of two vectors cannot be equal to the magnitude of the sum of the same two vectors.
You blow across the open mouth of an empty test tube and producethe fundamental standing wave of the air column inside the testtube. The speed of sound in air is 344 m/s and the test tube actsas a stopped pipe.
(a) If the length of the air column in the test tubeis 11.0 cm, what is thefrequency of this standing wave?
kHz
(b) What is the frequency of the fundamental standing wave in theair column if the test tube is half filled with water?
kHz
Answer:
(a). The frequency of this standing wave is 0.782 kHz.
(b). The frequency of the fundamental standing wave in the air is 1.563 kHz.
Explanation:
Given that,
Length of tube = 11.0 cm
(a). We need to calculate the frequency of this standing wave
Using formula of fundamental frequency
[tex]f_{1}=\dfrac{v}{4l}[/tex]
Put the value into the formula
[tex]f_{1}=\dfrac{344}{4\times0.11}[/tex]
[tex]f_{1}=781.81\ Hz[/tex]
[tex]f_{1}=0.782\ kHz[/tex]
(b). If the test tube is half filled with water
When the tube is half filled the effective length of the tube is halved
We need to calculate the frequency
Using formula of fundamental frequency of the fundamental standing wave in the air
[tex]f_{1}=\dfrac{v}{4(\dfrac{L}{2})}[/tex]
Put the value into the formula
[tex]f_{1}=\dfrac{344}{4\times\dfrac{0.11}{2}}[/tex]
[tex]f_{1}=1563.63\ Hz[/tex]
[tex]f_{1}=1.563\ kHz[/tex]
Hence, (a). The frequency of this standing wave is 0.782 kHz.
(b). The frequency of the fundamental standing wave in the air is 1.563 kHz.
An ice cube at 0.00 ∘C with a mass of 21.5 g is placed into 500.0 g of water, initially at 31.0 ∘C, in an insulated container. Part A Assuming that no heat is lost to the surroundings, what is the temperature of the entire water sample after all of the ice has melted?
Answer:The temperature of the entire water sample after all of the ice has melted is [tex]30.3^0C[/tex]
Explanation:
As we know that,
[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex].......(1)
where,
q = heat absorbed or released
[tex]m_1[/tex] = mass of ice = 21.5 g
[tex]m_2[/tex] = mass of water = 500.0 g
[tex]T_{final}[/tex] = final temperature = ?
[tex]T1 =\text {temperature of ice}= 0^oC[/tex]
[tex]T_2[/tex] = temperature of water =[tex]31.0^oC[/tex]
[tex]c_1[/tex] = specific heat of ice= [tex]2.1J/g^0C[/tex]
[tex]c_2[/tex] = specific heat of water = [tex]4.184J/g^0C[/tex]
Now put all the given values in equation (1), we get
[tex]21.5\times 2.1\times (T_{final}-0)=-[500.0\times 4.184\times (T_{final}-31.0)][/tex]
[tex]T_{final}=30.3^0C[/tex]
Thus the temperature of the entire water sample after all of the ice has melted is [tex]30.3^0C[/tex]
A mixture of gaseous reactants is put into a cylinder, where a chemical reaction turns them into gaseous products. The cylinder has a piston that moves in or out, as necessary, to keep constant pressure on the mixture of 1 atm. The cylinder is also submerged in a large insulated water bath. The temperature of the water bath is monitored, and it is determined from this data that 133.0 kJ of heat flows into the system during the reaction. The position of the piston is also monitored, and it is determined from the data that the piston does 241.0 kJ of work on the system during the reaction.
(1) Does the temperature of the water bath go up or down?
(2) Does the piston move in or out?
(3) Does heat flow into or out of the gaseous mixture?
(4) How much heat flows?
Answer:
1) Temperature of the water bath rises
2) the piston moves out
3) 133 kJ of heat flows
Explanation:
Given:(Isobaric process)
Heat absorbed by the system,[tex]Q_{in}=133000\ J[/tex]work done on the system, [tex]W=241000\ J[/tex]pressure on the system, [tex]P=1\ atm=101325\ Pa[/tex]1)
Since the whole setup is isolated and the heat is absorbed by the system therefore the the temperature of the water will go down.
2)
When the system absorbs heat its pressure has to be constant so the piston needs to move up outwards giving the inside gases more volume.
As ideal gas law equation:
[tex]P.V=m.R.T[/tex]
[tex]P=[/tex]absolute pressure of the gas
[tex]V=[/tex] volume of the gas
[tex]m=[/tex] mass of the gas
[tex]R=[/tex] universal gas constant
[tex]T=[/tex] absolute temperature of the gas
Since pressure, mass and gas constants are the constant value we observe that: [tex]T\propto V[/tex]
3)
According to the given data the heat that flows is 133 kilo-joule in quantity.
If a 0.10−mL volume of oil can spread over a surface of water of about 47.0 m2 in area, calculate the thickness of the layer in centimeters. Enter your answer in scientific notation.
Answer:
[tex]2.1276595745\times 10^{-7}\ cm[/tex]
Explanation:
Volume of oil = 0.1 mL = 0.1 cm³
Area of oil = [tex]47\ m^2[/tex]
Converting to [tex]cm^2[/tex]
[tex]1\ m=10^2\ cm[/tex]
[tex]1\ m^2=10^4\ cm^2[/tex]
[tex]47\ m^2=47\times 10^4\ cm^2[/tex]
Volume is given by
[tex]V=At\\\Rightarrow t=\dfrac{V}{A}\\\Rightarrow t=\dfrac{0.1}{47\times 10^4}\\\Rightarrow t=2.1276595745\times 10^{-7}\ cm[/tex]
The thickness of the layer is [tex]2.1276595745\times 10^{-7}\ cm[/tex]
a silver ingot has a volume of 53.6 cm and weighs 500g. what is the desity in grams per cubic centimeter.
Q: a silver ingot has a volume of 53.6 cm³ and weighs 500g. what is the desity in grams per cubic centimeter.
Answer:
9.33 kg/m³
Explanation:
Density: This can be defined as the ratio of the mass of a substance and it's volume.
The Formula for Density is given as,
D = m/v ..................... Equation 1
Where D = Density of the silver ingot, m = mass of the silver ingot, v = volume of the silver ingot.
Given: m = 500 g, v = 53.6 cm³
Substitute into equation 1
D = 500/53.6
D = 9.33 kg/m³.
Hence the density of the ingot = 9.33 kg/m³.
Final answer:
The density of the silver ingot, which has a mass of 500 g and a volume of 53.6 cm³, is found by dividing the mass by the volume. The calculation yields a density of approximately 9.33 g/cm³ for the silver ingot.
Explanation:
The student is asking for the density of a silver ingot, which is a measurement of mass per unit volume of a substance. The formula for calculating density is mass divided by volume. Given that the silver ingot has a mass of 500 g and a volume of 53.6 cm³, we can determine its density using this formula.
To find the density of the silver ingot, you would:
Divide the mass of the ingot (500 g) by its volume (53.6 cm³).Perform the calculation: 500 g ÷ 53.6 cm³.The calculated density is approximately 9.33 g/cm³. This is the silver ingot's density.The height, h , in meters of a dropped object after t seconds can be represented by h ( t ) = − 4.9 t 2 + 136 . What is the instantaneous velocity of the object one second after it is dropped?
Answer:
The velocity of the object 1 s after it is dropped is -9.8 m/s.
Explanation:
Hi there!
The instantaneous velocity is defined by the change in height over a very small time. Mathematically, it is expressed as the derivative of the function h(t):
instantaneous velocity = dh/dt
dh/dt = h´(t) = -2 · 4.9 · t
h´(t) = -9.8 · t
Now we have to eveluate the function h´(t) at t = 1 s:
h´(1) = -9.8 · (1) = -9.8
The velocity of the object 1 s after it is dropped is -9.8 m/s.
Final answer:
The instantaneous velocity of the object one second after it is dropped is 9.8 m/s, moving downwards.
Explanation:
The question asks for the instantaneous velocity of a dropped object after a specific time has elapsed, which is a physics concept related to mechanics and motion. Given the height equation h(t) = -4.9t2 + 136, the instantaneous velocity at time t can be found by taking the derivative of the height equation with respect to time, which gives us the velocity as a function of time v(t) = dh/dt. At t = 1 second, the derivative of the height equation is v(1) = -9.8(1) = -9.8 m/s. The negative sign indicates that the object is moving downwards. However, when we speak about the speed or magnitude of the velocity (instantaneous velocity), we typically refer to the positive value, which would be 9.8 m/s.
When measuring weight on a scale that is accurate to the nearest 0.1 pound, what are the real limits for the weight of 145 pounds?
When measuring weight on a scale that is accurate to the nearest 0.1
pound, the real limits for the weight of 145 pounds is 144.9- 145.1
ScaleThis is an instrument which is used to measure the weight of objects. There
may be differences in the measurement as a result of air interference and
other factors.
We were told that the accuracy is to the nearest 0.1 pound which means
= 145± 0.1
= (145-0.1) - (145+0.1)
= 144.9 - 145.1
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The real limits for the weight of 145 pounds on a scale accurate to the nearest 0.1 pound are 144.95 to 145.05 pounds.
Explanation:When measuring weight on a scale that is accurate to the nearest 0.1 pound, the real limits for the weight of 145 pounds would be 144.95 to 145.05 pounds. This is because the scale is accurate to the tenths place, meaning it can measure weights up to one decimal place. In this case, the weight of 145 pounds would be rounded to 145.0 pounds. Therefore, the real limits would be 144.95 pounds (145.0 - 0.05) to 145.05 pounds (145.0 + 0.05).
Your bedroom has a rectangular shape, and you want to measure its area. You use a tape that is precise to 0.001 and find that the shortest wall in the room is 3.547 long. The tape, however, is too short to measure the length of the second wall, so you use a second tape, which is longer but only precise to 0.01 . You measure the second wall to be 4.79 long. Which of the following numbers is the most precise estimate that you can obtain from your measurements for the area of your bedroom?
If your bedroom has a circular shape, and its diameter measured 6.32 , which of the following numbers would be the most precise value for its area?
a)30 m^2
b) 31.4 m^2
c)31.37 m^2
d)31.371 m^2
Answer:
The question is incomplete.the complete question is giving below "College Physics 10+5 pts
Your bedroom has a rectangular shape, and you want to measure its area. You use a tape that is precise to 0.001 and find that the shortest wall in the room is 3.547 long. The tape, however, is too short to measure the length of the second wall, so you use a second tape, which is longer but only precise to 0.01 . You measure the second wall to be 4.79 long. Which of the following numbers is the most precise estimate that you can obtain from your measurements for the area of your bedroom?
A. 17.0m^2
B. 16.990m^2
C.16.99m^2
D.16.9m^2
E. .16.8m^2
b. If your bedroom has a circular shape, and its diameter measured 6.32 , which of the following numbers would be the most precise value for its area?
a)30 m^2
b) 31.4 m^2
c)31.37 m^2
d)31.371 m^2
Answers:
A. 17.0m²
B. 31.4m²
Explanation:
First, we determine the area of the room using the measured parameters I.e
Length=4.79 long
Breadth= 3.547 long
Hence area is calculated as
A=length *Breadth
A=4.79*3.547
A=16.99013m²
From the question the different measurements have different precision, the answer should match the number of significant figures of the least precisely known number in the calculation which is 3 significant digits.
Hence the correct answer will be 17.0m²
B. The are of a circle is express as
A=πd²/4
A=π(6.32)²/4
A=31.37069m²
The π and 4 are exact number they have no effect on the accuracy on the accuracy of the area. Hence we express the answer to the same significant figure as giving in the question..
The correct answer will be 31.4m²
The most precise estimate for the area of the rectangular bedroom is 17.004613 units. The most precise value for the area of the circular bedroom is 31.373968 units.
Explanation:To find the most precise estimate for the area of your rectangular bedroom, you need to consider the precision of the measurements. The first wall measurement is 3.547, which has a precision of 0.001. The second wall measurement is 4.79, which has a precision of 0.01. Since the second measurement has a lower precision, it will determine the precision of the final estimate. Therefore, the most precise estimate for the area of your bedroom is obtained by multiplying the two measurements: 3.547 x 4.79 = 17.004613.
To find the most precise value for the area of your circular bedroom, you need to consider the precision of the diameter measurement. The diameter is measured as 6.32, which has a precision of 0.01. The formula to calculate the area of a circle is A = πr^2, where r is the radius (half of the diameter). So, the radius is 6.32/2 = 3.16. Therefore, the most precise value for the area of your circular bedroom is obtained by calculating the area using the radius: A = π(3.16)^2 = 31.373968.
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A screen is placed 1.20m behind a single slit. The central maximum in the resulting diffraction pattern on the screen is 1.40cm wide-that is, the two first-order diffraction minima are separated by 1.40cm
What is the distance between the two second-order minima?
Answer:
2.8 cm
Explanation:
[tex]y_1[/tex] = Separation between two first order diffraction minima = 1.4 cm
D = Distance of screen = 1.2 m
m = Order
Fringe width is given by
[tex]\beta_1=\dfrac{y_1}{2}\\\Rightarrow \beta_1=\dfrac{1.4}{2}\\\Rightarrow \beta_1=0.7\ cm[/tex]
Fringe width is also given by
[tex]\beta_1=\dfrac{m_1\lambda D}{d}\\\Rightarrow d=\dfrac{m_1\lambda D}{\beta_1}[/tex]
For second order
[tex]\beta_2=\dfrac{m_2\lambda D}{d}\\\Rightarrow \beta_2=\dfrac{m_2\lambda D}{\dfrac{m_1\lambda D}{\beta_1}}\\\Rightarrow \beta_2=\dfrac{m_2}{m_1}\beta_1[/tex]
Distance between two second order minima is given by
[tex]y_2=2\beta_2[/tex]
[tex]\\\Rightarrow y_2=2\dfrac{m_2}{m_1}\beta_1\\\Rightarrow y_2=2\dfrac{2}{1}\times 0.7\\\Rightarrow y_2=2.8\ cm[/tex]
The distance between the two second order minima is 2.8 cm
The distance between the two second-order minima is 2.80 cm.
The central maximum's width given is 1.40 cm (this is the distance between the first-order minima, m = ±1). So, the distance from the center to the first-order minimum (m = ±1) is 0.70 cm.
Using the formula for the first-order minimum (m = 1):
a sin(θ₁) = λ
We know that the distance to the first minima (y₁) with the screen distance (L) is given by:
tan(θ₁) ≈ sin(θ₁) = y₁ / L
so, for the first-order minima:
y₁ = Lλ / a
We have y₁ = 0.70 cm, L = 1.20 m. Solving for aλ:
a = Lλ / 0.007
Next, for the second-order minima (m = 2), we use:
y₂ = 2Lλ / a
Thus, the distance between the second-order minima will be:
2y₂ = 2 × 2Lλ / a = 2 × 1.40 cm = 2.80 cm
So, the distance between the two second-order minima is 2.80 cm.
A stone is thrown vertically upward with a speed of 12m/s from the edge of a cliff 70 m high (a) How much later it reaches the bottom of the cliff? (b) What is its speed just before hitting? And (c) what total distance did it travel? (Ans: 5.2 sec, 38.94 m/s, 84.7 m)
Answer
given,
vertical speed of stone,v = 12 m/s
height of the cliff = 70 m
a) time taken by the stone to reach at the bottom of the cliff
We know that,
S = u t + 1/2 a t²
- 70 = 12 t - 0.5 x 9.8 t²
4.9 t² - 12 t - 70 = 0
solving the equation
t = 5.2 s (neglecting the negative value)
b) again using equation of motion
v = u + a t
v = 12 - 9.8 x 5.2
v = -38.96 m/s
ignoring the negative sign
magnitude of velocity is equal to 38.96 m/s
c) total distance travel by the stone
vertical distance covered by the stone
v² = u² + 2 g h
0 = 12² - 2 x 9.8 x h
h = 7.34 m
to reach the stone to the same level distance travel be doubled.
Total distance travel by the stone
H = h + h + 70
H = 7.34 x 2 + 70
H = 84.7 m.
Using the equations of motion, we calculated that the stone hits the ground approximately 7.44 seconds after being thrown. The speed at impact would be approximately 36.46 m/s, and the stone would travel a total distance of about 138 meters.
Explanation:To solve this question, we need to use the equations of motion which are concepts of Physics.
(a) The time it takes for the stone to reach the bottom can be found using the equation: h = ut + 0.5gt², where h is the height =70m, u is the initial velocity =12 m/s, g is the acceleration due to gravity ~= 9.8 m/s², and t is the time we need to find. Solving for 't' gives us approximately 3.72 seconds for the upward journey. The total time taken is twice this value (since the return journey takes the same amount of time), so the stone hits the bottom after approximately 7.44 seconds.
(b) The speed at impact can be calculated using the equation of motion v = u + gt. Using the initial speed u when the stone starts falling from the top = 0, g = 9.8 m/s², and t ~=3.72 seconds, we find the speed is approximately 36.46 m/s.
(c) The total distance traveled can be calculated as the sum of the upward journey (the maximum height the stone reached, which can be calculated using the equation h = ut + 0.5gt² where u = 12m/s, g = 9.8 m/s² and t = ~3.72 seconds) and the downward journey (the fall from the cliff, which is 70m). This gives us an approximate total distance of 70 m + 68 m = 138 m.
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A package falls out of an airplane that is flying in a straight line at a constant altitude and speed. If you ignore air resistance, what would be the path of the package as observed by the pilot? As observed by a person on the ground?
Answer:
Explanation:
If you ignore air resistant, then nothing affects the package horizontal motion. In Newton's first law it would keep the package at a constant speed, the speed of the airplane.
So to the eyes of the pilot who is also moving at the same horizontal speed, the lateral position of the package does not change. He can only perceive that the package is getting further away from him as it's dropping vertically.
To a person on the ground then the package is travelling in a parabolic path, where its horizontal speed is constant but vertical speed is increasing toward the ground at the rate of g.
The package would follow a straight line for the pilot and a parabolic curve for an observer on the ground.
Explanation:
The path of the package as observed by the pilot would be a straight line parallel to the airplane's motion. This is because the package falls with the same horizontal velocity as the airplane due to the absence of air resistance. As observed by a person on the ground, the path of the package would be a parabolic curve. This is because the package has an initial horizontal velocity but is acted upon by the force of gravity, causing it to follow a curved trajectory.
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At a certain time a particle had a speed of 87 m/s in the positive x direction, and 6.0 s later its speed was 74 m/s in the opposite direction. What was the average acceleration of the particle during this 6.0 s interval?
Answer:
The average acceleration during the 6.0 s interval was -27 m/s².
Explanation:
Hi there!
The average acceleration is defined as the change in velocity over time:
a = Δv/t
Where:
a = acceleration.
Δv = change in velocity = final velocity - initial velocity
t = elapsed time
The change in velocity will be:
Δv = final velocity - initial velocity
Δv = -74 m/s - 87 m/s = -161 m/s
(notice the negative sign of the velocity that is in opposite direction to the direction considered positive)
Then the average acceleration will be:
a = Δv/t
a = -161 m/s / 6.0 s
a = -27 m/s²
The average acceleration during the 6.0 s interval was -27 m/s².
A woman is riding on a train while watching the display on a GPS unit. She notices that both the "speed" and the "heading" readings are not changing. What can she conclude about the net force acting on the train car?
Answer:
There is no force acting on the body.
Explanation:
While riding a train when a person watches on a GPS that the speed is not changing and also 'heading' is not changing that means the train is moving with constant speed.Heading in GPS means the direction shown in the compass, when the heading is not changing then that means there is no change in the direction of the train.Therefore the person is in a train moving with constant velocity. When any object moves with constant velocity this implies that there is no force acting on the body. This is in accordance with the Newton's first law of motion which states that any particle continues to be in the state of rest or uniform motion until an external force acts upon it.Acceleration is defined in the Newton's second law as the rate of change in velocity. Here we have a constant velocity in terms of both the magnitude and the direction.A railroad car with mass m is moving with an initial velocity v when it collides and connects with a second railroad car with a mass of 3m, which is initially at rest. How do the speed and momentum of the connected car system compare with those of the car with mass m before the collision
Answer:
Same momentum but speed is reduced 4 times.
Explanation:
According to the law of momentum conservation, the total momentum of the system before and after the collision is the same
Before the collision, the bigger car is at rest, only the 1st car of mass m is moving at speed v
p = mv
After the collision, both cars are connected and moving at speed V
P = (m + 3m)V = 4mV
These 2 momentum are equal
p = P
mv = 4mV
V = v/4
So after the collision, they have the same momentum but the speed decreased 4 times
Final answer:
After the collision of a moving car with mass m and a stationary car with mass 3m, the final velocity of the connected system is v/4, a quarter of the original car's velocity. The momentum of the system remains unchanged due to the conservation of momentum, but the total mechanical energy is typically not conserved as some is lost to other forms of energy.
Explanation:
Conservation of Momentum and Collision of Railway Cars
When a railroad car of mass m moving with an initial velocity v collides with and connects to another car at rest with a mass of 3m, the conservation of momentum principle applies. Given that neither external forces nor friction are significant, the total system momentum before and after the collision remains constant. The initial momentum of the system, which is mv (since the second car is at rest), must equal the final momentum of the now combined mass (4m) moving at the new velocity v'. To find the final velocity, we use the conservation of momentum equation:
mv = (m + 3m)v'
Which simplifies to:
v' = (mv)/(4m) = v/4
This equation shows that the speed of the combined railroad cars after the collision is a quarter of the initial speed of the moving car. While the speed of the connected car system is reduced compared to the initial speed of the car with mass m, the total momentum remains the same because the mass has increased fourfold.
Energy Conservation After the Collision
As for energy conservation, the total mechanical energy may not be conserved in an inelastic collision (like when cars connect after collision). Although the momentum is conserved, some kinetic energy is usually lost as other forms of energy such as heat, sound, or deformation of the vehicles.
You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics, you ask one of the archers to shoot an arrow parallel to the ground. You find the arrow stuck in the ground 63.0 mm away, making a 2.8 ∘∘ angle with the ground. Ignore all possible aerodynamic effects on the motion of the arrow. Use 9.80 m/s22 for the acceleration due to gravity.
Answer:
[tex]v_0 = 3.53~{\rm m/s}[/tex]
Explanation:
This is a projectile motion problem. We will first separate the motion into x- and y-components, apply the equations of kinematics separately, then we will combine them to find the initial velocity.
The initial velocity is in the x-direction, and there is no acceleration in the x-direction.
On the other hand, there no initial velocity in the y-component, so the arrow is basically in free-fall.
Applying the equations of kinematics in the x-direction gives
[tex]x - x_0 = v_{x_0} t + \frac{1}{2}a_x t^2\\63 \times 10^{-3} = v_0t + 0\\t = \frac{63\times 10^{-3}}{v_0}[/tex]
For the y-direction gives
[tex]v_y = v_{y_0} + a_y t\\v_y = 0 -9.8t\\v_y = -9.8t[/tex]
Combining both equation yields the y_component of the final velocity
[tex]v_y = -9.8(\frac{63\times 10^{-3}}{v_0}) = -\frac{0.61}{v_0}[/tex]
Since we know the angle between the x- and y-components of the final velocity, which is 180° - 2.8° = 177.2°, we can calculate the initial velocity.
[tex]\tan(\theta) = \frac{v_y}{v_x}\\\tan(177.2^\circ) = -0.0489 = \frac{v_y}{v_0} = \frac{-0.61/v_0}{v_0} = -\frac{0.61}{v_0^2}\\v_0 = 3.53~{\rm m/s}[/tex]
Final answer:
The question is about projectile motion in Physics, requiring knowledge of equations of motion, range calculation, and trigonometry to find the release angle for an arrow to hit a target and resolve if it will clear an obstacle.
Explanation:
The subject of this question is Physics, specifically dealing with concepts related to projectile motion and the laws of motion. The grade level would likely be High School where students are typically introduced to these topics in a physics curriculum.
As an example, for the question where an archer shoots an arrow at a 75.0 m distant target with an initial speed of 35.0 m/s, one would need to use the equations of projectile motion to solve for the correct release angle. This involves breaking down the initial velocity into its horizontal and vertical components, using the range equation for projectile motion, and applying trigonometry. To determine whether the arrow will go over or under a branch 3.50 m above the release point, one would need to calculate the maximum height of the arrow's trajectory.
In scenarios such as these, understanding the principles of inertia, conservation of momentum, and the conversion of potential energy into kinetic energy are essential.
The maximum lift-to-drag ratio of the World War I Sopwith Camel was 7.7. If the aircraft is in flight at 5000 ft when the engine fails, how far can it glide in terms of distance measured along the ground?
The related concepts to solve this problem is the Glide Ratio. This can be defined as the product between the height of fall and the lift-to-drag ratio. Mathematically, this expression can be written as,
[tex]R = h (\frac{L}{D})_{max}[/tex]
Replacing,
[tex]R = 5000ft (7.7)[/tex]
[tex]R = 38500ft[/tex]
Converting this units to miles.
[tex]R = 38500ft (\frac{1mile}{5280ft})[/tex]
[tex]R = 7.2916miles[/tex]
Therefore the glide in terms of distance measured along the ground is 7.2916miles
The far that can it glide in terms of distance measured along the ground is 38,500 ft.
Given that,
World War I Sopwith Camel was 7.7. If the aircraft is in flight at 5000 ft when the engine fails.Based on the above information, the calculation is as follows:
[tex]\frac{lift}{drag} = \frac{Distance}{height}\\\\7.7 = Distance \div 5500\\\\[/tex]
So, the distance is 38,500 ft.
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Water flows at speed of 5.9 m/s through a horizontal pipe of diameter 3.1 cm . The gauge pressure P1 of the water in the pipe is 1.5 atm . A short segment of the pipe is constricted to a smaller diameter of 2.1 cm. What is the gauge pressure of the water flowing through the constricted segment?
Final answer:
Using Bernoulli's equation and the continuity equation, we find that as the diameter of a pipe decreases, the velocity of the water increases, and to conserve the total mechanical energy, gauge pressure in the constricted segment decreases.
Explanation:
The situation described in the question can be analyzed using the principle of conservation of energy, specifically Bernoulli's equation for incompressible fluid flow. We are given that water flows at a speed of 5.9 m/s through a horizontal pipe of diameter 3.1 cm, and the gauge pressure is 1.5 atm. With the constriction of the pipe's diameter to 2.1 cm, we want to find the new gauge pressure of the water.
According to Bernoulli's equation, the total mechanical energy per unit volume is the same at all points along the streamline, i.e.,
[tex]P + \(\frac{1}{2}\)\(\rho\)v^2 + \(\rho\)gh = constant[/tex]
where P is the pressure, \(\rho\) is the density of the fluid, v is the flow velocity, g is acceleration due to gravity, and h is the elevation. For the situation described, h remains constant (horizontal pipe), and there is no change in the gravitational potential energy. Therefore, we can simplify the equation to:
[tex]P1 + \(\frac{1}{2}\)\(\rho\)v1^2 = P2 + \(\frac{1}{2}\)\(\rho\)v2^2[/tex]
Since the fluid is incompressible and the flow rate must be conserved, the velocity v2 in the constricted segment is determined by the continuity equation:
[tex]A_1v_1 = A_2v_2[/tex]
To solve this problem, we'll use the principle of continuity, which states that the product of cross-sectional area and velocity is constant for an incompressible fluid flowing through a tube. Mathematically, it can be expressed as:
Step 1: Find [tex]\( v_2 \)[/tex]
Using the continuity equation:
[tex]\[ A_1 V_1 = A_2 V_2 \][/tex]
We can find [tex]\( v_2 \)[/tex] as:
[tex]\[ V_2 = \frac{\frac{\pi D_1^2}{4}}{\frac{\pi D_2^2}{4}} \times V_1 \]\[ V_2 = \frac{D_1^2}{D_2^2} \times V_1 \][/tex]
Since [tex]\( A = \frac{\pi D^2}{4} \)[/tex], we can rewrite the equation as:
[tex]\[ V_2 = \frac{\frac{\pi D_1^2}{4}}{\frac{\pi D_2^2}{4}} \times V_1 \]\[ V_2 = \frac{D_1^2}{D_2^2} \times V_1 \][/tex]
Now, let's plug in the values:
[tex]\[ V_2 = \frac{(0.031 \, \text{m})^2}{(0.021 \, \text{m})^2} \times 5.9 \, \text{m/s} \]\[ V_2 = \frac{0.000961 \, \text{m}^2}{0.000441 \, \text{m}^2} \times 5.9 \, \text{m/s} \]\[ V_2 = \frac{0.000961}{0.000441} \times 5.9 \, \text{m/s} \]\[ V_2 \approx 12.84 \, \text{m/s} \][/tex]
Step 2: Calculate [tex]\( P_2 \)[/tex]
We'll use Bernoulli's equation, which states that the sum of pressure energy, kinetic energy, and potential energy per unit volume is constant along any streamline of flow. In this case, we'll ignore changes in height (assuming horizontal flow), and the equation simplifies to:
[tex]\[ P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2 \][/tex]
Where:
- [tex]\( P_1 \)[/tex] is the initial pressure
- [tex]\( P_2 \)[/tex] is the pressure in the constricted segment
- [tex]\( \rho \)[/tex]is the density of the fluid (we'll assume water,[tex]\( \rho = 1000 \, \text{kg/m}^3 \))[/tex]
-[tex]\( v_1 \) and \( v_2 \)[/tex] are initial and final velocities, respectively.
We rearrange this equation to solve for [tex]\( P_2 \)[/tex]:
[tex]\[ P_2 = P_1 + \frac{1}{2} \rho v_1^2 - \frac{1}{2} \rho v_2^2 \][/tex]
Now, let's plug in the values:
[tex]\[ P_2 = P_1 + \frac{1}{2} \times 1000 \times (5.9)^2 - \frac{1}{2} \times 1000 \times (12.84)^2 \]\[ P_2 = P_1 + 17421.5 - 82626.24 \]\[ P_2 = P_1 - 65204.74 \][/tex]
Step 3: Convert [tex]\( P_2 \)[/tex] to atmospheres (gauge pressure)
To express the pressure in atmospheres and considering atmospheric pressure as the baseline, we need to subtract the atmospheric pressure from [tex]\( P_2 \)[/tex]. Assuming atmospheric pressure is [tex]\( 101.3 \, \text{kPa} \) (or \( 101300 \, \text{Pa} \))[/tex]:
[tex]\[ P_{\text{gauge}} = \frac{P_2 - \text{atmospheric pressure}}{\text{atmospheric pressure}} \]\[ P_{\text{gauge}} = \frac{-65204.74 - 101300}{101300} \]\[ P_{\text{gauge}} \approx -1.64 \][/tex]
So, the gauge pressure in the constricted segment is approximately [tex]\( -1.64 \)[/tex] atmospheres.
A uniformly charged, straight filament 6 m in length has a total positive charge of 3 µC. An uncharged cardboard cylinder 1 cm in length and 5 cm in radius surrounds the filament using the filament as its axis of symmetry, with the filament as the central axis of the cylinder. Find the total electric flux through the cylinder. The permittivity of free space is 8.8542 × 10−12 C 2 /N · m2 . 1. 501.959 2. 4356.29 3. 1452.1 4. 1355.29 5. 564.704 6. 250.979 7. 4065.87 8. 141.176 9. 3049.4 10. 847.056 Answer in units of N · m2 /C. 009 (part 2 of 2) 10.0 points What is the electric field at the surface of the cylinder? 1. 224688.0 2. 22468.8 3. 503302.0 4. 209709.0 5. 161776.0 6. 462216.0 7. 719003.0 8. 171191.0 9. 114127.0 10. 179751.0 Answer in units of V/m
Answer:
Part A : 5.) = 564.704 N*m²/C
Part B: 10.) = 179751.0 V/m
Explanation:
A) Applying Gauss'Law to the straight filament, using a cylindrical gaussian surface with the filament as the central axis of the surface, assuming that the electric field is normal to the surface (which means that no flux exist through the lids of the cylinder) and is constant at any point of the surface (except the lids where is 0), we can find the electric flux, as follows:
[tex]E*2*\pi *r*l = \frac{Qenc}{\epsilon 0} (1)[/tex]
where:
r is the radius of the cylinder = 0.05 m
l is the length of the cylinder = 0.01 m
Qenc, is the net charge on the filament enclosed by the gaussian surface
ε₀ = 8.8542*10⁻¹² C²/N*m²
In order to find the value of Qenc, we need to find first the linear charge density, as follows:
[tex]\lambda = \frac{Q}{L} =\frac{+3e-6C}{6m} = 5e-7 C/m[/tex]
The net charge enclosed by the gaussian surface will be just the product of the linear change density λ times the length of the gaussian surface:
[tex]Qenc = \lambda * l = 5e-7C/m * 0.01 m = 5e-9 C[/tex]
According Gauss ' Law, the net flux through the gaussian surface must be equal to the charge enclosed by the surface, divided by the permittivity of free space (in vacuum or air), as follows:
[tex]Flux = \frac{Qenc}{\epsilon 0} = \frac{5e-9C}{8.8542e-12 C2/N*m2} = 564.704 (N*m2)/C[/tex]
which is the same as the option 5.
B) Repeating the equation (1) from above:
[tex]E*2*\pi *r*l = \frac{Qenc}{\epsilon 0} (1)[/tex]
we can solve for E, as follows:
[tex]E = \frac{Qenc}{2*\pi*r*l* \epsilon 0} = \frac{5e-9C}{2*\pi*(0.05m)*(0.01m)*8.8542e-12 C2/N*m2} = 179751.0 V/m[/tex]
which is the same as the option 10 of part B.
A rectangular dam is 101 ft long and 54 ft high. If the water is 35 ft deep, find the force of the water on the dam (the density of water is 62.4 lb/ft3).
Final answer:
The force of water on a rectangular dam is given by the formula F = pgh²L/2, where p is the density of water, h is the depth at the dam, and L is the length of the dam. Using the given values, we can calculate the force of the water on the dam to be approximately 1,130,946 lb.
Explanation:
The force of water on a rectangular dam is given by the formula F = pgh²L/2, where p is the density of water, h is the depth at the dam, and L is the length of the dam. We are given the values p = 62.4 lb/ft³, h = 35 ft, and L = 101 ft.
Substituting these values into the formula, we have F = (62.4 lb/ft³) * (35 ft)² * (101 ft) / 2.
Simplifying the expression, we get F = 1,130,946 lb. Therefore, the force of the water on the dam is approximately 1,130,946 lb.
Some plants disperse their seeds when the fruit splits and contracts, propelling the seeds through the air. The trajectory of these seeds can be determined with a high-speed camera. In an experiment on one type of plant, seeds are projected at 20 cm above ground level with initial speeds between 2.3 m/s and 4.6 m/s. The launch angle is measured from the horizontal, with + 90° corresponding to an initial velocity straight up and – 90° straight down.
The experiment is designed so that the seeds move no more than 0.20 mm between photographic frames. What minimum frame rate for the high-speed camera is needed to achieve this? (a) 250 frames/s; (b) 2500 frames/s; (c) 25,000 frames/s; (d) 250,000 frames/s.
Answer:
trajectory
Explanation:
according to issac newton plants are important
A bimetallic strip consists of two 1-mm thick pieces of brass and steel bonded together. At 20C, the strip is completely straight and has a length of 10 cm. At 30C, the strip bends into a circular arc. What is the radius of curvature of this arc?
Answer:
R = 16.67 m
Explanation:
Given:
- Initial Temperature T_i = 20 C
- Thickness of both strips t = 0.001 m
- Final Temperature T_f = 30 C
- Length of the strip L = 0.1 m
- coefficient of linear expansion for brass a_b = 19*10^-6
- coefficient of linear expansion for steel a_s = 13*10^-6
Find:
What is the radius of curvature of this arc R?
Solution:
- The radius of curvature R in relation to dT and a_b, a_s:
R = t / dT*(a_b - a_s)
R = 0.001 / (30-20)*(19-13)*10^-6
R = 16.67 m
Waves travel along a 100-m length of string which has a mass of 55 g and is held taut with a tension of 75 N. What is the speed of the waves?
Answer:
[tex]v=369.27\frac{m}{s}[/tex]
Explanation:
The speed of the waves in a string is related with the tension and mass per unit length of the string, as follows:
[tex]v=\sqrt\frac{T}{\mu}[/tex]
First, we calculate the mass per unit length:
[tex]\mu=\frac{m}{L}\\\mu=\frac{55*10^{-3}kg}{100m}\\\mu=5.5*10^{-4}\frac{kg}{m}[/tex]
Now, we calculate the speed of the waves:
[tex]v=\sqrt\frac{75N}{5.5*10^{-4}\frac{kg}{m}}\\v=369.27\frac{m}{s}[/tex]
Given values,
Mass, [tex]m = 55 \ g[/tex]or, [tex]= 55\times 10^{-3} \ kg[/tex]
Length, [tex]L = 100 \ m[/tex]As we know,
→ [tex]v = \sqrt{\frac{T}{\mu} }[/tex]
or,
→ [tex]\mu = \frac{m}{L}[/tex]
By putting the values,
[tex]= \frac{55\times 10^{-3}}{100}[/tex]
[tex]= 5.5\times 10^{-4} \ kg/m[/tex]
hence,
The speed of wave:
→ [tex]v = \sqrt{\frac{75}{5.5\times 10^{-4}} }[/tex]
[tex]= 369.27 \ m/s[/tex]
Thus the above response is correct.
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Estimate the number of atoms in your body.
An average adult of weigh 70 kg has approximately [tex]7.0 \times 10^{27}[/tex].
Explanation:
Estimation of Atoms in Human Body
We generally evaluate the amount of elements, liquid, or solid bones in a human body but, we can relate these counts to the atomic level as well. Talking about the approximate amount of Hydrogen, Oxygen and Carbon that we carry in our body, they all make 99% of it neglecting 1% of trace elements.
On the basis of this calculation, we can say that a human body of weight 70 kg has around [tex]7.0 \times 10^{27}[/tex] atoms that are differentiated as 2/3 of hydrogen atoms, 1/4 of Oxygen and only 1/3 of Carbon atoms in the total count.
a 10.0 kg block on a smooth horizontal surface is acted upon by two forces: a horizontal force of 30 N acting to the right and a horizontal force of 60N to the left. What acceleration will the block experience?
Answer:
[tex]a=-3\ m/s^2[/tex]
Explanation:
Second Newton's Law
It allows to compute the acceleration of an object of mass m subject to a net force Fn. The relation is given by
[tex]F_n=m.a[/tex]
The net force is the sum of all vector forces applied to the object. The block has two horizontal forces applied (in absence of friction): The 30 N force acting to the right and the 60 N force to the left. The positive horizontal direction is assumed to the right, so the net force is
[tex]F_n=30\ N-60\ N=-30\ N[/tex]
Thus, the acceleration can be computed by
[tex]\displaystyle a=\frac{F_n}{m}=\frac{-30}{10}=-3\ m/s^2[/tex]
[tex]\boxed{\displaystyle a=-3\ m/s^2}[/tex]
The negative sign indicates the block is accelerated to the left
The block will experience an acceleration of 3.0 m/s² to the left, as calculated by using the net force (30 N leftward) divided by the mass of the block (10.0 kg).
Explanation:To determine the acceleration of a 10.0 kg block on a smooth horizontal surface that is being acted upon by two forces, one must first find the net force acting on the block. In this scenario, there is a 30 N force acting to the right and a 60 N force acting to the left. The net force can be found by subtracting the smaller force from the larger one, which gives us a result of 60 N - 30 N = 30 N acting to the left.
Now, using Newton's second law of motion which states that F = ma (where F is the net force, m is the mass of the object, and a is the acceleration), we can solve for acceleration by rearranging the equation to a = F/m. Plugging the net force (F = 30 N) and the mass of the block (m = 10.0 kg) into the equation yields:
a = 30 N / 10.0 kg = 3.0 m/s²
The acceleration of the block will be 3.0 m/s², and since the net force is to the left, the acceleration will also be directed to the left.
Calculate the electric field (in V/m) between two parallel plates connected to a 45 V battery if the plates are separated by a) 3 cm and b) 5cm. What is the resulting force on an electron in each of these fields?
Answer:
Explanation:
Given
Voltage of battery [tex]V=45\ V[/tex]
electric field(E) is given by
[tex]E=|-\frac{dV}{dx}|[/tex]
i.e. Change in Potential over a distance x
(a)When Plate are separated by 3 cm apart
[tex]E=|\frac{45}{3\times 10^{-2}}|[/tex]
[tex]E=1500\ V/m[/tex]
(b)When Plates are 5 cm apart
[tex]E=|-\frac{45}{5\times 10^{-2}}|[/tex]
[tex]E=900\ V/m[/tex]
Force on electron is given by
[tex]F=charge\times Electric\ Field[/tex]
[tex]F=q\times E[/tex]
For case a
[tex]F=1.6\times 10^{-19}\times 1500[/tex]
[tex]F=24\times 10^{-17}\ N[/tex]
(b)[tex]F=1.6\times 10^{-19}\times 900=14.4\times 10^{-17}\ N[/tex]