Final answer:
The change in momentum of the basketball after impacting the floor and rebounding at the same speed is 2.268 kg·m/s upward.
Explanation:
You asked what the momentum change was when a basketball with a mass of 567 g hits the floor at a speed of 2 m/s and rebounds at nearly the same speed. To find the change in momentum, we need to consider the initial and final momentums of the basketball. The momentum of an object is given by the product of its mass and velocity. Given the initial velocity (vi) is 2 m/s downward and the final velocity (vf) is 2 m/s upward, and the direction is important in momentum calculations.
Momentum before impact = mass × velocity = 0.567 kg × (-2 m/s) = -1.134 kg·m/s (negative sign indicates downward direction)
Momentum after impact = mass × velocity = 0.567 kg × (2 m/s) = 1.134 kg·m/s (positive sign indicates upward direction)
To find the momentum change, Δp, we subtract the initial momentum from the final momentum:
Δp = pf - pi = 1.134 kg·m/s - (-1.134 kg·m/s) = 1.134 kg·m/s + 1.134 kg·m/s = 2.268 kg·m/s.
Therefore, the momentum change is 2.268 kg·m/s in the upward direction.
Newton's law of motions worksheet answer solutions What is the mass of an object that needs a force of 6 600 N to increase its speed from rest to 107 m/s in 2.3 seconds?
Answer:
141.87 kg.
Explanation:
Deduction From Newton's second law of motion.
F = ma....................... Equation 1
Where F = Force acting on the object, m = mass of the object, a = acceleration of the object.
Making m the subject of the equation,
m = F/a .................. Equation 2
But
a = (v-u)/t............... Equation 3
Where v = final velocity, u = initial velocity, t = time.
Given: v = 107 m/s, u = 0 m/s ( fro rest), t = 2.3 s.
Substituting into equation 3
a = (107-0)/2.3
a = 107/2.3
a = 46.52 m/s².
Also Given, F = 6600 N
Substitute into equation 2
m = 6600/46.52
m = 141.87 kg.
Hence the mass of the object = 141.87 kg.
If the distance between two charges is doubled, by what factor is the magnitude of the electric force changed? F_e final/F_e, initial =____
To solve this problem we will apply the concepts related to Coulomb's law for which the Electrostatic Force is defined as,
[tex]F_{initial} = \frac{kq_1q_2}{r^2}[/tex]
Here,
k = Coulomb's constant
[tex]q_{1,2}[/tex] = Charge at each object
r = Distance between them
As the distance is doubled so,
[tex]F_{final} = \frac{kq_1q_2}{( 2r )^2}[/tex]
[tex]F_{final} = \frac{ kq_1q_2}{ 4r^2}[/tex]
[tex]F_{final} = \frac{1}{4} \frac{ kq_1q_2}{r^2}[/tex]
[tex]F_{final} = \frac{1}{4} F_{initial}[/tex]
[tex]\frac{F_{final}}{ F_{initial}} = \frac{1}{4}[/tex]
Therefore the factor is 1/4
A 5.00 kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force F(t) is applied to the end of the rope, and the height of the crate above its initial position is given by y(t)=(2.80 m/s)t +(0.61 m/s^3)t^3.
a. What is the magnitude of the force F when 4.10s ?b. is the magnitude's unit N but the system doesn't accept it?
Answer
F = 124 N
Explanation:
given,
mass, m = 5 Kg
time, t = 4.1 s
displacement = y(t)=(2.80 m/s)t +(0.61 m/s³)t³
velocity
[tex]\dfrac{dy(t)}{dt}=2.80 + 1.83 t^2[/tex]
[tex]v=2.80 + 1.83 t^2[/tex]
again differentiating to get the equation of acceleration
[tex]\dfrac{dv}{dt}= 3.66 t[/tex]
[tex]a= 3.66 t[/tex]
force at time t = 4.10 s
F = m a
F = 5 x 3.66 x 4.1
F = 75 N
the net force when crate is moving upward
F = Mg + Ma
F = 5 x 9.8 + 75
F = 124 N
the magnitude of force is equal to 124 N
An athlete at high performance inhales 4.0L of air at 1 atm and 298 K. The inhaled and exhaled air contain 0.5% and 6.2% by volume of water,respectively. For a respiration rate of 40 breaths per minute, how many moles of water per minute are expelled from the body through the lungs?
To solve this problem we will calculate the total volume of inhaled and exhaled water. From the ideal gas equation we will find the total number of moles of water.
An athlete at high performance inhales 4.0L of air at 1atm and 298K.
The inhaled and exhaled air contain 0.5% and 6.2% by volume of water, respectively.
During inhalation, volume of water taken is
[tex]V_i = (4L)(0.5\%)[/tex]
[tex]V_i = 0.02L[/tex]
During exhalation, volume of water expelled is
[tex]V_e = (4L)(6.2\%)[/tex]
[tex]V_e = 0.248L[/tex]
During 40 breathes, total volume of water taken is
[tex]V_{it} = (40L)(0.02L) = 0.8L[/tex]
During 40 breathes, total volume of water expelled out is
[tex]V_{et} = (40L)(0.248L) = 9.92L[/tex]
Therefore resultant volume of water expelled out from the lung is
[tex]\Delta V = 9.92L-0.8L = 9.12[/tex]
From the body through the lung we have that
[tex]n = \frac{PV}{RT}[/tex]
Here,
P = Pressure
R= Gas ideal constant
T= Temperature
V = Volume
Replacing,
[tex]n = \frac{(1atm)(9.12L)}{(8.314J/mol \cdot K)(298K)}[/tex]
[tex]n = 0.373mol/min[/tex]
Therefore the moles of water per minute are expelled from the body through the lungs is 0.373mol/min
The athlete expels 8.89 x 10^-2 moles of water per minute through the lungs.
Explanation:To calculate the number of moles of water per minute expelled by the athlete through the lungs, we need to determine the amount of water evaporated with each breath and then multiply it by the respiration rate. According to the information provided, an average breath is about 0.5 L, and each breath evaporates 4.0 x 10^-2 g of water. We can convert grams of water to moles by dividing by the molar mass of water (18.02 g/mol). So, the moles of water evaporated with each breath are (4.0 x 10^-2 g)/(18.02 g/mol) = 2.22 x 10^-3 mol/breath.
Next, we can calculate the number of breaths per minute multiplied by the moles of water evaporated per breath to find the moles of water expelled per minute. The respiration rate is given as 40 breaths per minute. Therefore, the moles of water expelled per minute are (2.22 x 10^-3 mol/breath) x 40 breaths/minute = 8.89 x 10^-2 mol/minute.
A block with velocity v>0 slides along the floor (with no friction). It hits an ideal spring at time t=0 (configuration #1). The spring starts to compress until the block comes to a (momentary) stop (configuration #2). (Figure 1) (Later, the spring will of course expand, pushing the block back). Here we show you some plots relating to the motion of the block and spring. You will need to identify what these plots represent. In each plot, the point we label as "1" refers to configuration #1 (when the block first comes in contact with the spring). The point we label "2" refers to configuration #2 (which is the moment the block comes to rest, with the spring fully compressed). Here, "force" refers to the x-component of the force of the spring on the block and "position" (and "velocity") refer to the x-components of the position (and velocity) of the block. In all cases, consider the origin to be (0,0); that is, the x-axis represents y=0 and the y-axis represents x=0.
Part A
Look first a t graph A. (Figure 2)
Which of the choices given could this graph represent?
1. position (x) vs. time
2. velocity (v) vs. time
3. force (F) vs. time
4. force (F) vs. position
Part B
Now look at graph B. (Figure 3)
Which of the choices given could this graph represent?
1. position (x) vs. time
2. velocity (v) vs. time
3. force (F) vs. time
4. force (F) vs. position
Part C
Next look at graph C. (Figure 4)
Which of the choices given could this graph represent?
1. position (x) vs. time
2. velocity (v) vs. time
3. force (F) vs. time
4. force (F) vs. position
Answer:
(A) position vs time
(B) Force vs position
(C) velocity vs time
Explanation:
Part A
This graph shows that the position of the block increases with time along the x-axis exponentially (that is it increases in unequal amounts in equal time intervals). This is because the velocity of the block is changing with time and as a result the position changes in unequal amounts per time
PartB
The force on the spring increases in a negative direction going from zero to a negative value. This is because the spring is being compressed from configuration 1 to 2. The force of compression on a spring is usually taken to have a negative sign and expansion to have a positive sign. So in this case force becomes increasingly negative with time.
Part C
The velocity of the block decreases from a positive nonzero value (v>0) to zero because the spring resists the motion of the block. As a result the block comes to a stop momentarily. The velocity decreases exponentially because the acceleration of the block is also changing with time since the force of the block is decreasing with time.
Thank you for reading.
A uniform line charge that has a linear charge density λ = 3.3 nC/m is on the x axis between x = 0 to x = 5.0 m.
(a) What is its total charge?
.... nC
(b) Find the electric field on the x axis at x = 6 m.
.... N/C
(c) Find the electric field on the x axis at x = 10.0 m.
.... N/C
(d) Find the electric field on the x axis at x = 270 m.
.... N/C
(e) Estimate the electric field at x = 270 m, using the approximation that the charge is a point charge on the x axis at x = 2.5 m.
.... N/C
(f) Compare your result with the result calculated in part (d) by finding the ratio of the approximation to the exact result. To do this, you will need to assume that the values given in this problem statement are valid to more than two significant figures.
.....
(g) Is your approximate result greater or smaller than the exact result?
... greater
... smaller
so what do we want our website for the best of all I can get to see
A uniform line charge with 16.5 nC total charge creates an electric field that weakens with distance from the line. The exact electric field strength can be calculated using the formula for an infinitely long line charge, while approximating the line as a point charge underestimates the true field strength.
Solving for the Electric Field of a Uniform Line Charge
(a) Total Charge:
The total charge (Q) of a line charge can be found by integrating the linear charge density (λ) over the length (L) of the charge distribution:
Q = ∫λ dx (from x = 0 to x = 5.0 m)
Here, λ = 3.3 nC/m and L = 5.0 m.
Q = (3.3 nC/m) * (5.0 m) = 16.5 nC
Therefore, the total charge is 16.5 nC.
(b) Electric Field at x = 6 m (Exact):
Because the line charge is infinitely long, we can apply the electric field formula:
E = (λ / (2πε₀)) * ln(2a / b)
Plugging in the values:
E = (3.3 x 10⁻⁹ C/m) / (2π * 8.854 x 10⁻¹² C²/N∙m²) * ln(2 * 6 m / 0)
E ≈ 47.7 N/C (rounded to two significant figures)
(c) Electric Field at x = 10.0 m (Exact):
Following the same method as part (b):
E ≈ (3.3 x 10⁻⁹ C/m) / (2π * 8.854 x 10⁻¹² C²/N∙m²) * ln(2 * 10.0 m / 0)
E ≈ 33.1 N/C (rounded to two significant figures)
(d) Electric Field at x = 270 m (Exact):
Using the same formula:
E ≈ (3.3 x 10⁻⁹ C/m) / (2π * 8.854 x 10⁻¹² C²/N∙m²) * ln(2 * 270 m / 0)
E ≈ 0.012 N/C (rounded to three significant figures)
(e) Electric Field at x = 270 m (Approximation):
Assuming the charge is a point charge at x = 2.5 m (center of the line charge):
E ≈ k * Q / (x - 2.5 m)²
(f) Ratio of Approximation to Exact Result:
Ratio = (Approximate Electric Field) / (Exact Electric Field)
Ratio ≈ (2.02 x 10⁻⁴ N/C) / (0.012 N/C) ≈ 0.0017
(g) Comparison of Results:
Since the ratio is less than 1, the approximate result (2.02 x 10⁻⁴ N/C) is smaller than the exact result (0.012 N/C). This is reasonable because approximating the finite line charge as a point charge weakens the effect of the charge, leading to a lower electric field value.
A block of wood is floating in water; it is depressed slightly and then released to oscillate up and down. Assume that the top and bottom of the block are parallel planes which remain horizontal during the oscillations and that the sides of the block are vertical. Show that the period of the motion (neglecting friction) is 2π ph/g, where h is the vertical height of the part of the block under water when it is floating at rest. Hint: Recall that the buoyant force is equal to the weight of displaced water.
Explanation:
Equilibrium position in y direction:
W = Fb (Weight of the block is equal to buoyant force)
m*g = V*p*g
V under water = A*h
hence,
m = A*h*p
Using Newton 2nd Law
[tex]-m*\frac{d^2y}{dt^2} = Fb - W\\\\-m*\frac{d^2y}{dt^2} = p*g*(h+y)*A - A*h*p*g\\\\-A*h*p*\frac{d^2y}{dt^2} = y *p*A*g\\\\\frac{d^2y}{dt^2} + \frac{g}{h} * y =0[/tex]
Hence, T time period
T = 2*pi*sqrt ( h / g )
Common transparent tape becomes charged when pulled from a dispenser. If one piece is placed above another, the repulsice force can be great enough to support the top pieces weight. Assuming equal point charges, calculate the magnitide of the charge if electrostatic force is great enough to support the weight of a 11.0 mg piece of tape held 1.00 cm above another.
Answer:
Q = 1.095 x 10^-9 C
Let the force experienced by the top piece of tape be F
F = kQ²/r²
r = distance between the two pieces tape = 1.00cm = 1.00 x 10^ -2 m
1/4(pi)*Eo = k = 8.99 x 10^9 Nm²/C²
The electric force of repulsion between the two charges and the weight of the top piece of tape are equal so
F = kQ²/r² = mg
Where m is the mass of the top piece of tape and g is the acceleration due to gravity
On re-arranging the equation above,
Q² = mgr²/k
Q² = ((11.0 x 10^-6) x 9.8 x (1.00x10^-2)²)/(8.99 x 10^9)
Q = 1.095x10^-9 C
Explanation:
The charge Q on both pieces of tape are equal and both act with a force of repulsion on each other.
The force of repulsion between both tapes pushes the top piece of tape upwards. The weight of the top piece of tape acts vertically downward. Since the top tape is in a position of equilibrium, the two forces acting on the top piece of tape must be equal to each other. This assumption is backed up by newton's first law of motion which states that the summation of all forces acting on a body at rest must be equal to zero. That is
Fe (electric force) - Fg (gravitational force) = 0
Fe = Fg
kQ²/r² = mg
On substituting the respective values for all variables except Q and rearranging the equation Q = 1.09 x 10^-9
If the gap between C and the rigid wall at D is initially 0.15 mm, determine the magnitudes of the support reactions at A and D when the force P
Answer / Explanation
The question in the narrative is incomplete.
Kindly find the complete question below:
If the gap between C and the rigid wall at D is initially 0.15 mm, determine the support reactions at A and D when the force is applied. The assembly is made of A36 steel
Procedure
Recalling the the equation of equilibrium and referencing the free body diagram of the assembly,
Therefore, ∑fₓ = 0 ,
where, 20 ( 10³) - Fₐ - Fₙ = 0 --------------equation (1)
Now, recalling the compatibility equation, while utilizing the superposition method,
Therefore, δₓ - δfₓ
= 0.15 = 200(10³)(600) ÷ π/4 (0.05²)(200)(10⁹) - [ Fₐ (600) / π/4 (0.05²)(200)(10⁹) + Fₐ (600) π / 4 (0.05²)(200)(10⁹) ]
Solving this further,
We get: Fₐ = 20365.05 N
Which is equivalent to = 20.4 kN.
Now, substituting the answer (Fₐ) into equation (1)
Fₙ = 179634.95 N
= 180 kN
A rocket is launched upward with a constant acceleration of 165 m/s^2. After 8.00 seconds of ascension a passenger on the rocket drops a rock out of it. How long does it take for this rock to hit the ground?
To solve this problem we will apply the concepts related to the linear kinematic movement. We will start by finding the speed of the body from time and the acceleration given.
Through the position equations we will calculate the distance traveled.
Finally, using this same position relationship and considering the previously found speed, we can determine the time to reach your goal.
For time (t) and acceleration (a) we have to,
[tex]t = 8s, a = 165m/s^2[/tex]
The velocity would be,
[tex]u = a*t \\u = 165*8\\u = 1320m/s[/tex]
Now the position is,
[tex]h= \frac{1}{2} at^2[/tex]
[tex]h = \frac{1}{2} 165*8^2[/tex]
[tex]h = 5280m[/tex]
Now with the initial speed and position found we will have the time is,
[tex]h=ut +\frac{1}{2} at^2[/tex]
[tex]-5280=1320t - \frac{1}{2} 9.8t^2[/tex]
[tex]4.9t^2-1320t-5280=0[/tex]
Solving the polynomian we have,
[tex]t = 273.33s = 4.56minutes[/tex]
Therefore the rocket will take to hit the ground around to 4.56min
First, we need to determine the velocity of the rocket at the time the rocket is dropped after 8 seconds of powered ascent.
Given the constant acceleration of 165 m/s^2, the velocity (v) at 8 seconds can be found using the formula
v = at
where 'a' is the acceleration and 't' is the time.
Therefore, v = 165 m/s^2 x 8 s = 1320 m/s.
Now that the rocket is dropped, it will initially have the velocity of the rocket at that instant, which is 1320 m/s upward. To find out how long it takes for the rock to reach the highest point, we use the formula
v = u + at
where 'u' is initial velocity, 'v' is final velocity (0 m/s at the highest point), 'a' is the acceleration due to gravity (which is negative since it is in the opposite direction to the initial motion), and 't' is the time.
Solving for time, we get
t = -u/g. With g ≈ 9.81 m/s^2, the time to reach the highest point is
t ≈ -1320 m/s / -9.81 m/s^2 ≈ 134.6 s.
After reaching the highest point, the rocket will start falling back to the ground. Since the rocket starts from rest at the highest point, we can use the formula
s = 0.5gt^2
where 's' is the distance and 't' is time, to calculate the time it takes to fall to the ground.
But since we already have the time to reach the highest point, we can simply double that time to find the total time taken for the round trip, because the time to go up is the same as the time to come down in free fall. So the total time taken for the rocket to hit the ground is
134.6 s up + 134.6 s down = 269.2 s.
If you must do positive work to bring a charged balloon toward a negatively charged sphere, is the charge on the balloon positive or negative? Or can it have any sign?
Let's start from the definition of attraction and repulsion. Similar charges tend to repel each other, while different charges attract. When work is done due to the force of attraction its value will be negative, while if work is done due to the force of repulsion its value will be positive.
Given this the sphere has a negative charge.
In other words, if the balloon has a positive charge, it will be attracted by the sphere with negative charge. In this case, you would not have to do a positive job to bring them together. If the balloon has a negative charge, it will be repelled by the sphere with a negative charge. In this case, you will do a positive job to unite them.
Therefore, the load on the balloon is negative.
The charge on the ballon will be Negative
What do you understand by Columbus's law of attraction and repulsion forces?The answer to this question is based on the column's law of forces of attraction and repulsion between two charged atoms.
The formula for calculating the magnitude of the forces is given as
[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]
Here [tex]q_1 & \ q_2[/tex] are the two charged atoms
r = Distance between the two particles
k is constant
Now if two charged particles are having like charges that are both charged particles carrying positive or both carrying a negative charge then there will be Force of repulsion created between them
If both the Particles have the opposite charges on them then there will be a force of attraction between them.
The charge on the ballon will be negative that's why there is positive work needed to bring the ballon towards the negatively charged sphere
Thus the charge on the ballon will be Negative
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A rigid tank internal energy of fluid 800kJ. Fluid loses 500kJ of heat and padle wheel does 100kJ of work. Find final internal energy in tank.
Answer:
U₂ = 400 KJ
Explanation:
Given that
Initial energy of the tank ,U₁= 800 KJ
Heat loses by fluid ,Q= - 500 KJ
Work done on the fluid ,W= - 100 KJ
Sign -
1.Heat rejected by system - negative
2.Heat gain by system - Positive
3.Work done by system = Positive
4.Work done on the system-Negative
Lets take final internal energy =U₂
We know that
Q= U₂ - U₁ + W
-500 = U₂ - 800 - 100
U₂ = -500 +900 KJ
U₂ = 400 KJ
Therefore the final internal energy = 400 KJ
An expensive vacuum system can achieve a pressure as low as 1.83 ✕ 10−7 N/m2 at 28°C. How many atoms are there in a cubic centimeter at this pressure and temperature?
Answer:[tex]44.1\times 10^6[/tex] atoms
Explanation:
According to the ideal gas equation:
[tex]PV=nRT[/tex]
P = Pressure of the gas = [tex]1.83\times 10^{-7}N/m^2=1.81\times 10^{-12}atm[/tex] [tex]1N/m^2=9.87\times 10^{-6}atm[/tex]
V= Volume of the gas = [tex]1cm^3=1ml=0.001L[/tex] (1L=1000ml)
T= Temperature of the gas = 28°C = 301 K [tex]0^0C=273K[/tex]
R= Gas constant = 0.0821 atmL/K mol
n= moles of gas= ?
[tex]n=\frac{PV}{RT}=\frac{1.81\times 10^{-12}atm\times 0.001L}0.0821Latm/Kmol\times 301K}=7.32\times 10^{-17}moles[/tex]
Number of atoms =[tex]moles\times {\text {avogadro's number}}=7.32\times 10^{-17}mol\times 6.023\times 10^{23}mol^{-1}=44.1\times 10^6atoms[/tex]
Thus there are [tex]44.1\times 10^6[/tex] atoms in a cubic centimeter at this pressure and temperature.
A device called a parallel-plate capacitor consists of two large, flat, metal plates held parallel to each other and separated by a small gap. One plate is positively charged and the other plate is negatively charged. A positive point charge is placed in the gap between the two plates and near the center of each plate. Does the charge experience an electric force, and if so, in which direction does the force on the charge point?a. No, there is no force on the negative charge. b. Yes, but electric force on the negative charge points equally in both directions, towards the positive plate and towards the negative plate. c. Yes, the electric force on the negativc charge is directed away from negative plate to and towards the positive plate. d. Yes, the electric force on the negative charge is directed parallel to both plates. e. Yes, the electric force on the negative charge is directed away from the positive plate and towards the negative plate.
Answer:
Yes, there will a force acting on the positive point charge. The options provided are that of negative point charge, rather than a positive point charge stated in the question
Explanation:
Here is the explanation:
When a capacitor is charged, one plate is positively charge and the other is negatively charged. In between the oppositely charged plates, there exist a potential difference. A positive point charge placed in the distance between the two plate will experience an electric force due to the potential difference. The direction of the force will be directed away from the positvely charged plate and towards the negatively charged plate. The reason is due to the law of electrostatic force, which states: like charges repel and unlike charges attract.
A parallel-plate capacitor consists of two large, flat, metal plates held parallel to each other. A positive point charge placed in the gap between the plates and near the center will experience an electric force directed away from the positive plate and towards the negative plate.
Explanation:A device called a parallel-plate capacitor consists of two large, flat, metal plates held parallel to each other and separated by a small gap. When a positive point charge is placed in the gap between the two plates and near the center of each plate, it experiences an electric force.
The direction of the electric force on the negative charge is from the positive plate towards the negative plate. This means the electric force is directed away from the positive plate and towards the negative plate.
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Masses are stacked on top of the block until the top of the block is level with the waterline. This requires 20 g of mass. What is the mass of the wooden block
Answer:
Mass of the wooden Block is 20g.
Explanation:
The buoyant force equation will be used here
Buoyant Force= ρ*g*1/2V Here density used is of water
m*g= ρ*g*1/2V
Simplifying the above equation
2m= ρ*V Eq-1
Also we know from the question that
ρ*V = m + 0.020 Eq-2 ( Density = (Mass+20g)/Volume )
Equating Eq-1 & Eq-2 we get
2m = m+0.020
m = 0.020kg
m = 20g
An object is undergoing SHM with period 0.300 s and amplitude 6.00 cm. At t = 0 the object is instantaneously at rest at x = 6.00 cm. Calculate the time it takes the object to go from x = 6.00 cm to x = -1.50 cm
Answer:
Time taken is 0.087 s
Solution:
As per the question:
Time period, T = 0.300 s
Amplitude, A = 6.00 cm
Now,
To calculate the time taken:
For SHM, we know that:
[tex]x = Acos\omega t[/tex] (1)
At x = 6.00 cm, the object comes to rest instantaneously at times t = 0.00 s
Thus from eqn (1), for x = 6.00 cm:
[tex]6.00 = 6.00cos\omega t[/tex]
[tex]cos\omega t = 1[/tex]
[tex]\omega t = cos^{- 1}(1)[/tex]
[tex]\omega t = 0[/tex]
Thus at t = 0.00 s, x = 6.00 cm
Now,
Using eqn (1) for x = - 1.50 cm:
[tex]- 1.50 = 6.00cos\omega t'[/tex]
[tex]cos\omega t' = -0.25[/tex]
We know that:
[tex]\omega = \frac{2\pi}{T}[/tex]
Thus
[tex]\frac{2\pi}{0.300} t' = cos^{- 1}(0.25)[/tex]
[tex]t' = 0.087\ s[/tex]
Time taken by the object in moving from x = 6.00 cm to x = 1.50 cm:
t' - t = 0.087 - 0.00 = 0.087 s
To find the time it takes for the object to go from x = 6.00 cm to x = -1.50 cm, we use the equation for simple harmonic motion. The time it takes is half of the period, which is 0.150 s.
Explanation:To find the time it takes for the object to go from x = 6.00 cm to x = -1.50 cm, we need to use the equation for simple harmonic motion (SHM). The equation is given by x = A * cos(2π/T * t + φ), where x is the position, A is the amplitude, T is the period, t is the time, and φ is the phase shift.
First, we need to determine the phase shift. At t = 0, the object is at rest at x = 6.00 cm. This means the phase shift is 0, because the cosine function is at a maximum at t = 0.
Next, we can plug in the values into the equation. The amplitude A is 6.00 cm and the period T is 0.300 s. We want to find the time it takes for the object to go from x = 6.00 cm to x = -1.50 cm. In SHM, the object goes from -A to A and back in one period, so the time it takes to go from x = 6.00 cm to x = -1.50 cm is half of the period. Therefore, the time is 0.300 s / 2 = 0.150 s.
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A parallel-plate capacitor is connected to a battery. What happens to the stored energy if the plate separation is doubled while the capacitor remains connected to the battery?
(a) It remains the same
(b) It is doubles
(c) It decreases by a factor of 2
(d) It decreases by a factor of 4
(e) It increases by a factor of
Answer:
(c) As 'd' becomes doubled, energy decreases by the factor of 2
Explanation:
Energy stored in a parallel plate capacitor is given by:
[tex]U=\frac{1}{2}CV^2\\\\C=\frac{A\epsilon_{o}}{d}\\\\then\\\\U=\frac{1}{2}\frac{A\epsilon_{o}}{d}V^2--(1)\\\\[/tex]
As capacitor remains connected to the battery so V remains constant. As can be seen from (1) that energy is inversely proportional to the separation between the plates so as 'd' becomes doubled, energy decreases by the factor of 2.
Answer:
(c) It decreases by a factor of 2
Explanation:
Since the capacitor is still connected to the power source, the potential difference remain the same even when the distance is a doubled.
The energy stored in a capacitor can be written as:
E = (1/2)CV^2 .....1
And the capacitance of a capacitor is inversely proportional to the distance between the two plates of the capacitor.
C = kA/d ....2
Therefore, when d doubles, and every other determinant of capacitance remains the same, the capacitance is halved.
Cf = kA/2d = C/2
Cf = C/2
Since the capacitance has been halved and potential difference remains the same, the energy stored would also be halved since the energy stored in the capacitor is directly proportional to the capacitance.
Ef = (1/2)(Cf)V^2
Ef = (1/2)(C/2)V^2 = [(1/2)CV^2]/2
Ef = E/2
Where;
E and Ef are the initial and final energy stored in the capacitor respectively
C and Cf are the initial and final capacitance of the capacitor.
d is the distance between the plates
A is the area of plates
k is the permittivity of dielectrics
Therefore the energy stored in the capacitor is decreased by a factor of 2, when the distance is doubled.
A rod of length 30.0 cm has linear density (mass per length) given by l 5 50.0 1 20.0x where x is the distance from one end, measured in meters, and l is in grams/meter. (a) What is the mass of the rod? (b) How far from the x 5 0 end is its center of mass?
Answer:
(a). The mass of the rod is 15.9 g.
(b). The center of mass is 0.153 m.
Explanation:
Given that,
Length = 30.0 cm
Linear density [tex]\labda=50.0+20.0x[/tex]
We need to calculate the mass of rod
Using formula of mass
[tex]M=\int{dm}[/tex]
[tex]M=\int{(50.0+20.0x)dx}[/tex]
[tex]M=50.0x+10x^2[/tex]
Put the value of x
[tex]M=50.0\times0.30+10\times(0.30)^2[/tex]
[tex]M=15.9\ g[/tex]
We need to calculate center of mass
The center of mass has an x coordinate is given by
[tex]x_{cm}=\dfrac{\int{xdm}}{\int{dm}}[/tex]
We need to calculate the value of [tex]\int{xdm}[/tex]
[tex]\int{xdm}=\int{(50.0x+20.0x^2)dx}[/tex]
[tex]\int{xdm}=25x^2+\dfrac{20}{3}x^3[/tex]
Put the value into the formula
[tex]\int{xdm}=25\times0.3^2+\dfrac{20}{3}\times(0.3)^3[/tex]
[tex]\int{xdm}=2.43[/tex]
Put the value into the formula of center of mass
[tex]x_{cm}=\dfrac{2.43}{15.9}[/tex]
[tex]x_{cm}=0.153\ m[/tex]
Hence, (a). The mass of the rod is 15.9 g.
(b). The center of mass is 0.153 m.
To find the mass of the rod, integrate the linear density function. To find the center of mass, set up an integral to find the position x such that the total mass on one side is equal to the total mass on the other side.
Explanation:(a) To find the mass of the rod, we need to integrate the linear density function over the length of the rod. The linear density function is given by ℓ(x) = l + 20x, where x is the distance from one end measured in meters and l is in grams/meter. We can integrate this function from 0 to 0.3 meters (corresponding to a length of 30.0 cm) to find the total mass:
M = ∫(0 to 0.3) (l + 20x) dx
M = ∫(0 to 0.3) l dx + ∫(0 to 0.3) 20x dx
(b) To find the center of mass of the rod, we need to find the position x such that the total mass on one side of it is equal to the total mass on the other side. We can set up an integral to find this position:
x_cm = ∫(0 to x_cm) (l + 20x) dx - ∫(x_cm to 0.3) (l + 20x) dx
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You are 9.0 m from the door of your bus, behind the bus, when it pulls away with an acceleration of 1.0 m/s2. You instantly start running toward the still-open door at 5.7 m/s.
How long does it take for you to reach the open door and jump in?
What is the maximum time you can wait before starting to run and still catch the bus?
Final answer:
To determine the time it takes for you to reach the open door and jump in, we can use the equations of motion. We know that you are initially 9.0 m away from the door, and the bus is accelerating at 1.0 m/s². The final velocity of the bus is not given, so we can't find the exact time it takes for you to reach the door. However, we can find the maximum time you can wait before starting to run and still catch the bus.
Explanation:
To determine the time it takes for you to reach the open door and jump in, we can use the equations of motion. We know that you are initially 9.0 m away from the door, and the bus is accelerating at 1.0 m/s². The final velocity of the bus is not given, so we can't find the exact time it takes for you to reach the door. However, we can find the maximum time you can wait before starting to run and still catch the bus.
To find the maximum time you can wait, we need to calculate when the distance between you and the door is equal to 0. Since you are moving towards the door at a constant speed of 5.7 m/s and the bus is accelerating away from you, the distance between you and the door will decrease over time. Let's call the time you wait before starting to run as 't'.
The distance traveled by the bus can be calculated using the equation:
S = ut + (1/2)at^2
Where S is the distance traveled, u is the initial velocity which is 0 m/s, a is the acceleration which is 1.0 m/s², and t is the time.
The distance traveled by you can be calculated using the equation:
S = vt
Where S is the distance traveled, v is your constant velocity which is 5.7 m/s, and t is the time.
After the time 't', both the bus and you will be at the same position which is the door. So the total distance traveled by you and the bus will be equal. We can set up the equation:
ut + (1/2)at^2 = vt
Simplifying this equation, we get:
(1/2)at^2 - vt = 0
Since we know that you can wait a maximum of 't' seconds, we need to solve this quadratic equation for 't'.
Using the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac))/2a
Where a = (1/2)a, b = (-v), and c = 0
Substituting the values, we get:
t = (-(-v) ± sqrt((-v)^2 - 4(1/2)a(0)))/2(1/2)a
t = v + sqrt(v^2)/a
Now we can substitute the values of 'v' and 'a' to find the maximum time you can wait before starting to run:
t = 5.7 + sqrt((5.7)^2)/1.0
t = 5.7 + sqrt(32.49)/1.0 ≈ 5.7 + 5.7 = 11.4 seconds
Therefore, the maximum time you can wait before starting to run and still catch the bus is approximately 11.4 seconds.
The pressure rise p associated with wind hitting a window of a building can be
estimated using the formula p = rho(V2
/2), where rho is density of air and V is the speed of the
wind. Apply the grid method to calculate pressure rise for rho = 1.2 kg/m3
and V = 100 km/h.
(20%)
a. Express your answer in pascals.
b. Express your answer in meters of water column (m-H2O).
Answer:
a)P=462.70 Pa
b)h = 0.047 m of water
Explanation:
Given that
Pressure ,[tex]P=\dfrac{1}{2}\rho V^2[/tex]
[tex]\rho = 1.2\ kg/m^3[/tex]
V= 100 km/h
[tex]V=100\times \dfrac{1000}{3600}\ m/s[/tex]
V=27.77 m/s
The pressure P
[tex]P=\dfrac{1}{2}\rho V^2[/tex]
[tex]P=\dfrac{1}{2}\times 1.2\times 27.77^2\ Pa[/tex]
P=462.70 Pa
We know that density of the water [tex]\rho=1000\ kg/m^3[/tex]
Lets height of the water column = h m
We know that
[tex]_P=\rho _w g h[/tex]
462.70 = 1000 x 9.81 h
[tex]h=\dfrac{462.7}{1000\times 9.81}\ m[/tex]
h = 0.047 m of water
a)P=462.70 Pa
b)h = 0.047 m of water
a)The presuure rise will be P=462.70 Pa
b) The height of the water column h = 0.047 m of water
What will be the pressure rise and the height of the water column of the fluid?It is given that
Pressure,
[tex]p= \dfrac{1}{2} \rho v^2[/tex]
Here [tex]\rho =1.2 \ \dfrac{kg}{m^3}[/tex]
[tex]V=100\ \frac{km}{h} =\dfrac{100\times 1000}{3600} =27.77 \ \dfrac{m}{s}[/tex]
Now to calculate the pressure P
[tex]P=\dfrac{1}{2} \rho v^2[/tex]
[tex]P= \dfrac{1}{2}\times 1.2\times (27.77)^2[/tex]
[tex]P=462.70 \ \frac{N}{m^2}[/tex]
As we know that the density of water
[tex]\rho = 1000\ \frac{kg}{m^3}[/tex]
Lets height of the water column = [tex]h_m[/tex]
As We know that
[tex]P= \rho_w gh[/tex]
[tex]462.70=1000\times 9.81\times h_m[/tex]
[tex]h_m=0.047\ m \ of \ water[/tex]
Thus
a)The presuure rise will be P=462.70 Pa
b) The height of the water column h = 0.047 m of water
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Solar energy strikes earth’s atmosphere at 343 W m-2 . About 30% is reflected and the rest is absorbed
Answer:
The black body temperature of Earth T_e= 180.4 K
Explanation:
Assuming we have to find Black body temperature of the earth.
[tex]T_e =(\frac{s_o(1-\alpha)}{4\sigma})^{0.25}[/tex]
S0= solar energy striking the earth= 343 Wm^{-2}
\alpaha = 30% = 0.3
\sigma = stephan boltsman constant.= 5.67×10^{-8} Wm^{-2}K^4
[tex]T_e =(\frac{343(1-0.3)}{4\times5.67×10^{-8}})^{0.25}[/tex]
T_e= 180.4 K
A hunter is aiming horizontally at a monkey who is sitting in a tree. The monkey is so terrified when it sees the gun that it falls off the tree. At that very instant, the hunter pulls the trigger. What will happen?
a) The bullet will miss the monkey because the monkey falls down while the bullet speeds straight forward.
b) The bullet will hit the monkey because both the monkey and the bullet are falling downward at the same rate due to gravity.
c) The bullet will miss the monkey because although both the monkey and the bullet are falling downward due to gravity, the monkey is falling faster.
d) It depends on how far the hunter is from the monkey.
Answer:a) The bullet will miss the monkey because the monkey falls down while the bullet speeds straight forward.
Explanation: The bullet keeps as it aim( the monkey) unless it is redirected by an external force that could redirect it. Hence, the bullet speeds straight forward.
Final answer:
(b) The bullet will hit the monkey because both the monkey and the bullet are subject to gravity's acceleration equally upon being released or fired; both will fall downward at the same rate. Hence, (b) is the correct option.
Explanation:
When a hunter aims horizontally at a monkey in a tree and the monkey drops at the moment the gun is fired, the outcome is determined by Newtonian physics.
According to Newton's laws, the bullet and the monkey are both subject to gravity and will start to fall toward the ground at the same rate, regardless of any horizontal motion. Therefore, the correct answer is:
b) The bullet will strike the monkey because gravity causes both the bullet and the monkey to fall at the same speed.
This scenario illustrates the principle that horizontal and vertical motions are independent of each other. When the gun is fired, the bullet travels forward while also accelerating downward due to gravity.
Since the monkey begins to fall at the same moment the bullet is fired, both the bullet and the monkey undergo the same downward acceleration, meaning they will fall together.
Earth travels around the Sun at an average speed of 29.783 km/skm/s. Convert this speed to miles per hour. Express your answer using five significant figures.
The given speed of Earth in km/sec is converted first to miles/sec and then to miles/hour, resulting in an average speed of Earth around the Sun of approximately 66,661.6 miles/hour.
Explanation:To solve this, we need to convert kilometers to miles and seconds to hours. First, we should know that 1 kilometer is approximately 0.621371 miles, and 1 hour has 3600 seconds.
Given that, we can first convert Earth's speed from kilometers per second to miles per second by multiplying by the conversion factor:
29.783 km/sec * 0.621371 mile/km = 18.5171 miles/sec.
Next, we convert seconds to hours:
18.5171 miles/sec * 3600 sec/hour = 66,661.6 miles/hour.
So, the average speed of the Earth around the Sun, in miles per hour, to five significant figures is 66,661.6 miles/hour.
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Final answer:
The Earth travels around the Sun at an average speed of 29.783 km/s, which is approximately 66,636.7 miles per hour when converted using the steps of multiplying by 3600 to get km/hr and then by the conversion factor for km to mi.
Explanation:
The student's question is about converting the speed of the Earth's orbit around the Sun from kilometers per second to miles per hour. The given speed is 29.783 km/s. To convert this to miles per hour, we can follow these steps:
Multiply the kilometres per second by 3600, which is the number of seconds in an hour, to get the kilometres per hour.
Convert kilometers per hour to miles per hour by multiplying by the conversion factor (1 kilometer is approximately 0.621371 miles).
Performing these calculations:
29.783 km/s × 3600 s/hr = 107218.8 km/hr
107218.8 km/hr × 0.621371 mi/km = 66636.7 miles per hour
Therefore, Earth travels around the Sun at an average speed of 66,636.7 miles per hour, expressed with five significant figures.
When the spring, with the attached 275.0 g mass, is displaced from its new equilibrium position, it undergoes SHM. Calculate the period of oscillation, T , neglecting the mass of the spring itself.
Answer:
The period of oscillation is 1.33 sec.
Explanation:
Given that,
Mass = 275.0 g
Suppose value of spring constant is 6.2 N/m.
We need to calculate the angular frequency
Using formula of angular frequency
[tex]\omega=\sqrt{\dfrac{k}{m}}[/tex]
Where, m = mass
k = spring constant
Put the value into the formula
[tex]\omega=\sqrt{\dfrac{6.2}{275.0\times10^{-3}}}[/tex]
[tex]\omega=4.74\ rad/s[/tex]
We need to calculate the period of oscillation,
Using formula of time period
[tex]T=\dfrac{2\pi}{\omega}[/tex]
Put the value into the formula
[tex]T=\dfrac{2\pi}{4.74}[/tex]
[tex]T=1.33\ sec[/tex]
Hence, The period of oscillation is 1.33 sec.
The position of a particle moving along the x-axis varies with time according to x(t) = 5.0t2-4.0t3m. Find (a) the displacement, average velocity, average acceleration between 0.0 s and 1.0,
Answer:
[tex]\Delta x=1-0=1\ m[/tex]
[tex]\Delta v=-2-0=-2\ m.s^{-1}[/tex]
[tex]\Delta a=-14-10=-24\ m.s^{-1}[/tex]
Explanation:
The equation governing the position of the particle moving along x-axis is given as:
[tex]x=5\times t^2-4\times t^3[/tex]
we know that the time derivative of position gives us the velocity:
[tex]\frac{d}{dt} x=v[/tex]
[tex]v=10\ t-12\ t^2[/tex]
and the time derivative of of velocity gives us the acceleration:
[tex]\frac{d}{dt} v=a[/tex]
[tex]a=10-24\ t[/tex]
Now, when t = 0
[tex]x=0\ m[/tex]
[tex]v=0\ m.s^{-1}[/tex]
[tex]a=10\ m.s^{-2}[/tex]
When t=1 s
[tex]x_1=5\times 1^2-4\times 1^3=1\ m[/tex]
[tex]v_1=10\times 1-12\times 1^2=-2\ m.s^{-1}[/tex]
[tex]a_1=10-24\times 1=-14\ m.s^{-2}[/tex]
Hence,
Displacement between the stipulated time:
[tex]\Delta x=x_1-x[/tex]
[tex]\Delta x=1-0=1\ m[/tex]
Velocity between the stipulated time:
[tex]\Delta v=v_1-v[/tex]
[tex]\Delta v=-2-0=-2\ m.s^{-1}[/tex]
Acceleration between the stipulated time:
[tex]\Delta a=a_1-a[/tex]
[tex]\Delta a=-14-10=-24\ m.s^{-1}[/tex]
Here negative sign indicates that the vectors are in negative x direction.
One pipe has successive harmonics of 165 Hz, 275 Hz, and 385 Hz, and another pipe has successive harmonics of 165 Hz, 220 Hz, and 275 Hz. (a) Which pipe is closed at one end and which is open at both ends
Answer:
165 Hz, 220 Hz, and 275 Hz belongs to pipe open at both ends
165 Hz, 275 Hz, and 385 Hz belongs to pipe closed at one end
Explanation:
Open ended pips have harmonic frequencies that are multiple of the fundamental frequency
Find the fundamental frequency for each of the samples:
165Hz, 275Hz, 385Hz
(275-165)=110
(385-275)= 110
165 Hz, 220 Hz, and 275 Hz
(220-165)=55
(275-220)=55
F= 55
Note that 165 =3f
220=4f
275=5f
SO these frequencies are multiples of the fundamental frequency
You measure the intensity of a sound wave to be 9.80 W/m2 . The power output of the signal is 75 W and the signal is emitted in all directions.
Part A
How far away from the source are you?
r = ?
Answer:
r = 0.78 m
Explanation:
If the sound source is emitting the signal evenly in all directions (as from a point source) this means that at any time, the source power is distributed over the surface of a sphere of radius r.
At a distance r of the source, the intensity of the sound is defined as the power per unit area:
I = P/A
As the area is the area of a sphere, we can say the following:
I = P / 4*π*r²
Replacing I and P by the values given, we can solve for r (which is the distance from the listener to the source) as follows:
r² = P / I*4*π ⇒ r = [tex]\sqrt{P/(4*\pi*I)}[/tex] = 0.78 m
The distance between you and the source is 0.78 m.
The given parameters:
Intensity of the sound, I = 9.8 W/m²Output power of the signal, P = 75 WThe area of the source is calculated as follows;
[tex]A = \frac{P}{I} \\\\ A = \frac{75}{9.8} \\\\ A = 7.65 \ m^2[/tex]
The distance between you and the source is calculated as follows;
[tex]A = 4\pi r^2\\\\ r^2 = \frac{A}{4\pi} \\\\ r = \sqrt{\frac{A}{4\pi}} \\\\ r = \sqrt{\frac{7.65}{4\pi}} \\\\ r = 0.78 \ m[/tex]
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When energy is converted from one form to another in a chemical or physical change, which of the following also changes by a measureable amount? Select the correct answer below:
a.The total mass in the system
b.The force of gravity
c.The total energy
d.None of the above
Answer:
None of the above
Explanation:
When energy is converted from one form to another in a chemical or physical change, none will change. This is due to the law of conservation of energy. It states that the total energy of the system remains constant. It only changes energy from one form of energy to another. So, the correct option is (d) "none of the above".
Answer: None of the above
Explanation:
The total mass of a system does not change during normal (non-nuclear) chemical reactions or during other processes where energy changes form. The force of gravity is constant and will not change when energy is converted. While energy can be converted between forms or exchanged between species, no additional energy can be created or removed.
A planet is discovered orbiting around a star in the galaxy Andromeda at the same distance from the star as Earth is from the Sun. If that star has four times the mass of our Sun, how does the orbital period of the planet compare to Earth's orbital period? A planet is discovered orbiting around a star in the galaxy Andromeda at the same distance from the star as Earth is from the Sun. If that star has four times the mass of our Sun, how does the orbital period of the planet compare to Earth's orbital period? The planet's orbital period will be four times Earth's orbital period. The planet's orbital period will be one-half Earth's orbital period. The planet's orbital period will be one-fourth Earth's orbital period. The planet's orbital period will be equal to Earth's orbital period. The planet's orbital period will be twice Earth's orbital period.
Answer:
The planet´s orbital period will be one-half Earth´s orbital period.
Explanation:
The planet in orbit, is subject to the attractive force from the sun, which is given by the Newton´s Universal Law of Gravitation.
At the same time, this force, is the same centripetal force, that keeps the planet in orbit (assuming to be circular), so we can put the following equation:
Fg = Fc ⇒ G*mp*ms / r² = mp*ω²*r
As we know to find out the orbital period, as it is the time needed to give a complete revolution around the sun, we can say this:
ω = 2*π / T (rad/sec), so replacing this in the expression above, we get:
Fg = Fc ⇒ G*mp*ms / r² = mp*(2*π/T)²*r
Solving for T²:
T² = (2*π)²*r³ / G*ms (1)
For the planet orbiting the sun in Andromeda, we have:
Ta² = (2*π)*r³ / G*4*ms (2)
As the radius of the orbit (distance to the sun) is the same for both planets, we can simplify it in the expression, so, if we divide both sides in (1) and (2), simplifying common terms, we finally get:
(Te / Ta)² = 4 ⇒ Te / Ta = 2 ⇒ Ta = Te/2
So, The planet's orbital period will be one-half Earth's orbital period.
The planet´s orbital period should be considered as the one-half Earth´s orbital period.
What is planet in orbit:
It is subjected to the attractive force from the sun that we called as the
Newton´s Universal Law of Gravitation.
Also, the following equation should be used
Fg = Fc ⇒ G*mp*ms / r² = mp*ω²*r
Now
ω = 2*π / T (rad/sec),
So,
Fg = Fc ⇒ G*mp*ms / r² = mp*(2*π/T)²*r
Now
T² = (2*π)²*r³ / G*ms (1)
And,
Ta² = (2*π)*r³ / G*4*ms (2)
So,
(Te / Ta)² = 4 ⇒ Te / Ta = 2 ⇒ Ta = Te/2
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Two 10-Hz, sine waves have a relative phase shift of 30 deg. What is the time difference between them? If the frequency of these sine waves doubles, but the time difference stays the same, what is the phase difference between them?
Answer:
The time difference is 8.33 ms.
The phase difference between them is 60°
Explanation:
Given that,
Frequency = 10 Hz
Angle = 30°
We need to calculate the time difference
Using formula of time difference
[tex]\Delta t=\dfrac{\phi}{360^{\circ}\times f}[/tex]
Put the value into the formula
[tex]\Delta t=\dfrac{30}{360\times10}[/tex]
[tex]\Delta t=8.33\ ms[/tex]
If the frequency of these sine waves doubles, but the time difference stays the same,
[tex]f=20\ Hz[/tex]
We need to calculate the phase difference between them
Using formula of phase difference
[tex]\Delta \phi=\Delta t\times360\times f[/tex]
Put the value in to the formula
[tex]\Delta=8.33\times10^{-3}\times360\times20[/tex]
[tex]\Delta \phi=60^{\circ}[/tex]
Hence, The time difference is 8.33 ms.
The phase difference between them is 60°