A) Angular speed: 0.42 rev/s
B) Frequency: 0.42 Hz
C) Tangential speed: 26.4 cm/s
D) Distance travelled: 528 cm
Explanation:
A)
In this problem, the ladybug is rotating together with the record.
The angular velocity of the ladybug, which is defined as the rate of change of the angular position of the ladybug, in this problem is
[tex]\omega = 25 rpm[/tex]
where here it is measured in revolutions per minute.
Keeping in mind that
1 minute = 60 seconds
We can rewrite the angular speed in revolutions per second:
[tex]\omega = 25 \frac{rev}{min} \cdot \frac{1}{60 s/min}=0.42 rev/s[/tex]
B)
The relationship between angular speed and frequency of revolution for a rotational motion is given by the equation
[tex]\omega = 2 \pi f[/tex] (1)
where
[tex]\omega[/tex] is the angular speed
f is the frequency of revolution
For the ladybug in this problem,
[tex]\omega=0.42 rev/s[/tex]
Keeping in mind that [tex]1 rev = 2\pi rad[/tex], the angular speed can be rewritten as
[tex]\omega = 0.42 \frac{rev}{s} \cdot 2\pi = 2\pi \cdot 0.42[/tex]
And re-arranginf eq.(1), we can find the frequency:
[tex]f=\frac{\omega}{2\pi}=\frac{(2\pi)0.42}{2\pi}=0.42 Hz[/tex]
And the frequency is the number of complete revolutions made per second.
C)
For an object in circular motion, the tangential speed is related to the angular speed by the equation
[tex]v=\omega r[/tex]
where
[tex]\omega[/tex] is the angular speed
v is the tangential speed
r is the distance of the object from the axis of rotation
For the ladybug here,
[tex]\omega = 2\pi \cdot 0.42 rad/s[/tex] is the angular speed
r = 10 cm = 0.10 m is the distance from the center of the record
So, its tangential speed is
[tex]v=(2\pi \cdot 0.42)(0.10)=0.264 m/s = 26.4 cm/s[/tex]
D)
The tangential speed of the ladybug in this motion is constant (because the angular speed is also constant), so we can find the distance travelled using the equation for uniform motion:
[tex]d=vt[/tex]
where
v is the tangential speed
t is the time elapsed
Here we have:
v = 26.4 cm/s (tangential speed)
t = 20 s
Therefoe, the distance covered by the ladybug is
[tex]d=(26.4)(20)=528 cm[/tex]
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An object of mass 2.5 kg has a momentum < 3, 6, 7 > kg m/s. At this instant the object is acted on a by a force < 50, 50, 100 > N for 5 x 10-3 s . What is the momentum of the object at the end of this time interval?
Answer:
The momentum of the object at the end is [tex](3.25i+6.25j+7.5k)\ kg-m/s[/tex]
Explanation:
Given that,
Mass of object = 2.5 kg
Momentum [tex]p= 3i+6j+7k[/tex]
Force [tex]F=50i+50j+100k[/tex]
Time [tex]t=5\times10^{-3}\ s[/tex]
We need to calculate the momentum of the object at the end
Using formula of impulse
[tex]J=\Delta p[/tex]
[tex]J=m\Delta v[/tex]...(I)
[tex]J=F\times\Delta t[/tex]....(II)
From equation (I) and (II)
[tex]F\times\Delta t=m\Delta v[/tex]
[tex]F\times \Delta t=m(v_{f}-v_{i})[/tex]
[tex]mv_{f}=F\times \Delta t+mv_{i}[/tex]
Put the value into the formula
[tex]p_{f}=(50i+50j+100k)\times5\times10^{-3}+3i+6j+7k[/tex]
[tex]p_{f}=0.25i+0.25j+0.5k+3i+6j+7k[/tex]
[tex]p_{f}=(3.25i+6.25j+7.5k)\ kg m/s[/tex]
Hence, The momentum of the object at the end is [tex](3.25i+6.25j+7.5k)\ kg-m/s[/tex]
To determine the object's momentum after the force is applied, calculate the change in momentum from the applied force over the time interval and add it to the initial momentum. The final momentum of the object is <3.25, 6.25, 7.5> kg·m/s.
Explanation:The student asked: What is the momentum of the object at the end of this time interval? To calculate the final momentum, we must use the formula p = p_0 + F ∙ Δt, where p_0 is the initial momentum, F is the force applied, and Δt is the time interval.
With the given values:
Initial momentum: p_0 = <3, 6, 7> kg·m/sForce: F = <50, 50, 100> NTime: Δt = 5 × 10^-3 sCalculating the change in momentum due to the force:
Change in momentum: Δp = F ∙ Δt = <50, 50, 100> N ∙ 5 × 10^-3 s = <0.25, 0.25, 0.5> kg·m/sAdding the initial momentum to the change in momentum gives the final momentum:
Final momentum: p = <3, 6, 7> + <0.25, 0.25, 0.5> = <3.25, 6.25, 7.5> kg·m/sTherefore, the object's momentum at the end of the time interval is <3.25, 6.25, 7.5> kg·m/s.
A capacitor consists of a set of two parallel plates of area A separated by a distance d. This capacitor is connected to a battery that maintains a constant potential difference V across the plates. If the separation between the plates is doubled, the electrical energy stored in the capacitor will be:_______
a- doubled
b- unchanged
c- quadrupled
d- quartered
e- halved
Answer:e
Explanation:
Given
Area of parallel plates is A
distance between plates is d
Potential difference between Plates is V
Capacitance is given by
[tex]C=\frac{\epsilon _0A}{d}[/tex]
If separation is doubled then capacitance become half
[tex]C'=\frac{\epsilon _0A}{2d}[/tex]
[tex]C'=\frac{C}{2}[/tex]
Electrical energy stored in the capacitor is given by
[tex]E=\frac{1}{2}CV^2[/tex]
When distance is doubled
[tex]E'=\frac{1}{2}\times \frac{C}{2}\times V^2[/tex]
[tex]E'=\frac{E}{2}[/tex]
Therefore Energy is halved
If the separation between the plates is doubled, the electrical energy stored in the capacitor will be: d. The electrical energy stored in the capacitor will be quartered.
To understand why the energy stored in the capacitor is quartered when the separation between the plates is doubled, let's consider the formula for the capacitance of a parallel plate capacitor and the energy stored in a capacitor.
The capacitance C of a parallel plate capacitor is given by:
[tex]\[ C = \frac{\varepsilon_0 A}{d} \][/tex]
The energy U stored in a capacitor is given by:
[tex]\[ U = \frac{1}{2} C V^2 \][/tex]
[tex]\[ C' = \frac{\varepsilon_0 A}{2d} = \frac{1}{2} \frac{\varepsilon_0 A}{d} = \frac{1}{2} C \][/tex]
[tex]\[ U' = \frac{1}{2} C' V^2 = \frac{1}{2} \left(\frac{1}{2} C\right) V^2 = \frac{1}{4} C V^2 = \frac{1}{4} U \][/tex]
So, when the separation between the plates is doubled, the capacitance is halved, and since the energy is directly proportional to the capacitance, the energy stored in the capacitor is quartered.
Based on the data thomson collected in his experiments using cathode rays, the concept of atomicc structure was modified. What were the four things validated by his cathode ray expirement?
Answer:
Based on the data thomson collected in his experiments using cathode rays, the concept of atomicc structure was modified. What were the four things validated by his cathode ray expirement?
Explanation:
Cathode Ray tubes' Experiment of J.J. Thomson, showed that atoms contain electrons or tiny negative charged subatomic particles.
Based on the Thomson's cathode ray tubes experiment, He validated four things which are as following
1. Cathode rays have mass.
2. Matter contains positive and negative charge.
3. Particles of cathode rays are fundamental to all matter.
4. An atom is divisible.
A localized electron has been polarized so that its spin is oriented in the positive z- direction. It is now subject to the application of a constant uniform magnetic field B = Bî along x over a period of time of duration T. After that, it is subject to the application of another magnetic field of the same magnitude B but along y: B2 = Bý, also with duration T. (a) What is the probability P that the spin-flip would occur as a result? That is, what is the probability that the spin of the electron would be found oriented in the negative z-direction after the application of the magnetic fields is over? (b) Is it possible to find such duration that the spin-flip would occur with certainty: P = 1? If yes, what would be that time r?
Answer: The answer is attached
Explanation:
Final answer:
In quantum mechanics, the probability of a spin-flip and the potential for achieving a spin flip with certainty are explored through the application of magnetic fields in different directions.
Explanation:
P(a): The probability that the spin-flip would occur can be calculated using quantum mechanics. When applying magnetic fields B along x and y, the probability of finding the spin of the electron oriented in the negative z-direction after is over is determined by the transition probabilities.
P(b): The time to ensure a spin flip with certainty (P = 1) can be calculated by adjusting the duration of the magnetic field application. By manipulating the duration, the spin can be controlled to achieve a certain outcome at a specific time.
Sabe-se que um alqueire paulista equivale a 24200 metros quadrados. Uma chácara retangular tem um alqueire e mede 100m de frente. Quanto ela mede de fundo?
Answer:
b = 242 m
Explanation:
A = 24200 m²
a = 100 m
b = ?
A seguinte fórmula é aplicada
A = a*b
⇒ b = A / a
⇒ b = (24200 m²) / (100 m)
⇒ b = 242 m
A space vehicle is traveling at 5320 km/h relative to Earth when the exhausted rocket motor is disengaged and sent backward. The relative speed between the motor and the command module is then 98 km/h. The mass of the motor is four times the mass of the module. What is the speed of the command module relative to Earth just after the separation ?
Answer:
5398.4km/h
Explanation:
IN THIS CASE THE MOMENTUM IS CONSERVED. THE VALUE OF MOMENTUM OF ONE COMBINED ROCKET WILL BE SAME AS OF TWO COMBINED.
Let mass of module be m
then
mass of motor = 4m (four times the mass of rocket module)
total mass = m + 4m = 5m
combined velocity = V = 5320kph
Let
absolute (relative to earth)motor velocity after disengagement = v
then
rocket module velocity (relative to earth) after disengagement = v+98 (relative velocity = 98)
momentum conservation equation
combined momentum = module momentum + motor momentum
(m+4m)V = m(v+98) + 4m*v
5mV = 98m+mv + 4mv
5V = 98+v + 4v (m cancels out)
5V - 98 = 5v
((5*5320)-98)/5 = v
v = 5300.4 km/h
velocity of rocket module relative to earth = v +98
= 5300.4 + 98
= 5398.4km/h
The height of a typical playground slide is about 6 ft and it rises at an angle of 30 ∘ above the horizontal.
a.)Some children like to slide down while sitting on a sheet of wax paper. This makes the friction force exerted by the slide very small. If a child starts from rest and we take the friction force to be zero, what is the speed of the child when he reaches the bottom of the slide?
b.)If the child doesn't use the wax paper, his speed at the bottom is half the value calculated in part A. What is the coefficient of kinetic friction between the child and the slide when wax paper isn't used? μk = ?
I found the answer to part a, it is 3m/s, but I need part b. Please show all steps, thank you!
Answer:
What is the coefficient of kinetic friction = 0.432
Explanation:
The detailed steps and derivation with appropriate substitution is as shown in the attached file.
This question involves the concepts of the law of conservation of energy and frictional energy.
a) The speed of the child when he reaches the bottom of the slide is "6 m/s".
b) The coefficient of kinetic friction between the child and slide when the wax paper isn't used is "0.432".
a)
According to the law of conservation of energy in this situation:
Loss in Potential Energy = Gain in Kinetic Energy
[tex]mgh = \frac{1}{2}mv^2\\\\2gh=v^2\\v=\sqrt{2gh}\\\\[/tex]
where,
v = velocity = ?
g = acceleration due to gravity = 9.81 m/s²
h = height lost = 6 ft = 1.83 m
Therefore,
[tex]v=\sqrt{2(9.81\ m/s^2)(1.83\ m)}[/tex]
v = 6 m/s
b)
Now, the velocity becomes half and the friction comes into action. So in this case the law of conservation of energy will be written as:
Loss of Potential Energy = Gain of Kinetic Energy + Frictional Energy
[tex]mgh=\frac{1}{2}m(\frac{v}{2})^2+\mu_kRl\\\\mgh=\frac{1}{2}m(\frac{v}{2})^2+\mu_kmgCos\theta l\\\\gh-\frac{1}{2}(\frac{v}{2})^2+\mu_kgCos\theta l[/tex]
where,
l = length of slide = [tex]\frac{h}{sin\theta}=\frac{1.83\ m}{sin30^o}=3.66\ m[/tex]
[tex]\mu_k[/tex] = coefficient of kinetic friction = ?
Therefore,
[tex](9.81\ m/s^2)(1.83\ m)-\frac{1}{2}(\frac{6\ m/s}{2})^2=\mu_k(9.81\ m/s^2)(Cos30^o)(3.66\ m)\\\\\mu_k=\frac{13.45\ m^2/s^2}{31.09\ m^2/s^2}\\\\\mu_k=0.432[/tex]
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The attached picture explains the law of conservation of energy.
According to the cartoon video on physics, electrons sent through 2 slits at once without measuring what slit is goes through, will produce what?
A) An interference pattern
B) A single line
C) 4 lines
D) A blank screen
Answer:
A) An interference pattern
Explanation:
the two slit experiment is key to understand the microscopic world. The wave-like properties of light were demonstrated by the famous experiment first performed by Thomas Young in the early nineteenth century. In original experiment, a point source of light illuminates two narrow adjacent slits in a screen, and the image of the light that passes through the slits is observed on a second screen.
Key Points
waves can interfere, for light this will make a series of light and dark bands matter particles, such as electrons, also produce interference patterns due to their wave-like nature so with a high flux of either photons or electrons, the characteristic interference pattern is visibleA light-year is _________. a. about 10 trillion kilometers the time it takes light to travel around the Sun b. about 300,000 kilometers per second the time it takes light to reach the nearest star
Answer:
about 10 trillion kilometers
Explanation:
c = Speed of light = [tex]3\times 10^8\ m/s[/tex]
t = Seconds in one year = [tex]365.25\times 24\times 60\times 60[/tex]
1 Light year
[tex]1\ ly=ct\\\Rightarrow 1\ ly=3\times 10^8\times 365.25\times 24\times 60\times 60=9.46728\times 10^{15}\ m\\ =9.46728\times 10^{15}\times 10^{-3}\\ =9.46728\times 10^{12}\ km\approx 10\ trillion\ km[/tex]
The answer is a. about 10 trillion kilometers
The water molecule has a dipole with the negative portion
A) localized on one of the hydrogens
B) localized between the hydrogen atoms
C) pointing toward the oxygen atom
D) pointing from the oxygen through the hydrogen atoms
In a water molecule, the dipole's negative portion is located on the oxygen atom and points towards the hydrogen atoms due to the difference in electronegativity between these elements.
Explanation:The water molecule, H2O, is a polar molecule, meaning it has a net dipole due to the presence of polar bonds, which result from a significant difference in electronegativities of the atoms involved. In a water molecule, the oxygen atom is more electronegative than hydrogen and therefore pulls the shared electrons toward itself. This creates a charge separation where the oxygen side of the molecule becomes partially negative, and the hydrogen side becomes partially positive. Looking at the provided options, the most accurate statement would be that the negative portion of the dipole is D) pointing from the oxygen through the hydrogen atoms
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A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof a time of 1.19 s s later. You may ignore air resistance. Part A Part complete If the height of the building is 21.0 m m , what must the initial speed be of the first ball if both are to hit the ground at the same time? v v = 9.53 m/s m/s SubmitPrevious Answers Correct Part B Part complete Consider the same situation, but now let the initial speed v 0 v0 of the first ball be given and treat the height h h of the building as an unknown. What must the height of the building be for both balls to reach the ground at the same time for v 0 v0v_0 = 8.70 m/s m/s . h h = 6.51 m m SubmitPrevious Answers CorrectPart C Part complete If v 0 v0 is greater than some value v max vmax , a value of h h does not exist that allows both balls to hit the ground at the same time. Solve for v max vmax . v max vmax = 11.7 m/s m/s SubmitPrevious Answers Correct Part D Part complete If v 0 v0 is less than some value v min vmin , a value of h h does not exist that allows both balls to hit the ground at the same time. Solve for v min vmin . v min vmin = 5.83 m/s m/s SubmitPrevious Answers Correct Provide Feedback Next
A) Initial velocity of ball 1: 9.53 m/s upward
B) Height of the building: 6.48 m
C) Maximum velocity: 11.7 m/s
D) Minimum velocity: 5.83 m/s
Explanation:
A)
The y-position of the 1st ball at time t is given by the equation for free fall motion:
[tex]y_1 = h + v_0 t - \frac{1}{2}gt^2[/tex] (1)
where
h = 21.0 m is the initial height of the ball, the height of the building
[tex]v_0[/tex] is the initial velocity of the ball, upward
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
The y-position of the 2nd ball instead, dropped from the roof 1.19 s later, is given by
[tex]y_2 = h-\frac{1}{2}g(t-1.19)^2[/tex]
where
h = 21.0 m is the initial height of the ball, the height of the building
t' = 1.19 s is the delay in time of the 2nd ball (we can verify that at t = 1.19 s, then [tex]y_2=h[/tex], so the ball is still on the roof
The 2nd ball reaches the ground when [tex]y_2=0[/tex], so:
[tex]0=h-\frac{1}{2}g(t-1.19)^2\\0=(21.0)-4.9(t^2-2.38t+1.42)\\4.9t^2-11.66t-14.04=0[/tex]
Which has two solutions:
t = -0.88 s (negative, we discard it)
t = 3.26 s (this is our solution)
The 1st ball reaches the ground at the same time, so we can substitute t = 3.26 s into eq.(1) and [tex]y_1=0[/tex], so we find the initial velocity:
[tex]0=h+v_0 t -\frac{1}{2}gt^2\\v_0 = \frac{1}{2}gt-\frac{h}{t}=\frac{1}{2}(9.8)(3.26)-\frac{21.0}{3.26}=9.53 m/s[/tex]
B)
In this case, the height of the building h is unknown, while the initial velocity of ball 1 is known:
[tex]v_0 = 8.70 m/s[/tex]
When the two balls reach the ground at the same time, there position is the same, so we can write:
[tex]y_1=y_2\\h+v_0 t - \frac{1}{2}gt^2 = h-\frac{1}{2}g(t-1.19)^2[/tex]
Solving the equation, we find:
[tex]v_0t=1.19gt-\frac{1}{2}g(1.19)^2\\t=\frac{0.5g(1.19)^2}{1.19g-v_0}=2.34 s[/tex]
This is the time at which both balls reache the ground; and substituting into the eq. of ball 2, we find the height of the building:
[tex]0=h-\frac{1}{2}g(t-1.19)^2\\h=0.5g(t-1.19)^2=0.5(9.8)(2.34-1.19)^2=6.48 m[/tex]
C)
If [tex]v_0[/tex] is greater than some value [tex]v_{max}[/tex], then there is no value of h such that the two balls hit the ground at the same time. This situation occurs when the demoninator of the formula found in part b:
[tex]t=\frac{0.5g(1.19)^2}{1.19g-v_0}[/tex]
becomes negative: in that case, the time becomes negative, so no solution is possible.
The denominator becomes negative when
[tex]1.19g-v_0 < 0[/tex]
Therefore when
[tex]v_0>1.19g=(1.19)(9.8)=11.7 m/s[/tex]
So, if the initial velocity of ball 1 is greater than 11.7 m/s, the two balls cannot reach the ground at the same time.
D)
There is also another condition that must be true in order for the two balls to reach the ground at the same time: the time at which ball 1 reaches the ground must be larger than 1.19 s (because ball 2 starts its motion after 1.19 s). This means that the following condition must be true
[tex]t=\frac{0.5g(1.19)^2}{1.19g-v_0}>1.19[/tex]
Solving the equation for [tex]v_0[/tex], we find:
[tex]0.5g(1.19)^2>1.19(1.19g-v_0)\\6.94>13.88-1.19v_0\\1.19v_0>6.94[/tex]
Which gives
[tex]v_0>5.83 m/s[/tex]
Therefore, the minimum speed of ball 1 at the beginning must be 5.83 m/s.
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Water behind a dam has a certain amount of stored energy that can be released as the water falls over the top of the dam. It may be enough energy to turn a mill wheel or an electricity-generating turbine. Choose the term that best describes the type of energy stored in the water at the top of the dam.
Answer:
Gravitational potential energy
Explanation:
Gravitational potential energy is the type of energy an object has due to its position in a gravitational field. Water behind a dam possesses gravitational potential energy due to it being at a higher level than the water on the other side of the dam. When the water falls the gravitational potential energy is converted to kinetic energy, leading to the turning of the turbines to generate electricity.
When you jump from an elevated position you usually bend your knees upon reaching the ground. By doing this, you make the time of the impact about 10 times as great as for a stiff-legged landing. In this way the average force your body experiences is ________.a. less than 1/10 as great. b. more than 1/10 as great. c. about 1/10 as great.d. about 10 times as great.
Answer:
c. about 1/10 as great.
Explanation:
While jumping form a certain height when we bend our knees upon reaching the ground such that the time taken to come to complete rest is increased by 10 times then the impact force gets reduced to one-tenth of the initial value when we would not do so.
This is in accordance with the Newton's second law of motion which states that the rate of change in velocity is directly proportional to the force applied on the body.
Mathematically:
[tex]F\propto\frac{d}{dt} (p)[/tex]
[tex]\Rightarrow F=\frac{d}{dt} (m.v)[/tex]
since mass is constant
[tex]F=m\frac{d}{dt}v[/tex]
when [tex]dt=10t[/tex]
then,
[tex]F'=m.\frac{v}{10\times t}[/tex]
[tex]F'=\frac{1}{10} \times \frac{m.v}{t}[/tex]
[tex]F'=\frac{F}{10}[/tex] the body will experience the tenth part of the maximum force.
where:
[tex]\frac{d}{dt} =[/tex] represents the rate of change in dependent quantity with respect to time
[tex]p=[/tex] momentum
[tex]m=[/tex] mass of the person jumping
[tex]v=[/tex] velocity of the body while hitting the ground.
In one contest at the county fair, a spring-loaded plunger launches a ball at a speed of 3.2m/s from one corner of a smooth, flat board that is tilted up at a 20 degree angle. To win, you must make the ball hit a small target at the adjacent corner, 2.40m away. At what angle theta should you tilt the ball launcher?
Answer:
Explanation:
Given
Speed of ball [tex]u=3.2\ m/s[/tex]
Plane is inclined at an angle [tex]20^{\circ}[/tex]
To win the Game we need to hit the target at [tex]x=2.4\ m[/tex] away
Launch angle of ball [tex]\theta [/tex]
Motion of ball can be considered in two planes i.e. Vertical to the plane and horizontal to the plane
So Net acceleration in vertical plane is [tex]g\sin 20[/tex]
Range of Projectile is given by
[tex]R=\frac{u^2\sin 2\theta }{g}[/tex]
for [tex]R=2.4\ m[/tex]
[tex]2.4=\frac{3.2^2\times sin 2\theta }{g\sin 20}[/tex]
[tex]\sin 2\theta =\frac{2.4\times 9.8\times \sin 20}{3.2^2}[/tex]
[tex]\sin 2\theta =0.7855[/tex]
[tex]2\theta =51.77[/tex]
[tex]\theta =25.88^{\circ}[/tex]
so ball must be launched at an angle of [tex]25.88^{\circ}[/tex]
To answer this, we use projectile motion principles with an equation specifically structured for the problem. Inserting the given values into the equation derived from the horizontal and vertical equations of motion, we can find the required launching angle to hit the target.
Explanation:To solve this problem, we can apply the concepts found in projectile motion physics. Given the initial speed or velocity (3.2 m/s) and the horizontal distance of the target (2.4 m) we can find the necessary angle (theta) to hit the target. The angle θ can be given by the equation of motion for a projectile which can be derived from the horizontal and vertical equations of motion, θ = atan[(vf² ± sqrt(vf⁴ - g*(g*x² + 2*y*vf²)) / (g*x)] where g = 9.8 m/s² is the acceleration due to gravity, x = 2.4 m is the horizontal distance, y = 0 (height difference), and vf = 3.2 m/s is the final velocity, the speed at which the ball is launched.
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A pitcher throws a 0.15 kg baseball so that it crosses home plate horizontally with a speed of 20 m/s. It is hit straight back at the pitcher with a final speed of 25 m/s. Assume the direction of the initial motion of the baseball to be positive.(a) What is the impulse delivered to the ball? (b) Find the average force exerted by the bat on the ball if the two are in contact for 2.0?
Answer:
a.-6.75 kgm/s
b.[tex]3375 N[/tex]
Explanation:
We are given that
Mass of baseball=0.15 kg
Initial speed=[tex]u=20m/s[/tex]
Final speed=[tex]v=-25m/s[/tex]
a.We know that
Impulse=Change in momentum=[tex]\Delta p=mv-mu=m(v-u)[/tex]
Momentum=[tex]mass\times velocity[/tex]
Using the formula
Impulse=[tex]0.15(-25-20)=-6.75 kgm/s[/tex]
b.Time=[tex]2\times 10^{-3} s[/tex]
Force=[tex]\frac{Impulse}{time}[/tex]
Using the formula
Average force exerted by the bat on the ball=[tex]\frac{-6.75}{2\times 10^{-3}}[/tex] N
Average force exerted by the bat on the ball=[tex]3375N[/tex]
A particle moves along the x-axis according to x(t)=10t−2t²m. (a) What is the instantaneous velocity at t = 2 s and t = 3 s? (b) What is the instantaneous speed at these times? (c) What is the average velocity between t = 2 s and t = 3 s?
Answer:
a) v(2) = 2m/s, v(3) = -2m/s
b) speed at t = 2s is 2m/s
speed at t = 3s is 2m/s
c) 0 m/s
Explanation:
We can take the derivative of x(t) to find the equation of velocity
v(t) = x'(t) = 10 - 4t
(a) v(2) = 10 - 4*2 = 10 - 8 = 2 m/s
v(3) = 10 - 4*3 = 10 - 12 = -2 m/s
(b) The speed would be the same as velocity without the direction
speed at t = 2s is 2m/s
speed at t = 3s is 2m/s
(c) The average velocity between t = 2s and t = 3s is distance it travels over period of time
[tex]v_a = \frac{s(3) - s(2)}{\Delta t} = \frac{10*3 - 2*3^2 - (10*2 - 2*2^2)}{3 - 2}[/tex]
[tex]v_a = \frac{12 - 12}{1} = 0/1 = 0 m/s[/tex]
Final answer:
The instantaneous velocity at t = 2 s is 2 m/s and at t = 3 s is -2 m/s. The instantaneous speed at both times is 2 m/s. The average velocity between t = 2 s and t = 3 s is 12 m/s.
Explanation:
(a) To find the instantaneous velocity, we need to find the derivative of the position function x(t) with respect to time. The derivative of x(t) = 10t - 2t² is v(t) = 10 - 4t. Substituting t = 2 and t = 3 into v(t), we get v(2) = 10 - 4(2) = 2 m/s and v(3) = 10 - 4(3) = -2 m/s.
(b) The instantaneous speed is the magnitude of the instantaneous velocity. Since speed is always positive, the speed at t = 2 s and t = 3 s is 2 m/s for both.
(c) The average velocity between t = 2 s and t = 3 s is given by the change in position divided by the change in time. The change in position is x(3) - x(2) = (10(3) - 2(3)²) - (10(2) - 2(2)²) = 12 m, and the change in time is 3 s - 2 s = 1 s. Therefore, the average velocity is 12 m/1 s = 12 m/s.
The volume of gas in a container is 125,000 liters, and the pressure is 1.2 atmospheres. Suppose the temperature remains constant, and the pressure changes to 1.6 atmospheres. What is the new volume of the gas in liters?
Answer:
the new volume =93,750 Liters
Explanation:
From Boyles law
P1V1 = P2V2
1.2 X 125000 = 1.6 X V2
V2 = 93,750 Liters
A friend of yours who takes her astronomy class very seriously challenges you to a contest to find the thinnest crescent moon you can find just after new moon? What time of day is best for looking for this very thin crescent?
Answer:
after the sun sets or just as it is setting
Explanation:
a crescent moon is thin and reflects less sunlight during the daylight sky so it becomes difficult to spot, but can be spotted when the sun is setting or just sets.
The best time to find the thinnest crescent moon just after the new moon is in the early evening, just after sunset, when the moon starts to reflect sunlight towards the earth, showing the crescent shape. The appearance of the Moon's surface can vary significantly with its phase and this can be better viewed through binoculars.
To find the thinnest crescent moon just after new moon, the best time to look is usually in the early evening just after sunset. The Moon, moving eastward each day in its 30 days cycle around the Earth, moves roughly 12° in the sky each day. A day or two after the new phase, the thin crescent first appears, as we begin to see a small part of the Moon's illuminated hemisphere reflecting a little sunlight toward us.
Because the Moon is moving eastward away from the Sun, it rises later and later each day. Therefore, after a new moon, the thin crescent will be seen in the west just after sunset.
Keep in mind that the bright crescent increases in size on successive days as the Moon moves farther and farther around the sky away from the direction of the Sun.
Bear in mind that the brighter the Moon is in the night sky, the harder it is to see the faint flashes of meteors. Furthermore, as seen through a good pair of binoculars, the appearance of the Moon's surface changes dramatically with its phase, revealing more topographic details when sunlight streams in from the side, causing topographic features to cast sharp shadows.
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An object moves 15.0 m north and then 11.0 m south. Find both the distance traveled and the magnitude of the displacement vector.
Answer:
Distance = 26.0m Displacement = 4.0m
Explanation:
Distance specifies only how far an object has traveled while displacement is the distance traveled in a specified direction.
Total distance traveled by the object will be distance travelled through north + distance travelled through south i.e 15.0m + 11.0m = 26.0m
Displacement is gotten by using the Pythagoras theorem. Since the object traveled in the same vertical direction (15.0m through north which is upward i.e positive y direction and 11.0m through south i.e in the negative y direction), the displacement will be 15.0m - 11.0m = 4.0m
The distance traveled is 26.0 m and the magnitude of the displacement vector is 4.0 m
First, we will define the terms distance and displacement
Distance is the total movement of an object without any regard to direction.
Displacement is the difference between the original and final position of a path taken by an object.
Since, the object moves 15.0 m north and then 11.0 m south,
Then,
Distance traveled = 15.0 m + 11.0 m
Distance traveled = 26.0 m
For the magnitude of the displacement,
The object moves 15.0 m north and then 11.0 m south, which is in the opposite (negative) direction
Then,
Magnitude of displacement = 15.0 m - 11.0 m
Magnitude of displacement = 4.0 m
Hence, the distance traveled is 26.0 m and the magnitude of the displacement vector is 4.0 m
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In a population of Mendel's garden peas, the frequency of dominant yellow-flowered plants is 50%. The population is in Hardy-Weinberg equilibrium. What is the frequency of the homozygous recessive genotype in the population? 0.71 0.25 0.5 The frequency cannot be determined from the data provided.
The frequency of the homozygous recessive genotype in the population is approximately 0.71.
Explanation:The frequency of the homozygous recessive genotype in the population can be calculated using the Hardy-Weinberg equation. In this case, the frequency of the dominant yellow-flowered plants is given as 50% or 0.5. To calculate the frequency of the homozygous recessive genotype (qq), we need to find the value of q. The equation for the Hardy-Weinberg equilibrium is p² + 2pq + q² = 1.
Given that the frequency of the dominant allele (p) is 0.5, we can substitute the value of p into the equation and solve for q:
0.5² + 2(0.5)(q) + q² = 1
Simplifying the equation gives:
0.25 + q + q² = 1
Combining like terms:
q² + q - 0.75 = 0
Using the quadratic formula to solve for q, we find that q ≈ 0.71 or -1.21. Since the frequency of a trait cannot be negative, we can disregard the negative value. Therefore, the frequency of the homozygous recessive genotype in the population is approximately 0.71.
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A flying dragon is rising vertically at a constant speed of 6.0m/s. When the dragon is 30.0m above the ground, the rider on its back drops a small golden egg which, subsequently, is in free fall.
a) What is the maximum height above the ground reached by the egg?
b) How long after its release does the egg hit the ground?
c) What is the egg’s velocity immediately before it hits the ground?
d) Sketch, qualitatively, position, velocity, and acceleration of the egg as functions of time.
Final answer:
The detailed response covers the maximum height reached by the egg, the time it takes to hit the ground, and its velocity just before landing.a)60.0m,b)2.45 seconds,c)-6.0m/s.
Explanation:
a) Maximum Height: The maximum height above the ground reached by the egg can be calculated using the kinematic equation. It will be twice the initial height. In this case, it would be 60.0m.
b) Time to Hit Ground: You can determine the time it takes for the egg to hit the ground by using the kinematic equation for vertical motion. The time would be approximately 2.45 seconds after its release.
c) Velocity Before Landing: The velocity of the egg immediately before hitting the ground would be the same as the initial velocity when it was thrown, but in the opposite direction, which is -6.0m/s.
Calculate the speed of a proton having a kinetic energy of 0.995 × 10−19 J and a mass of 1.673 × 10−27 kg. Answer in units of m/s.
Answer:
1.19×10²² m/s
Explanation:
Kinetic Energy: This can be defined as the the energy of a body due to motion.
The formula for kinetic energy is given as,
Ek = 1/2mv²................... Equation 1
Where Ek = Kinetic energy, m = mass of proton. v = velocity of proton.
Making v the subject of the equation,
v = √(2Ek/m).................. Equation 2.
Given: Ek = 0.995×10⁻⁵ J, m = 1.673×10⁻²⁷ kg.
Substitute into equation 2
v = √(2×0.995×10⁻⁵/1.673×10⁻²⁷ )
v = 1.19×10²² m/s.
Therefore, the velocity of proton = 1.19×10²² m/s
If an astronaut goes on a space walk outside the Space Station, she will quickly float away from the station unless she has a tether holding her to the station. Part A Choose the correct explanation why does the statement make sense (or is clearly true) or does not make sense (or is clearly false). Choose the correct explanation why does the statement make sense (or is clearly true) or does not make sense (or is clearly false).
a. This statement is true. She and the Space Station have different orbits at the beginning and will move apart.
b. This statement is true. She and the Space Station cannot share the same orbit and will move apart quickly.
c. This statement is false. She and the Space Station have different orbits at the beginning but will stay together due to mutual gravity.
d. This statement is false. She and the Space Station share the same orbit and will stay together unless they are pushed apart.
Answer:
d. This statement is false. She and the Space Station share the same orbit and will stay together unless they are pushed apart.
Explanation:
In astronomy, orbit is simply a path of an object around another object in a space. That is, orbit is a path of a body that revolves around a gravitating center of mass. Examples of an orbit is are satellite around a planet, orbit around a center of galaxy, planet around the sun, and among others.
On the other hand, space station refers to a spacecraft that can support a group of human for long time in the orbit. Another names for space stations are orbital space station and orbital station.
Therefore, an astronaut goes on a space walk outside the Space Station shares the same orbit with the space station and they will stay together unless they are pushed apart.
The statement is true that an astronaut would float away during a spacewalk if not tethered. The astronaut and ISS, although influencing each other gravitationally to a small extent, they would continue along their separate paths under the influence of Earth's gravity and microgravity.
Explanation:
This statement is true. When an astronaut steps out for a spacewalk, she and the International Space Station (ISS) are both in orbit around Earth but the astronaut will not stay in place relative to the ISS without a tether holding her to the station.
The reason for this is due to microgravity and the properties of motion in space. The astronaut and the ISS are both in free-fall around the Earth, so in the absence of other forces, they would continue along their separate paths. Without a tether, even a small force (like the push off the astronaut does to get away from the airlock) can cause her to drift away from the station.
While the ISS and the astronaut do influence each other gravitationally, this effect is extremely small compared to the force of Earth's gravity. So, without a tether, the astronaut can float away from the ISS.
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write a statement that warns people about the presence of peroxide in hair dyes
Answer:
Hydrogen peroxides are known to be an active ingredient in hair dyes. Hydrogen peroxide is a damaging chemical although it has been added in diluted amounts in hair dyes. Some hair dye companies have already started looking for alternatives for hydrogen peroxide to use in their products, concerning the damaging effects of the chemical.
Hydrogen peroxide can cause hair loss, dermatitis and scalp burns.
Concerning the harmful chemicals in dyes, many people consider it is best to use henna if you really want a colour change for your hair.
___ twisted pair wire is used in environments that have a noticeable amount of electromagnetic interference.
Answer:
shielded
Explanation:
shielded twisted pair wire is used in environments that have a noticeable amount of electromagnetic interference.
Connor is a basketball player who makes, on average, 65% of her free throws. Asume each shot is independent. SHOW ALL WORK FOR FULL CREDIT. A) What is the probability Connor will miss three straight free throws before she makes one? (If you use your calculator to get your answer, make sure to show the key strokes in which you got your answer). Write your answer in Standard notation. B) During a season, Connor takes 150 free throws. What is the probability that he will make at least 100 out of 150 of these throws? C) What is the probability that he will make no more than 110 out of 150 free throws?
Answer:
DUNNO
Explanation:
BIT BAD FOR 65% WOULD GET -100%
Consider a spherical Gaussian surface and three charges: q1 = 1.60 μC , q2 = -2.61 μC , and q3 = 3.67 μC . Find the electric flux through the Gaussian surface if it completely encloses (a) only charges q1 and q2, (b) only charges q2 and q3, and (c) all three charges.
Answer:
Explanation:
Guass Law: Also known as "Gauss's flux theorem" is the total of the electric flux "φ" out of a closed surface is equal to the charge "Q" enclosed divided by the permittivity εο. Solution is attached.The electric flux through a Gaussian surface can be calculated using Gauss's law. (a) Calculate electric flux for q1 and q2, (b) Calculate electric flux for q2 and q3, (c) Calculate electric flux for all three charges.
Explanation:The electric flux through a Gaussian surface can be calculated using Gauss's law. Gauss's law states that the total electric flux through a closed surface is equal to the net charge enclosed by that surface divided by the permittivity of free space.
(a) To find the electric flux through the Gaussian surface enclosing only charges q1 and q2, we need to calculate the net charge enclosed by the surface, which is the sum of the two charges. Then, we divide this sum by the permittivity of free space to obtain the electric flux.
(b) Following the same procedure, we can find the electric flux through the Gaussian surface enclosing only charges q2 and q3.
(c) To find the electric flux through the Gaussian surface enclosing all three charges, we calculate the net charge enclosed by the surface, which is the sum of all three charges. Again, we divide this sum by the permittivity of free space to obtain the electric flux.
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Juanita lifts a round box and a square box to a shelf. The gravitational potential energy (GPE) for the round box increases by 50 J. The GPE for the square box increases by 100 J. On which box did Juanita do more work? Explain your reasoning.
Juanita did more work on the square box
Explanation:
According to the law of conservation of energy, the work done in lifting an object is equal to the increase in gravitational potential energy of the object. This is due to the fact that the work done on the object is converted into potential energy.
The potential energy of an object (GPE) is given by
[tex]GPE=mgh[/tex]
where
m is the mass of the object
g is the acceleration of gravity
h is the height of the object
In this problem, we have two objects:
- The roud box is lifted and its GPE increases by 50 J --> this means that the work done by Juanita on the box is 50 J
- Thr square box is lifted and its GPE increases by 100 J --> this means that the work done by Juanita on the box is 100 J
Therefore, Juanita did more work on the square box.
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The square box received more work by Juanita than the round box based on the increase in gravitational potential energy.
The square box received more work done by Juanita compared to the round box. When the gravitational potential energy increase for the square box is 100 J, which is greater than the 50 J increase in the round box, it indicates that Juanita did more work on the square box.
An old manuscript reveals that a landowner in the time of King Arthur held 3.00 acres of plowed land plus a livestock area of 25.0 perches by 4.00 perches. What was the total area in (a) the old unit of roods and (b) the more modern unit of square meters? Here, 1 acre is an area of 40 perches by 4 perches, 1 rood is an area of 40 perches by 1 perch, and 1 perch is the length 16.5 ft.
Answer:
(a) Total area is 14.5 roods
(b) Total area is 14674.522 square meters
Explanation:
Area occupied by land = 3 acres
1 acre = 40 perches by 4 perches = 160 square perches
3 acres = 3×160 = 480 square perches
Area occupied by livestock = 25 perches by 4 perches = 100 square perches
Total area = 480 + 100 = 580 square perches
1 rood = 4 perches by 1 perch = 4 square perches
580 square perches = 580/4 = 14.5 roods
(b) Total area = 580 square perches
1 perch = 16.5ft = 16.5/3.2808 = 5.03 meters
580 square perches × (5.03 meters/1 perch)^2 = 580 ×25.3009 square meters = 14674.522 square meters
The total area owned by the landowner is 580 perches, which is equivalent to 14.5 roods. When converted to the modern unit of measurement, this totals to approximately 17516 square meters.
Explanation:To solve this problem, we first need to convert each area to the same unit. In this case, we can convert all areas to perches. According to the old manuscript, the landowner had 3 acres of plowed land. Given that 1 acre is equivalent to 160 perches (40 perches by 4 perches), the plowed land was 480 perches (3 x 160).
The livestock area is 25 perches by 4 perches, which totals to 100 perches. Therefore, the total area owned by the landowner is 580 perches (480 perches + 100 perches).
(a) To convert this to roods, we divide by 40 (since 1 rood is 40 perches by 1 perch) which equals to 14.5 roods
(b) To convert perches to square meters, we need to know that 1 perch is 16.5 ft. and 1 square foot is approximately 0.0929 square meters. So, 1 perch is 325.125 sq ft (16.5 ft * 16.5 ft). Therefore, 1 perch is about 30.2 square meters (325.125 ft * 0.0929), and 580 perches equate to approximately 17516 square meters.
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A distant star is moving toward the earth at a speed of 1/4 the speed of light. Compared to the light from a flashlight on the earth, the speed of the light from the star would be__________
Answer: THE SAME
Explanation: visible light is an electromagnetic wave, which has the properties of both magnet and electrical. The speed of light in air has Generally been estimated to be
299,792 kilometers per second. For for a distant start moving towards the Earth at a speed one-fourth (1/4th) of 299,792kilometers per second compared to a flashing light from the Earth,the speed of the light from the star is expected to be the same.