Answer:
[tex]t_t=4.131\ s[/tex]
Explanation:
Given:
height above the horizontal form where the ball is hit, [tex]y=1\ m[/tex]
angle of projectile above the horizontal, [tex]\theta=30^{\circ}[/tex]
initial speed of the projectile, [tex]u=40\ m.s^{-1}[/tex]
Firstly we find the vertical component of the initial velocity:
[tex]u_y=u.\sin\theta[/tex]
[tex]u_y=40\times \sin30^{\circ}[/tex]
[tex]u_y=20\ m.s^{-1}[/tex]
During the course of ascend in height of the ball when it reaches the maximum height then its vertical component of the velocity becomes zero.
So final vertical velocity during the course of ascend:
[tex]v_y=0\ m.s^{-1}[/tex]Using eq. of motion:
[tex]v_y^2=u_y^2-2g.h[/tex] (-ve sign means that the direction of velocity is opposite to the direction of acceleration)
[tex]0^2=20^2-2\times 9.8\times h[/tex]
[tex]h=20.4082\ m[/tex] (from the height where it is thrown)
Now we find the time taken to ascend to this height:
[tex]v_y=u_y-g.t[/tex]
[tex]0=20-9.8t[/tex]
[tex]t=2.041\ s[/tex]
Time taken to descent the total height:
we've total height, [tex]h'=h+y[/tex] [tex]=20.4082+1[/tex][tex]h'=u_y'.t'+\frac{1}{2} g.t'^2[/tex]
during the course of descend its initial vertical velocity is zero because it is at the top height, so [tex]u_y'=0\ m.s^{-1}[/tex][tex]21.4082=0+4.9t'^2[/tex]
[tex]t'=2.09\ s[/tex]
Now the total time taken by the ball to hit the ground:
[tex]t_t=t'+t[/tex]
[tex]t_t=2.09+2.041[/tex]
[tex]t_t=4.131\ s[/tex]
Suppose that you are holding a pencil balanced on its point. If you release the pencil and it begins to fall, what will be the angular acceleration when it has an angle of 10.0 degrees from the vertical
The angular acceleration of the pencil is 17 rad/sec^2
Explanation:
When the object is balanced at its center of mass then it is said to be motionless because the net force acts through the center of mass. If we shift the location of the net force then which results in the rotation of the object about the center of mass. The angular acceleration possessed by the object is proportional to the torque generated by the net force about the center of mass.
τ [tex]= I \alpha[/tex]
[tex]I[/tex] is the moment of inertia of a body
α is the angular acceleration
Length of the pencil L = 15 c m = 0.15 m
Mass of the pencil = 10 g = 10 [tex]\times 10^{-3}[/tex] kg
Let α be the angular acceleration of pencil
θ be the inclination pencil from vertical
θ = 10 °
Assume the pencil as thin rod and moment of inertia of the thin rod with the axis of rotation at one end is
[tex]I = \frac{mL^{2} }{3}[/tex]
When the pencil is balanced then the net torque acting on the pencil is zero and when we release the pencil and begin to fall then net torque acts on the pencil due to the weight of the pencil.
τ [tex]= F \times d[/tex]
τ [tex]= mgsin\theta \times \frac{L}{2}[/tex]
τ [tex]= 10 \times 10^{-3} \times 9.81 \times sin10 \times \frac{0.15}{2}[/tex]
τ [tex]= 1.277 \times 10^{-3} Nm[/tex]
We know that relation between torque and angular acceleration
τ [tex]= I\alpha[/tex]
[tex]\alpha = \frac{\tau}{I}[/tex]
[tex]\alpha = \frac{\tau}{\frac{mL^{2} }{3} }[/tex]
[tex]\alpha = \frac{1.277 \times 10^{-3} \times 3 }{10 \times 10^{-3} \times 0.15^{2} }[/tex]
[tex]\alpha = 17.02[/tex]
[tex]\alpha \approx 17 rad/sec^{2}[/tex]
The angular acceleration of the pencil is 17 rad/sec^2
Two 20.0 g ice cubes at − 20.0 ∘ C are placed into 285 g of water at 25.0 ∘ C. Assuming no energy is transferred to or from the surroundings, calculate the final temperature, T f , of the water after all the ice melts.
Final answer:
To calculate the final temperature after placing ice cubes into water, principles of thermodynamics are used, involving calculations for warming the ice, melting it, and adjusting the water temperature. The steps highlight the application of heat transfer and phase change concepts within a closed system.
Explanation:
The student's question involves the calculation of the final temperature of the water after two 20.0 g ice cubes at -20.0 °C are placed into 285 g of water at 25.0 °C. The principles of thermodynamics, specifically the concept of heat transfer and the conservation of energy, are essential to solving this problem. However, with the given information, an exact numerical solution cannot be provided without knowing the specific heats of ice and water, as well as the heat of fusion of ice. Generally, the solution involves several steps:
Calculate the amount of energy required to warm the ice from -20.0 °C to 0 °C using the specific heat of ice.
Calculate the energy needed to melt the ice into water at 0 °C using the enthalpy of fusion.
Calculate the energy released by the water as it cools down to the new equilibrium temperature.
Set the energy gained by the ice equal to the energy lost by the water to find the final temperature of the mixture.
Without the numerical values, the step-by-step method demonstrates how principles of heat transfer and phase change are applied to predict changes in temperature and state within a closed system.
Mateo drew the field lines around the ends of two bar magnets but forgot to label the direction of the lines with arrows. At left a rectangular box with the long horizontal side labeled S. At right a rectangular box with the long horizontal side labeled N. There is a gap between the boxes. Curved lines labeled 1 emerge from the right end of S and reach the left end of N. In which direction should an arrow at position 1 point?
Answer:
left
Explanation:
The magnetic field lines always point outwards from the north pole and merges towards the south pole. Thus, the arrow, is from right to left.
What is bar magnet?A bar magnet is an object made of materials having permanent magnetism. They have two poles namely south poles and north poles. The like poles of magnets will repel and unlike pole attracts each other.
Conventionally, it is assumed that the magnet's field flows from its north pole inward to its south pole. Ferromagnetic materials can be used to create permanent magnets.
The magnetic field is strongest inside the magnetic substance, as seen by the magnetic field lines. Near the poles, there are the strongest external magnetic fields. A magnetic north pole will pull a magnet's south pole toward it while repelling another magnet's north pole. Hence, the arrow is pointing to left direction.
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A heavy steel ball is hung from a cord to make a pendulum. The ball is pulled to the side so that the cord makes a 4 ∘ angle with the vertical. Holding the ball in place takes a force of 20 N . If the ball is pulled farther to the side so that the cord makes a 8 ∘ angle, what force is required to hold the ball? Express your answer to two significant figures and include the
Answer:40.19 N
Explanation:
Given
Force needed to hold the ball at [tex]\theta =4^{\circ}\ is\ F_1=20\ N[/tex]
From FBD we can write
[tex]F\cos \theta =mg\sin \theta [/tex]
[tex]\tan \theta =\dfrac{F}{mg}[/tex]
For [tex]\theta =4^{\circ}[/tex]
[tex]F_1=20\ N[/tex]
[tex]\tan (4)=\dfrac{20}{mg}----1[/tex]
for [tex]\theta =8^{\circ}[/tex]
Force is [tex]F_2[/tex]
[tex]\tan (8)=\dfrac{F_2}{mg}---2[/tex]
Divide 1 and 2 we get
[tex]\dfrac{\tan (4)}{\tan (8)}=\dfrac{F_1}{F_2}[/tex]
[tex]F_2=20\times \dfrac{\tan 8}{\tan 4}[/tex]
[tex]F_2=20\times 2.009[/tex]
[tex]F_2=40.19\ N[/tex]
The specific heat of water is 4.184 J g—1 K—1. A piece of iron (Fe) weighing 40 g is heated to 80EC and dropped into 100 g of water at 25EC. (Specific heat of Fe is 0.446 J g—1 K—1) What is the temperature when thermal equilibrium has been reached
Explanation:
Below is an attachment containing the solution.
A 20.0 kg rock is sliding on a rough , horizontal surface at8.00 m/s and eventually stops due to friction .The coefficient ofkinetic fricction between the rock and the surface is 0.20
what average thermal power is produced as the rock stops?
Answer:
Therefore the average thermal power is = 19.61 W
Explanation:
Given that the mass of the rock = 20.0 kg
initial velocity (u) = 8.00m/s
Final velocity = 0
Kinetic fiction [tex](\mu_k)[/tex] = 0.20.
Friction force = Kinetic fiction× weight
=[tex]\mu_k mg[/tex]
Force = mass × acceleration
[tex]\Rightarrow acceleration =\frac{Force}{mass}[/tex]
[tex]=-\frac{\mu_kmg}{m}[/tex]
[tex]=-\mu_kg[/tex]
To find the time , we use the the following formula,
v=u+at
Here a [tex]=-\mu_kg[/tex]
[tex]\Rightarrow 0=8+(-\mu_kg)t[/tex]
[tex]\Rightarrow 0.20 \times 9.8\times t=8[/tex]
⇒t = 4.08 s
Now,
Thermal energy= work done by friction = change of kinetic energy
The change of kinetic energy
= [tex]\frac{1}{2} mu^2-\frac{1}{2} mv^2[/tex]
[tex]=\frac{1}{2}\times 20\times 8 -\frac{1}{2}\times 20\times 0[/tex]
=80 J
Thermal energy=80 J
Thermal power = Thermal energy per unit time
[tex]=\frac{80}{4.08}[/tex] w
=19.61 W
Therefore 19.61 W average thermal power is produced as the rock stops.
Suppose that in a lightning flash the potential difference between a cloud and the ground is 1.0*109 V and the quantity of charge of the charge transferred I s30 C? (a )What is the charge in energy of that transferred charge? (b) If all the energy released could be used to accelerate a 1000kg car from rest. What would be its final speed?
Answer:
a) [tex]U_{e} = 3 \times 10^{10}\,J[/tex], b) [tex]v \approx 7745.967\,\frac{m}{s}[/tex]
Explanation:
a) The potential energy is:
[tex]U_{e} = Q \cdot \Delta V[/tex]
[tex]U_{e} = (30\,C)\cdot (1.0\times 10^{9}\,V)[/tex]
[tex]U_{e} = 3 \times 10^{10}\,J[/tex]
b) Maximum final speed:
[tex]U_{e} = \frac{1}{2}\cdot m \cdot v^{2}\\v = \sqrt{\frac{2\cdot U_{e}}{m} }[/tex]
The final speed is:
[tex]v=\sqrt{\frac{2\cdot (3 \times 10^{10}\,J)}{1000\,kg} }[/tex]
[tex]v \approx 7745.967\,\frac{m}{s}[/tex]
The change in the energy transferred or potential energy is 3 x 10¹⁰ J.
The final speed of the charge released is 7745.97 m/s.
Change in energy transferredThe change in the energy transferred or potential energy is calculated as follows;
U = QV
U = 30 x 1 x 10⁹
U = 3 x 10¹⁰ J
Speed of the charge releasedThe speed of the charge released is calculated by applying law of conservation of energy.
[tex]K.E = U\\\\\frac{1}{2} mv^2 = U\\\\v= \sqrt{\frac{2U}{m} } \\\\v = \sqrt{\frac{2\times 3\times 10^{10} }{1000} }\\\\v = 7745.97 \ m/s[/tex]
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A physics student is driving home after class. The car is traveling at 14.7 m/s when it approaches an intersection. The student estimates that he is 20.0 m from the entrance to the intersection when the traffic light changes from green to yellow and the intersection is 10.0 m wide. The light will change from yellow to red in 3.00 s. The maximum safe deceleration of the car is 4.00 m/s2 while the maximum acceleration of the car is 2.00 m/s2. Should the physics student decelerate and stop or accelerate and travel through the intersection?
The student should accelerate and travel through the intersection.
Explanation:
As per the given problem, the velocity with which the car of the student is moving is 14.7 m/s and he is 20 m away from intersection. Also the width of intersection is 10 m. It is also stated that traffic lights will change from yellow to red in 3 s. So totally , the student has to cover a distance of 30 m in 3 s to avoid getting stopped in intersection.
The velocity or speed required by the car to cover 30 m in 3 s is 30/3 = 15 m/s.
And the car is moving with initial velocity of 14.7 m/s. The student has two options either to decelerate at 4m/s² rate or to accelerate at the rate of 2 m/s².
So the velocity or speed of the car in 3 s with acceleration of 2 m/s² will be
Velocity = Acceleration × Time = 2 × 3 =6 m/s.
Thus on accelerating the car by 2 m/s² in 3 s, it will have the speed of 6 m/s and it can cover a distance of 6 × 3 = 18 m.
And the time taken by the car to reach from 20 m to intersection point with speed of 14.7 m/s is 20/14.7 = 1.36 s.
So, the car will take 1.36 s to reach the entrance of intersection with the speed of 14.7 m/s. But if it gets accelerated to 2 m/s², then it will have an increase in speed which will make the car to cover the complete intersection without stopping. So the student should accelerate the car.
3) A person catches a ball with a mass of 145 g dropped from a height of 60.0 m above his glove. His hand stops the ball in 0.0100 s. What is the force exerted by his glove on the ball? Assume the ball slows down with constant acceleration.
Answer: 87KN
Explanation: F= m(v-u)/t
V= h/t
V= 60/0.0100
v= 6000m/s
M=145g/1000=0.145kg
F= 0.145*6000/0 .001
F= 87000N
F= 87KN
A 55.0 kg ice skater is moving at 4.07 m/s when she grabs the loose end of a rope, the opposite end of which is tied to a pole. She then moves in a circle of radius 0.808 m around the pole. (b) Compare this force with her weight by finding the ratio of the force to her weight.
Answer:
Explanation:
mass of ice skater, m = 55 kg
velocity, v = 4.07 m/s
radius, r = 0.808 m
The force is centripetal force, the formula for this force is
[tex]F_{c}=\frac{mv^{2}}{r}[/tex]
[tex]F_{c}=\frac{55\times 4.07\times 4.07}{0.808}[/tex]
Fc = 1127.56 N
Weight of the person, W = mg
W = 55 x 9.8 = 539 N
The ratio of force to he weight
[tex]\frac{F_{c}}{W}=\frac{1127.56}{539}[/tex]= 2.09
A 10-kg disk-shaped flywheel of radius 9.0 cm rotates with a rotational speed of 320 rad/s. Part A Determine the rotational momentum of the flywheel. Express your answer to two significant figures and include the appropriate units. Part B With what magnitude rotational speed must a 10-kg solid sphere of 9.0 cm radius rotate to have the same rotational momentum as the flywheel? Express your answer to two significant figures and include the appropriate units.
Answer:
(A). The rotational momentum of the flywheel is 12.96 kg m²/s.
(B). The rotational speed of sphere is 400 rad/s.
Explanation:
Given that,
Mass of disk = 10 kg
Radius = 9.0 cm
Rotational speed = 320 m/s
(A). We need to calculate the rotational momentum of the flywheel.
Using formula of momentum
[tex]L=I\omega[/tex]
[tex]L=\dfrac{1}{2}mr^2\omega[/tex]
Put the value into the formula
[tex]L=\dfrac{1}{2}\times10\times(9.0\times10^{-2})^2\times320[/tex]
[tex]L=12.96\ kg m^2/s[/tex]
(B). Rotation momentum of sphere is same rotational momentum of the flywheel
We need to calculate the magnitude of the rotational speed of sphere
Using formula of rotational momentum
[tex]L_{sphere}=L_{flywheel}[/tex]
[tex]I\omega_{sphere}=I\omega_{flywheel}[/tex]
[tex]\omega_{sphere}=\dfrac{I\omega_{flywheel}}{I_{sphere}}[/tex]
[tex]\omega_{sphere}=\dfrac{I\omega_{flywheel}}{\dfrac{2}{5}mr^2}[/tex]
Put the value into the formula
[tex]\omega_{sphere}=\dfrac{12.96}{\dfrac{2}{5}\times10\times(9.0\times10^{-2})^2}[/tex]
[tex]\omega_{sphere}=400\ rad/s[/tex]
Hence, (A). The rotational momentum of the flywheel is 12.96 kg m²/s.
(B). The rotational speed of sphere is 400 rad/s.
To find the rotational momentum of a flywheel, one must calculate the moment of inertia and then use it to ascertain the angular momentum by multiplying the moment of inertia with the angular speed. L = Iw. To compare this with another rotating object, for example a sphere, one would then utilize their common momentum and calculate the necessary angular speed the sphere would need in order to match the flywheel's momentum.
Explanation:Part A: To determine the rotational momentum of the flywheel, we first need to find its moment of inertia. For a disk, the moment of inertia (I) is given by I = 1/2 MR² where M is the mass and R is the radius. Substituting values, I becomes = 1/2 * 10 Kg * (0.09m)² = 0.0405 kg • m². The rotational momentum can then be found by multiplying moment of inertia by the rotational speed w. Therefore the rotational momentum (L) of the flywheel is L = Iw = 0.0405 kg • m² * 320 rad/s = 12.96 Kg • m²/s.
Part B: Let's now find the speed a spherical object needs to achieve the same momentum. The moment of inertia of a solid sphere is given by I=2/5 MR². The angular momentum L (which should be the same as the flywheel's) equals Iw, so we solve for w giving w = L/I = 12.96 kg•m²/s / (2/5*10 kg (0.09 m)²) = 180 rad/s. Thus, a 10-kg solid sphere of 9.0 cm radius would have to rotate at a speed of 180 rad/s to have the same rotational momentum as the flywheel.
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Two 4.0-Ω resistors are connected in parallel, and this combination is connected in series with 3.0 Ω. What is the effective resistance of this combination?
Answer
given,
R₁= 4 Ω
R₂ = 3 Ω
When two resistors are connected in series
R = R₁ + R₂
R = 4 + 3
R = 7 Ω
When two resistors are connected in series then their effective resistance is equal to 7 Ω .
When two resistors are connected in parallel.
[tex]\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}[/tex]
[tex]\dfrac{1}{R}=\dfrac{1}{3}+\dfrac{1}{4}[/tex]
[tex]\dfrac{1}{R}=\dfrac{7}{12}[/tex]
[tex]R = 1.714 \Omega[/tex]
Hence, the equivanet resistance in parallel is equal to [tex]R = 1.714 \Omega[/tex]
Answer:
5.0 Ω
Explanation:
[tex]R_{1}[/tex] = 4.0 Ω, [tex]R_{2}[/tex] = 4.0 Ω, [tex]R_{3}[/tex] = 3.0 Ω
[tex]\frac{1}{R_{parallel}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} \\\frac{1}{R_{parallel}} = \frac{1}{4.0} + \frac{1}{4.0}\\\frac{1}{R_{parallel}} = \frac{2}{4.0}\\\frac{1}{R_{parallel}} = 0.5\\R_{parallel} = (0.5)^{-1} \\R_{parallel} = 2.0[/tex]
[tex]R_{T} = R_{parallel} + R_{3} \\R_{T} = 2.0 + 3.0\\R_{T} = 5.0[/tex]
Therefore, the effective resistance of this combination is 5.0 Ω.
A box is moved 10 m across a smooth floor by a force making a downward angle with the floor, so that there is effectively a 10 N force acting parallel to the floor in the direction of motion and a 5 N force acting perpendicular to the floor. The work done is
Answer:
Work done = 100 J
Explanation:
Given:
Distance moved by the box (d) = 10 m
Parallel force acting on it [tex](F_1)[/tex] = 10 N
Perpendicular force acting on it [tex](F_2)[/tex] = 5 N
Work done is the product of force and displacement along the line of action of force.
So, the perpendicular force doesn't do any work and it's only the parallel force that does work as the displacement caused is along the parallel force direction.
So, work done is given as:
[tex]Work=F_1\times d\\\\Work=10\ N \times 10\ m \\\\W=100\ J[/tex]
Therefore, work done is 100 J.
Three identical balls are thrown from the top of a building, all with the same initial speed but at different angles. Neglecting air resistance, rank the speeds of the balls at the instant each hits the ground.
Answer:
The speed of each ball would be same at the instant each hits the ground.
Explanation:
From the principle of conservation of energy we know that
PE₁ + KE₁ = PE₂ + PE₂
where PE = mgh
The initial and final potential energies would be same for each ball since all the balls are thrown from the same height.
Now what about KE ?
We know that each ball is thrown with the same initial speed so their initial kinetic energies would be same and in turn their final kinetic energies must be same in order to satisfy th e conservation of energy principle. Therefore, we can conclude that the speed of each ball would be same at the instant they hit the ground. The initial angle of ball doesn't have any impact on the speed of the ball.
If an object in space is giving off a frequency of 10^13 wavelength of 10^-6 what will scientist be looking for?
Answer:
The scientist will be looking for the velocity of the wave in air which is equivalent to 10^7m/s
Explanation:
If an object in space is giving off a frequency of 10^13Hz and wavelength of 10^-6m then the scientist will be looking for the velocity of the object in air.
The relationship between the frequency (f) of a wave, the wavelength (¶) and the velocity of the wave in air(v) is expressed as;
v = f¶
Given f = 10^13Hz and ¶ = 10^-6m,
v = 10¹³ × 10^-6
v = 10^7 m/s
The value of the velocity of the object in space that the scientist will be looking for is 10^7m/s
Low-frequency vertical oscillations are one possible cause of motion sickness, with 0.30 Hz having the strongest effect. Your boat is bobbing in place at just the right frequency to cause you the maximum discomfort. The water wave that is bobbing the boat has crests that are 30 m apart. What will be the boat’s vertical oscillation frequency if you drive the boat at 5.0 m/s in the direction of the oncoming waves?.
Answer:
0.467 Hz
Explanation:
Wave properties are related thus
v = fλ
v = velocity of the wave = ?
f = 0.30 Hz
λ = wavelength = 30 m
v = 0.3×30 = 9.0 m/s
But if the boat is now moving at 5.0 m/s in the direction of the oncoming wave,
The speed of the wave relative to the boat = 9 - (-5) = 14.0 m/s
f = (v/λ) = (14/30) = 0.467 Hz
Hope this helps!
Answer:
The answer to the question is;
The boat’s vertical oscillation frequency if you drive the boat at 5.0 m/s in the direction of the oncoming waves moving at 9 m/s in the opposite direction is 0.467 Hz.
Explanation:
We note the frequency of the water wave = 0.3 Hz
Distance between crests or wavelength of water wave = 30 m
Speed v of a wave is given by Frequency f × Wavelength λ
Therefore the speed of the wave = f·λ = 0.3 × 30 = 9 m/s
(b) Distance between crests = 30 m = wavelength λ
Boat speed = 5.0 m/s v
Speed of the wave with respect to the boat = 5 + 9 = 14 m/s
From the wave speed relationship, we have
v = fλ f = [tex]\frac{v}{\lambda}[/tex] =[tex]\frac{14}{30} = \frac{7}{15}[/tex] = 0.467 Hz which is high.
1) Steven carefully places a 1.85 kg wooden block on a frictionless ramp, so that the block begins to slide down the ramp from rest. The ramp makes an angle of 59.3° up from the horizontal. Which forces below do non-zero work on the block as it slides down the ramp?a) gravityb) normalc) frictiond) spring2) How much total work has been done on the block after it slides down along the ramp a distance of 1.85 m? 3) Nancy measures the speed of the wooden block after it has gone the 1.85 m down the ramp. Predict what speed she should measure.4) Now, Steven again places the wooden block back at the top of the ramp, but this time he jokingly gives the block a big push before it slides down the ramp. If the block's initial speed is 2.00 m / s and the block again slides down the ramp 1.91 m , what should Nancy measure for the speed of the block this time?
The forces doing non-zero work as the block slides down a frictionless ramp are gravity and any applied forces. The work done can be calculated using the mass, gravitational acceleration, and height. For the pushed block, the initial kinetic energy is combined with gravitational work to predict the final speed.
The forces that do non-zero work on a block as it slides down a frictionless ramp include gravity and any applied forces (like a push or a pull), but not the normal force or friction since the ramp is frictionless. Gravity does positive work as it pulls the block down the plane, increasing its kinetic energy. To calculate the total work done on the 1.85 kg wooden block after sliding down a distance of 1.85 m on the ramp at a 59.3° angle, you can use the formula:
Work = mgh, where m is mass, g is acceleration due to gravity, and h is the height.To predict the speed of the block after sliding down, you can use the conservation of energy principle or kinematic equations that relate distance, acceleration, and velocity. For the scenario where the block is given an initial push, the final speed can be predicted using the same principles by accounting for the initial kinetic energy provided by the push.
Here's an example of calculating work done:
Problem: A block of mass 10 kg slides down through a length of 10 m over an incline of 30°. If the coefficient of kinetic friction is 0.5, then find the work done by the net force on the block. In this example, work done by the net force can be calculated by finding the components of the gravitational force parallel and perpendicular to the incline and applying the kinetic friction force in the opposite direction of motion.Laboratory measurements show hydrogen produces a spectral line at a wavelength of 486.1 nanometers (nm). A particular star's spectrum shows the same hydrogen line at a wavelength of 486.0 nm. What can we conclude
Answer: we can conclude that the wavelength is decreasing. This means that the star is moving towards the observer on earth.
Explanation:
Since light has a constant speed of 3 x 10^8m/s, and this speed is a product of its wavelength and its frequency c = f¥
Where f is the frequency and ¥ is the wavelnght.
For a decreasing wavelength, it is seen that the frequency is increasing.
According to the doppler's effect, a moving body that is a source of a wave of frequency f, moving relatively towards an observer, the frequency will increase as they move closer to a frequency f' which is greater than f. This is known as the doppler shift of light wave.
A boy and a girl are riding a merry- go-round which is turning ata constant rate. The boy is near the outer edge, while a girl is closer to the center. Who has the greater tangential acceleration? 1. the boy 2. the girl 3. both have zero tangential 25% try penalty acceleration Hints: 0,0 4. both have the same non-zero tangential acceleration
Answer:
The correct answer is
1. the boy 2.
Explanation:
Since Tangential acceleration = v²/r then the
v = ωr and v² = ω²r² then the tangential acceleration = ω²r²/r = ω²r
This shows that since ω is the same for both of them, the boy with greater r has the most acceleration.
The tangential acceleration [tex]a_t =[/tex] α·r
A force of 4.9 N acts on a 14 kg body initially at rest. Compute the work done by the force in (a) the first, (b) the second, and (c) the third seconds and (d) the instantaneous power due to the force at the end of the third second.
Explanation:
The work done on the body = force applied x displacement
in this case , the acceleration of body a = [tex]\frac{Force}{Mass}[/tex] = [tex]\frac{4.9}{14}[/tex] = 0.35 ms⁻²
The displacement in first second S₁ = u + [tex]\frac{1}{2}[/tex] a x t²
here u = 0 , because body was at rest
Thus S₁ = [tex]\frac{1}{2}[/tex] x 0.35 x ( 1 )² = = 0.175 m
The work done = 4.9 x 0.175 = 0.86 J
The displacement in 2 seconds = [tex]\frac{1}{2}[/tex] x 0.35 x ( 2 )² = 0.7 m
Work done in 2 seconds = 4.9 x 0.7 = 3.43 J
Work done in second second = 3.43 - 0.86 = 2.57 J
The displacement in three seconds = [tex]\frac{1}{2}[/tex] x 0.35 x ( 3 )² = 1.575 m
Work done in three seconds = 4.9 x 1.575 = 7.7 J
Work done in third second = 7.7 - 3.43 = 4.3 J
The power at the end of third second is 4.3 watt
Because 4.3 J of work is done in the last second .
in a phenomenon known as ________, many incoming signals produce progressively larger graded potentials in a cell−larger than any single impulse would produce alone.
Answer:
neurotransmitters
Explanation:
Neurotransmitters are endogenous chemicals that enable neurotransmission. It is a type of chemical messenger which transmits signals across a chemical synapse, such as a neuromuscular junction, from one neuron (nerve cell) to another "target" neuron, muscle cell, or gland cell.
The number density of an ideal gas at stp is called the loschmidt number. True or False
Answer:
True
Explanation:
The number density of an ideal gas at stp is called the loschmidt number, being named after the Austrian physicist Johann Josef Loschmidt who first estimated this quantity in 1865
n = [tex]\frac{P}{K_{B}T }[/tex]
where [tex]K_{B}[/tex] is the Boltzman constant
T is the temperature
P is the pressure
Loschmidt Number is the number of molecules of gas present in one cubic centimetre of it at STP conditions.Loschmidt constant is also used to define the amagat, which is a practical unit of number density for gases and other substances:
1 amagat = n = 2.6867811×1025 m−3,
Answer:
True.
Explanation:
The Loschmidt number, n is defined as the number of particles (atoms or molecules) of an ideal gas in a given volume (the number density).
It is also the number of molecules in one cubic centimeter of an ideal gas at standard temperature and pressure which is equal to 2.687 × 10^19.
A large steel water storage tank with a diameter of 20 m is filled with water and is open to the atmosphere (1 atm = 101 kPa) at the top of the tank. If a small hole rusts through the side of the tank, 5.0 m below the surface of the water and 20.0 m above the ground, assuming wind resistance and friction between the water and steel are not significant factors, how far from the base of the tank will the water hit the ground?
Answer:
Explanation:
Their explanation is: We first need to determine the velocity of the water that comes out of the hole, using Bernoulli's equation.
P1+ρgy1 + ½ρv1² = P2+ρgy2+½ρv2²
The atmospheric pressure exerted on the surface of the water at the top of the tank and at the hole are essentially the same.
Then, P1-P2 = 0
. Additionally, since the opening at the top of the tank is so large compared to the hole on the side, the velocity of water at the top of the tank will be essentially zero. We can also set y1 as zero, simplifying the equation to:
0 = ρgy2 + ½ρv2²
Divide through by density
-gy2 = ½ v2²
y2= 5m and g =-9.81
-•-9.81×5 = ½v2²
9.81×5×2 = v2²
98.1 = v2²
v2 = √98.1
v2 = 9.9m/s
Using equation of motion to know the time of fall
We assume that the initial vertical velocity of the water is zero and the displacement of the water is -20 m
y-yo = ut + ½gt²
0-20 = 0•t -½ ×9.81 t²
-20 = -4.905t²
t² = -20÷-4.905
t² = 4.07747
t =√4.07747
t = 2.02secs
Using range formula
Then, R=Voxt
R= 9.9 × 2.02
R =19.99m
R ≈20m
Answer:
The distance from the base of the tank to the ground is 20 m
Explanation:
From Bernoulli equation where P is pressure, V is velocity, ρ is density, g is acceleration due to gravity and y is the height
P₁ + 1/2ρV₁² + ρgy₁ = P₂ + 1/2ρV₂² + ρgy₂
The pressure on the surface of the water on the top of the tank and that in the hole is the same (P₁ = P₂). Since the opening at the top of the tank is large compared to the hole at the side of the tank, the velocity at the top of the tank is 0 and y₁ is 0
Therefore: 0 = 1/2ρV₂² + ρgy₂
1/2ρV₂² = -ρgy₂ g = -10 m/s²
-10(5) = -1/2V₂²
V₂ = 10 m/s
The time it would take the water to fall on the ground is given as:
d = ut + 1/2gt²
u is the initial velocity = 0 , g = -10 m/s² and the displacement (d) = - 20 m
Therefore: -20 = 1/2(10)t²
t = 2 sec
The horizontal displacement (d) can be gotten from
d = V₂t
d = 10(2) = 20 m
The distance from the base of the tank to the ground is 20 m
Three resistors connected in series have potential differences across them labeled /\V1 , /\V2 , and /\V3. What expresses the potential difference taken over the three resistors together
Answer:
[tex]\Delta V=\Delta V_1+\Delta V_2+\Delta V_3[/tex]
Explanation:
We are given that three resistors R1, R2 and R3 are connected in series.
Let
Potential difference across [tex]R_1=\Delta V_1[/tex]
Potential difference across [tex]R_2=\Delta V_2[/tex]
Potential difference across [tex]R_3=\Delta V_3[/tex]
We know that in series combination
Potential difference ,[tex]V=V_1+V_2+V_3[/tex]
Using the formula
[tex]\Delta V=\Delta V_1+\Delta V_2+\Delta V_3[/tex]
Hence, this is required expression for potential difference.
List the number of sigma bonds and pi bonds in a double bond.
Yo sup??
a double bond has
1 Sigma bond
1 pi bond
and in total 2 bonds
Hope this helps
Final answer:
In a double bond, there is one sigma bond, which allows for free rotation, and one pi bond, which restricts rotation due to side-to-side overlap of orbitals.
Explanation:
In a double bond, there is always one sigma bond (sigma bond) and one pi bond (pi bond). The sigma bond is formed by the head-on overlap of orbitals, such as those that occur between s-s orbitals, s-p orbitals, or p-p orbitals. This type of bond allows for the free rotation of atoms around the bond axis. On the other hand, a pi bond arises from the side-to-side overlap of p-orbitals, creating electron density above and below the plane of the nuclei of the bonding atoms. This pi bond restricts the rotation of atoms around the bond axis due to the electron cloud above and below the bond plane.
If a car is taveling with a speed 6 and comes to a curve in a flat road with radius ???? 13.5 m, what is the minumum value the coefficient of friction must be so the car doesn’t slide of the road?
Answer:
[tex]\mu_s \geq 0.27[/tex]
Explanation:
The centripetal force acting on the car must be equal to mv²/R, where m is the mass of the car, v its speed and R the radius of the curve. Since the only force acting on the car that is in the direction of the center of the circle is the frictional force, we have by the Newton's Second Law:
[tex]f_s=\frac{mv^{2}}{R}[/tex]
But we know that:
[tex]f_s\leq \mu_s N[/tex]
And the normal force is given by the sum of the forces in the vertical direction:
[tex]N-mg=0 \implies N=mg[/tex]
Finally, we have:
[tex]f_s=\frac{mv^{2}}{R} \leq \mu_s mg\\\\\implies \mu_s\geq \frac{v^{2}}{gR} \\\\\mu_s\geq \frac{(6\frac{m}{s}) ^{2}}{(9.8\frac{m}{s^{2}})(13.5m) }\\\\\mu_s\geq0.27[/tex]
So, the minimum value for the coefficient of friction is 0.27.
When we breathe, we inhale oxygen and exhale carbon dioxide plus water vapor. Which likely has more mass, the air that we inhale or the same volume of air we exhale? Does breathing cause you to lose or gain weight?
Answer:
If the same volume of air is inhaled and exhaled, the air we breathe out normally weighs more than the air we breathe in.
Since the output from the body normally exceeds the input, breathing leads to weight loss.
Explanation:
If equal volumes of gas is inhaled and exhaled, the exhaled gas is heavier.
The inhaled gas contains Oxygen and majorly Nitrogen.
The exhaled gas contains CO₂, H₂O and a very large fraction of the unused inhaled air that goes into the lungs.
So, basically, the body exchanges O₂ with CO₂ and H₂O (and some other unwanted gases in the body) in a composition that CO₂, the heavier gas of the ones mentioned here, is prominent.
So, because the mass leaving the body is more than the mass entering, breathing leads to a loss of weight. This is one of the reasons why we need food for sustenance. Breathing alone will wear one out.
Final answer:
The air we exhale likely has less mass than the air we inhale due to the replacement of dense oxygen with less dense carbon dioxide and water vapor.
Explanation:
When we inhale, we take in air that is rich in oxygen, and when we exhale, we expel air with a higher concentration of carbon dioxide and water vapor. Given that a liter of oxygen weighs more than the same volume of water vapor, and we replace some of the oxygen with the less dense carbon dioxide, the air we exhale likely has less mass than the air we inhale.
The breathing process does result in weight loss, but only in small amount attributable to the exhaled carbon dioxide which comes from the metabolic processes in the body. As for the volume and density of our bodies, taking a deep breath increases the volume, but because the density of air is much smaller than that of the body, the overall density decreases.
During gas exchange, oxygen flows from the bloodstream into the body's cells, while carbon dioxide flows out of the cells into the bloodstream to be expelled. This process is driven by the differing partial pressures of oxygen and carbon dioxide in the blood and the cells, facilitating diffusion across the cellular membrane.
We will use a video to analyze the dependence of the magnitude of the Coulomb force between two electrically-charged spheres on the distance between the centers of the spheres. The electrical interaction is one of the fundamental forces of nature and acts between any pair of charged objects, therefore it is important to understand how precisely the separation distance affects the corresponding force between them. Specifically, we will:______. A. Study conceptually the nature of electric charge and force.B. Take measurements of the force exerted between two electrically-charged spheres as the distance between them is varied.C. Determine graphically the relationship between electric force and distance.
Answer: A. Study conceptually the nature of the electric charge and force
Explanation: Since the electrons are the mobile charge carriers, they will therefore be the particles that are transferred. For this case, since the electrophorus is positively- charged, the electrons from the sphere will be attracted to the electrophorus and thus will be transfer there.
A power plant produces 1000 MW to supply a city 40 km away. Current flows from the power plant on a single wire of resistance 0.050Ω/km, through the city, and returns via the ground, assumed to have negligible resistance. At the power plant the voltage between the wire and ground is 115 kV.What is the current in the wire?What fraction of the power is lost in transmission?
Answer:
Current = 8696 A
Fraction of power lost = [tex]\dfrac{80}{529}[/tex] = 0.151
Explanation:
Electric power is given by
[tex]P=IV[/tex]
where I is the current and V is the voltage.
[tex]I=\dfrac{P}{V}[/tex]
Using values from the question,
[tex]I=\dfrac{1000\times10^6 \text{ W}}{115\times10^3\text{ V}} = 8696 \text{ A}[/tex]
The power loss is given by
[tex]P_\text{loss} = I^2R[/tex]
where R is the resistance of the wire. From the question, the wire has a resistance of [tex]0.050\Omega[/tex] per km. Since resistance is proportional to length, the resistance of the wire is
[tex]R = 0.050\times40 = 2\Omega[/tex]
Hence,
[tex]P_\text{loss} = \left(\dfrac{200000}{23}\right)^2\times2[/tex]
The fraction lost = [tex]\dfrac{P_\text{loss}}{P}=\left(\dfrac{200000}{23}\right)^2\times2\div (1000\times10^6)=\dfrac{80}{529}=0.151[/tex]
In fireworks displays, light of a given wavelength indicates the presence of a particular element. What are the frequency and color of the light associated with each of the following?
Answer:
The four wavelengths of the problem are not given. Here they are:
a) [tex]Li^+,\lambda=671 nm[/tex]
b) [tex]Cs^+, \lambda=456 nm[/tex]
c) [tex]Ca^{2+}, \lambda=649 nm[/tex]
d) [tex]Na^+, \lambda=589 nm[/tex]
The relationship between wavelength and frequency of light wave is
[tex]f=\frac{c}{\lambda}[/tex]
where
f is the frequency
[tex]c=3.0\cdot 10^8 m/s[/tex] is the speed of light
[tex]\lambda[/tex] is the wavelength
For case a), [tex]\lambda=671 nm = 6.71\cdot 10^{-7}m[/tex] (corresponds to red color), so its frequency is
[tex]f=\frac{3\cdot 10^8}{6.71\cdot 10^{-7}}=4.47\cdot 10^{14}Hz[/tex]
For case b), [tex]\lambda=456 nm = 4.56\cdot 10^{-7}m[/tex] (corresponds to blue color), so its frequency is
[tex]f=\frac{3\cdot 10^8}{4.56\cdot 10^{-7}}=6.58\cdot 10^{14}Hz[/tex]
For case c), [tex]\lambda=649 nm = 6.49\cdot 10^{-7}m[/tex] (corresponds to red color), so its frequency is
[tex]f=\frac{3\cdot 10^8}{6.49\cdot 10^{-7}}=4.62\cdot 10^{14}Hz[/tex]
For case d), [tex]\lambda=589 nm = 5.89\cdot 10^{-7}m[/tex] (corresponds to yellow color), so its frequency is
[tex]f=\frac{3\cdot 10^8}{5.89\cdot 10^{-7}}=5.09\cdot 10^{14}Hz[/tex]
In fireworks displays, the color of light is determined by its wavelength and frequency. The color red has the longest wavelength and the lowest frequency, while the color violet has the shortest wavelength and the highest frequency. Therefore, the order of the given colors from shortest wavelength to longest wavelength is blue, yellow, and red. Similarly, the order of the given colors from lowest frequency to highest frequency is also blue, yellow, and red.
Explanation:The colors of light in a fireworks display indicate the presence of different elements. The wavelength and frequency of the light determines its color. Within the visible range, our eyes perceive radiation of different wavelengths as light of different colors. The color red corresponds to the longest wavelength and the lowest frequency, while the color violet corresponds to the shortest wavelength and the highest frequency. Therefore, the order of the given colors from shortest wavelength to longest wavelength would be blue, yellow, and red. Similarly, the order of the given colors from lowest frequency to highest frequency would also be blue, yellow, and red.
Learn more about Colors of light in fireworks displays here:https://brainly.com/question/30183969
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