Answer with Explanation:
We are given that
Mass , m=372 g=[tex]\frac{372}{1000}=0.372 Kg[/tex]
1 kg=1000g
Maximum acceleration, a=[tex]17.6 m/s^2[/tex]
Maximum speed ,v=1.75 m/s
a.We know that
Maximum acceleration, a=[tex]A\omega^2[/tex]
Maximum speed, v=[tex]\omega A[/tex]
[tex]17.6=A\omega^2[/tex]
[tex]1.75=A\omega[/tex]
[tex]\frac{17.6}{1.75}=\frac{A\omega^2}{A\omega}=\omega[/tex]
Angular frequency,[tex]\omega=10.06 rad/s[/tex]
b.Substitute the value of angular frequency
[tex]1.75=A(10.06)[/tex]
[tex]A=\frac{1.75}{10.06}=0.17 m[/tex]
Hence, the amplitude=0.17 m
c.Spring constant,k=[tex]m\omega^2[/tex]
Using the formula
[tex]k=0.372\times (10.06)^2[/tex]
Hence, the spring constant,k=37.6 N/m
Final answer:
The angular frequency is approximately 7.686 rad/s, the amplitude is approximately 0.227 m, and the spring constant is approximately 6.5472 N/m.
Explanation:
To determine the angular frequency, we can use the formula:
ω = √(k/m)
where ω is the angular frequency in radians per second, k is the spring constant in Newtons per meter, and m is the mass in kilograms.
Given that the maximum acceleration is 17.6 m/s^2 and the mass is 372 g (or 0.372 kg), we can calculate the spring constant:
k = m * a
k = 0.372 kg * 17.6 m/s^2 = 6.5472 N/m
Now we can find the angular frequency:
ω = √(6.5472 N/m / 0.372 kg) ≈ 7.686 rad/s
To determine the amplitude, we can use the formula:
A = vmax / ω
where A is the amplitude, vmax is the maximum speed, and ω is the angular frequency.
Given that the maximum speed is 1.75 m/s and the angular frequency is 7.686 rad/s, we can calculate the amplitude:
A = 1.75 m/s / 7.686 rad/s ≈ 0.227 m
Therefore, the angular frequency is approximately 7.686 rad/s, the amplitude is approximately 0.227 m, and the spring constant is approximately 6.5472 N/m.
Two plastic spheres each with radius 1.30 cm and mass 9.75 g, have uniform charge densities of 0.100 C/m^3 and -0.100 C/m^3, respectively. The spheres are released from rest when their centers are 80.0 cm apart. What is the speed of each sphere at the moment they collide? Ignore polarization of the spheres.
To develop this problem we will proceed to use the principle of energy conservation. For this purpose we will have that the change in the electric potential energy and kinetic energy at the beginning must be equal at the end. Our values are given as shown below:
[tex]m = 9.75g = 0.00975kg[/tex]
[tex]r = 1.3cm = 0.013m[/tex]
[tex]q_1 = 0.1 C/m^3 * \frac{4}{3} \pi r^3[/tex]
[tex]q_1 = 9.2*10^{-7}C[/tex]
[tex]q_2 = -9.2*10^{-7}C[/tex]
Applying energy conservation equations
[tex]U_1+K_1 = U_2+K_2[/tex]
[tex]\frac{k q_1q_2}{d} +0 = \frac{kq_1q_2}{2r}+ \frac{1}{2} (2m)v^2[/tex]
Replacing,
[tex]9*10^{9} (9.2*10^{-7})^2(\frac{1}{0.026}-\frac{1}{0.8}) = v^2[/tex]
Solving for v,
[tex]v = 9.2*10^{-7} (\frac{9*10^9}{0.00975}(\frac{1}{0.026}-\frac{1}{0.8}))^{1/2}[/tex]
[tex]v = 5.4 m/s[/tex]
Therefore the speed of each sphere at the moment they collide is 5.4m/s
Two small spheres with charges q1 = +9 nC and q2 = -3 nC are placed at marks x1 = -2 cm and x2 = 13 cm. Where along the x-axis should you place a third sphere with charge q3 so that the sum of the forces exerted by the first two spheres on the third one is zero?
To find the position along the x-axis where the sum of the forces exerted by the first two spheres on the third one is zero, we can use Coulomb's law. The forces between the charges are given by F = k * (|q1*q2| / r^2). By setting the forces equal to zero and solving for x, we can determine the desired position.
Explanation:To find the location along the x-axis where a third sphere with charge q3 should be placed so that the sum of the forces exerted by the first two spheres on the third one is zero, we need to consider the forces between the charges. According to Coulomb's law, the force between two point charges is given by:
F = k * (|q1*q2| / r^2)
Where k is the Coulomb's constant, q1 and q2 are the charges of the two spheres, and r is the distance between them. In this case, the force exerted by q1 on q3 is:
F1 = k * (|q1*q3| / (x - x1)^2)
And the force exerted by q2 on q3 is:
F2 = k * (|q2*q3| / (x - x2)^2)
For the sum of the forces to be zero, we must have:
F1 + F2 = 0
Substituting the values of q1, q2, x1, and x2, we can solve for x to find the position where the sum of the forces is zero.
Here, a=2, b=−82, and
c=437. Plug these values into the formula to find x_{3} . You should find two possible positions for x_{3}
Pluto's atmosphere. As recently observed by the New Horizons mission, the surface pressure of Pluto is about 11 microbar. The surface temperature is about 37 K.
(a) What is the number density (in units of number per cubic centimeter) of molecules at Pluto's surface (Hint: use ideal gas law)? The radius of Pluto is about 1187 km and the surface gravity is about 0.62 m s. What is the total mass of the atmosphere in terms of Kg?
(b) Calculate the saturation vapor pressure (in units of Pa) of ethane at Pluto's surface. The saturated vapor pressure of ethane can be assumed as: log1o(P)-10.01-1085.0/(T-0.561). T is temperature in K and the vapor pressure (P) in units of millimeters of Hg (~133.32 Pa).
(c) If the volume mixing ratio of ethane on Pluto is about 1%, what is mass mixing ratio of ethane (assume the mean molecular weight is 28 g mol')? What is the partial pressure of ethane at the surface? (Hint: should you use volume mixing ratio or mass mixing ratio to calculate the partial pressure? Think about the physical meaning of gas pressure.) Finally, is ethane condensable at Pluto's surface)
Answer:
a) The number density is 3.623 × 10⁻³ [tex]\frac{mol}{m^{3} }[/tex]
The mass of the atmosphere is 1.3 × 10²²Kg
b) The pressure is 10⁻²⁰ Millimeter of mercury
c) The mass mixing ratio is 0.0107
The partial pressure of ethane is 0.01114 Pa
Yes it is condensable because it boiling point is -88.5 C which is equivalent to 184.5 K i.e is adding 273 to -88.5C and the temperature of the atmosphere is 37 K.
Explanation:
The explanation is on the first and second uploaded image
Describe the relationship between the direction of the velocity vector and the direction of the acceleration or a body moving in a circle at constant speed.
In the centripetal movement, what happens with velocity is that it will remain constant, always pointing in its tangential direction of the trajectory. Said speed, although constant, will have a constant direction that will generate an acceleration that will always point towards the center of the circle radius. Both vectors as the turn is performed will always be perpendicular to each other.
The direction of the acceleration is longitudinal to the direction of velocity in for a body moving in a circle at constant speed.
Circular motion is the motion of an object in a circle at a constant speed. As the object is constantly changing its direction as it moving in a circle at constant speed.
At all instances, the object is moving tangent to the circle. The direction of velocity vector is tangent to the circle, as the direction of velocity vector is towards the direction of object's motion.
An object moving in a circle is accelerating due to its change in direction. The direction of the acceleration is inwards.
Therefore, the direction of the acceleration is longitudinal to the direction of velocity in for a body moving in a circle at constant speed.
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A telephone cable is clamped at A to the pole AB. Knowing that the tension in the right-hand portion of the cable is T2 = 1000 lb, determine by trigonometry (a) the required tension T1 in the left-hand portion if the resultant R of the forces exerted by the cable at A is to be vertical, (b) the corresponding magnitude of R.
Answer:
(a) T1 = 938.3lb
(b) R = 665.5lb
The detail solution to this problem can be found in the attachment below.
This problem was solved by resolving the forces along the vertical and horizontal and equating to Rx and Ry respectively. Rx = 0 because the resultant is directed along the vertical.
Explanation:
See attachment below for the full solution.
Thank you for reading.
Final answer:
To find the tension T1 and the vertical resultant R, use trigonometry, considering that T1 must have an equal and opposite horizontal force component to T2 and a vertical component that combines with T2's vertical component.
Explanation:
Understanding the Tension in Two Portions of a Cable
The student's question pertains to the tensions in a telephone cable clamped at point A to a pole. We are given that the tension in the right-hand portion of the cable (T2) is 1000 lb and are asked to determine two things:
The required tension T1 in the left-hand portion if the resultant R of the forces exerted by the cable at A is to be vertical.
The corresponding magnitude of R.
To solve for the required tension T1 in the left-hand portion so that the resultant force R is vertical, we need to use trigonometry. Given that the resultant R must be vertical and that we already have a horizontal force component from T2, T1 must provide an equal and opposite horizontal force component to cancel it out, as well as contribute to the resultant vertical force. The magnitude of the horizontal component of T1 must equal that of T2, and the vertical component of T1 will combine with T2 to form the vertical resultant R.
To determine the magnitude of the resultant force R, which is vertical, we will sum up the vertical components of T1 and T2. This magnitude can be found using the Pythagorean theorem if necessary, or simply as the sum of the vertical components if T1 is already calculated and has the proper direction to ensure a vertical R.
For this specific scenario, without additional information such as angles or the tension division between T1 and T2, we cannot provide numerical answers but can describe the method to find the solution.
A bottle with a volume of 193 U. S. fluid gallons is filled at the rate of 1.9 g/min. (Water has a density of 1000 kg/m3, and 1 U.S. fluid gallon = 231 in.3.) How long does the filling take?
Answer:
0.732 years
Explanation:
We are given that
Volume of fluid filled in bottle=193 U.S
1 U.S fluid gallon=231 cubic inch
193 U.S gallon=[tex]231\times 193=44583in^3[/tex]
[tex]1 in^3=1.63871\times 10^{-5}m^3[/tex]
[tex]44583in^3=44583\times 1.63871\times 10^{-5}=0.731m^3[/tex]
Density of water=[tex]1000kg/m^3[/tex]
Mass=[tex]Volume\times density[/tex]
Using the formula
Mass=[tex]0.731\times 1000=731kg[/tex]
1kg =1000g
[tex]731kg=731\times 1000=731\times 10^3g[/tex]
By using [tex]1000=10^3[/tex]
1.9 g of fluid takes time to fill in the bottle=1 min
1 g of fluid takes time to fill in the bottle=[tex]\frac{1}{1.9}min[/tex]
[tex]731\times 10^3[/tex]g of fluid takes time to fill in the bottle=[tex]\frac{1}{1.9}\times 731\times 10^3[/tex]min
Time=[tex]384.7\times 10^3[/tex]min
[tex]1min=\frac{1}{60\times 24\times 365}[/tex]years
Time=[tex]384.7\times 10^3\times \frac{1}{60\times 24\times 365}years[/tex]
Time =0.732 years
Hence, it takes to filling the 193 U.S fluid gallons in bottle=0.732 years
A spelunker is surveying a cave. She follows a passage 120 mm straight west, then 250 mm in a direction 45ââ east of south, and then 280 mm at 30ââ east of north. After a fourth unmeasured displacement, she finds herself back where she started.
(A) Use a scale drawing to determine the magnitude of the fourth displacement. Express your answer using two significant figures.
(B) Determine the direction of the fourth displacement.Express your answer using two significant figures.
Answer:
R = 207.45 mm , θ_return = 18.47 south west
Explanation:
This vector addition exercise is schematized in the attachment where the displacements are
1 d1 = 120 mm west
2 d2 = 250mm at 45 south east
3 d3 = 280 mm at 30 east of nort.
R is the final displacement that takes the goat to its initial point (origin)
The analytical way to perform this exercise is to find the components of each displacement and add them
Decompose the displacement using trigonometry
Displacement d1
d1ₓ = 120 cos 180 = -120 mm
Displacement d2, with the angle measured from the axis this θ = 270 + 45
sin 45 = [tex]d2_{y}[/tex] / d2
cos 45 = d2ₓ / d2
[tex]d2_{y}[/tex] = d2 sin45
[tex]d2_{y}[/tex] = 250 sin (270 + 45)
[tex]d2_{y}[/tex] = -176.77 mm
d2ₓ = d2 cos (270 + 45)
d2ₓ = 176.77 mm
displacement d3, for half the angle from the east axis θ = 90-30 = 60
sin 60 = [tex]d3_{y}[/tex] / d3
cos 60 = d3ₓ / d3
[tex]d3_{y}[/tex] = d3 sin 60
d3ₓ = d3 cos 60
[tex]d3_{y}[/tex] = 280 sin 60 = 242.49 mm
d3ₓ = 280 cos 60 = 140 mm
Having all the displacement components we can find the total displacement
Rₓ = d1ₓ + d2ₓ + d3ₓ
Ry = [tex]d1_{y}[/tex] + [tex]d2_{y}[/tex] + [tex]d3_{y}[/tex]
Rₓ = -120 + 176.77 +140
Rₓ = 196.77 mm
Ry = 0 -176.77 +242.49
Ry = 65.72 mm
Therefore the displacement you must make to return to the starting point is
R = RA Rx2 + Ry2)
R = RA (196.77 2 + 65.72 2)
R = 207.45 mm
We used trigonometry
tan tea = RY / Rx
tea = tan-1 Ry / Rx
ea = tan-1 (65.72 / 196.77)
tea = 18.47
This is the point where the girl is, to return to its origin this path must be serial, but in the opposite direction,
θ_return = 18.47 south west
The problem involves adding vectors to find the total displacement, then finding the negative of this displacement to calculate the magnitude and direction of the fourth displacement.
Explanation:This problem involves the mathematical concept of vectors, particularly in determining the resultant vector, which in this context represents the spelunker's path. First, we convert the movements into rectangular coordinates: going 120 mm west is -120i, going 250 mm 45° east of south is -250 cos(45)i - 250 sin(45)j, and going 280 mm 30° east of north is 280 cos(30)i + 280 sin(30)j. Adding these vectors together, we get the spelunker's total displacement vector. The negatives of this sum will represent the fourth displacement needed to get the spelunker back to where she started.
(A) The magnitude of the fourth displacement is the sum of these vectors taken as negatives, because she has to go back, which is equivalent to the length of the path taken. This can be calculated using Pythagoras' theorem for two dimensions.
(B) The direction of the fourth displacement is calculated by finding the angle made by the resulting vector with respect to one of the axes (for instance, the x-axis). For this, we take the inverse tangent of the y-coefficient over the x-coefficient of the vector.
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If you punched thousands of holes in the aluminum foil of the scope (so there were more "holes" than "foil"), how many imageswould you see in the viewer?
The correct answer is (a.) thousands of small images. Each hole in the aluminum foil acts as a tiny pinhole camera, allowing light to pass through and create an image on the other side. Since there are numerous holes, each one forms a separate image, resulting in a multitude of small, overlapping pictures. This phenomenon is known as a pinhole array or pinhole sieve.
If you punched thousands of holes in the aluminum foil of the scope (so there were more “holes” than “foil”), the resulting images in the viewer would be thousands of small images. Each hole acts as a pinhole camera, allowing light to pass through and create an image on the other side. Since there are numerous holes, each one would form a separate image, resulting in a multitude of small, overlapping pictures. This phenomenon is known as a pinhole array or pinhole sieve.
complete question;
If you punched thousands of holes in the aluminum foil of the scope (so there were more "holes" than "foil"), how many images would you see in the viewer?
Choices are:
a. thousands of small images
b. a few bright images
c. one large blurry image
d. no images at all, the light waves would cancel
When a moving car is brought to a stop with the brakes, its acceleration vector is 29. (a) in the same direction as its velocity vector b) in the opposite direction as its velocity vector c) equal to zero (d) directed downward
To develop this problem we will apply the concepts related to the kinematic equations of motion, specifically that of acceleration. Acceleration can be defined as the change of speed in an instant of time, mathematically this is
[tex]\vec{a} = \frac{\vec{v_2}-\vec{v_1}}{\Delta t}[/tex]
If a mobile is decreasing its speed (it is slowing down), then its acceleration is in the opposite direction to the movement. This would imply that the acceleration vector is opposite to the velocity vector.
Therefore the correct answer is B.
Two balloons (m = 0.012 kg) are separated by a distance of d = 15 m. They are released from rest and observed to have an instantaneous acceleration of a = 1.9 m/s2 toward each other at the moment they are released. Assume the two balloons are point charges and have equal but opposite charges.
Answer:
1.492*10^14 electrons
Explanation:
Since we know the mass of each balloon and the acceleration, let’s use the following equation to determine the total force of attraction for each balloon.
F = m * a = 0.012 * 1.9 = 0.0228 N
Gravitational forces are negligible
Charge force = 9 * 10^9 * q * q ÷ 225
= 9 * 10^9 * q^2 ÷ 225 = 0.0228
q^2 = 5.13 ÷ 9 * 10^9
q = 2.387 *10^-5
This is approximately 2.387 *10^-5 coulomb of charge. The charge of one electron is 1.6 * 10^-19 C
To determine the number of electrons, divide the charge by this number.
N =2.387 *10^-5 ÷ 1.6 * 10^-19 = 1.492*10^14 electrons
Work is done on an object as you drop it from the second floor of a building to the ground floor. True False.
Answer:
True, the work is done by the gravitational force on the object being dropped from the second floor of a building.
Explanation:
When an object is dropped from the second floor of a building then no force is applied by the person on the object but under the influence of gravity the object experience a force equal to its own weight and travels a displacement from the second floor to the ground.
Hence in this case the work is done by the gravitational force but not by the person dropping the object.
As we know that the work done is given as:
[tex]W=F.s[/tex]
where:
F = force on the object getting displaced by a length s
Here:
[tex]F=m.g[/tex]
where, m = mass of the object
g = acceleration due to gravity
A small metal sphere has a mass of 0.19 g and a charge of -21.0 nC. It is 10.0 cm directly above an identical sphere that has the same charge. This lower sphere is fixed and cannot move. If the upper sphere is released, it will begin to fall.
What is the magnitude of its initial acceleration?
The magnitude of the initial acceleration is 11.52 m/s^2.
Acceleration is a fundamental concept that describes how an object's velocity changes over time. Velocity includes both speed and direction, so acceleration can involve changes in speed, direction, or both. When an object speeds up, slows down, or changes direction, it experiences acceleration.
To find the magnitude of the initial acceleration, we need to use Newton's Law of Gravitation and Coulomb's Law. The weight and electric force acting on the sphere will determine its acceleration. The weight of the sphere is given by the formula W = mg, where m is the mass and g is the acceleration due to gravity. The electric force is given by the formula[tex]F = k(q1q2/r^2)[/tex], where k is the Coulomb's constant, q1 and q2 are the charges, and r is the distance between the spheres.
First, let's calculate the weight of the sphere:
[tex]W = (0.19 g)(9.8 m/s^2) = 1.862 N[/tex]
Next, let's calculate the electric force:
[tex]F = (8.99 x 10^9 Nm^2/C^2)(21.0 x 10^-9 C)^2 / (0.10 m)^2 = 0.4083 N[/tex]
Now, we can calculate the acceleration using Newton's second law:
F = ma, so a = F/m
a = (1.862 N + 0.4083 N) / 0.19 g
a = 11.52 m/s^2
Therefore, the magnitude of the initial acceleration is [tex]11.52 m/s^2.[/tex]
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A boat is able to move throught still water at 20m/s. It makes a round trip to a town 3.0km upstrea. If the river flows at 5m/s, the time required for this round trip is
Answer:
t=320s
Explanation:
Given Data
Boat speed=20 m/s
River flows=5 m/s
Total trip of distance d=3.0km = 3000m
To find
Total time taken
Solution
As
[tex]Velocity=distance/time\\time=distance/velocity\\[/tex]
Here we have two conditions
First when boat moves upward and the river pushing back.then velocity is given as
velocity=20m/s-5m/s
velocity=15 m/s
Time for that velocity
[tex]t_{1} =distance/velocity\\t_{1}=\frac{3000m}{15m/s}\\ t_{1}=200s[/tex]
Now for second condition when river flows and boat speed on same direction
velocity=20m/s+5m/s
velocity=25 m/s
Time taken for that velocity
[tex]t_{2}=distance/velocity\\t_{2}=\frac{3000m}{25m/s}\\ t_{2}=120m/s[/tex]
Now the total time
[tex]t=t_{1}+t_{2}\\t=(200+120)s\\t=320s[/tex]
For projectile motion on level ground assuming negligible air resistance (the initial angle being neither 0o nor 90o): Can the speed ever be the same as the initial speed at a time other than at t = 0 ?
a. No
b. Yes, at the end of the arch (end of trajectory)
c. Yes, at its maximum height
d. Yes, at half of its maximum height
Answer:
B. Yes, at the end of the trajectory.
Explanation:
By the conservation of energy, in order the object to have the same speed, it must have the same potential and kinetic energy, therefore, it must be at the same height as its initial height.
Answer:
b. Yes, at the end of the arch (end of trajectory)
Explanation:
according to conservation of energy and considering no air resistance:
initially the object has some kinetic energy due to speed which gets converted to gravitational potential energy till its maximum height.
since there is no energy loss, the object again gets to same amount of kinetic energy and hence the same speed.
An object has a mass of 20 kg. Determine its weight, in N, at a location where the acceleration of gravity is 9.78 m/s2.
Answer:
Weight of the object will be 195.6 N
Explanation:
We have given mass of the object m = 20 kg
And acceleration due to gravity at any location is given [tex]g=9.78m/sec^2[/tex]
We have to find the weight of the object in N at the location where value of acceleration due to gravity is [tex]g=9.78m/sec^2[/tex]
Weight of the object is given by W = mg , here W is weight , m is mass and g is acceleration due to gravity
So weight W = 20×9.78 = 195.6 N
So weight of the object will be 195.6 N
The weight of an object is calculated by multiplying its mass with the acceleration due to gravity. Given a mass of 20 kg and gravitational acceleration of 9.78 m/s2, the weight of the object is 195.6 N.
Explanation:The subject of this question is Physics, specifically dealing with the concept of weight, which is a force. The weight of an object can be calculated by multiplying its mass by the acceleration due to gravity. Given in this problem, the object's mass is 20 kg and the place's gravitational acceleration is 9.78 m/s2.
To find the object's weight, we use the formula: Weight = mass x gravity. Applying the values, we get: Weight = 20 kg x 9.78 m/s2 = 195.6 N.
Therefore, the weight of the object at the given location is 195.6 Newtons.
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An object is thrown upwards with a speed of 14 m/s. How long does it take to reach a height of 5.0 m above the projection point while descending?
Answer:
2.43 s
Explanation:
Using newton's equation of motion.
T = (v-u)/g
Where T = time taken for the object to return to the point of projection , u = initial velocity, v = final velocity, g = acceleration due to gravity.
Given: v =-14 m/s, u = 14 m/s, g = -9.8 m/s²
T = (-14-14)/-9.81
T = 2.85 s
Note: We look for the object's speed at 5.0 m.
using
v² = u²+2gs.................................... Equation 1
Where v = final velocity, u = initial velocity, g = acceleration due to gravity, s = distance.
Given: u = 14 m/s, g = -9.81 m/s², s = 5.0 m
Substitute into equation 1
v² = 14²+(-9.81×5×2)
v² = 196-98.1
v = √97.9
v = 9.89
We look for the time taken for the velocity to decrease from 14 m/s to 9.89 m/s.
using
v = u+gt
t =(v-u)/g........................... Equation 2
Where t = time taken for the object to decrease it velocity from 14 m/s to 9.89 m/s
Given: v = 9.89 m/s, u =14 m/s g = -9.81 m/s²
t = (14-9.89)/-9.81
t = -4.11/-9.81
t = 0.42 s
Thus,
Time taken to reach 5.0 m above projection point = T-t
=2.85-0.42
2.43 s
It takes approximately 1.03 seconds for the object to reach a height of 5.0 m above the projection point while descending.
The time it takes for an object to reach a certain height when thrown upwards can be determined using equations of projectile motion.
In this case, with an initial speed of 14 m/s and a height of 5.0 m, we can use the equation :
[tex]$$y = y_0 + v_{oy}t - \frac{1}{2}gt^2$$[/tex]
Substituting the known values, we get:
[tex]$$5.0 = 0 + 14t - \frac{1}{2}(9.8)t^2$$[/tex]
By rearranging the equation and solving it,
we find that it takes approximately 1.03 seconds for the object to reach a height of 5.0 m above the projection point while descending.
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Two tiny, spherical water drops, with identical charges of −8.00 ✕ 10^(−17) C, have a center-to-center separation of 2.00 cm.
What is the magnitude of the electrostatic force acting between them?
Answer:
F=1.4384×10⁻¹⁹N
Explanation:
Given Data
Charge q= -8.00×10⁻¹⁷C
Distance r=2.00 cm=0.02 m
To find
Electrostatic force
Solution
The electrostatic force between between them can be calculated from Coulombs law as
[tex]F=\frac{kq^{2} }{r^{2} }[/tex]
Substitute the given values we get
[tex]F=\frac{(8.99*10^{9} )*(-8.00*10^{-17} )^{2} }{(0.02)^{2} }\\ F=1.4384*10^{-19} N[/tex]
Light illuminating a pair of slits contains two wavelengths, 600 nm and an unknown wavelength. The 10th bright fringe of the unknown wavelength overlaps the 9th bright fringe of the 600 nm light.
1. What is the unknown wavelength?
Answer:
unknown wavelength is 540 nm
Explanation:
given data
wavelengths = 600 nm
10th bright fringe overlap = 9th bright fringe of 600 nm
to find out
unknown wavelength
solution
we get here first nth bright fringe from the central maximum at distance y
y = n ×[tex]\frac{\lambda D }{d}[/tex] ...............1
and we have given 10th bright fringe overlap = 9th bright fringe of 600 nm
so here
10 × [tex]\frac{\lambda D }{d}[/tex] = 9 × 600 × [tex]\frac{D}{d}[/tex]
solve it we get [tex]\lambda[/tex]
[tex]\lambda[/tex] = 9 × [tex]\frac{600}{10}[/tex]
[tex]\lambda[/tex] = 540 nm
so unknown wavelength is 540 nm
The unknown wavelength is equal to 540 nanometers.
Let the unknown wavelength be W.Given the following data:
Wavelength = 600 nmTo determine the unknown wavelength:
At the maximum distance, the nth bright fringe is given by the formula:
[tex]Y = \frac{n\lambda D}{d}[/tex]
From the question, we were told that the 10th bright fringe of the unknown wavelength overlaps the 9th bright fringe of the 600 nm light.
Mathematically, this can be expressed as follows:
[tex]Y_{10} = Y_9\\\\\frac{10WD}{d} =\frac{9\times 600D}{d} \\\\10W=5400\\\\W=\frac{5400}{10}[/tex]
W = 540 nanometers.
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Dan is gliding on his skateboard at 4.00m/s . He suddenly jumps backward off the skateboard, kicking the skateboard forward at 6.00m/s . Dan's mass is 60.0kg and the skateboard's mass is 7.00kg .
How fast is Dan going as his feet hit the ground?
To solve this problem we will apply the concepts related to the conservation of the Momentum. For this purpose we will define the momentum as the product between mass and velocity, and by conservation the initial momentum will be equal to the final momentum. Mathematically this is,
[tex]m_1u_1+m_2u_2 = m_1v_1+m_2v_2[/tex]
Here,
[tex]m_{1,2}[/tex] = Mass of Dan and Skateboard respectively
[tex]u_{1,2}[/tex] = Initial velocity of Dan and Skateboard respectively
[tex]v_{1,2}[/tex] = Final velocity of Dan and Skateboard respectively
Our values are:
Dan's mass
[tex]m_1 = 60kg[/tex]
Mass of the skateboard
[tex]m_2 = 7.0kg[/tex]
Both have the same initial velocity, then
[tex]u_1= u_2 = 4m/s[/tex]
Final velocity of Skateboard is
[tex]v_2 = 6m/s[/tex]
Rearranging to find the final velocity of Dan we have then,
[tex]m_1u_1+m_2u_2 = m_1v_1+m_2v_2[/tex]
[tex]m_1v_1+m_2v_2 = (m_1+m_2)u_1[/tex]
[tex]v_1 = \frac{ (m_1+m_2)u_1 -m_2v_2}{m_1}[/tex]
Replacing,
[tex]v_1 = \frac{(60+7)(4)-(7)(6)}{60}[/tex]
[tex]v_1 = 3.76m/s[/tex]
Therefore Dan will touch the ground at a speed of 3.76m/s
A mass m at the end of a spring oscillates with a frequency of 0.83 Hz.When an additional 730 gmass is added to m, the frequency is 0.65 Hz.What is the value of m?
Answer:
m will be equal to 1158.73 gram
Explanation:
We have given mass m when frequency is 0.83 Hz
So mass [tex]m_1=m[/tex] and frequency [tex]f_!=f[/tex] let spring constant of the spring is KK
Frequency of oscillation of spring is given by [tex]f=\frac{1}{2 \pi }\sqrt{\frac{k}{m}}[/tex]
From above relation we can say that [tex]{\frac{f_1}{f_2}}=\sqrt{\frac{m_2}{m_1}}[/tex]
It is given that when an additional 730 gram is added to m then frequency become 0.65 Hz , [tex]f_2=0.65Hz[/tex]
So [tex]m_2=m+730[/tex]
So [tex]\frac{0.93}{0.65}=\sqrt{\frac{m+730}{m}}[/tex]
[tex]\frac{m+730}{m}=1.63[/tex]
[tex]0.63m=730[/tex]
m= 1158.73 gram
To find the value of m, we can set up a proportion using the formula for frequency and solve for m. The value of m is 0.93 kg.
Explanation:To solve this problem, we can use the formula for the frequency of an object in simple harmonic motion:
f = 1 / T
Where f is the frequency and T is the period. Let's denote the mass of the object as m, and the original frequency as f1. When the additional mass is added, the frequency becomes f2. We can set up a proportion to solve for the value of m:
f1 / f2 = m / (m + 0.73)
Solving for m, we have:
m = (f1 / f2) * 0.73
Substituting the values f1 = 0.83 Hz and f2 = 0.65 Hz, we can find the value of m:
m = (0.83 / 0.65) * 0.73 = 0.93 kg
Therefore, the value of m is 0.93 kg.
A thread is being pulled off a spool at the rate of 57.3 cm per sec. Find the radius of the spool if it makes 141 revolutions per min. Round to two decimal places as needed.
Answer: r= 0.04meter
Explanation:
The velocity V and angular speed w is related to the radio by
V=w*r
r=V/w
But the standard unit for velocity is m/s so we convert 57.3cm/sec to m/s
57.3/100= 0.57m/s
Also Angular speed w is rad/secs
So we convert rev/min to rad/sec
0.1rad/sec equals 1 rev/mins
141rev/mins * 0.1rad/secs
=14.1rad/sec
r = V/w
= 0.573/14.1
= 0.04m to 2d.p
The radius of the spool is 12.3 cm.
Explanation:To find the radius of the spool, we need to use the formula for the linear speed of a point on the edge of a rotating object: v = ωr. Here, v is the linear speed, ω is the angular velocity, and r is the radius. By converting the given information, we can calculate the linear speed:
Linear speed (v) = (57.3 cm/sec) / (100 cm/m) = 0.573 m/sec
The angular velocity can be found by converting the rate of revolution to radians per second:
Angular velocity (ω) = (141 rev/min) * (2 π rad/rev) / (60 sec/min) = 4.676 rad/sec
Now we can rearrange the formula to solve for the radius:
Radius (r) = v / ω = 0.573 m/sec / 4.676 rad/sec = 0.123 m = 12.3 cm
A heat engine uses two containers held at different temperatures. One container is at 294 K 294 K , while the other is kept at 552 K 552 K . What is the maximum possible efficiency for this engine?
Answer:
Explanation:
Given
Lower Temperature [tex]T_L=294 \K[/tex]
Higher Temperature [tex]T_H=522 \K[/tex]
Maximum Possible efficiency is achieved when the engine works as carnot Engine
i.e. [tex]\eta _{max}=1-\frac{T_L}{T_H}[/tex]
[tex]\eta_{max}=1-\frac{294}{522}[/tex]
[tex]\eta _{max}=\frac{228}{522}=0.436[/tex]
[tex]\eta _{max}=43.64\ %[/tex]
The maximum possible efficiency of a heat engine can be determined using the temperatures of the hot and cold reservoirs in the formula Effc = 1 - Tc / Th. Applying this formula to the given temperatures of 294 K and 552 K results in a maximum efficiency of 46.8%.
Explanation:The maximum possible efficiency of a heat engine (also known as the Carnot efficiency) can be calculated using the temperatures of the heat source (hot reservoir) and the heat sink (cold reservoir). This efficiency can be determined by using the formula Effc = 1 - Tc / Th, where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir. Both these temperatures should be in Kelvin.
In the given problem, Tc is 294 K and Th is 552 K. Substituting these values into the formula gives the maximum possible efficiency:
Effc = 1 - 294 / 552 = 0.468. Thus, the maximum possible efficiency of this engine is approximately 46.8%.
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A golf ball is hit with an angle of elevation 30∘ and speed 20????????/????. Find the horizontal and vertical components of the velocity vector.
The horizontal and vertical components of the velocity vector for the golf ball hit with an angle of elevation of 30° and speed of 20 m/s are determined to be 17.32 m/s and 10 m/s, respectively.
Explanation:Velocity Vector Components
The components of the velocity vector, which are horizontal and vertical, are calculated by multiplying the speed of the object by cos(θ) for the horizontal component and sin(θ) for the vertical component, where θ is the angle of elevation.
Given that the angle of elevation is 30 degrees and the speed is 20 m/s, we can use these equations to find:
Horizontal Component: Vx = speed * cos(θ) = 20 * cos(30) = 17.32 m/sVertical Component: Vy = speed * sin(θ) = 20 * sin(30) = 10 m/sLearn more about Velocity Vector Components here:https://brainly.com/question/37190332
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If you double the radius, for a fixed angular velocity, the tangential velocity will:a. Decrease by halfb. Stay the samec. Increase by a factor of twod. Increase by a factor of four
Answer:
C. Increase by a factor of two.
Explanation:
Lets take angular speed = [tex]\omega \ rad/s[/tex]
Initially radius = r m
The tangential velocity = v m/s
We know that
[tex]v=\omega r[/tex]
If the radius becomes double ,lets say r'
r'= 2 r
Then
The tangential velocity = v' m/s
[tex]v'=\omega r'[/tex]
[tex]v'=\omega (2r)[/tex]
[tex]v'=2\omega r[/tex]
v' = 2 v
Then we can say that velocity will become double.
Therefore the answer will be C.
C. Increase by a factor of two.
Final answer:
Doubling the radius while maintaining a fixed angular velocity results in the tangential velocity increasing by a factor of two.
Explanation:
When the radius of an object is doubled while keeping the angular velocity constant, the tangential velocity will increase by a factor of two. This is because the tangential velocity (v) is calculated by multiplying the angular velocity (
ω) by the radius (r) of the circle, resulting in v = r x
ω. Therefore, if we double the radius (2r), the new tangential velocity becomes 2r x
ω, which is twice the original tangential velocity.
For a constant angular velocity, the relationship between tangential velocity and radius is direct; as one increases, so does the other. So, if you double the radius, the tangential velocity also doubles, not decreases. This explains why the correct answer to the student's question is c. Increase by a factor of two.
The electric force is exerted by a 40-nC charged particle located at the origin of a Cartesian coordinate system on a 15-nC charged particle located at (2.0 m, 2.0 m ).
Determine the direction of the force.
a. ?y direction
b. +x direction
c. ?x direction
d. 45? clockwise from the +x axis.
e. 45? counterclockwise from the +x axis.
f. +y direction
Answer:
Explanation:
Check the attachment for solution
The direction of the electric force is exerted by a 40-nC charged particle located at the origin of a cartesian coordinate system on a 15-nC charged particle located at (2.0 m, 2.0 m ) is 45° counterclockwise from the +x axis.
What is electric force?Electric force is the attracting or repulsive interaction between any two charged things. Similar to any force, Newton's laws of motion describe how it affects the target body and how it does so. One of the many forces that affect objects is the electric force.
The given two charges are: 40 nC and 15 nC. As both of them are positive, the electric force is repulsive in nature.
15-nC charged particle located at (2.0 m, 2.0 m ).
So, the direction of repulsive force = tan⁻¹(2.0/2.0) = 45⁰.
Hence, The direction of the electric force is 45° counterclockwise from the +x axis. Option (e) is correct.
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You are hiking in the mountains and find a shiny gold nugget. It might be the element gold, or it might be "fool's gold ," which is a nickname for iron pyrite,
FeS2
Which of the following physical properties do you think would help determine if the shiny nugget is really gold?
a. appearance,
b. melting point,
c.. density, or
d. physical state
Final answer:
Option C.) To determine if the nugget is gold, you can consider its density, melting point, and appearance.
Explanation:
To determine if the shiny nugget is really gold, one physical property that would be helpful is its density. Gold has a high density of approximately 19 g/cm³, which means it is much denser than most other minerals. Another physical property that could be useful is its melting point. Gold has a relatively low melting point of 1,064 degrees Celsius, while iron pyrite has a much higher melting point. Lastly, the appearance of the nugget can also provide some clues. Gold has a distinctive yellow color, while fool's gold (iron pyrite) has a brassy or pale yellow color.
Jupiter's moon Io has active volcanoes (in fact, it is the most volcanically active body in the solar system) that eject material as high as 500 km (or even higher) above the surface. Io has a mass of 8.93×1022 kg and a radius of 1821 km . For this calculation, ignore any variation in gravity over the 500-km range of the debris.A. How high would this material go on earth if it were ejected with the same speed as on Io?
The height that an object reaches in a gravitational field is dependent on its initial speed and the strength of the field. With the same initial speed, an object in a weaker gravitational field will reach a greater height. Therefore, material ejected with the same speed on Io and Earth will reach a greater height on Io due to its weaker gravitational field.
Explanation:To understand how high material would go on Earth if ejected at the same speed as on Io, we first need to consider that the height that an object reaches in a gravitational field (ignoring air resistance) depends on its initial speed and the strength of the gravitational field. Specifically, the maximum height is given by the equation H = (v^2)/(2g), where v is the initial speed, and g is the acceleration due to gravity.
When an object is thrown upwards, it slows down under the force of the Earth's gravity until its speed decreases to zero. At that point, it begins to fall back down. Therefore, if two objects are thrown upwards with the same speed but under the influence of different gravitational fields (one stronger and one weaker), the object in the weaker gravitational field will reach a greater height.
In the case of Io, Jupiter's moon, the strength of the gravitational field is much less than that on Earth because its mass is much smaller. That's why volcanic material can be ejected to such high altitudes. Now, if we were to eject material with the same speed on Earth, which has a stronger gravitational field compared to Io, it would not reach the same height.
To calculate the specific height that material would reach on Earth assuming the same ejection speed as on Io, we would need to know that ejection speed. Unfortunately, the question does not provide this information.
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material ejected from Io with the same speed on Earth would reach a height of approximately [tex]\( 3.41 \times 10^8 \, \text{m} \) or \( 341 \, \text{km} \)[/tex] above the surface of Earth.
To determine how high material ejected from Io would go on Earth if it were ejected with the same speed, we can use the concept of escape velocity and the energy conservation principle.
The escape velocity [tex](\( v_e \))[/tex] of a celestial body is the minimum speed an object must have to break free from its gravitational pull and escape into space. It is given by the formula:
[tex]\[ v_e = \sqrt{\frac{2GM}{r}} \][/tex]
Where:
- [tex]\( v_e \)[/tex] is the escape velocity,
- ( G ) is the universal gravitational constant[tex](\( 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)),[/tex]
- ( M \ is the mass of the celestial body,
- ( r ) is the distance from the center of the celestial body.
Let's calculate the escape velocity of Io first:
[tex]\[ v_e = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \times 8.93 \times 10^{22} \, \text{kg}}{1821 \times 10^3 \, \text{m}}} \]\[ v_e = \sqrt{\frac{2 \times 6.67 \times 8.93}{1821}} \times 10^{11} \, \text{m/s} \]\[ v_e = \sqrt{2 \times 3.34} \times 10^{11} \, \text{m/s} \]\[ v_e \approx 8.19 \times 10^4 \, \text{m/s} \][/tex]
Now, let's assume that material ejected from Io is given this speed of [tex]\( 8.19 \times 10^4 \, \text{m/s} \)[/tex] on Earth. We can calculate how high it would go using energy conservation principles.
The kinetic energy ( KE ) of the material is given by:
[tex]\[ KE = \frac{1}{2} m v^2 \][/tex]
Where:
- \( m \) is the mass of the material, and
- \( v \) is the velocity.
The potential energy (\( PE \)) gained by the material at height \( h \) above the surface of Earth is given by:
[tex]\[ PE = mgh \][/tex]
Where:
- ( g ) is the acceleration due to gravity on Earth[tex](\( 9.81 \, \text{m/s}^2 \))[/tex],
- ( h ) is the height above the surface of Earth.
At the maximum height, the kinetic energy is converted entirely into potential energy. So we have:
[tex]\[ \frac{1}{2} m v^2 = mgh \]\[ \frac{1}{2} v^2 = gh \]\[ h = \frac{v^2}{2g} \][/tex]
Substitute the values:
[tex]\[ h = \frac{(8.19 \times 10^4 \, \text{m/s})^2}{2 \times 9.81 \, \text{m/s}^2} \]\[ h = \frac{6696.61 \times 10^8 \, \text{m}^2/\text{s}^2}{19.62 \, \text{m/s}^2} \]\[ h \approx 3.41 \times 10^8 \, \text{m} \][/tex]
So, material ejected from Io with the same speed on Earth would reach a height of approximately [tex]\( 3.41 \times 10^8 \, \text{m} \) or \( 341 \, \text{km} \)[/tex] above the surface of Earth.
A particle is moving along a straight line such that its' acceleration is defined as
a(v) = (-2v) m/s^2 where v is in meters per second.
If v = 20 m/s when s= 0 and t=0 find:
1. The particles position as a function of time
2. The particles velocity as a function of time
3. The particles acceleration as a function of time
Final answer:
To find the particle's dynamics, we need to integrate the given acceleration function in terms of velocity and apply the initial conditions. The velocity function v(t) is √(400 - 2t), the position function s(t) would result from further integration, and the acceleration a(t) is -2√(400 - 2t).
Explanation:
A particle is moving in a straight line with its acceleration a(v) given as a(v) = (-2v) m/s², where v is the velocity in meters per second. To find the particle's position, velocity, and acceleration as functions of time, we'll need to integrate the acceleration function.
Finding the velocity as a function of time:
We have the acceleration in terms of velocity; thus, we can write:
'(v) = a(v) = -2v
(v) = -∫ 2v dv = -v² + C
When t=0, v = 20 m/s:
C = 20² = 400
v(t) = √(400 - 2t)
Finding the particle's position as a function of time:
s(t) = ∫ v(t) dt
s(t) = ∫ √(400 - 2t) dt
The integration will give us the position, s(t), in terms of t, which needs to be integrated carefully using appropriate techniques such as substitution.
Finding the acceleration as a function of time:
a(t) can be found by substituting the expression for v(t) into a(v), which gives us a(t) = -2√(400 - 2t).
A basketball has a mass of 567 g. Heading straight downward, in the direction, it hits the floor with a speed of 2 m/s and rebounds straight up with nearly the same speed. What was the momentum change ?
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.
A 5.60 g bullet moving at 501.8 m/s penetrates a tree trunk to a depth of 4.59 cm. a) Use work and energy considerations to find the magnitude of the force that stops the bullet. Answer in units of N
Answer:
Explanation:
Given
mass of bullet [tex]m=5.6\ gm[/tex]
velocity of bullet [tex]v=501.8\ m/s[/tex]
Depth of penetration [tex]d=4.59\ cm[/tex]
According to the work energy theorem work done by all the force will be equal to change in kinetic energy of Particle
Suppose F is the force which is opposing the bullet motion
change in kinetic Energy [tex]\Delta K=\frac{1}{2}mv^2-0[/tex]
[tex]\Delta K=\frac{1}{2}mv^2=\frac{1}{2}\times 5.6\times 10^{-3}\times (501.8)^2[/tex]
[tex]\Delta K=705.049\ J[/tex]
[tex]\Delta K=F\cdot d[/tex]
[tex]F=\frac{\Delta K}{d}[/tex]
[tex]F=\frac{705.049}{4.59\times 10^{-2}}[/tex]
[tex]F=15,360.54\ N[/tex]
[tex]F=1.536\times 10^4\ N[/tex]