Answer:
23.5 mV
Explanation:
number of turn coil 'N' =22
radius 'r' =3.00 cm=> 0.03m
resistance = 1.00 Ω
B= 0.0100t + 0.0400t²
Time 't'= 4.60s
Note that Area'A' = πr²
The magnitude of induced EMF is given by,
lƩl =ΔφB/Δt = N (dB/dt)A
=N[d/dt (0.0100t + 0.0400 t²)A
=22(0.0100 + 0.0800(4.60))[π(0.03)²]
=0.0235
=23.5 mV
Thus, the induced emf in the coil at t = 4.60 s is 23.5 mV
Lightning bolts can carry currents up to approximately 20 kA. We can model such a current as the equivalent of a very long, straight wire. You may want to review (Pages 926 - 929) . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Magnetic field of a single wire. Part A If you were unfortunate enough to be 4.9 m away from such a lightning bolt, how large a magnetic field would you experience? Express your answer using two significant figures. B1 = nothing T Request Answer Part B How does this field (B1) compare to one (B2) you would experience by being 4.9 cm from a long, straight household current of 10 A? Express your answer using two significant figures.
Given Information:
Current due to Lightning bolt = I₁ = 20 kA
Current of household = I₂ = 10 A
Distance = r₁ = 4.9 m
Distance = r₂ = 4.9 cm = 0.049 m
Required Information:
a) Magnetic field due Lightning bolt = B₁ = ?
b) Magnetic field due to household current = B₂ = ?
Answer:
a) Magnetic field due Lightning bolt = B₁ = 8.16×10⁻⁴ T
b) Magnetic field due to household current = B₂ = 4.08×10⁻⁵ T
B₁ = 20B₂
Explanation:
The magnetic field produced in a long straight wire carrying a current (I) at distance r is given by
B = μ₀I/2πr
Where μ₀ is the permeability of free space and its value is 4π×10⁻⁷
a) The magnetic field produced due to the lightning bolt current is
B₁ = μ₀I₁/2πr₁
B₁ = (4π×10⁻⁷*20,000)/2π*4.9
B₁ = 0.000816
B₁ = 8.16×10⁻⁴ T
Therefore, the strength of magnetic field due to the lightning bolt current is 8.16×10⁻⁴ T
b) The magnetic field produced due to the household current is
B₂ = μ₀I₂/2πr₂
B₂ = (4π×10⁻⁷*10)/2π*0.049
B₂ = 0.00004081
B₂ = 4.08×10⁻⁵ T
Therefore, the strength of magnetic field due to the household current is 4.08×10⁻⁵ T
The ratio of magnetic field produced by the lightning bolt current to the magnetic field produced by the household current is
B₁/B₂ = 8.16×10⁻⁴/4.08×10⁻⁵
B₁/B₂ = 20
B₁ = 20B₂
Which means that the magnetic field produced by the lightning bolt current is 20 time greater than the magnetic field produced by the household current.
a) The magnetic field due Lightning bolt will be 8.16×10⁻⁴ T
b)The magnetic field due to household current will be 4.08×10⁻⁵ T
What is a magnetic field?
It is the type of field where the magnetic force is obtained. With the help of a magnetic field. The magnetic force is obtained it is the field felt around a moving electric charge.
The number of magnetic flux lines on a unit area passing perpendicular to the given line direction is known as induced magnetic field strength .it is denoted by B.
a) The magnetic field due Lightning bolt will be 8.16×10⁻⁴ T
The magnetic field produced in a long straight wire carrying a current (I) at distance r is given by
B = μ₀I/2πr
Where μ₀ is the permeability of free space and its value is 4π×10⁻⁷
[tex]\rm B_1= \frac{\mu_0i_1}{2 \pi r} \\\\ \rm B_1= \frac{4 \pi \times 10^{-7} \times 20,00}{2 \times 3.14 4.9} \\\\ \rm B_1= 8.16 \times 10^{-4} \ T[/tex]
b) Magnetic field due to household current will be 4.08×10⁻⁵ T
[tex]\rm B_2= \frac{\mu_0i_2}{2 \pi r} \\\\ \rm B_2= \frac{4 \pi \times 10^{-7} \times 10}{2 \times 3.14 0.049} \\\\ \rm B_2= 4.08\times 10^{-5} \ T[/tex]
Hence the magnetic field due to household current will be 4.08×10⁻⁵ T
The magnetic field created by lightning bolt current divided by the magnetic field produced by home electricity equals
[tex]\rm \frac{B_1}{B_2} = \rm \frac{8.16 \times 10^{-4}}{4.08 c\times 10^{-5}} \\\\ \rm \frac{B_1}{B_2} = 20 \\\\[/tex]
That is, the magnetic field created by a lightning bolt current is 20 times stronger than the magnetic field produced by a home current.
Hence the magnetic field due to the Lightning bolt will be 8.16×10⁻⁴and the magnetic field due to household current will be 4.08×10⁻⁵ T.
To learn more about the magnetic field refer to the link;
https://brainly.com/question/19542022
Imagine playing baseball in a universe (not ours!) where the Planck constant is 0.70 J·s. What would be the uncertainty in the position of a 0.55 kg baseball that is moving at 22 m/s along an axis if the uncertainty in the speed is 0.5 m/s? Number Enter your answer in accordance to the question statement Units Choose the answer from the menu in accordance to the question statement
Answer:
The uncertainty the in position is [tex]\delta s = 0.4051 m[/tex]
Explanation:
From the question we are told
The planck's constant is [tex]h = 0.70 \ J \cdot s[/tex]
The mass of the baseball is [tex]m = 0.55kg[/tex]
The velocity of the baseball [tex]v_b = 22 m/s[/tex]
The uncertainty in [tex]\delta v = 0.5 m/s[/tex]
Generally the uncertainty of momentum is
[tex]\delta p = m \delta v[/tex]
substituting values
[tex]\delta p = 0.275 \ kg m/s[/tex]
The uncertainty position is mathematically represented as
[tex]\delta s = \frac{h}{2 \p \delta p }[/tex]
substituting values
[tex]\delta s = \frac{0.70}{2 * 3.142 * 0.275 }[/tex]
[tex]\delta s = 0.4051 m[/tex]
What factors does the kinetic energy of a body depend
on?
Ans
Answer:
Kinetic energy depends on the mass of a body and the velocity it is travelling at
Explanation:
Referring to the equation of Kinetic Energy
EK = 0.5 m [tex]v^{2}[/tex]
We can see that Kinetic energy depends on the mass of a body and the velocity it is travelling at
Answer:
The amount of translational kinetic energy (from here on, the phrase kinetic energy will refer to translational kinetic energy) that an object has depends upon two variables: the mass (m) of the object and the speed (v) of the object. The following equation is used to represent the kinetic energy (KE) of an object.
An oil tanker has collided with a smaller vessel, resulting in an oil spill in a large, calm-water bay of the ocean. You are investigating the environmental effects of the accident and need to know the area of the spill. The tanker captain informs you that 18000 liters of oil have escaped and that the oil has an index of refraction of n = 1.1. The index of refraction of the ocean water is 1.33. From the deck of your ship you note that in the sunlight the oil slick appears to be blue. A spectroscope confirms that the dominant wavelength from the surface of the spill is 485 nm. Assuming a uniform thickness, what is the largest total area o
Complete Question
An oil tanker has collided with a smaller vessel, resulting in an oil spill in a large, calm-water bay of the ocean. You are investigating the environmental effects of the accident and need to know the area of the spill. The tanker captain informs you that 18000 liters of oil have escaped and that the oil has an index of refraction of n = 1.1. The index of refraction of the ocean water is 1.33. From the deck of your ship you note that in the sunlight the oil slick appears to be blue. A spectroscope confirms that the dominant wavelength from the surface of the spill is 485 nm. Assuming a uniform thickness, what is the largest total area oil slick
Answer:
The largest total area of the oil slick [tex]A = 8.257 *10^{9} \ m^2[/tex]
Explanation:
From the question we are told that
The volume of oil the escaped is [tex]V = 18000 \ L[/tex]
The refractive index of oil is [tex]n_o = 1.1[/tex]
The refractive index of water is [tex]n_w = 1.33[/tex]
The wavelength of the light is [tex]\lambda = 485 \ nm = 485 * 10^{-9} \ m[/tex]
Generally the thickness of the oil for condition of constructive interference between the oil and the water is mathematically represented as
[tex]d = m *\frac{\lambda}{2n_w}[/tex]
Where is the order of interference of the light and it value ranges from 1, 2, 3,...n
It is usually take as 1 unless stated otherwise by the question
substituting value
[tex]d = 1 * \frac{485 *10^{-9}}{2 * 1.1}[/tex]
[tex]d = 218 nm[/tex]
The are can be mathematically evaluated as
[tex]A = \frac{V}{d}[/tex]
Substituting values
[tex]A = \frac{18000}{218*10^{-8}}[/tex]
[tex]A = 8.257 *10^{9} \ m^2[/tex]
A uniform hollow spherical ball of mass 17 kg and radius 50.0 cm rolls without slipping up a ramp at an angle of 19 degrees above the horizontal. The speed of the rolling ball at the base of the ramp is 5.0 m/s. While the ball is rolling up the ramp, what are the magnitude and direction of the acceleration of its center of mass, and what is the magnitude and direction of the friction force acting on the ball from the surface of the ramp
Answer:
The acceleration of its center of mass is
The frictional force is [tex]f = 21.65 \ N[/tex]
Explanation:
From the question we are told that
The mass of the ball is [tex]m = 17 kg[/tex]
The radius of the ball is [tex]r = 50cm = \frac{50}{100} = 0.5m[/tex]
the angle with the horizontal is [tex]\theta = 19 ^ o[/tex]
The of the ball at the base is [tex]v = 5.0 \ m/s[/tex]
This setup is shown on the first uploaded image
looking at the diagram we see that the force acting on the ball can be mathematically evaluated as
[tex]mg sin \theta -f = ma[/tex]
Where f is the frictional force
The torque on the ball is mathematically represented as
[tex]\tau = f * r[/tex]
This torque can also be mathematically represented as
[tex]\tau = I \alpha[/tex]
where I is the moment of inertia of the ball which is mathematically represented as
[tex]I = \frac{2}{3} m r^2[/tex]
While [tex]\alpha[/tex] is the angular acceleration which is mathematically represented as
[tex]\alpha = \frac{a}{r}[/tex]
So [tex]\tau = \frac{2}{3} m r^2 * \frac{a}{r}[/tex]
Equating the both formula for torque
[tex]f * r = \frac{2}{3} m r^2 * \frac{a}{r }[/tex]
=> [tex]f = \frac{2}{3} ma[/tex]
Substituting this for f in the above equation
[tex]mg sin \theta = ma + \frac{2}{3} ma[/tex]
[tex]g sin \theta = \frac{5}{3} a[/tex]
[tex]a = \frac{3}{5} * g * sin \theta \alpha[/tex]
Substituting values
[tex]a = 1.91 m/s^2[/tex]
Now substituting into the equation frictional force equation
[tex]f = \frac{2}{3} * 17 * 1.91[/tex]
[tex]f = 21.65 \ N[/tex]
Answer:
[tex]a=-1.92 m/s^{2}[/tex]
[tex]F_{f}=-21.76 N[/tex]
Explanation:
We can use the definition of the torque:
[tex]\tau=I\alpha[/tex]
When:
I is the inertia of a uniform hallow sphere [tex]I = (2/3)mR^{2}[/tex]α is the angular acceleration (α = a/R)Now, torque is the product of the friction force times the radius.
[tex]F_{f}*R=\frac{2}{3}mR^{2}*\frac{a}{R}[/tex]
[tex]F_{f}=\frac{2}{3}ma[/tex] (1)
Now, let's analyze the force acting over the sphere using the Newton's second law.
[tex]F=ma[/tex]
[tex]-mgsin(\theta)-F_{f}=ma[/tex] (2)
Let's put F(f) of the equation (1) into the equation (2):
[tex]-mgsin(\theta)-\frac{2}{3}ma=ma[/tex]
[tex]a=-\frac{3}{5}gsin(\theta)[/tex]
[tex]a=-\frac{3}{5}*9.81*sin(19)[/tex]
[tex]a=-1.92 m/s^{2}[/tex]
Hence: [tex]F_{f}=\frac{2}{3}ma=\frac{2}{3}*17*(-1.92)[/tex]
[tex]F_{f}=-21.76 N[/tex]
I hope it helps you!
A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 16.0 m/s when the hand is 1.90 m above the ground. You may want to review (Pages 49 - 51) . For help with math skills, you may want to review: Quadratic Equations For general problem-solving tips and strategies for this topic, you may want to view a Video Tutor Solution of Time in the air for a tossed ball. Part A How long is the ball in the air before it hits the ground
Answer:
t = 3.38 s
Explanation:
We have,
Initial speed of the ball that leaves the student's hand is 16 m/s
Initially, the hand is 1.90 m above the ground.
It is required to find the time for which the ball in the air before it hits the ground. We can use the equation of kinematics as :
[tex]y_f=y_i+ut+\dfrac{1}{2}at^2[/tex]
Here, [tex]y_f=-1.9\ m, y_i=0[/tex] and a=-g
The equation become:
[tex]-1.9=16t-\dfrac{1}{2}\times 9.8t^2[/tex]
After rearranging we get the above equation as :
[tex]4.9t^2-16t-1.9=0[/tex]
It is a quadratic equation, we need to find the value of t. On solving the above equation, we get :
t = -0.115 s and t = 3.38 s (ignore t = -0.115 s )
So, the ball is in air for 3.38 seconds before it hits the ground.
٠Light bulb A is rated at 60 W and light bulb B is rated at 100 W. Both are designed to operate at 110 V. Which statement is correct?
A-The 60 W bulb has a greater resistance and greater current than the 100 W bulb.
B-The 60 W bulb has a greater resistance and smaller current than the 100 W bulb.
C-We need to know the resistivities of the filaments to answer this question.
D-The 60 W bulb has a smaller resistance and greater current than the 100 W bulb.
F-The 60 W bulb has a smaller resistance and smaller current than the 100 W bulb
Answer:B
Explanation:
Power=p
Voltage=v
Resistance=r
Voltage for both:110v
For 60 watts bulb:
Resistance=v^2/p
Resistance=110^2/60
Resistance=(110x110)/60
Resistance=12100/60
Resistance=201.7 ohms
Current=power/voltage
Current=60/110
Current=0.55 amperes
For 100watts bulb:
Resistance=v^2/p
Resistance=110^2/100
Resistance=(110 x 110)/100
Resistance=(12100)/100
Resistance =121 ohms
Current=power/voltage
Current=100/110
Current=0.91
The 60W bulb has greater resistance and smaller current than the 100W bulb, based on calculations from the power, voltage and Ohm's Law.
Explanation:In the case of light bulbs, power (P) is given by the formula P = IV, where I is the current and V is the voltage. Bulb A is rated at 60W and Bulb B at 100W, both functioning at a voltage (V) of 110V. To find the current (I) for each bulb, you would divide the power (P) by the voltage (V). This shows that Bulb B, with a higher wattage, has a greater current.
The resistance (R) of the bulbs can be found using Ohm's Law, which states R = V/I. This shows that Bulb A, with a smaller current, has a greater resistance.
Therefore, option B-'The 60W bulb has a greater resistance and smaller current than the 100 W bulb' is the correct statement.
Learn more about Electricity and Resistance here:https://brainly.com/question/33719176
#SPJ2
A merry-go-round spins freely when Diego moves quickly to the center along a radius of the merry-go-round. As he does this, it is true to say that Group of answer choices the moment of inertia of the system increases and the angular speed increases. the moment of inertia of the system increases and the angular speed decreases. the moment of inertia of the system decreases and the angular speed increases. the moment of inertia of the system decreases and the angular speed decreases.
Answer:
The moment of inertia of the system decreases and the angular speed increases.
Explanation:
This very concept might not seem to be interesting at first, but in combination with the law of the conservation of angular momentum, it can be used to describe many fascinating physical phenomena and predict motion in a wide range of situations.
In other words, the moment of inertia for an object describes its resistance to angular acceleration, accounting for the distribution of mass around its axis of rotation.
Therefore, in the course of this action, it is said that the moment of inertia of the system decreases and the angular speed increases.
When Diego moves quickly to the center along a radius of the merry-go-round, the moment of inertia of the system increases and the angular speed decreases.
Explanation:When Diego moves quickly to the center along a radius of the merry-go-round, the moment of inertia of the system increases and the angular speed decreases.
The moment of inertia is a measure of how resistant an object is to changes in its rotation. As Diego moves closer to the center, the distribution of mass in the system changes, resulting in an increase in moment of inertia.
According to the conservation of angular momentum, if the moment of inertia increases, the angular speed must decrease in order to maintain a constant angular momentum.
Learn more about Moment of inertia and angular speed here:https://brainly.com/question/14190791
#SPJ3
Match the descriptions to the feature,
principal axis
Center of the spherical mirror from which a
curved mirror was cut
center of curvature
Distance from the center of a mirror to the
focal point
Line that runs through the center of
curvature to a mirror
focal point
Point where reflected light converges or
appears to diverge
focal length
Point where the principal axis and mirror
vertex
meet
Answer:
Vertex: Point where the principal axis and mirror meet
Focal point: Point we are reflected light converges or appears to diverge
Focal length: distance from the center of a mirror to the focal point
Principal axis: line that runs to the center of curvature to a mirror
Center of curvature: sensor of spherical mirror from which a curved mirror was cut
Explanation:
Just did the assignment on Edge.
A spherical mirror is a mirror which has the shape of a piece cut out of a spherical surface. There are two types of spherical mirrors: concave, and convex.
What is a spherical mirror?A spherical mirror is a mirror which has the shape of a piece cut out of a spherical surface. There are two types of spherical mirrors: concave, and convex. These are illustrated in.
The most commonly occurring examples of concave mirrors are shaving mirrors and makeup mirrors.
As is well-known, these types of mirrors magnify objects placed close to them. The most commonly occurring examples of convex mirrors are the passenger-side wing mirrors of cars.
These types of mirrors have wider fields of view than equivalent flat mirrors, but objects which appear in them generally look smaller
The answers to the given questions are:
Vertex: Point where the principal axis and mirror meet
Focal point: Point we are reflected light converges or appears to diverge
Focal length: distance from the centre of a mirror to the focal point
Principal axis: line that runs to the centre of curvature of a mirror
Center of curvature: sensor of spherical mirror from which a curved mirror was cut
To know more about Spherical mirrors follow
https://brainly.com/question/7512320
Your mental health is a reflection of your...
O
Thoughts
Emotions
Behaviors
O
All of the above
Answer:all of the above
Explanation:
because thoughts relate to your emotions and your behaviors reflect to your emotions
Mental health is a reflection of one's thoughts, emotions, and behaviors. Positive thinking, managed emotions and constructive behaviors contribute to a healthy mental state.
Explanation:Your mental health is indeed a reflection of your thoughts, emotions and behaviors. Accumulative thoughts and feelings can shape our mental well-being to a great extent. Similarly, our behaviors, influenced by these thoughts and feelings, can either nurture or deteriorate our mental health. A balance in all three – positive thinking, managing emotions and constructive behaviors lead to a healthy mental state.
Learn more about Mental Health here:https://brainly.com/question/33043544
#SPJ6
Since sinusoidal waves are cyclical, a particular phase difference between two waves is identical to that phase difference plus a cycle. For example, if two waves have a phase difference of π4, the interference effects would be the same as if the two waves had a phase difference of π4+2π. The complete criterion for constructive interference between two waves is therefore written as follows: phase difference=0+2πnfor any integer n Write the full criterion for destructive interference between two waves
Final answer:
Destructive interference between two waves occurs when their phase difference is [tex]\pi + 2\pi n[/tex], for any integer n, resulting in a minimized or zero resultant amplitude.
Explanation:
The full criterion for destructive interference between two waves is when the phase difference between the waves is [tex]\pi + 2\pi n[/tex] for any integer n. This occurs because two waves are exactly half a wavelength out of phase [tex]\pi[/tex] radians phase difference), resulting in the peaks of one wave aligning with the troughs of the other, canceling each other out. Unlike constructive interference, where the waves reinforce each other leading to increased amplitude, destructive interference leads to a decrease in amplitude, potentially down to zero in the case of perfect destructive interference with identical waves.
Expressed mathematically, for pure destructive interference, the phase difference must fulfill the condition: phase difference = [tex]\pi + 2\pi n[/tex] for any integer n. This means that the path length difference between the two waves must be an odd multiple of half the wavelength. The result is that at the points of destructive interference, the amplitude of the resultant wave is minimized or even becomes zero, eliminating the wave at that point.
The hydraulic oil in a car lift has a density of 8.53 x 102 kg/m3. The weight of the input piston is negligible. The radii of the input piston and output plunger are 5.43 x 10-3 m and 0.135 m, respectively. What input force F is needed to support the 22600-N combined weight of a car and the output plunger, when (a) the bottom surfaces of the piston and plunger are at the same level, and (b) the bottom surface of the output plunger is 1.20 m above that of the input plunger
Answer:
(a) the input force is 36.56 N
(b) the input force is 37.49 N
Explanation:
Given;
density of hydraulic oil, ρ = 8.53 x 10² kg/m³
radius of plunger, r₁ = 0.135 m
radius of piston, r₂ = 5.43 x 10⁻³ m
Part (a) The input force needed to support 22600-N weight, when the bottom surfaces of the piston and plunger are at the same level;
[tex]P =\frac{F}{A}[/tex]
Where;
P is pressure
F is force
A is circular area = πr²
[tex]\frac{F_1}{A_1} =\frac{F_2}{A_2} \\\\F_2 = \frac{F_1*A_2}{A_1} =\frac{F_1* \pi r_2^2}{\pi r_1^2} = \frac{F_1* r_2^2}{ r_1^2} \\\\F_2 = \frac{22600*(5.43*10^{-3})^2 }{(0.135)^2}\\\\F_2 = 36.56 \ N[/tex]
Part (b) The input force needed to support 22600-N weight, when the bottom surface of the output plunger is 1.20 m above that of the input plunger
[tex]P_2 = P_1 + \rho gh[/tex]
But, F = PA and A = πr²
[tex]F_2 = F_1(\frac{A_2}{A_1} ) + \rho gh*A_2\\\\F_2 = F_1(\frac{r_2^2}{r_1^2} )+\rho gh(\pi r_2^2)\\\\F_2 = 22600(\frac{5.43*10^{-3}}{0.135})^2 \ + 853*9.8*1.2*\pi (5.43*10^{-3})^2\\\\F_2=36.56 + 0.93\\\\F_2 = 37.49 \ N[/tex]
The solution of this problem involves the use of Pascal's principle and a concept known as hydrostatic pressure.
First, let's identify the given parameters:
- The density (ρ) of the hydraulic oil is 8.53 x 10ˆ2 kg/m³
- The radius of the input piston is 5.43 x 10ˆ-3 m
- The radius of the output plunger is 0.135 m
- The total weight (W) being lifted is 22600 N
- The acceleration due to gravity (g) is approximately 9.81 m/s²
- The height difference (Δh) between the pistons is 1.2 m
We will use these in our calculations.
We begin by determining the areas of the input piston and the output plunger using the formula for the area of a circle A = πr². We denote the area of the input piston as Ain and the area of the output plunger as Aout.
Let's now focus on when the pistons are at the same level.
Part (a) requires us to find the input force (F) needed to support the weight of the car and output plunger. This is governed by Pascal's principle, which states that for an incompressible, non-viscous fluid in a hydraulic system, pressure is transmitted undiminished throughout the fluid and acts with equal force on all areas. Using Pascal's principle, the force required on the input piston is equal to the weight divided by the ratio of the areas. F = W * (Ain / Aout). After performing the calculation, we find that F equivales to approximately 36.562893827160494 N.
Next, let's turn to when the bottom surface of the output plunger is 1.2 m above the input piston.
Part (b) of the problem asks us to calculate the input force needed in this scenario. We must consider not only Pascal's principle but the additional hydrostatic pressure due to the height difference between the pistons. Firstly, we calculate the hydrostatic pressure difference (ΔP): ΔP = ρ * g * Δh. We then add this pressure difference to the force required when pistons were at the same level. Remember that pressure is force divided by area, thus when we add pressure we need to multiply it by the area of the input piston to maintain the units consistent. F = F_same_level + ΔP * Ain. The result of this calculation yields an input force of approximately 37.49208673165364 N.
In conclusion, the input force required to support the specified weight when the bottom surfaces of the pistons are at the same level is about 36.56 N. This force increases to roughly 37.49 N when the output plunger is pushed 1.2 m above the input piston due to the added hydrostatic pressure.
https://brainly.com/question/20876092
#SPJ11
The ballistic pendulum is an apparatus used to measure the speed of a projectile. An 8.0 g bullet is fired into a 2.5 kg ballistic pendulum bob, which is initially at rest, and becomes embedded in the bob. The pendulum then rises to a vertical distance of 6.0 cm. What was the initial speed of the bullet (in m/s)?
The initial speed of the bullet was determined by utilizing the principles of conservation of energy and momentum. The final velocity of the bullet and bob was first determined from the given height and the known conversion of kinetic to potential energy. This final velocity was then input into the conservation of momentum equation to find the initial speed of the bullet.
Explanation:The ballistic pendulum is a classic example of a problem in physics which can be solved either by using the principles of work and energy or the principles of impulse and momentum. For this problem, we will utilize conservation of momentum. We know that the momentum before the collision is equal to the momentum after the collision.
Therefore, we can create the following equation: (m1×v1) + (m2×v2) = (m1+m2)×V, where m1 and v1 represent the mass and velocity of the bullet, m2 and v2 represent the mass and velocity of the block, and V is the final velocity of the bullet-block system. We know that the block was initially at rest, hence v2 = 0.
After the bob reaches maximum height, there's no kinetic energy (because the speed is zero), so all that original kinetic energy has been converted into potential energy, m×g×h, where m is the total mass (mass of bullet + mass of bob), g is the acceleration due to gravity, and h is the height the bob was raised. From here we can solve for V, then substitute back into the momentum equation to solve for v1, which represents the initial speed of the bullet.
Learn more about Ballistic pendulum here:https://brainly.com/question/35573030
#SPJ12
Using conservation of momentum and energy, the initial speed of an 8.0 g bullet fired into a 2.5 kg ballistic pendulum bob, which rises 6.0 cm, is approximately 339.9 m/s. The key steps involve converting units, applying conservation laws, and solving for the initial speed. Therefore, the bullet's speed is determined to be 339.9 m/s.
For solving this problem, we will use the principle of conservation of momentum and the principle of conservation of energy.
Step-by-Step Solution
Convert the masses into common units:
mass of bullet, [tex]m_{bullet[/tex] = 8.0 g = 0.008 kg
mass of pendulum bob = 2.5 kg.
Convert the rise height into meters:
h = 6.0 cm = 0.06 m.
Determine the final velocity of the combined bullet and pendulum system at the lowest point (right after the collision) using energy conservation:
At maximum height, all kinetic energy is converted to potential energy:
mgh = 0.5 (M + m) v²
Using v = √(2gh) since M includes the bullet:
v = √(2 * 9.8 m/s² * 0.06 m)
v = √(1.176)
v ≈ 1.084 m/s
Use conservation of momentum to find the initial speed of the bullet:
Initial momentum = [tex]m_{bullet[/tex] * [tex]v_{bullet[/tex]
Final momentum = (M + m) * v = (2.5 kg + 0.008 kg) * 1.084 m/s
So, p = (2.508 kg) * 1.084 m/s
p = 2.719 kg·m/s
Initial speed of the bullet:
[tex]v_{bullet[/tex] = p / [tex]m_{bullet[/tex]
[tex]v_{bullet[/tex] = 2.719 kg·m/s / 0.008 kg
[tex]v_{bullet[/tex] ≈ 339.9 m/s
Therefore, the initial speed of the bullet is approximately 339.9 m/s.
How do magnets act when they are near each other?
Answer:
Either repel or attract.
Explanation:
A horizontal platform in the shape of a circular disk rotates on a frictionless bearing about a vertical axle through the center of the disk. The platform has a radius of 2.97 m and a rotational inertia of 358 kg·m2 about the axis of rotation. A 69.5 kg student walks slowly from the rim of the platform toward the center. If the angular speed of the system is 1.96 rad/s when the student starts at the rim, what is the angular speed when she is 1.06 m from the center?
Answer:
4.36 rad/s
Explanation:
Radius of platform r = 2.97 m
rotational inertia I = 358 kg·m^2
Initial angular speed w = 1.96 rad/s
Mass of student m = 69.5 kg
Rotational inertia of student at the rim = mr^2 = 69.5 x 2.97^2 = 613.05 kg.m^2
Therefore initial rotational momentum of system = w( Ip + Is)
= 1.96 x (358 + 613.05)
= 1903.258 kg.rad.m^2/s
When she walks to a radius of 1.06 m
I = mr^2 = 69.5 x 1.06^2 = 78.09 kg·m^2
Rotational momentuem of system = w(358 + 78.09) = 436.09w
Due to conservation of momentum, we equate both momenta
436.09w = 1903.258
w = 4.36 rad/s
Select the three correct parts that make up the cell theory.
A
All living things are composed of cells.
B
All cells are microscopic.
C
Cells are the basic unit of structure and function in animals only.
D
All cells are produced from other cells.
E
Cells are the basic unit of structure and function in all living things.
F
Cells are made of organelles that perform functions necessary for the cell to do its job.
Plsss help me!! :/
Answer:
Options A, D and E....make up cell theory
Rudolf Virchow was the first to propose the concept of Omnis cellula-e cellula in relation to cell division.
What is the cell theory?It is a scientific theory that was proposed in the mid-nineteenth century.
Who proposed it?It was proposed by Theodore Schwann a British Zoologist and Matthias Schleiden a German botanist.
What are the three principles of the cell theory?The three principles are:
All living things are composed of cells.Cells are the basic unit of structure and function in all living things.All cells are produced from pre-existing cells.Hence options A, D, E are correct.
To learn more about cell theory and cells here,
https://brainly.com/question/12129097
#SPJ2
A 12.0 g wad of sticky clay is hurled horizontally at a 100 g wooden block initially at rest on a horizontal surface. The clay sticks to the block. After the impact, the block slides 7.5 m before coming to rest. If the coefficient of friction between the block and the surface is 0.650, what was the speed of the clay immediately before the impact
Answer:
The speed of the clay immediately before the impact is 91.23 m/s
Explanation:
Given;
mass of clay, m₁ = 12g = 0.012 kg
mass of wooden block, m₂ = 100g = 0.1 kg
initial velocity of the wooden block, u₂ = 0
distance moved by the wooden block, d = 7.5 m
coefficient of friction, μk = 0.65
Apply the principle of conservation of linear momentum;
Total momentum before collision = Total momentum after collision
m₁u₁ + m₂u₂ = v(m₁ +m₂)
where;
u₁ is the initial velocity of the clay immediately before the impact
v is the common velocity of clay-block system after impact
u₂ = 0
m₁u₁ = v(m₁ +m₂)
[tex]U_1 = \frac{(m_1 + m_2)V}{m_1}[/tex] ------- equ. (i)
Apply the principle of conservation of energy after the impact
ΔK + ΔU = 0
where;
ΔK is change in kinetic energy
ΔU is change in internal energy of the system due to frictional force
[tex](K_f -K_i) + (F_k*d) = 0\\\\-K_i +F_k*d = 0\\\\K_i = F_k*d \\\\\frac{1}{2} (m_1+m_2)v_i^2 = \mu_k (m_1 +m_2)gd\\\\\frac{1}{2}v_i^2 = \mu_kgd\\\\v_i^2 = 2 \mu_kgd\\\\v_i = \sqrt{2 \mu_kgd}[/tex]
[tex]v_i[/tex] is the common velocity of the clay-block system immediately after the impact, which is equal to V in equation (i)
[tex]U_1 = \frac{(m_1+m_2)V}{m_1} = \frac{(m_1+m_2)v_i}{m_1}\\\\U_1 = \frac{m_1+m_2}{m_1}(\sqrt{2 \mu_kgd})\\\\U_1 = \frac{0.012+0.1}{0.012}(\sqrt{2 *0.65*9.8*7.5})\\\\U_1 = 9.3333(9.77497)\\\\U_1 = 91.23 \ m/s[/tex]
Therefore, the speed of the clay immediately before the impact is 91.23 m/s
A 2 kg object is subjected to three forces that give it an acceleration −→a = =(8.00m/s2 )ˆi + (6.00m/s2 )ˆj. If two of the three forces, are −→F1 = (30.0N)ˆi + (16.0N)ˆj and −→F2 = =(12.0N)ˆi + (8.00N)ˆj, find the third force.
Answer:
The third force = [tex]-(26.0 \, N \, \hat{i} + 12.00 \, N \, \hat{j})[/tex]
Explanation:
Here we note that, the formula for the resultant force is as follows;
[tex]\Sigma F = \Sigma F_x + \Sigma F_y[/tex]
Also
∑F = m×a
Where:
[tex]F_x[/tex] = Force components in the x direction
[tex]F_y[/tex] = Force components in the y direction
∑F = Resultant force vector
m = Mass of the object
a = Acceleration vector ob the object =
[tex]a = 8.00 \, m/s^2 \, \hat{i} + 6.00 \, m/s^2 \, \hat{j}[/tex]
[tex]F_1 = 30.0 \, N \, \hat{i} + 16.0 \, N \, \hat{j}[/tex]
[tex]F_2 = 12.0 \, N \, \hat{i} + 8.00 \, N \, \hat{j}[/tex]
Therefore, since ∑F = m×a, we have;
[tex]\Sigma F = 2 kg \times (8.00 \, m/s^2 \, \hat{i} + 6.00 \, m/s^2 \, \hat{j})[/tex]
[tex]\Sigma F = (16.00 \, m/s^2 \, \hat{i} + 12.00 \, m/s^2 \, \hat{j})[/tex]
Hence from [tex]\Sigma F = \Sigma F_x + \Sigma F_y[/tex], we have;
[tex]F_{x1} + F_{x2} + F_{x3} = 16[/tex]
That is 30 + 12 + [tex]F_{x3}[/tex] = 16
∴ [tex]F_{x3}[/tex] = 16 - (30 + 12) = -26
Similarly,
[tex]F_{y1} + F_{y2} + F_{y3} = 12[/tex]
Therefore, [tex]F_{y3}[/tex] = 12 - (16 + 8) = -12
Hence, [tex]F_3 = -26.0 \, N \, \hat{i} - 12.00 \, N \, \hat{j} = -(26.0 \, N \, \hat{i} + 12.00 \, N \, \hat{j})[/tex]
The third force, [tex]F_3, = -(26.0 \, N \, \hat{i} + 12.00 \, N \, \hat{j})[/tex]
The third force applied on the object calculated using Newton's second law provides a magnitude of (-26.00 N î - 12.00 N ĵ).
Explanation:The subject of this problem is Newton's second law (F=ma), where F is the total force, m is the mass of the object, and a is acceleration. Each of these elements (i, j) represent a vector quantity with both a magnitude and a direction, and in this case, they represent different directions in two-dimensional space. The total acceleration of the object is a vector sum of these accelerations, as is the total force. Therefore, you can calculate the third force by subtracting the first two forces from the total force (which is found by multiplying mass and acceleration).
Total Force (F) = m*a = (2 kg)*(8.00m/s² î + 6.00m/s² ĵ) = 16.00 N î + 12.00 N ĵ.
We know two forces, F1 = 30.00 N î + 16.00 N ĵ and F2 = 12.00 N î + 8.00 N ĵ. Adding them, we get F1 + F2 = 42.00 N î + 24.00 N ĵ.
The third Force (F3) is the total force (F) subtracted from (F1 + F2), thus, F3 = F - (F1 + F2) = (-26.00 N î - 12.00 N ĵ).
Learn more about Newton's second law here:
https://brainly.com/question/13447525
#SPJ3
An alpha particle (a helium nucleus, consisting of two protons and two neutrons) has a radius of approximately 1.6 × 10-15 m. A certain heavy nucleus contains 79 protons in addition to all its neutrons and has a radius of approximately 5.8 × 10-15 m. An alpha particle is shot directly from a large distance at such a resting heavy nucleus.
What is the initial momentum of the alpha particle?
To find the initial momentum of an alpha particle, we can use the equation: momentum = mass x velocity. The mass of an alpha particle is approximately 4 atomic mass units (amu). The velocity can be determined by considering the initial kinetic energy of the alpha particle.
Explanation:The initial momentum of an alpha particle can be calculated using the equation:
momentum = mass x velocity
The mass of an alpha particle is approximately 4 atomic mass units (amu). The velocity of the alpha particle can be determined by considering its initial kinetic energy. Since the alpha particle is shot directly at a resting heavy nucleus, we can assume that its initial kinetic energy is equal to the energy of the system.
Using the equation:
kinetic energy = (1/2) x mass x velocity^2
we can solve for velocity. Then, using the calculated velocity, we can find the momentum of the alpha particle.
Learn more about Initial momentum of an alpha particle here:https://brainly.com/question/14863785
#SPJ3
Determine how fast the length of an edge of a cube is changing at the moment when the length of the edge is 5cm and the volume of the edge is decreasing at the rate of 100cm^3/sec
Answer:
1.333 cm/s
Explanation:
The formula for the volume of the cube V in term of its edge s is:
[tex]V = s^3[/tex]
By using chain rule we have the following equation between the rate of change of the volume and the rate of change of the edge:
[tex]\frac{dV}{dt} = \frac{dV}{ds}\frac{ds}{dt}[/tex]
[tex]100 = \frac{d(s^3)}{ds}\frac{ds}{dt}[/tex]
[tex]100 = 3s^2\frac{ds}{dt}[/tex]
[tex]\frac{ds}{dt} = \frac{100}{3s^2}[/tex]
We can substitute s = 5 cm:
[tex]\frac{ds}{dt} = \frac{100}{3*5^2} = 100 / 75 = 1.333 cm/s[/tex]
Jeff's body contains about 5.31 L of blood that has a density of 1060 kg/m3. Approximately 45.0% (by mass) of the blood is cells and the rest is plasma. The density of blood cells is approximately 1125 kg/m3, and about 1% of the cells are white blood cells, the rest being red blood cells. The red blood cells are about 7.50µm across (modeled as spheres). What is the mass of the blood mblood in Jeff's body?
Answer:
The mass of the blood is 5.6286 kg
Explanation:
Given:
V = volume of blood = 5.31 L = 0.00531 m³
ρ = density = 1060 kg/m³
Question: What is the mass of the blood mblood in Jeff's body, m = ?
To calculate the mass of the blood you just have to multiply the density of the blood by the volume occupied by it:
[tex]m=\rho *V=1060*0.00531=5.6286kg[/tex]
At the bottom, deep bodies of water always measure 4°C, because at that temperature water has its highest [...] (fill in the physical quantity in question)
Answer:density
Explanation:
At the bottom, deep bodies of water always measure [tex]4^{\circ}C[/tex] because at that temperature water has it highest density .
In water bodies like lake warm water rises up due to convection and at higher depths there is cold water, which is found to be at highest density of water.
When water molecules acquired heat energy it become widely spread at the same volume and thus posses low density but at low temperature water molecule occupy less space therefore posses maximum density at [tex]4^{\circ}C[/tex]
A moving van collides with a sports car in a high-speed head-on collision. During the impact, the truck exerts a force Ftruck on the car and the car exerts a force Fcar on the truck. Which of the following statements about these forces is true?A. The force exerted by the truth on the car is the same size as the force exerted by the car on the truck: Ftruck = Fcar B. Ftruck < Fcar C. Ftruck > Fcar
Answer:
A. The force exerted by the truth on the car is the same size as the force exerted by the car on the truck: Ftruck = Fcar
Explanation:
Both vehicles will experience equal both opposite forces one on the other.
Specimen of steel has a rectangular cross section 20 mm wide and 40 mm thick, an elastic modulus of 207 GPa, and a Poisson’s ratio of 0.30. If this specimen is pulled in tension with a force of 60,000 N, what is the change in width if deformation is totally elastic? (A) Increase in width of 3.62 10−6 m (B) Decrease in width of 7.24 10−6 m (C) Increase in width of 7.24 10−6 m (D) Decrease in width of 2.18 10−6 m
Answer:
(D) Decrease in width of 2.18 x [tex]10^{-6[/tex]m
Explanation:
Given:
force 'F'= 60,000 N
elastic modulus 'E' = 207 GPa => 2.07 x [tex]10^{11[/tex]N/m²
cross section area ' [tex]A_{0}[/tex]'= 20 mm x 40 mm => 800mm² =>8 x [tex]10^{-4}[/tex] m²
∈z = б/E => (F/ [tex]A_{0}[/tex])/E => F/ [tex]A_{0}[/tex]E
∈z = 60,000/(8 x [tex]10^{-4}[/tex] x 2.07 x [tex]10^{11[/tex])
∈z =3.62 x [tex]10^{-4}[/tex]
Lateral strain is given by,
∈x= -v∈z => -(0.30)(3.62 x [tex]10^{-4}[/tex])
∈x=1.09 x [tex]10^{-4}[/tex]
Next is to calculate the change in width
ΔW= Wo x ∈x =>20 x 1.09 x [tex]10^{-4}[/tex]
ΔW= -2.18 x [tex]10^{-6[/tex] m
Therefore, the correct option is 'D'
The steel specimen, when pulled in tension, will experience a decrease in width. This decrement is calculated to be about 1.75*10^-7 m, closest to option (D) decrease in width of 2.18 10−6 m.
Explanation:The change in the width of a steel specimen being pulled in tension can be found using the formula for strain in the lateral direction = -Poisson's ratio * (stress/Young's modulus). Here, the stress (force/area) is 60,000N/(20mm*40mm), the Young's modulus is 207 GPa, and the Poisson's ratio is 0.30. Plugging these values in, we get a lateral strain of -8.74*10^-6. The change in width, which is this lateral strain times the original width (20mm), will thus be a decrease of about 1.75*10^-4 mm, or 1.75*10^-7 m. Amongst the given options, this is closest to (D) decrease in width of 2.18 10−6 m.
Learn more about Elastic Deformation here:https://brainly.com/question/31577326
#SPJ3
what is the only part of a motor through which a current does not flow
Answer:
An electric motor
Explanation:
Answer:
the only part it doesn't flow through is an insulator
Explanation:
A conductor is material through which current flows easily; an insulator is material through which current does not flow easily.
A 1.9 kg ball drops vertically onto a floor, hitting with a speed of 25 m/s. It rebounds with an initial speed of 9.0 m/s. (a) What impulse acts on the ball during the contact? (b) If the ball is in contact with the floor for 0.0158 s, what is the magnitude of the average force on the floor from the ball?
Answer:
(a) 64.6 kg/ms
(b) 4088.61 N
Explanation:
(a)
I = mΔv.................. Equation 1
Where I = impulse acting on the ball, Δv = change in velocity.
But,
Δv = v-u.............. Equation 2
Where v = final velocity, u = initial velocity.
Substitute equation 2 into equation 1
I = m(v-u).................. Equation 3
Assuming: upwards to be positive
Given: m = 1.9 kg, v = 9 m/s, u = -25 m/s (downward)
Substitute into equation 3
I = 1.9[9-(25)]
I = 1.9(9+25)
I = 1.9(34)
I = 64.6 kgm/s.
(b)
F = I/t................... Equation 4
Where F = Average force on the floor from the ball, t = time of contact of the floor with the ball.
Given: I = 64.6 kgm/s, t = 0.0158 s
Substitute into equation 4
F = 64.6/0.0158
F = 4088.61 N
Answer:
A) Impulse = 64.6 N.s
B) Force = 4088.6 N
Explanation:
We are given;
Mass of ball;m = 1.9 kg
Initial velocity of ball;u = -25 m/s (negative value because ball was drop downward)
Final velocity of ball;v = 9 m/s
From Newton's second law of motion, we know that;
Impulse = Change in momentum
Thus;
Impulse = final momentum - initial momentum = mv - mu
Thus;
Impulse = m(v - u)
Plugging in the relevant values, we have;
Impulse = 1.9(9 - (-25)
Impulse = 1.9(9 + 25)
Impulse = 1.9(34)
Impulse = 64.6 N.s
B) As said earlier,
impulse = 64.6 N.s
Now, we are looking for the magnitude of the average force.
Now, Impulse is also expressed as Force x time.
Thus,
Force x time = 64.6 N.s
We are given time as 0.0158 s
Thus, making Force the subject of the formula, we now have;
Force = 64.6/0.0158
Force = 4088.6 N
What is the gravitational potential energy of a 2.5kg object that is 300m above the surface of the earth? g=10m/s
Answer:
7350 J
Explanation:
Gravitational Potential Energy: This is defined as the energy possessed by a body due to it's position in the gravitational field. The S.I unit is Joules(J).
Applying,
E.p = mgh..................... Equation 1
Where E.p = Gravitational potential Energy, m = mass of the object, h = height of the object above the surface of the earth, g = acceleration due to gravity.
Given: m = 2.5 kg, h = 300 m
Constant: g = 9.8 m/s²
Substitute these values into equation 1
E.p = 2.5(300)(9.8)
E.p = 7350 J.
To practice Problem-Solving Strategy 16.1 Standing waves. An air-filled pipe is found to have successive harmonics at 800 HzHz , 1120 HzHz , and 1440 HzHz . It is unknown whether harmonics below 800 HzHz and above 1440 HzHz exist in the pipe. What is the length of the pipe
Answer:
Length of the pipe = 53.125 cm
Explanation:
given data
harmonic frequency f1 = 800 Hz
harmonic frequency f2 = 1120 Hz
harmonic frequency f3 = 1440 Hz
solution
first we get here fundamental frequency that is express as
2F = f2 - f1 ...............1
put here value
2F = 1120 - 800
F = 160 Hz
and
Wavelength is express as
Wavelength = Speed ÷ Fundamental frequency ................2
here speed of waves in air = 340 m/s
so put here value
Wavelength =340 ÷ 160
Wavelength = 2.125 m
so
Length of the pipe will be
Length of the pipe = 0.25 × wavelength ......................3
put here value
Length of the pipe = 0.25 × 2.125
Length of the pipe = 0.53125 m
Length of the pipe = 53.125 cm
Final answer:
The length of the pipe, which resonates at odd harmonics and has successive frequencies of 800 Hz, 1120 Hz, and 1440 Hz, is calculated to be approximately 32.2 centimeters. This calculation assumes that the pipe is closed at one end, and uses the relationship between the harmonics and the fundamental frequency.
Explanation:
The student is inquiring about determining the length of a pipe based on the frequencies of its successive harmonics. Since the successive harmonics are at 800 Hz, 1120 Hz, and 1440 Hz, and these frequencies are not direct multiples of each other, it suggests that we're dealing with a pipe that is closed at one end. Such pipes produce odd harmonics only. The given frequencies thus correspond to the fundamental (first harmonic), the third harmonic, and the fifth harmonic, respectively.
The frequency of the nth harmonic in a pipe closed at one end is given by:
fn = n(f1), where n is an odd integer and f1 is the fundamental frequency, which in this case is 800 Hz. So, the third harmonic would be 3(800 Hz) = 2400 Hz, which is incorrect given our second frequency is 1120 Hz. The provided frequencies imply that 800 Hz is, in fact, the third harmonic (800 Hz = 3f1). Hence, the fundamental frequency (f1) is 800 Hz / 3 = 266.67 Hz.
The wavelength (λ1) of the fundamental frequency in a tube closed at one end is given by λ1 = 4L. Using the formula for frequency (
f = v / λ), where v is the velocity of sound in air (approximately 343 m/s), we can calculate the length of the pipe by rearranging it to L = v / (4f1).
Therefore, L = 343 m/s / (4 * 266.67 Hz) = 0.322 meters or 32.2 cm.
Polly is pushing a box across the floor with a force of 30 N. The force of gravity is -8 N, and the normal force is 8 N. Which
value could describe the force of friction if Polly could not move the box?
-30 N
8N
8N
30 N
Mark this and retum
Save and Exit
Submit
Answer:
Force of friction is (-30 N).
Explanation:
The force applied on the box across the floor is 30 N.
The force of gravity is (-8 N) and the the normal force is 8 N.
It is based on Newton's third law of motion. Newton's third law of motion states that the force acting on object 1 to object 2 is equal in magnitude of the force from object 2 to 1 but in opposite direction.
Here there is force of 30 N is applied in horizontal direction. The frictional force act in opposite direction. So, the force of friction is -30 N so that box across the floor.
If the amount of work done on a book was 10 J and the force required to move the book was 2.5 N, what was the distance the book was moved?
Answer:
Explanation:4meters
Work=10J
Force=2.5N
Distance=work ➗ force
Distance=10 ➗ 2.5
Distance=4meter
Final answer:
Given that the work done on a book was 10 J and the force required to move the book was 2.5 N, we use the formula Work = Force x Distance to find that the book was moved a distance of 4 meters.
Explanation:
The question asks to calculate the distance a book was moved given that the work done on the book was 10 J and the force required to move the book was 2.5 N.
To find the distance, we can use the formula for work, which is:
Work (W) = Force (F) * Distance (d)
From the formula, we can solve for distance (d) by rearranging the equation:
Distance (d) = Work (W) / Force (F)
Plugging in the given values:
d = 10 J / 2.5 N
This calculation yields:
d = 4 m
Therefore, the book was moved a distance of 4 meters.