Answer:
Work is defined as a force doing a movement, for example, if with a force F we move an object a distance D, the work done is:
W = F*D
(note, the force is causing the movement, the product here is a dot product, this means that if the force and the displacement are perpendiculars, the product is zero)
So the examples where there are not work being done may be:
We do not have any movement:
For example, you can go to a wall in your house and push it really hard.
There is a force, but the wall will not move, so we have D = 0
and W = F*D = F*0 = 0
Because we have no motion.
Another case is where the force and the direction of motion are perpendiculars.
If we have for example a car, moving at a constant speed, and you push it sidewise (perpendicular to the direction of movement) we have a force being applied and movement, but those are in different directions (so the force does not cause the movement) so we dont have work being done.
Two cannonballs are dropped from a second-floor physics lab at height h above the ground. Ball B has four times the mass of ball A. When the balls pass the bottom of a first-floor window at height above the ground, the relation between their kinetic energies, KA and KB, is:_______.A) KA- 4KB B) KA 2KB C) KA KB D) KB 4KA.
Formula of kinetic energy
[tex]e = \frac{1}{2} m {v}^{2} [/tex]
Therefore Kinetic energy of ball B is 4 times more than ball A.
ans is KB=4KA
Consider the previous situation. Under what condition would the acceleration of the center of mass be zero? Keep in mind that F1x and F2x represent the components, of the corresponding forces. Consider the previous situation. Under what condition would the acceleration of the center of mass be zero? Keep in mind that and represent the components, of the corresponding forces. F1x=−F2x F1x=F2x m1=m2 m1≪m2
Answer:
a) m₁ = m₂ F₁ₓ = F₂ₓ
b) m₁ << m₂ F₂ₓ =0
Explanation:
This interesting exercise is unclear your statement, so that in a center of mass system has an acceleration of zero it is necessary that the sum of the forces on each axis is zero, to see this we write Newton's second law
∑ F = m a
for acceleration to be zero implies that the net force is zero.
we must write the expression for the center of mass
[tex]x_{cm}[/tex] = 1 / M (m₁ x₁ + m₂ x₂)
now let's use the derivatives
[tex]a_{cm}[/tex] = d² x_{cm}/dt² = 1 / M (m₁ a₁ + m₂a₂)
where M is the total mass M = m₁ + m₂
so that the acceleration of the center of mass is zero
0 = 1 / M (m₁ a₁ + m₂a₂)
m₁ a₁ = - m₂ a₂
In the case that we have components on the x axis, the modulus of the two forces are equal and their direction is opposite, therefore
F₁ₓ = -F₂ₓ
b)r when the two masses are equal , in the case of a mass greater than the other m₁ << m₂
acm = d2 xcm / dt2 = 1 / M (m1 a1 + m2a2)
so that the acceleration of the center of mass is zero
0 = 1 / M (m1 a1 + m2a2)
m1 a1 = - m 2 a2
with the initial condition, we can despise m₁, therefore
0 = m₂a₂
if we use Newton's second law
F₂ = 0
I tell you that in this case with a very high mass difference the force on the largest mass must be almost zero
An aluminium bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. The hole is 30mm in diameter and is 100mm long. If the modulus of elasticity for the aluminium is 85GN/m2, calculate the total contraction on the bar due to a compressive load of 180kN.
Answer:
[tex]\delta = 0.385\,m[/tex] (Compression)
Explanation:
The aluminium bar is experimenting a compression due to an axial force, that is, a force exerted on the bar in its axial direction. (See attachment for further details) Under the assumption of small strain, the deformation experimented by the bar is equal to:
[tex]\delta = \frac{P\cdot L}{A \cdot E}[/tex]
Where:
[tex]P[/tex] - Load experimented by the bar, measured in newtons.
[tex]L[/tex] - Length of the bar, measured in meters.
[tex]A[/tex] - Cross section area of the bar, measured in square meters.
[tex]E[/tex] - Elasticity module, also known as Young's Module, measured in pascals, that is, newtons per square meter.
The cross section area of the bar is now computed: ([tex]D_{o} = 0.04\,m[/tex], [tex]D_{i} = 0.03\,m[/tex])
[tex]A = \frac{\pi}{4}\cdot (D_{o}^{2}-D_{i}^{2})[/tex]
Where:
[tex]D_{o}[/tex] - Outer diameter, measured in meters.
[tex]D_{i}[/tex] - Inner diameter, measured in meters.
[tex]A = \frac{\pi}{4}\cdot [(0.04\,m)^{2}-(0.03\,m)^{2}][/tex]
[tex]A = 5.498 \times 10^{-4}\,m^{2}[/tex]
The total contraction of the bar due to compresive load is: ([tex]P = -180\times 10^{3}\,N[/tex], [tex]L = 0.1\,m[/tex], [tex]E = 85\times 10^{9}\,Pa[/tex], [tex]A = 5.498 \times 10^{-4}\,m^{2}[/tex]) (Note: The negative sign in the load input means the existence of compressive load)
[tex]\delta = \frac{(-180\times 10^{3}\,N)\cdot (0.1\,m)}{(5.498\times 10^{-4}\,m^{2})\cdot (85\times 10^{9}\,Pa)}[/tex]
[tex]\delta = -3.852\times 10^{-4}\,m[/tex]
[tex]\delta = -0.385\,mm[/tex]
[tex]\delta = 0.385\,m[/tex] (Compression)
What is a major criticism of Maslow's hierarchy of needs? Select one: a. It is subjective. b. It does not take gender differences into account. c. It is humanistic. d. It only accounts for the objective world.
The correct answer is A. It is subjective
Explanation:
In 1943, the recognized psychologist Abraham Maslow proposed a theory to understand and classify human needs. The work of Maslow included five different categories to classify all basic needs, psychological needs, and self-esteem needs; additionally, in this, Maslow proposed individuals need to satisfy the needs of previous levels to satisfy more complex needs. For example, the first level includes physiological needs such as hunger and these are necessary to get to more complex needs such as the need for safety or self-satisfaction.
This hierarchy is still used all around the world to understand human needs; however, it was been widely criticized because the classification itself is related to Maslow's perspective as this was mainly based on Maslow's ideas about needs, which makes the hierarchy subjective. Also, due to its subjectivity, the hierarchy may apply only in some individuals or societies.
A spaceship travels toward the Earth at a speed of 0.97c. The occupants of the ship are standing with their torsos parallel to the direction of travel. According to Earth observers, they are about 0.50 m tall and 0.50 m wide. Calculate what the occupants’ height and width according to the others on the spaceship?
Answer:
Explanation:
We shall apply length contraction einstein's relativistic formula to calculate the length observed by observer on the earth . For the observer , increased length will be observed for an observer on the earth
[tex]L=\frac{.5}{\sqrt{1-(\frac{.97c}{c})^2 } }[/tex]
[tex]L=\frac{.5}{.24}[/tex]
L= 2.05
The length will appear to be 2.05 m . and width will appear to be .5 m to the observer on the spaceship. . It is so because it is length which is moving parallel to the direction of travel. Width will remain unchanged.
Hellppppppppppp please thanks
Point C would the greatest
Consider a situation in which you are moving two point charges such that the potential energy between them decreases. (NOTE: ignore gravity).
This means that you are moving the charges:
a) Closer to each other
b) Farther apart
c) Either A or B
Answer: Option A
Explanation:
The potential energy decreases in the case when the charges are opposite and they attract each other.
In this case there is no external energy required in order to put the charges together.
This is so because the charges are opposite and they will attract each other. Yes, the only condition should be that the charges should be alike.
Example: a negative charge and a positive charge.
Two stones are thrown vertically upward from the ground, one with three times the initial speed of the other. Assume free fall. Part A If the faster stone takes 12.0 s to return to the ground, how long will it take the slower stone to return
Answer:
36s
Explanation:
Let the objects be A and B.
Let the initial velocity of A be U and the initial velocity of B be 3U
The height sustain by A will be;
The final velocity would be zero
V2 = U2-2gH
Hence
0^2= U2 -2gH
H = U^2/2g
Similarly for object B, the height sustain is;
V2 = (3U)^2-2gH
Hence
0^2= 3U^2 -2gH
U2-2gH
Hence
0^2= U2 -2gH
H = 3U^2/2g
By comparism. The object with higher velocity sustains more height and so should fall longer than object A.
Now object A would take;
From V=U+gt as the object falls freely, the initial velocity is zero hence and the final velocity of the object is;
V=10×12=120m/s let g be 10m/S2
Similarly for object B,
The final velocity for B when it's falling it should be 3×that of A
Meaning
3V= gt
t =3V/g = 3× 120/10 = 36s
A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 3.60 m/s. The car is a distance d away. The bear is 39.3 m behind the tourist and running at 5.00 m/s. The tourist reaches the car safely. What is the maximum possible value for d
Answer:
101 meters
Explanation:
Distance traveled by the tourist:
d = 3.60 m/s × t
Distance traveled by the bear:
d + 39.3 m = 5.00 m/s × t
Substitute:
3.6 t + 39.3 = 5 t
39.3 = 1.4 t
t = 28.1
d = 101
If the mass of the ladder is 12.0 kgkg, the mass of the painter is 55.0 kgkg, and the ladder begins to slip at its base when her feet are 70% of the way up the length of the ladder, what is the coefficient of static friction between the ladder and the floor
Answer:
μ = 0.336
Explanation:
We will work on this exercise with the expressions of transactional and rotational equilibrium.
Let's start with rotational balance, for this we set a reference system at the top of the ladder, where it touches the wall and we will assign as positive the anti-clockwise direction of rotation
fr L sin θ - W L / 2 cos θ - W_painter 0.3 L cos θ = 0
fr sin θ - cos θ (W / 2 + 0,3 W_painter) = 0
fr = cotan θ (W / 2 + 0,3 W_painter)
Now let's write the equilibrium translation equation
X axis
F1 - fr = 0
F1 = fr
the friction force has the expression
fr = μ N
Y Axis
N - W - W_painter = 0
N = W + W_painter
we substitute
fr = μ (W + W_painter)
we substitute in the endowment equilibrium equation
μ (W + W_painter) = cotan θ (W / 2 + 0,3 W_painter)
μ = cotan θ (W / 2 + 0,3 W_painter) / (W + W_painter)
we substitute the values they give
μ = cotan θ (12/2 + 0.3 55) / (12 + 55)
μ = cotan θ (22.5 / 67)
μ = cotan tea (0.336)
To finish the problem, we must indicate the angle of the staircase or catcher data to find the angle, if we assume that the angle is tea = 45
cotan 45 = 1 / tan 45 = 1
the result is
μ = 0.336
From the edge of a cliff, a 0.41 kg projectile is launched with an initial kinetic energy of 1430 J. The projectile's maximum upward displacement from the launch point is 150 m. What are the (a) horizontal and (b) vertical components of its launch velocity
Answer:
v₀ₓ = 63.5 m/s
v₀y = 54.2 m/s
Explanation:
First we find the net launch velocity of projectile. For that purpose, we use the formula of kinetic energy:
K.E = (0.5)(mv₀²)
where,
K.E = initial kinetic energy of projectile = 1430 J
m = mass of projectile = 0.41 kg
v₀ = launch velocity of projectile = ?
Therefore,
1430 J = (0.5)(0.41)v₀²
v₀ = √(6975.6 m²/s²)
v₀ = 83.5 m/s
Now, we find the launching angle, by using formula for maximum height of projectile:
h = v₀² Sin²θ/2g
where,
h = height of projectile = 150 m
g = 9.8 m/s²
θ = launch angle
Therefore,
150 m = (83.5 m/s)²Sin²θ/(2)(9.8 m/s²)
Sin θ = √(0.4216)
θ = Sin⁻¹ (0.6493)
θ = 40.5°
Now, we find the components of launch velocity:
x- component = v₀ₓ = v₀Cosθ = (83.5 m/s) Cos(40.5°)
v₀ₓ = 63.5 m/s
y- component = v₀y = v₀Sinθ = (83.5 m/s) Sin(40.5°)
v₀y = 54.2 m/s
An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.Find the current I through the conductor at time 5.0 seconds.
Answer:
I = 4.75 A
Explanation:
To find the current in the wire you use the following relation:
[tex]J=\frac{E}{\rho}[/tex] (1)
E: electric field E(t)=0.0004t2−0.0001t+0.0004
ρ: resistivity of the material = 2.75×10−8 ohm-meters
J: current density
The current density is also given by:
[tex]J=\frac{I}{A}[/tex] (2)
I: current
A: cross area of the wire = π(d/2)^2
d: diameter of the wire = 0.205 cm = 0.00205 m
You replace the equation (2) into the equation (1), and you solve for the current I:
[tex]\frac{I}{A}=\frac{E(t)}{\rho}\\\\I(t)=\frac{AE(t)}{\rho}[/tex]
Next, you replace for all variables:
[tex]I(t)=\frac{\pi (d/2)^2E(t)}{\rho}\\\\I(t)=\frac{\pi(0.00205m/2)^2(0.0004t^2-0.0001t+0.0004)}{2.75*10^{-8}\Omega.m}\\\\I(t)=4.75A[/tex]
hence, the current in the wire is 4.75A
You have two balloons, one that has a spherical core of radius 4 cm and the other that is tubular with a radius of 0.5 cm and a length of 8 cm. Knowing that the force that you can initially apply to trying to expand each balloon is directly proportional to the volume of the balloon, show that the higher initial stress is achievable with the spherical balloon.
Answer:
Explanation:
Given That:
radius of spherical core r₁ = 4cm
radius of tubular r₂ = 0.5cm
length of tubular l = 8cm
Volume of spherical V₁
[tex]=\frac{4}{3} \pi r_1^3[/tex]
[tex]=\frac{4}{3} \pi(4)^3\\\\=\frac{4}{3} \pi 64\\\\=268.1cm^3[/tex]
Volume of tabular V₂
[tex]=\pi r ^2_2h[/tex]
[tex]=\pi(0.5)^2\times 8\\\\ =\pi 90.250\times8\\\\ =\pi 2\\\\=6.283cm^3[/tex]
F ∝ V
[tex]F_1 \propto V_1[/tex] and [tex]F_2 \propto V_2[/tex]
As V₁ is greater than V₂
⇒ F₁ is greater than F₂
F is force
V is volume
This is the required answer
Motion maps for two objects, Y and Z, are shown.
A motion map. The position line is a long black arrow pointing right with x as the reference point at left. Above the line are three dots, each with a vector pointed away from x back to back in a line labeled B. Above B, there are four dots, each with a shorter vector pointing away from x in a line labeled A starting closer to x .
Object Z passes object Y after how many seconds?
2
3
4
5
Answer: it takes 3 seconds (b)
Explanation:
Answer: B. 3
Explanation:
Each black point on the map represents one second. There are three black points with vectors representing Z's movement before Y begins to move.
When light with a wavelength of 225 nm is incident on a certain metal surface, electrons are ejected with a maximum kinetic energy of 2.98 × 10-19 J. Determine the wavelength (in nm) of light that should be used to double the maximum kinetic energy of the electrons ejected from this surface.
Answer:
The new wavelength is 112.5 nm.
Explanation:
It is given that,
When light with a wavelength of 225 nm is incident on a certain metal surface, electrons are ejected with a maximum kinetic energy of 2.98 × 10⁻¹⁹ J. We need to find the wavelength (in nm) of light that should be used to double the maximum kinetic energy of the electrons ejected from this surface.
The energy of incident electron is given by :
[tex]E=\dfrac{hc}{\lambda}[/tex]
New energy is 2 E and new wavelength is [tex]\lambda'[/tex]. So,
[tex]\dfrac{E}{2E}=\dfrac{\lambda'}{\lambda}\\\\\dfrac{1}{2}=\dfrac{\lambda'}{\lambda}\\\\\lambda'=\dfrac{\lambda}{2}\\\\\lambda'=\dfrac{225}{2}\\\\\lambda=112.5\ nm[/tex]
So, the new wavelength is 112.5 nm.
Which of the following is NOT a type of electromagnetic wave?
Seismic waves
Visible light
Radio waves
Microwave
Answer:
Seismic waves
Explanation:
seismic waves are not represented by electromagnetic graphs, nor can they be reflected on an electromagnetic spectrum. Visible light, radio waves, and microwaves are all electromagnetic waves, which are represented by graphs and electromagnetic spectrums.
Answer:
Siesmic waves
Explanation:
How many times can a three-dimensional object that has a radius of 1,000 units fit something with a radius of 10 units inside of it? How many times can something with a radius of 2,000 units fit something with a radius of 1 unit?
Answer:
# _units = 1000
Explanation:
This exercise we can use a direct proportion rule.
If a volume of radius r = 1 is one unit, how many units can fit in a volume of radius 10?
# _units = V₁₀ / V₁
The volume of a body of radius 1 is
V₁ = 4/3 π r₁³
V₁ = 4/3π
the volume of a body of radius r = 10
V₁₀ = 4/3 π r₂³
V10 = 4/3 π 10³
the number of times this content is
#_units = 4/3 π 1000 / (4/3 π 1)
# _units = 1000
Mention 4
applications of thermal expansion
thermak expansion is for cocluding the elctric value of solar panels
Answer:
thermometerscombustion enginesfitting or loosening of metallic partsproviding lift for hot-air balloonsExplanation:
The thermal expansion properties of liquids and metals are used in thermometers of many types. The liquid in a bulb thermometer expands to provide indication on a calibrated scale. Thermal expansion of a metal coil is used in dial thermometers and in thermostats of many kinds.
The thermal expansion of hot gas drives the cylinders or turbines in combustion engines, jet engines, and thermal cycle motors.
Thermal expansion of metal relative to glass can help remove a stuck jar lid. Similarly, machine parts can be expanded by heating to facilitate assembly or disassembly.
The expansion of warmer air in the atmosphere gives rise to updrafts and thermals that can be used by birds and bugs and people for gaining altitude. The expansion of air in the envelope of a hot-air balloon drives its lift as well.
HELP ASAP!
There is a lever with 5 m long. The fulcrum is 2 m from the right end. Each end hangs a box. The whole system is in balance. If the box hung to the right end is 12 kg, then what is the mass of the box hung to the left end?
Answer:
8kg
Explanation:
For the box to be in equilibrium. The clockwise moment ensued by the box on the right should be same as that ensued by the one on the right. Hence :
M ×3 = 12 ×2
M = 24/3 = 8kg
Note mass is used because trying to compute the weight by multiplying by the acceleration of free fall due to gravity on both sides will cancel out.
You have two charges; Q1 and Q2, and you move Q1 such that the potential experienced by Q2 due to Q1 increases.
Gravity should be ignored.
Then, you must be:
a) Moving Q1 further away from Q2.
b) Moving in the opposite direction to that of the field due to Q1
c) Moving Q1 closer to Q2.
d) Moving in the same direction as the field due to Q1.
e) Any of the above
Given that,
First charge = Q₁
Second charge = Q₂
The potential experienced by Q2 due to Q1 increases
We know that,
The electrostatic force between two charges is defined as
[tex]F=\dfrac{kQ_{1}Q_{2}}{r^2}[/tex]
Where,
k = electrostatic constant
[tex]Q_{1}[/tex]= first charge
[tex]Q_{2}[/tex]= second charge
r = distance
According to given data,
The potential experienced by Q₂ due to Q₁ increases.
We know that,
The potential is defined from coulomb's law
[tex]V=\dfrac{Q_{1}}{4\pi\epsilon_{0}r}[/tex]
[tex]V\propto\dfrac{1}{r}[/tex]
If r decrease then V will be increases.
If V decrease then r will be increases.
Since, V is increases then r will decreases that is moving Q₁ closer to Q₂.
Hence, Moving Q₁ closer to Q₂.
(c) is correct option.
John and Tau take their toys down to a nearby pond to play. John has a toy boat made from wood. He places the wooden boat on the pond and it floats. Tau, being a cheeky little boy, wants to make John’s wooden boat sink. So he attaches a piece of lead on one end of the string, and ties the other loose end onto John’s boat and gently places the lead into the water. Determine the smallest amount of lead (mass) that will be enough to sink John’s boat, assuming the specific gravity of wood is 0.5 and the density of the pond water is equal to the density of pure water
Answer:
The smallest amount of lead that needs to be attached to John's boat in order to sink it has to have a mass slightly greater than the mass of the boat.
Explanation:
A body floats in a fluid when its density is less than the density of the fluid.
The body sinks when its density is more than the density of the fluid.
Density of the pond = density of pure water = 1 g/cm³
Specific gravity of an object = (density of object)/(density of water)
0.5 = (density of John's boat)/1
Density of the John's boat = 0.5 g/cm³
And density is given as
Density = (mass/volume)
0.5 = (mass of John's boat)/(volume of John's boat)
Let the volume of John's boat be v and the mass of John's boat be m
0.5 = (m/v)
v = (m/0.5) = 2m
To sink the boat, we need the total mass on the boat to increase the density to a value greater than 1.
Let the minimum mass of lead required for this to be M
The volume of the boat remains the same, but the total mass on the boat is now (m+M)
1 < (m + M)/v
M + m > v
Recall, v = 2m
M + m > 2m
M > 2m - m
M > m
Hence, the smallest amount of lead that needs to be attached to John's boat in order to sink it has to have a mass slightly greater than the mass of the boat.
Hope this Helps!!!!
Two forces are acting on an object as shown in Fig. on the right. What is the magnitude of the resultant force?
A) 47.5 N
B) 185 N
C) 198 N
D) 200 N
Answer:
185 N
Explanation:
Sum of forces in the x direction:
Fₓ = -(80 N cos 75°) + (120 N cos 60°)
Fₓ = 39.3 N
Sum of forces in the y direction:
Fᵧ = (80 N sin 75°) + (120 N sin 60°)
Fᵧ = 181.2 N
The magnitude of the net force is:
F = √(Fₓ² + Fᵧ²)
F = √((39.3 N)² + (181.2 N)²)
F = 185 N
We have that for the Question "Two forces are acting on an object as shown in Fig. on the right. What is the magnitude of the resultant force?" it can be said that the magnitude of the resultant force is
R=200N
From the question we are told
Two forces are acting on an object as shown in Fig. on the right. What is the magnitude of the resultant force?
A) 47.5 N
B) 185 N
C) 198 N
D) 200 N
Generally the equation for the Resultant force is mathematically given as
For x axis resolution
[tex]Fx=80cos75+120cos60\\\\Fx=80.7N[/tex]
For y axis resolution
[tex]Fx=80sin75+120sin60\\\\Fx=181.2N[/tex]
Therefore
[tex]R=\sqrt{80.7^2+181.2N^2}\\\\R=200N[/tex]
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A frog hops at 2.45 m/s a distance of 2.11 m. How long does it take?
Answer:
Explanation:I don't say u must have to mark my ans as brainliest but if it has really helped u plz don't forget to thnk me...
Answer:
0.86s
Explanation:
How long it takes is the time required.
Time = distance /speed
Time =2.11/2.45=0.86s
Bats use sound to sense objects by sending out short ultrasound pulses and listening for the echo off the object. A. Sketch the ultrasound pulse leaving the bat, reflecting off the object and returning to the bat. B. If a stationary bat is 43 m from an object, how much time elapses between when the bat emits the pulse and it hears the echo
Answer:
B. t = 0.250s
Explanation:
A. An image with the sketch of the bat emitting a sound, which reflects on a surface and return to the bat is attached below.
B. In order to calculate the time that the pulse emitted by the bat, return to the bat, you first calculate the time that pulse takes to arrive to the object.
You use the following formula:
[tex]x=vt[/tex] (1)
x: distance to the object = 43m
t: time = ?
v: speed of sound beat = 343 m/s
You solve the equation (1) for t:
[tex]t=\frac{x}{v}=\frac{43m}{343m/s}=0.125s[/tex]
The time on which the bat hears the echo is twice the value of t, that is:
[tex]t'=2(0.125s)=0.250s[/tex]
The time on which bat heart the echo of its sound, from the moment on which bat emitted it, is 0.250s
A very long, straight solenoid with a cross-sectional area of 2.22 cm2 is wound with 85.6 turns of wire per centimeter. Starting at t = 0, the current in the solenoid is increasing according to i(t)=(0.177A/s2)t2. A secondary winding of 5.0 turns encircles the solenoid at its center, such that the secondary winding has the same cross-sectional area as the solenoid. What is the magnitude of the emf induced in the secondary winding at the instant that the current in the solenoid is 3.2A?
Answer:
The induce emf is [tex]\epsilon = 1.7966*10^{-5} V[/tex]
Explanation:
From the question we are told that
The cross-sectional area is [tex]A = 2,22 cm^3 = \frac{2.22}{10000} = 2.22*10^{-4} \ m^2[/tex]
The number of turn is [tex]N = 85.6 \ turns/cm = 85.6 \ \frac{turns }{\frac{1}{100} } = 8560 \ turns / m[/tex]
The starting time is [tex]t_o[/tex] = 0 s
The current increase is [tex]I(t) = (0.177A/s^2) t^2[/tex]
The number of turn of secondary winding is [tex]N_s = 5 \ turn s[/tex]
The current at the solenoid is [tex]I_(t) = 3.2 \ A[/tex]
at [tex]I_(t) = 3.2 \ A[/tex]
[tex]3.2 = 0.177* t^2[/tex]
=> [tex]t = \sqrt{ \frac{3.2}{0.177} }[/tex]
[tex]t = 4.25 s[/tex]
Generally Faraday's law of induction is mathematically represented as
[tex]\epsilon = A\mu_o N_s N * \frac{di}{dt}[/tex]
[tex]\epsilon = A\mu_o N_s N * \frac{d (0.177 t^2)}{dt}[/tex]
[tex]\epsilon = A\mu_o N_s N * (0.177)(2t)[/tex]
substituting values
[tex]\epsilon = (2.22*10^{-4}) * ( 4\pi * 10^{-7}) * 5 * [8560]* 0.177 * 2 * 4.25[/tex]
[tex]\epsilon = 1.7966*10^{-5} V[/tex]
What is the impulse on a car (750 kg) that accelerates from rest to 5.0 m/s in 10 seconds
Explanation:
impulse J = m × (v2-v1) =750 × ( 5 - 0 ) =3750( N×s)
Answer:
3750Ns
Explanation:
Impulse is defined as Force × time
Force = mass × acceleration,
Hence impulse is;
mass × acceleration × time.
From Newton's second law
Force × time = mass × ∆velocity
750× 5 = 3750Ns
∆velocity = Vfinal-Vinitial ; the initial velocity is zero since the body starts from rest.
Two identical point charges q=71.0 pCq=71.0 pC are separated in vacuum by a distance of 2d=29.0 cm.2d=29.0 cm. Calculate the total electric flux ΦΦ through the infinite surface placed at a distance dd from each charge, perpendicular to the line on which the point charges are located.
Answer:
The electric flux at the infinite surface is ZERO
Explanation:
From the question we are told that
The point charge are identical and the value is [tex]q = 71.0 pC = 71 * 10^{-12} \ C[/tex]
The distance of separation is [tex]D = 29.0 \ cm = 0.29 \ m[/tex]
The distance of both from the infinite surface is d
Generally the electric force exerted by each of the charge on the infinite surface is
[tex]\phi = \frac{q}{\epsilon_o}[/tex]
Now given from the question that they are identical, it then means that the electric flux of the first charge on the infinite surface will be nullified by the electric flux of the second charge hence the electric flux at that infinite surface due to this two identical charges is ZERO
A particle initially located at the origin has an acceleration of = 1.00ĵ m/s2 and an initial velocity of i = 6.00î m/s. (a) Find the vector position of the particle at any time t (where t is measured in seconds). ( t î + t2 ĵ) m (b) Find the velocity of the particle at any time t. ( î + t ĵ) m/s (c) Find the coordinates of the particle at t = 4.00 s. x = m y = m (d) Find the speed of the particle at t = 4.00 s.
Answer:
a) d = (6.00 t i ^ + 0.500 t²) m , b) v = (6.00 i ^ + 1.00 t j ^) m / s
c) d = (24.00 i ^ + 8.00 j^ ) m , d) v = (6.00 i ^ + 5 j^ ) m/s
Explanation:
This exercise is about kinematics in two dimensions
a) find the position of the particle on each axis
X axis
Since there is no acceleration on this axis, we can use the relation of uniform motion
v = x / t
x = v t
we substitute
x = 6.00 t
Y Axis
on this axis there is an acceleration and there is no initial speed
y = v₀ t + ½ a t²
y = ½ at t²
we substitute
y = ½ 1.00 t²
y = 0.500 t²
in vector position is
d = x i ^ + y j ^
d = (6.00 t i ^ + 0.500 t²) m
b) x axis
as there is no relate speed is concatenating
vₓ = v₀
vₓ = 6.00 m / s
y Axis
there is an acceleration and the initial speed is zero
[tex]v_{y}[/tex] = v₀ + a t
v_{y} = a t
v_{y} = 1.00 t
the velocity vector is
v = vₓ i ^ + v_{y} j ^
v = (6.00 i ^ + 1.00 t j ^) m / s
c) the coordinates for t = 4 s
d = (6.00 4 i ^ + 0.50 4 2 j⁾
d = (24.00 i ^ + 8.00 j^ ) m
x = 24.0 m
y = 8.00 m
d) the velocity of for t = 4 s
v = (6 i ^ + 1 5 j ^)
v = (6.00 i ^ + 5 j^ ) m/s
(III) A baseball is seen to pass upward by a window with a vertical speed of If the ball was thrown by a person 18 m below on the street, (a) what was its initial speed, (b) what altitude does it reach, (c) when was it thrown, and (d) when does it reach the street again? Giancoli, Douglas C.. Physics (p. 45). Pearson Education. Kindle Edition.
Answer:
Assuming that the vertical speed of the ball is 14 m/s we found the given values:
a) V₀ = 23.4 m/s
b) h = 27.9 m
c) t = 0.96 s
d) t = 4.8 s
Explanation:
a) Assuming that the vertical speed is 14 m/s (founded in the book) the initial speed of the ball can be calculated as follows:
[tex] V_{f}^{2} = V_{0}^{2} - 2gh [/tex]
Where:
[tex]V_{f}[/tex]: is the final speed = 14 m/s
[tex]V_{0}[/tex]: is the initial speed =?
g: is the gravity = 9.81 m/s²
h: is the height = 18 m
[tex] V_{0} = \sqrt{V_{f}^{2} + 2gh} = \sqrt{(14 m/s)^{2} + 2*9.81 m/s^{2}*18 m} = 23.4 m/s [/tex]
b) The maximum height is:
[tex] V_{f}^{2} = V_{0}^{2} - 2gh [/tex]
[tex] h = \frac{V_{0}^{2}}{2g} = \frac{(23. 4 m/s)^{2}}{2*9.81 m/s^{2}} = 27.9 m [/tex]
c) The time can be found using the following equation:
[tex] V_{f} = V_{0} - gt [/tex]
[tex] t = \frac{V_{0} - V_{f}}{g} = \frac{23.4 m/s - 14 m/s}{9.81 m/s^{2}} = 0.96 s [/tex]
d) The flight time is given by:
[tex] t_{v} = \frac{2V_{0}}{g} = \frac{2*23.4 m/s}{9.81 m/s^{2}} = 4.8 s [/tex]
I hope it helps you!
Consider the binding energy of two stable nuclei, one with 60 nucleons and one with 200 nucleons. a. Is the total binding energy of the nucleus with 200 nucleons more than, less than, or equal to the total binding energy of the nucleus with 60 nucleons. Justify your reasoning.
Answer:
The total binding energy of the nucleus with 200 nucleons more than the total binding energy of the nucleus with 60 nucleons
Explanation:
Binding energy can be given by the formula:
Binding energy = Binding energy per nucleon * Number of nucleons
This means that if the binding energy per nucleon = x MeV
Where x is a positive real number
The nucleus with 60 nucleons will have Binding energy = 60x MeV
The nucleus with 200 nucleons will have binding energy = 200x MeV
For a +ve x, 200x > 60x
Binding energy is proportional (directly) to nucleon volume. A further explanation is provided below.
Binding energy
Binding energy involving 200 nucleons would've been greater than 60 nucleons because so many more nucleons result in what seems like a stronger reaction.
It makes absolutely no difference out whether nucleon seems to be a proton as well as a neutron because they just have a similar strong coupling relatively steady.
Thus the response above is correct.
Find out more information about binding energy here:
https://brainly.com/question/23942204