Answer:
remains the same, but the apparent brightness is decreased by a factor of four.
Explanation:
A star is a giant astronomical or celestial object that is comprised of a luminous sphere of plasma, binded together by its own gravitational force.
It is typically made up of two (2) main hot gas, Hydrogen (H) and Helium (He).
The luminosity of a star refers to the total amount of light radiated by the star per second and it is measured in watts (w).
The apparent brightness of a star is a measure of the rate at which radiated energy from a star reaches an observer on Earth per square meter per second.
The apparent brightness of a star is measured in watts per square meter.
If the distance between us (humans) and a star is doubled, with everything else remaining the same, the luminosity remains the same, but the apparent brightness is decreased by a factor of four (4).
Some of the examples of stars are;
- Canopus.
- Sun (closest to the Earth)
- Betelgeuse.
- Antares.
- Vega.
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.
Two masses, each weighing 1.0 × 103 kilograms and moving with the same speed of 12.5 meters/second, are approaching each other. They have a head-on collision and bounce off away from each other. Assuming this is a perfectly elastic collision, what will be the approximate kinetic energy of the system after the collision?
A. 1.6 × 105 joules
B. 2.5 × 105 joules
C. 1.2 × 103 joules
D. 2.5 × 103 joules
Answer:
A. 1.6 × 105 joules
Explanation:
As per the question, the data given in the question is as follows
Number of masses = 2
Each weightage = [tex]1.0 \times 2.3[/tex]
Speed = 12.5 meters/ second
Based on the above information, the approximate kinetic energy after the collision is
A perfectly elastic collision is described as one where the collision does not cause any loss of kinetic energy.
So we sum the kinetic energy of each kind of system which is given below:
Kinetic energy is
[tex]= 0.5(1.0 \times 10^3) (12.5)^2 + 0.5 (1.0 \times 10^3) (12.5 )^2[/tex]
= 156250 J
[tex]= 1.6 \times 10^5 J[/tex]
100picometer is equal to
Answer:
100picometer is equal to:
0.1 NanometreExplanation:
Hope this works !!A cycler leaves home and rides due south for 65 km. She returns home on the same road. What is the cycler's
displacement?
Y
It doesn't matter what direction she travels, or how far, or whether she uses the same road in both directions.
If she ends up at the same place she started from, then her displacement for the ride is zero.
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
An uncharged, nonconducting, hollow sphere of
radius 10.0 cm surrounds a 10.0-uC charge located
at the origin of a Cartesian coordinate system. A drill
with a radius of 1.00 mm is aligned along the z axis,
and a hole is drilled in the sphere. Calculate the elec-
tric flux through the hole.
Answer:
Explanation:
The whole surface of hollow sphere = 4π r²
= 4 x 3.14 x (10 x 10⁻²)²
= 12.56 x 10⁻² m²
Area of the hole ( both side ) = 2 x π r²
= 2 x 3.14 x (10⁻³)²
= 6.28 x 10⁻⁶ m²
flux coming out of given charge at the centre as per Gauss's theorem
= q / ε₀ where q is charge at the centre and ε₀ is permittivity of the medium .
= 10 x 10⁻⁶ / 8.85 x 10⁻¹²
= 1.13 x 10⁶
This flux will pass through the surface of sphere so flux passing through per unit area
= 1.13 x 10⁶ / 12.56 x 10⁻²
= 8.99 x 10⁶ weber per m²
flux through area of hole
= 8.99 x 10⁶ x 6.28 x 10⁻⁶
= 56.45 weber .
A turtle and a rabbit are in a 150 meter race. The rabbit decides to give the turtle a 1 minute head start. The turtle moves at a constant speed of 0.500 m/s through the whole race (in fact the turtle even starts at a velocity of 0.500 m/s as while he was still approaching the starting line he was allowed to continue to keep going without stopping). The rabbit starts the race from rest and accelerates at a rate or 1.50 m/s2 until she reaches her top speed of 10 m/s. She then finishes the race running at a constant speed of 10 m/s. a) What is the turtle’s position when the rabbit starts to run (1 minute into the race)? b) How long does it take the turtle to finish the race? c) How long does it take the rabbit to reach max speed? d) What is the rabbit’s position when she reaches max speed? e) How long does it take the rabbit to finish the race? f) Who won?
Answer:
a) [tex]s_{T} = 30\,m[/tex], b) [tex]t = 5\,min[/tex], c) [tex]\Delta t = 6.667\,s[/tex], d) [tex]\Delta s_{R} = 33.333\,m[/tex], e) [tex]t' = 11.667\,s[/tex], f) The rabbit won the race.
Explanation:
a) As turtle moves at constant speed, its position is determined by the following formula:
[tex]s_{T} = v_{T}\cdot t[/tex]
Where:
[tex]t[/tex] - Time, measured in seconds.
[tex]v_{T}[/tex] - Velocity of the turtle, measured in meters per second.
[tex]s_{T}[/tex] - Position of the turtle, measured in meters.
Then, the position of the turtle when the rabbit starts to run is:
[tex]s_{T} = \left(0.5\,\frac{m}{s} \right)\cdot (60\,s)[/tex]
[tex]s_{T} = 30\,m[/tex]
The position of the turtle when the rabbit starts to run is 30 meters.
b) The time needed for the turtle to finish the race is:
[tex]t = \frac{s_{T}}{v_{T}}[/tex]
[tex]t = \frac{150\,m}{0.5\,\frac{m}{s} }[/tex]
[tex]t = 300\,s[/tex]
[tex]t = 5\,min[/tex]
The time needed for the turtle to finish the race is 5 minutes.
c) As rabbit experiments a constant acceleration until maximum velocity is reached and moves at constant speed afterwards, the time required to reach such speed is:
[tex]v_{R} = v_{o,R} + a_{R}\cdot \Delta t[/tex]
Where:
[tex]v_{R}[/tex] - Final velocity of the rabbit, measured in meters per second.
[tex]v_{o,R}[/tex] - Initial velocity of the rabbit, measured in meters per second.
[tex]a_{R}[/tex] - Acceleration of the rabbit, measured in [tex]\frac{m}{s^{2}}[/tex].
[tex]\Delta t[/tex] - Running time, measured in second.
[tex]\Delta t = \frac{v_{R}-v_{o,R}}{a_{R}}[/tex]
[tex]\Delta t = \frac{10\,\frac{m}{s}-0\,\frac{m}{s}}{1.50\,\frac{m}{s^{2}} }[/tex]
[tex]\Delta t = 6.667\,s[/tex]
The time taken by the rabbit to reach maximum speed is 6.667 s.
d) On the other hand, the position reached by the rabbit when maximum speed is reached is determined by the following equation of motion:
[tex]v_{R}^{2} = v_{o,R}^{2} + 2\cdot a_{R}\cdot \Delta s_{R}[/tex]
[tex]\Delta s_{R} = \frac{v_{R}^{2}-v_{o,R}^{2}}{2\cdot a_{R}}[/tex]
[tex]\Delta s_{R} = \frac{v_{R}^{2}-v_{o,R}^{2}}{2\cdot a_{R}}[/tex]
Where [tex]\Delta s_{R}[/tex] is the travelled distance of the rabbit from rest to maximum speed.
[tex]\Delta s_{R} = \frac{\left(10\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot \left(1.50\,\frac{m}{s^{2}} \right)}[/tex]
[tex]\Delta s_{R} = 33.333\,m[/tex]
The distance travelled by the rabbit from rest to maximum speed is 33.333 meters.
e) The time required for the rabbit to finish the race can be determined by the following expression:
[tex]t' = \frac{\Delta s_{R}}{v_{R}}[/tex]
[tex]t' = \frac{150\,m-33.333\,m}{10\,\frac{m}{s} }[/tex]
[tex]t' = 11.667\,s[/tex]
The time required for the rabbit from rest to maximum speed is 11.667 seconds.
f) The animal with the lowest time wins the race. Now, each running time is determined:
Turtle:
[tex]t_{T} = 300\,s[/tex]
Rabbit:
[tex]t_{R} = 60\,s + 6.667\,s + 11.667\,s[/tex]
[tex]t_{R} = 78.334\,s[/tex]
The rabbit won the race as [tex]t_{R} < t_{T}[/tex].
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.
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
f a curve with a radius of 97 m is properly banked for a car traveling 75 km/h , what must be the coefficient of static friction for a car not to skid when traveling at 100 km/h ?
Answer:
The coefficient of static friction is 0.26
Explanation:
Given;
radius of the road, R = 97 m
banking velocity, V₁ = 75 km/h = 20.83 m/s
velocity of the moving car, V₂ = 100 km/h = 27.78 m/s
Car in a banked circular turn:
[tex]\theta = tan^{-1}(\frac{V_1^2}{gR} )[/tex]
where;
θ is the angle between the horizontal ground and road in which the car move on
[tex]\theta = tan^{-1}(\frac{V_1^2}{gR} ) \\\\\theta = tan^{-1}(\frac{20.83^2}{9.8*97} ) \\\\\theta = 24.5^o[/tex]
During this type of motion, the body acquires some acceleration which tends to retain the circular motion towards its center, known as centripetal acceleration.
There are two components of this acceleration;
Parallel acceleration, [tex]a_|_|[/tex] = a*Cosθ
Perpendicular acceleration, a⊥ = a * Sinθ
Parallel acceleration, [tex]a_|_|[/tex] [tex]= \frac{V^2*Cos \theta}{R}[/tex]
Perpendicular acceleration, a⊥ [tex]= \frac{V^2*Sin \theta}{R}[/tex]
Apply Newton's second law of motion;
sum of perpendicular forces acting on the car;
ma⊥ [tex]= F_N - mg*cos \theta[/tex]
[tex]m(\frac{V^2*Sin \theta}{R} ) = F_N - mg*Cos \theta\\\\F_N = mg*Cos \theta + m(\frac{V^2*Sin \theta}{R} )[/tex] --------equation (1)
sum of parallel forces acting on the car
m[tex]a_|_|[/tex] [tex]= mg*Sin \theta - F_s[/tex]
[tex]m(\frac{V^2*Cos \theta}{R} ) = mg*Sin \theta - F_s\\\\F_s = mg*Sin \theta - m(\frac{V^2*Cos \theta}{R} )[/tex] ---------equation (2)
Coefficient of static friction is given as;
[tex]\mu = \frac{F_s}{F_N}[/tex]
Thus, divide equation (2) by equation (1)
[tex]\frac{F_s}{F_N} = \frac{mg*Sin \theta - m(\frac{V^2*Cos \theta}{R}) }{mg*Cos \theta + m(\frac{V^2*Sin \theta}{R}) } \\\\\frac{F_s}{F_N} = \frac{g*Sin \theta - (\frac{V^2*Cos \theta}{R}) }{g*Cos \theta + (\frac{V^2*Sin \theta}{R}) }[/tex]
V = V₂ = 27.78 m/s
θ = 24.5°
R = 97 m
g = 9.8 m/s²
Substitute in these values and solve for μ
[tex]\frac{F_s}{F_N} = \frac{9.8*Sin(24.5)+ (\frac{27.78^2*Cos (24.5)}{97}) }{9.8*Cos (24.5) + (\frac{27.78^2*Sin (24.5)}{97}) }\\\\\frac{F_s}{F_N} = \frac{4.0641 \ - \ 7.2391}{8.91702 \ + \ 3.299} = -0.26\\\\| \mu| = 0.26[/tex]
Therefore, the coefficient of static friction is 0.26
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 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.
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Point C would the greatest
. An object whose mass is 375 lb falls freely under the influence of gravity from an initial elevation of 253 ft above the surface of Earth. The initial velocity is downward with a magnitude of 10 ft/s. The effect of air resistance is negligible. (a) Determine the speed, in ft/s, of the object just before it strikes the surface of Earth. (b) Calculate the kinetic energy of the object, in Btu, before it strikes the surface of Earth. Assume that the gravitational acceleration is g
Answer:
(a) Vf = 128 ft/s
(b) K.E = 122.8 Btu
Explanation:
(a)
In order to find the velocity of the object just before striking the surface of earth or the final velocity, we use 3rd equation of motion:
2gh = Vf² - Vi²
where,
g = 32.2 ft/s²
h = height = 253 ft
Vf = Final Velocity = ?
Vi = Initial Velocity = 10 ft/s
Therefore,
(2)(32.2 ft/s²)(253 ft) = Vf² - (10 ft/s)²
16293.2 ft²/s² + 100 ft²/s² = Vf²
Vf = √(16393.2 ft²/s²)
Vf = 128 ft/s
(b)
The kinetic energy of the object before it hits the surface of earth is given by:
K.E = (0.5)(m)(Vf)²
where,
m = mass of object = 375 lb
K.E = Kinetic energy of object before it strikes the surface of earth = ?
Therefore,
K.E = (0.5)(375 lb)(128 ft/s)²
K.E = 3073725 lb.ft²/s²
Now, converting this to Btu:
K.E = (3073725 lb.ft²/s²)(1 Btu/25037 lb.ft²/s²)
K.E = 122.8 Btu
A cab driver heads south with a steady speed of V,22.0 m/s tor t 3.00 min, then makes a right turn and travels at v225.0 m/s for t2-2.40 min, and then drives northwest at v 30.0 m/s for 1.00 min. For this 6.40-min trip, calculate the following. Assume +x is In the eastward direction
(a) total vector displacement (Enter the magnitude in n and the direction in degrees south of west.) direvlion o south of west
(b) average speed (in m/s) m/s
(c) average velocity (Enter the magnitude in m/s and the direction in degrees south of west.) rmagnitude direction Im/'s o south of west
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
Which statement best explains why the overall charge on an atom is zero?
ОООО
The positive charge of the neutrons in the nucleus equals the negative charge in the electron cloud.
The positive charge of the protons in the nucleus equals the negative charge in the electron cloud.
The negative charge of the neutrons in the nucleus equals the positive charge in the electron cloud.
The negative charge of the protons in the nucleus equals the positive charge in the electron cloud.
Answer:
B) The positive charge of the protons in the nucleus equals the negative charge in the electron cloud.
Explanation:
For every negative charge of an electron, there is an equal positively charged proton in the nucleus of the atom. This is why the overall charge on an atom is zero.
Answer:
B
Explanation:
This is described by Gauss a scientist.
The positive charge is found in the proton in the nucleus.
The neutron has no charge.
The positive charge radiates in all directions and a counter negative charge ensues.
The equation = [(8.0 )(4.0)]cos(1.0) describes a wave generated by a sonometer. State whether or not the wave is a stationary wave and find its amplitude, with given that = 2.0 cm
Answer:
Yes. the wave is a stationary wave
Amplitude=7.91 cm
Explanation:
We are given that an equation
[tex]y=[8.0sin(4.0x)]cos(1.0t)[/tex]
We have to find that the given wave is a stationary wave or not and find its amplitude at x=2.0 cm
We know that equation of stationary wave
[tex]y(x,t)=2Asin(kx)cos(\omega t)[/tex]
Where
Amplitude=[tex]2Asin(kx)[/tex]
The given equation is of the form of stationary wave equation.
By comparing we get
[tex]2Asin(kx)=8sin(4.0x)[/tex]
Substitute the value of x
Therefore, the amplitude of the wave
Amplitude=[tex]8sin(4.0\times 2.0)=7.91 cm[/tex]
A baton twirler is twirling her aluminum baton in a horizontal circle at a rate of 2.33 revolutions per second. A baton held horizontally. Calculate the induced emf between the center and one end of the baton if the magnetic field of the Earth is 0.500 gauss and is oriented at 14.42 with respect to the horizontal. Assume the baton is 60.1 cm in length.
Answer:
Explanation:
Given that;
horizontal circle at a rate of 2.33 revolutions per second
the magnetic field of the Earth is 0.500 gauss
the baton is 60.1 cm in length.
the magnetic field is oriented at 14.42°
we wil get the area due to rotation of radius of baton is
[tex]\Delta A = \frac{1}{2} \Delta \theta R^2[/tex]
The formula for the induced emf is
[tex]E = \frac{\Delta \phi}{\Delta t}[/tex]
[tex]\phi = \texttt {magnetic flux}[/tex]
[tex]E=\frac{\Delta (BA) }{\Delta t}[/tex]
[tex]=B\frac{\Delta A}{\Delta t}[/tex]
B is the magnetic field strength
substitute
[tex]\texttt {substitute}\ \frac{1}{2} \Delta \theta R^2 \ \ for \Delta A[/tex]
[tex]E=B\frac{(\Delta \theta R^3/2)}{\Delta t} \\\\=\frac{1}{2} BR^2\omega[/tex]
The magnetic field of the earth is oriented at 14.42
[tex]\omega =2.33\\\\L=60.1c,\\\\\theta=14.42\\\\B=0.5[/tex]
we plug in the values in the equation above
so, the induce EMF will be
[tex]E=\frac{1}{2} \times (B\sin \theta)R^2\omega\\\\E=\frac{1}{2} \times (B\sin \theta)(\frac{L}{2} )\omega[/tex]
[tex]=\frac{1}{2} \times0.5gauss\times\frac{0.0001T}{1gauss} \times\sin 14.42\times(\frac{60.1\times10^-^2m}{2} )^2(2.33rev/s)(\frac{2\pi rad}{1rev} )\\\\=2.5\times10^-^5\times0.2490\times0.0903\times14.63982\\\\=2.5\times10^-^5\times0.32917\\\\=8.229\times10^-^6V[/tex]
A thin, metallic spherical shell of radius 0.347 m0.347 m has a total charge of 7.53×10−6 C7.53×10−6 C placed on it. A point charge of 3.65×10−6 C3.65×10−6 C is placed at the center of the shell. What is the electric field magnitude EE a distance 0.795 m0.795 m from the center of the spherical shell?
Answer:
E = 12640.78 N/C
Explanation:
In order to calculate the electric field you can use the Gaussian theorem.
Thus, you have:
[tex]\Phi_E=\frac{Q}{\epsilon_o}[/tex]
ФE: electric flux trough the Gaussian surface
Q: net charge inside the Gaussian surface
εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2
If you take the Gaussian surface as a spherical surface, with radius r, the electric field is parallel to the surface anywhere. Then, you have:
[tex]\Phi_E=EA=E(4\pi r^2)=\frac{Q}{\epsilon_o}\\\\E=\frac{Q}{4\pi \epsilon_o r^2}[/tex]
r can be taken as the distance in which you want to calculate the electric field, that is, 0.795m
Next, you replace the values of the parameters in the last expression, by taking into account that the net charge inside the Gaussian surface is:
[tex]Q=7.53*10^{-6}C+3.65*10^{-6}C=1.115*10^{-5}C[/tex]
Finally, you obtain for E:
[tex]E=\frac{1.118*10^{-5}C}{4\pi (8.85*10^{-12C^2/Nm^2})(0.795m)^2}=12640.78\frac{N}{C}[/tex]
hence, the electric field at 0.795m from the center of the spherical shell is 12640.78 N/C
A stiff wire 32.5 cm long is bent at a right angle in the middle. One section lies along the z axis and the other is along the line y=2x in the xy plane. A current of 25.0 A flows in the wire-down the z axis and out the line in the xy plane. The wire passes through a uniform magnetic field given by B =(0.318i)T.
Determine the magnitude and the direction of the total force on the wire.
Answer: i personally think it would be 789
Explanation:
Hopefully I am right
An airplane flies in a loop (a circular path in a vertical plane) of radius 200 m . The pilot's head always points toward the center of the loop. The speed of the airplane is not constant; the airplane goes slowest at the top of the loop and fastest at the bottom
(a) What is the speed of the airplane at the top of the loop, where the pilot feels weightless?
(b) What is the apparent weight of the pilot at the bottom of the loop, where the speed of the air-plane is 280 km/h? His true weight is 700 N.
Answer:
Explanation:
a ) for the pilot to feel weightless , his weight will provide the centripetal force . The reaction force from airplane will be zero.
mg = mv² / r
g = v²/r
v² = gr
= 9.8 x 200
v = 44.27 m / s .
b )
v = 280 km / h
v = 77.77 m /s
If R be the reaction force of floor of airplane on him
R - mg = mv² / r
R = mg + m v² / r
= 700 + 70 x 77.77² / 200 ( m = mg / g = 700/ 10 = 70 )
= 700 + 2116.86
= 2816.86 N .
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 woman is driving her truck with speed 45.0 mi/h on a horizontal stretch of road. (a) When the road is wet, the coefficient of static friction between the road and the tires is 0.105. Find the minimum stopping distance (in m). m (b) When the road is dry, μs = 0.602. Find the minimum stopping distance (in m).
Answer:
Explanation:
45 mi /h = 45 x 1.6 x 1000 / (60 x 60) m /s
= 20 m /s
Maximum frictional force possible on wet road
= μs x mg where μs is coefficient of static friction and m is mass of body
Applying work energy theorem
work done by friction = kinetic energy of truck
μs x mg x d = 1/2 m v ² where v is velocity of body and d is stopping distance
2xμs x g x d = v²
2 x .105 x 9.8 x d = 20 x 20
d = 194.36 m
b )
In this case
μs = 0.602
Inserting this value in the relation above
2xμs x g x d = v²
2 x .602 x 9.8 x d = 20 x 20
d = 33.9 m .
When the road is wet, the minimum stopping distance is 196.7 m
When the road is dry, the minimum stopping distance is 34.3 m.
The given parameters;
speed of the truck, v = 45 mi/hThe speed of the truck in m/s is calculated as follows;
[tex]v = \frac{45 \ mi}{h} \times \frac{1609.34 \ m}{1 \ mile} \times \frac{1 \ h}{3600 \ s} \\\\v = 20.12 \ m/s[/tex]
The work done by friction is calculated by applying work-energy theorem;
[tex]F_k d = \frac{1}{2} mv^2\\\\\mu_k mg d = \frac{1}{2} mv^2\\\\ 2 \mu_k g d = v^2\\\\d = \frac{v^2}{2\mu_k g} \\\\[/tex]
When the road is wet, the minimum stopping distance is calculated as follows;
[tex]d = \frac{20.12^2}{2 \times 0.105 \times 9.8} \\\\d = 196.7 \ m[/tex]
When the road is dry, the minimum stopping distance is calculated as follows;
[tex]d = \frac{(20.12)^2}{2\times 0.602 \times 9.8} \\\\d = 34.3 \ m[/tex]
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The angular velocity of a flywheel obeys the equation ωz(t)=A+Bt2, where t is in seconds and A and B are constants having numerical values 2.65 (for A) and 1.60 (for B).
Required:
a. What are the units of A and B if ωz is in rad/s?
b. What is the angular acceleration of the wheel at (i) t = 0 and (ii) t = 5.00 s?
c. Through what angle does the flywheel turn during the first 2.00 s?
Answer:
Explanation:
a )
Iz( t ) = A + B t²
Iz( t ) = angular velocity
putting dimensional formula
T⁻¹ = A + Bt²
A = T⁻¹
unit of A is rad s⁻¹
BT² = T⁻¹
B = T⁻³
unit of B is rad s⁻³
b )
Iz( t ) = A + B t²
dIz / dt = 2Bt
angular acceleration = 2Bt
at t = 0
angular acceleration = 0
at t = 5
angular acceleration = 2 x 1.6 x 5
= 16 rad / s²
Iz( t ) = A + B t²
dθ / dt = A + B t²
integrating ,
θ = At + B t³ / 3
when t = 0 , θ = 0
when t = 2
θ = At + B t³ / 3
= 2.65 x 2 + 1.6 x 2³ / 3
= 5.3 + 4.27
= 9.57 rad .
Flywheel turns by 9.57 rad during first 2 s .
(a) The unit of A is rad/s and the unit of B is rad/s³.
(b) The angular acceleration of the flywheel at 0 s is 0 and at 5 s is 16 rad/s²
(c) The angular displacement of the flywheel during the first 2 seconds, is 9.57 rad.
The given parameters;
z(t) = A + Bt²A = 2.65 and B = 1.6The units of A and B if z(t) is in radian per second (rad/s), are calculated as follows;
[tex]z(t)[\frac{rad}{s} ] = A[\frac{rad}{s} ] \ + \ Bt^2[\frac{rad}{s^3} ][/tex]
Thus, the unit of A is rad/s and the unit of B is rad/s³.
The angular acceleration of the flywheel is calculated as follows;
[tex]a = \frac{d\omega }{dt} =2Bt[/tex]
when, t = 0
a = 2(1.6)(0) = 0
when t = 5 s
a = 2(1.6)(5) = 16 rad/s²
The angular displacement of the flywheel during the first 2 seconds, is calculated as follows;
[tex]\theta = \int\limits {z(t)} \, dt\\\\\theta = At \ + \ \frac{Bt^3}{3} \\\\when, \ t = 2\ s;\\\\\theta = (2.65\times 2) \ + \ (\frac{1.6\times 2^3}{3} )\\\\\theta = 9.57 \ rad[/tex]
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A plane accelerates from rest at a constant rate of 5.00 m/s2 along a runway that is 1800 m long. Assume that the plane reaches the required takeoff velocity at the end of the runway. What is the time tTO needed to take off? Express your answer in seconds using three significant figures.
Answer:
26.8 seconds
Explanation:
To solve this problem we have to use 2 kinematics equations: *I can't use subscripts for some reason on here so I am going to use these variables:
v = final velocity
z = initial velocity
x = distance
t = time
a = acceleration
[tex] {v}^{2} = {z}^{2} + 2ax[/tex]
[tex]v = z + at[/tex]
First let's find the final velocity the plane will have at the end of the runway using the first equation:
[tex] {v}^{2} = {0}^{2} + 2(5)(1800)[/tex]
[tex]v = 60 \sqrt{5} [/tex]
Now we can plug this into the second equation to find t:
[tex]60 \sqrt{5} = 0 + 5t[/tex]
[tex]t = 12 \sqrt{5} [/tex]
Then using 3 significant figures we round to 26.8 seconds
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
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