(a) The magnitude of the wheels angular acceleration is 0.0088 rad/s².
(b) The magnitude of the tangential acceleration after the maximum operational speed is reached is 0.153 m/s².
Angular acceleration of the wheel
The angular acceleration of the wheel is calculated as follows;
α = ω/t
ω = v/r
α = v/(rt)
α = (2.3)/(17.5 x 15)
α = 0.0088 rad/s²
Tangential acceleration of the wheela = v/t
a = (2.3)/15
a = 0.153 m/s²
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A solenoid that is 93.9 cm long has a cross-sectional area of 17.3 cm2. There are 1270 turns of wire carrying a current of 7.80 A. (a) Calculate the energy density of the magnetic field inside the solenoid. (b) Find the total energy in joules stored in the magnetic field there (neglect end effects).
Answer:
[tex]65.6\ \text{J/m}^3[/tex]
[tex]0.11\ \text{J}[/tex]
Explanation:
B = Magnetic field = [tex]\mu_0 \dfrac{N}{l}I[/tex]
[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi10^{-7}\ \text{H/m}[/tex]
N = Number of turns = 1270
[tex]l[/tex] = Length of solenoid = 93.9 cm = 0.939 m
[tex]I[/tex] = Current = 7.8 A
A = Area of solenoid = [tex]17.3\ \text{cm}^2[/tex]
Energy density of a solenoid is given by
[tex]u_m=\dfrac{B^2}{2\mu_0}\\\Rightarrow u_m=\dfrac{(\mu_0 \dfrac{N}{l}I)^2}{2\mu_0}\\\Rightarrow u_m=\dfrac{\mu_0N^2I^2}{2l^2}\\\Rightarrow u_m=\dfrac{4\pi\times 10^{-7}\times 1230^2\times 7.8^2}{2\times 0.939^2}\\\Rightarrow u_m=65.6\ \text{J/m}^3[/tex]
The energy density of the magnetic field inside the solenoid is [tex]65.6\ \text{J/m}^3[/tex]
Energy is given by
[tex]U_m=u_mAl\\\Rightarrow U_m=65.6\times 17.3\times 10^{-4}\times 0.939\\\Rightarrow U_m=0.11\ \text{J}[/tex]
The total energy in joules stored in the magnetic field is [tex]0.11\ \text{J}[/tex].
What is a transfer of energy called?
A. Displacement
B. Acceleration
C. Work
D. Torque
A 0.70-kg disk with a rotational inertia given by MR 2/2 is free to rotate on a fixed horizontal axis suspended from the ceiling. A string is wrapped around the disk and a 2.0-kg mass hangs from the free end. If the string does not slip then as the mass falls and the cylinder rotates the suspension holding the cylinder pulls up on the mass with a force of______
Answer:
The force will be "9.8 N".
Explanation:
The given values are:
mass,
m = 0.7 kg
M = 2
g = 9.8
Now,
⇒ [tex]\tau = T \alpha[/tex]
then,
⇒ [tex]\frac{1}{2}mR^2(\frac{1}{R}\frac{dv}{dt}) =M(g-a_t)R[/tex]
⇒ [tex]\frac{1}{2}m \ a_t=m(g-a_t)[/tex]
⇒ [tex]a_t=\frac{2g}{(\frac{m}{M} +2)}[/tex]
On substituting the values, we get
⇒ [tex]=\frac{2\times 9.8}{\frac{0.7}{2} +2}[/tex]
⇒ [tex]=8.34 \ m/s[/tex]
hence,
⇒ [tex]T=mg+M(g-a_t)[/tex]
On substituting the values, we get
⇒ [tex]=0.7\times 9.8+2(9.8-8.34)[/tex]
⇒ [tex]=6.86+2(1.46)[/tex]
⇒ [tex]=6.86+2.92[/tex]
⇒ [tex]=9.8 \ N[/tex]
An object, with mass 64 kg and speed 14 m/s relative to an observer, explodes into two pieces, one 2 times as massive as the other; the explosion takes place in deep space. The less massive piece stops relative to the observer. How much kinetic energy is added to the system during the explosion, as measured in the observer's reference frame
Answer:
K_f = 1881.6 J
Explanation:
To solve this exercise, let's start by finding the velocities of the bodies.
We define a system formed by the initial object and its parts, with this the forces during the explosion are internal and the moment is conserved
initial instant. Before the explosion
p₀ = M v₀
final instant. After the explosion
p_f = m₁ v + m₂ 0
the moeoto is preserved
p₀ = p_f
M v₀ = m₁ v
v = [tex]\frac{m_1}{M}[/tex] v₀
in the exercise they indicate that the most massive part has twice the other part
M = m₁ + m₂
M = 2m₂ + m₂ = 3 m₂
m₂ = M / 3
so the most massive part is worth
m₁ = 2 M / 3
we substitute
v = ⅔ v₀
with the speed of each element we can look for the kinetic energy
initial
K₀ = ½ M v₀²
Final
K_f = ½ m₁ v² + 0
K_f = ½ (⅔ M) (⅔ v₀)²
K_f = [tex]\frac{8}{27}[/tex] (½ M v₀²)
K_f = [tex]\frac{8}{27}[/tex] K₀
the energy added to the system is
ΔK = Kf -K₀
ΔK = (8/27 - 1) K₀
ΔK = -0.7 K₀
K_f = K₀ + ΔK
K_f = K₀ (1 -0.7)
K_f = 0.3 K₀
let's calculate
K_f = 0.3 (½ 64 14²)
K_f = 1881.6 J
Please answer this for 15 points please don’t put in a link.
Answer:
c. Double Replacement
Explanation:
As in Double Replacement reaction exchanges the cations (or the anions) of two ionic compounds.
Here, in BaCl2 , Ba has replaced with NO3 to form Ba(NO3)2
and in 2AgNo3 , Ag has replaced with Cl to form 2AgCl.
Which of the following happens to
density as air pressure decreases?
С C
A. Density increases.
B. Density stays the same.
C. Density decreases.
D. There is no correlation between air pressure and
density.
Explanation:
As pressure increases, with temperature constant, density increases. Conversely when temperature increases, with pressure constant, density decreases. Air density will decrease by about 1% for a decrease of 10 hPa in pressure or 3 °C increase in temperature.
types of wave interactions include
Calculate the terminal velocity of
the following nain drops faning
through air (a) one with a diameter
of 0.3cm 6 one with a a diameter
of o. Olm. Take the density of
water to be looo Kym3 and the
eis cosity of air to be ixlos pas.
The buoyancy effect of the air
may be ignored)
Which of the
following
DECREASES
as you go UP a
mountain?
A. climate
B. altitude
C. amount of oxygen
D. buoyancy
Answer:
C. Amount of oxygen
Explanation:
Options A and D are invalid as they aren't affecting factors.
Option B is false as the altitude increases as you go up a mountain.
Option C is true as the air pressure (atmospheric pressure) is inversely proportional to the height/altitude of the mountain.
Find the wavelength of light which is capable of ionizing a hydrogen atom?
Answer:
The correct answer is - 91.4 nm
Explanation:
According to Bohr's model, the minimum wavelength to ionize Hydrogen atom from n= 1 state is expressed as:
(h×c)/λ=13.6eV
here,
h - Planck constant
c - the speed of light
λ - wavelength
Placing the value in the formula for the wavelength
(6.626×10^−34J.s × 3×10^8 m/s)/λ = 13.6 ×1.6 × 10^−19 J
λ≈91.4nm
Thus, the correct answer would be = 91.4 nm
Why is it harder to breathe on a
mountain?
A. The air pressure is so high the lungs can't expand.
B. The air is denser and oxygen can't flow easily into the
lungs.
C. The denser oxygen molecules sink below the
surrounding air.
D. The air is less dense so there are fewer oxygen
molecules.
A scientist notices that an oil slick floating on water when viewed from above has many different colors reflecting off the surface, making it look rainbow-like (an effect known as iridescence). She aims a spectrometer at a particular spot and measures the wavelength to be 750 nm (in air). The index of refraction of water is 1.33.
Part A: The index of refraction of the oil is 1.20. What is the minimum thickness of the oil slick at that spot? t= 313nm
Part B: Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now? t=125nm
Answer:
a) The minimum thickness of the oil slick at the spot is 313 nm
b) the minimum thickness be now will be 125 nm
Explanation:
Given the data in the question;
a) The index of refraction of the oil is 1.20. What is the minimum thickness of the oil slick at that spot?
t[tex]_{min[/tex] = λ/2n
given that; wavelength λ = 750 nm and index of refraction of the oil n = 1.20
we substitute
t[tex]_{min[/tex] = 750 / 2(1.20)
t[tex]_{min[/tex] = 750 / 2.4
t[tex]_{min[/tex] = 312.5 ≈ 313 nm
Therefore, The minimum thickness of the oil slick at the spot is 313 nm
b)
Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now?
minimum thickness of the oil slick at the spot will be;
t[tex]_{min[/tex] = λ/4n
given that; wavelength λ = 750 nm and index of refraction of the oil n = 1.50
we substitute
t[tex]_{min[/tex] = 750 / 4(1.50)
t[tex]_{min[/tex] = 750 / 6
t[tex]_{min[/tex] = 125 nm
Therefore, the minimum thickness be now will be 125 nm
Light containing two different wavelengths passes through a diffraction grating with 1,250 slits/cm. On a screen 17.5 cm from the grating, the third-order maximum of the shorter wavelength falls midway between the central maximum and the first side maximum for the longer wavelength. If the neighboring maxima of the longer wavelength are 8.44 mm apart on the screen, what are the wavelengths in the light
Answer:
[tex]\lambda_s =6.43*10^-4m[/tex]
Explanation:
From the question we are told that:
Diffraction grating [tex]N=1250slits/cm[/tex]
Distance b/w Screen and grating length [tex]d_{sg}=17.5 cm[/tex]
Distance b/w neighboring maxima and Screen [tex]d_{ms}=8.44[/tex]
Generally the equation for grating space is mathematically given by
[tex]d(g)=\frac{1}{N}[/tex]
[tex]d(g)=\frac{100}{1250}[/tex]
[tex]d(g)=0.08[/tex]
Generally the equation for small angle approximation is mathematically given by
[tex]\triangle y=\frac{\lambda d}{L}[/tex]
Therefore for longest wavelength
[tex]\lambda _l=\frac{8.44*10^{-3}*(0.08)}{0.175m}[/tex]
[tex]\lambda _l=3.858*10^{-3}[/tex]
Therefore the third order maximum equation for the shorter wavelength as
[tex]\lambda_s =\frac{1}{6} \lambda_l[/tex]
[tex]\lambda_s =\frac{1}{6} (3.858*10^-^3)[/tex]
[tex]\lambda_s =6.43*10^-4m[/tex]
The wavelengths in the light is given as
[tex]\lambda_s =6.43*10^-4m[/tex]
What is the importance of using locally available resources in creating art?
Answer:
please give me brainlist and follow
Explanation:
Using locally available resources for art help in the preservation of environment. A significant and practical aspects of art is material significance. The items used by artists while making an art piece affects both the form and the material. Every material delivers something special in the creative process.
Can someone tell me anything useful about energy management in the human body?
Answer:
The human body carries out its main functions by consuming food and turning it into usable energy. Immediate energy is supplied to the body in the form of adenosine triphosphate (ATP). Since ATP is the primary source of energy for every body function, other stored
Explanation:
this what teacher explain to us
true or false
The Total electric potential due to two or more charges is equal to the algebraic sum of the potentials due to the individual charges.
Answer:
i guess the answer is false
Scientists have concluded that the uppermost part of the mantle is partially-molten. Which observation helped them reach this conclusion?
Answer:
P and S waves slow down when they reach this layer. The asthenosphere, also known as the magma chamber, is the uppermost component of the mantle. This layer is partially molten and is a ductile zone in a tectonically poor state.
It's almost hard and seismic waves move through the asthenosphere at a slow rate. The fragile lithosphere and the uppermost portion of the asthenosphere are assumed to be rigid.
seismic waves travel more quickly through denser materials and therefore generally travel more quickly with the depth it moves more slowly through a liquid than a solid. Molten areas within the Earth slow down P waves and stop S waves because their shearing motion cannot be transmitted through a liquid. Partially molten areas may slow down the P waves and attenuate or weaken S waves.
hope this helps...
S and P wave slow down and stop in the uppermost part of the mantle. - For this, scientists have concluded that the uppermost part of the mantle is partially-molten.
What is mantle?A planetary body's mantle is a layer that is surrounded by the crust on top and the core underneath. The largest and most substantial layer of a planetary body, mantles are often comprised of rock or ice. Planetary bodies that have undergone density differentiation typically have mantles. Mantles are found on all terrestrial planets (including Earth), many asteroids, and a few planetary moons.
Between the crust and the outer core, there is a silicate rock layer known as the Earth's mantle. Despite being mostly solid, it behaves like a viscous fluid over geological time. Oceanic crust is created by the partial melting of the mantle at mid-ocean ridges, and continental crust is created by the partial melting of the mantle at subduction zones.
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What voltage would be measured across the 15 ohm resistor?
A)
2.5 volts
B)
5.0 volts
C)
7.5 volts
D)
10 volts
Answer:
7.5 volts
Explanation:
I did it on USA Testprep
A hand dryer blows heated air downwards out of the exit duct at a velocity of 4 m/s. The temperature and density of the ambient air at the inlet are 15 C and 1.23 kg/m3, while at the outlet it has temperature 35 C and density 1.15 kg/m3 The blower power is 10.0 W and the heater power is 715 W. Consider the inlet to be at the large mass of ambient air which has negligible velocity.
a) What is the pressure at the outlet? 4 m/s, 35 C
b) You will be applying the energy equation. Why can you ignore any height differences in this situation?
c) If the specific heat of air C-1000 J/(kg K), where Δυ-C Δ T, find the change in internal energy per unit mass from the inlet to outlet.
d) Find the mass flow rate through the dryer.
e) What is the power loss in the system?
f) What is the loss in the system?
g) What is the head loss in the system?
h) What is the total loss coefficient of the system, referred to the outlet velocity?
i) If there were no heater, would the temperature of gas at the outlet be higher, the same, or lower than the inlet? Explain why.
Answer:
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Explanation:
isos
What is the order of the events for the water cycle on a typical warm day?
А
rain, snow, sleet
B
precipitation, evaporation, rain
с
evaporation, condensation, precipitation
D
condensation, evaporation, precipitation
A 64.0 cm long cord is vibrating in such a manner that it forms a standing wave with two antinodes. (The cord is fixed at both ends.) Which harmonic does this wave represent
Answer:
the wave represents the second harmonic.
Explanation:
Given;
length of the cord, L = 64 cm
The first harmonic of a cord fixed at both ends is given as;
[tex]f_o = \frac{V}{2L}[/tex]
The wavelength of a standing wave with two antinodes is calculated as follows;
L = N---> A -----> N + N ----> A -----> N
Where;
N is node
A is antinode
L = N---> A -----> N + N ----> A -----> N = λ/2 + λ/2
L = λ
The harmonic is calculated as;
[tex]f = \frac{V}{\lambda} \\\\f = \frac{V}{L} = 2(\frac{V}{2L} ) = 2(f_o) = 2^{nd} \ harmonic[/tex]
Therefore, the wave represents the second harmonic.
L = λ
At noon, ship A is 110 km west of ship B. Ship A is sailing east at 20 km/h and ship B is sailing north at 15 km/h. How fast is the distance between the ships changing at 4:00 PM
Answer:
[tex]4.47\ \text{km/h}[/tex]
Explanation:
[tex]\dfrac{da}{dt}[/tex] = Rate at which the distance between A and starting point of B is changing = -20 km/h
[tex]\dfrac{db}{dt}[/tex] = Rate at which the distance of B is changing = 15 km/h
[tex]\dfrac{dc}{dt}[/tex] = Rate at which the distance between A and B is changing
Time after which the rate at which the distance between A and B is changing is 4 hours
Distance covered by A in 4 hours = [tex]20\times 4=80\ \text{km}[/tex]
a = Distance remaining to the start point of B = [tex]110-80=30\ \text{km}[/tex]
b = Distance covered by B in 4 hours = [tex]15\times 4=60\ \text{km}[/tex]
Distance between A and B after 4 hours
[tex]c=\sqrt{a^2+b^2}\\\Rightarrow c=\sqrt{30^2+60^2}\\\Rightarrow c=67.08\ \text{km}[/tex]
[tex]c^2=a^2+b^2[/tex]
Differentiating with respect to time we get
[tex]c\dfrac{dc}{dt}=a\dfrac{da}{dt}+b\dfrac{db}{dt}\\\Rightarrow \dfrac{dc}{dt}=\dfrac{a\dfrac{da}{dt}+b\dfrac{db}{dt}}{c}\\\Rightarrow \dfrac{dc}{dt}=\dfrac{30\times -20+60\times 15}{67.08}\\\Rightarrow \dfrac{dc}{dt}=4.47\ \text{km/h}[/tex]
The rate at which the distance between the ships is changing at 4 PM is [tex]4.47\ \text{km/h}[/tex].
How many gallons of water does it take to produce the following:
a. Cheeseburger
b. Pound of butter
c. A pair of jeans
Answer:
a. 660 gallons
b.665 gallons
c. 1,800
a disk of a radius 50 cm rotates at a constant rate of 100 rpm. what distance in meters will a point on the outside rim travel during 30 seconds of rotation?
Each minute, the disk completes 100 revolutions, so a point on the rim traverses a distance of 100 times the circumference of the disk and would have a linear speed of
100 rev/min
= (100 rev/min) × (2π × 50 cm/rev) × (1/100 m/cm) × (1/60 min/s)
= 5π/3 m/s ≈ 5.236 m/s
Then after 30 s of rotation, the point would have traveled a distance of
(5π/3 m/s) × (30 s) = 50π m ≈ 157.08 m
Which device converts electric energy into mechanical energy?
O A. An electromagnet
O B. A motor
O C. A transformer
O D. A generator
Answer:
B motor
Explanation:
One end of a meter stick is pinned to a table, so the stick can rotate freely in a plane parallel to the tabletop. Two forces, both parallel to the tabletop, are applied to the stick in such a way that the net torque is zero. The first force has a magnitude of 2.00 N and is applied perpendicular to the length of the stick at the free end. The second force has a magnitude of 6.00 N and acts at a 42.9o angle with respect to the length of the stick. Where along the stick is the 6.00-N force applied? Express this distance with respect to the end of the stick that is pinned.
Answer:
x = 0.455 L
Explanation:
For this exercise we must use the rotational equilibrium condition
Σ τ = 0
it has two forces, the first is perpendicular to the rod, so its stub is
τ₁ = F₁ L
the second force is applied with an angle, so we can use trigonometry to find its components
sin θ = F_parallel / F₂
cos θ = F_perpendicular / F₂
F_parallel = F₂ sin θ
F _perpendicular = F₂ cos θ
torque is
τ₂ = F_perpendicular x + F_parallel 0
the parallel force is on the rod therefore its distance is zero
we apply the equilibrium equation
τ₁ - τ₂ = 0
F₁ L = F₂ cos θ x
x = [tex]\frac{L}{cos \theta} \ \frac{F_1}{F_2}[/tex]
let's calculate
x = [tex]\frac{L}{cos \ 42.9} \ \frac{2.00}{6.00}[/tex]
x = 0.455 L
would it be m/s or kg?
Answer:
m.s
Explanation:
2. Plastic is a great conductor of charge so it moves quicker.
True
False
Answer:
the answer is false
Explanation:
plastic doesnt conduct anything
A baby carriage is sitting at the top of a hill that is 21 m high. The carriage with the baby weighs 20
kg. The carriage has
energy. Calculate it
Answer:
Energy in carriage (Potential energy) = 4,116 J
Explanation:
Given:
Mass of baby = 20 kg
Height = 21 m
Find:
Energy in carriage (Potential energy)
Computation:
The energy accumulated in an object as a result of its location relative to a neutral level is known as potential energy.
In carriage accumulated energy is potential energy.
Energy in carriage (Potential energy) = mgh
Energy in carriage (Potential energy) = (20)(9.8)(21)
Energy in carriage (Potential energy) = 4,116 J
The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have studied the change in length of the knee extensor tendon in sprinters and nonathletes. They find (on average) that the sprinters' tendons stretch 41 mm , while nonathletes' stretch only 33 mm .
Hello. Your question is incomplete. However, I managed to find it completely on the internet and I realized that you forgot to mention that the question asks you for the maximum energy difference between velovistas and non-athletes, considering that the spring constant for the tendon of the two groups is equal to 33n/mm.
To make this calculation you will need to use Hooke's law, using the formula: ¹/2*K*x², where "K" will be the value of the spring constant for the tendon and "X" will be the value of the sprinter and non-athlete terms.
So for the sprinter we will have the calculation:
¹/2*33*41² -------> 0,5*33*1681 = 27736. 5 Nmm
(To facilitate the calculation, first solve the division of ¹/2 and then multiply 41 by 41, lastly, just multiply all the results.)
For the non-athlete we will have the calculation:
¹/2*33*33² -------> 0,5*33*1089 = 17968. 5 Nmm
(To facilitate the calculation, first solve the division of ¹/2 and then multiply 41 by 41, lastly, just multiply all the results.)
Now, to reach the final result, you only need to subtract the two values presented by the sprinter and the non-athlete.
27736.5 - 17968.5 = 9768 Nmm