Answer:
d
Explanation:
Ya gon find the Kenitic Energy first
K=½mv²===> K=½×0.2×(0.2)²===> 0.1(0.04)===> 0.004
and now the replacement:
0.004=½×0.4V²====> v²=0.02===> V=0.14m/s
A ball is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s. How long will it take to get to the top of its trajectory? A. 3 seconds B. 4 seconds C. 2 seconds D. 6 seconds
Answer:
A (3 seconds)
Explanation:
Well here we have a type of motion called projectile motion and it is pretty similar to an upside down parabola. The top of the trajectory is the vertex of the parabola and is also when v=0.
Lets identify our givens.
Givens:
Horizontal speed= 30m/s
Vertical Speed= 30 m/s
Since the ball is in freefall after being launched ay=-g(take up to be positive) and ax=0
The ball is launched from the ground so y0=0
Final vertical velocity= 0
This problem is now relatively easy because we only need to find the vertical distance so we can ignore horizontal speed and use
vy=vy0+ayt
Plug in our givens
0=30-10t
solve for t
t=3 seconds
A 0.060 kg ball hits the ground with a speed of –32 m/s. The ball is in contact with the ground for 45 milliseconds and the ground exerts a +55 N force on the ball.
What is the magnitude of the velocity after it hits the ground?
Answer:
9.25 m/s
Explanation:
At an airport, luggage is unloaded from a plane into the three cars of a luggage carrier, as the drawing shows. The acceleration of the carrier is 0.12 mls2, and friction is negligible. The coupling bars have negligible mass. By how much would the tension in each of the coupling bars A, B, and C change if 39 kg of luggage were removed from car 2 and placed in (a) car I and (b) car 3
Answer:
a) ΔT₁ = -4.68 N, ΔT₂ = 4.68 N, b) ΔT₂ = 4.68 N, ΔT₁ = 4.68 N
Explanation:
In this exercise we will use Newton's second law.
∑F = m a
Let's start with the set of three cars
F_total = M a
F_total = M 0.12
where the total mass is the sum of the mass of each charge
M = m₁ + m₂ + m₃
This is the force with which the three cars are pulled.
Now let's write this law for each vehicle
car 1
F_total - T₁ = m₁ a
T₁ = F_total - m₁ a
car 2
T₁ - T₂ = m₂ a
T₂ = T₁ - m₂ a
car 3
T₂ = m₃ a
note that tensions are forces of action and reaction
a) They tell us that 39 kg is removed from car 2 and placed on car 1
m₂’= m₂ - 39
m₁'= m₁ + 39
m₃ ’= m₃
they ask how much each tension varies, let's rewrite Newton's equations
The total force does not change since the mass of the set is the same F_total ’= F_total
car 1
F_total ’- T₁ ’= m₁’ a
T₁ ’= F_total - m₁’ a
T₁ ’= (F_total - m₁ a) - 39 a
T₁ '= T₁ - 39 0.12
ΔT₁ = -4.68 N
car 2
T₁’- T₂ ’= m₂’ a
T₂ ’= T₁’- m₂’ a
T₂ '= (T₁'- m₂ a) + 39 a
T₂ '= T₂ + 39 0.12
ΔT₂ = 4.68 N
b) in this case the masses remain
m₁ '= m₁
m₂ ’= m₂ - 39
m₃ ’= m₃ + 39
we write Newton's equations
car 3
T₂ '= m₃' a
T₂ ’= (m₃ + 39) a
T₂ '= m₃ a + 39 a
T₂ '= T₂ + 39 0.12
ΔT₂ = 4.68 N
car 1
F_total - T₁ ’= m₁’ a
T₁ ’= F_total - m₁ a
car 2
T₁' -T₂ '= m₂' a
T₁ ’= T₂’- m₂’ a
T₁ '= (T₂'- m₂ a) + 39 a
T₁ '= T₁ + 39 0.12
ΔT₁ = 4.68 N
The tension in each of the coupling bars A, B, and C of the luggage carrier changes as,
When luggage were removed from car 2 and placed in car 1, the tension is A and C does not change and the tension in B is decreased by 4.68 N.When luggage were removed from car 2 and placed in car 3, the tension is A and B does not change and the tension in C is increased by 4.68 N.What is tension force?Tension is the pulling force carried by the flexible mediums like ropes, cables and string.
Tension in a body due to the weight of the hanging body is the net force acting on the body.
At an airport, luggage is unloaded from a plane into the three cars of a luggage carrier, as the drawing shows. The acceleration of the carrier is 0.12 m/s², and friction is negligible.
The acceleration is the same, Tension due to the horizontal component of the forces for car 1, 2 and 3 can be given as,
[tex]\sum F_{1h}=T_A-T_B=m_1a\\\sum F_{2h}=T_B-T_C=m_2a\\\sum F_{3h}=T_C=m_3a[/tex]
On solving the above 3 equation, we get the values of tension in each bar as,
[tex]T_A=(m_1+m_2+m_3)a\\T_B=(m_3+m_2)a\\T_C=m_3a[/tex]
Case 1- When 39 kg of luggage were removed from car 2 and placed in car IThe tension is A and C does not change for this case. The acceleration of the carrier is 0.12 m/s². Thus, the change in tension is B is,
[tex]\Delta T_B=39\times0.12\\\Delta T_B=4.68\rm \;N[/tex]
Case 2- When 39 kg of luggage were removed from car 2 and placed in car IIIThe tension is A and B does not change for this case. The acceleration of the carrier is 0.12 m/s². Thus, the change in tension is B is,
[tex]\Delta T_C=39\times0.12\\\Delta T_C=4.68\rm \;N[/tex]
Hence, the tension in each of the coupling bars A, B, and C of the luggage carrier changes as,
When luggage were removed from car 2 and placed in car 1, the tension is A and C does not change and the tension in B is decreased by 4.68 N.When luggage were removed from car 2 and placed in car 3, the tension is A and B does not change and the tension in C is increased by 4.68 N.
Learn more about the tension here;
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PLEASE HELP ME WITH THIS ONE QUESTION
The half-life of Barium-139 is 4.96 x 10^3 seconds. A sample contains 3.21 x 10^17 nuclei. What is the decay constant for this decay?
Explanation:
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A baseball of mass 0.145 kg is thrown at a speed of 40.0 m/s. The batter strikes the ball with a force of 15,000 N; the bat and ball are in contact for 0.500 ms. The force is exactly opposite to the original direction of the ball. Determine the final speed of the ball.
The final speed of the ball is 91.72 m/s.
Given the following data:
Mass of baseball = 0.145 kgInitial speed = 40.0 m/sForce = 15,000 NewtonTime = 0.500 milliseconds (ms) to seconds = 0.0005 seconds.To find the final speed of the ball, we would use the following formula:
[tex]F = \frac{M(V - U)}{t}[/tex]
Where:
F is the force applied. u is the initial speed. v is the final speed. t is the time measured in seconds.Substituting the parameters into the formula, we have;
[tex]15000 = \frac{0.145(V \;- \;40)}{0.0005}\\\\15000(0.0005) = 0.145(V \;- \;40)\\\\7.5 = 0.145V - 5.8\\\\0.145V = 7.5 + 5.8\\\\0.145V = 13.3\\\\V = \frac{13.3}{0.145}[/tex]
Final speed, V = 91.72 m/s
Therefore, the final speed of the ball is 91.72 m/s.
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Strategies for good health management involve:
A Avoiding stressful situations that may cause depression or moodiness insomnia, or lack motivation.
B) Denying, ignoring, or repressing feelings or problems, so that you don't have to face them.
Eating your favorite foods, imagining yourself working out (mind is power), sleeping a few hours a day, so as to make
the most of party time.
D Eating healthy, maintaining and ideal weight, resting, exercising, and establishing healthy relationships.
Answer:
D
Explanation:
This is a great way to manage health.
A would be avoiding everything which isnt good.
B. would be emotionally draining and damaging to bottle feelings and ignore them.
C. is unhealthy to not exercise and eat food while doing nothing.
How are Newton’s 1 and 2 law related?
A 2.0 kg wood block is launched up a wooden ramp that is inclined at a 30˚ angle. The block’s initial speed is 10 m/s. What vertical height does the block reach above its starting point? Use the coefficients μk=0.20 andμs=0.50.
Answer:
The wood block reaches a height of 4.249 meters above its starting point.
Explanation:
The block represents a non-conservative system, since friction between wood block and the ramp is dissipating energy. The final height that block can reach is determined by Principle of Energy Conservation and Work-Energy Theorem. Let suppose that initial height has a value of zero and please notice that maximum height reached by the block is when its speed is zero.
[tex]\frac{1}{2}\cdot m \cdot v^{2} = m \cdot g\cdot h + \mu_{k}\cdot m\cdot g\cdot s \cdot \sin \theta[/tex]
[tex]\frac{1}{2}\cdot v^{2} = g\cdot h + \mu_{k}\cdot g\cdot \left(\frac{h}{\sin \theta} \right)\cdot \sin \theta[/tex]
[tex]\frac{1}{2}\cdot v^{2} = g\cdot h +\mu_{k}\cdot g\cdot h[/tex]
[tex]\frac{1}{2}\cdot v^{2} = (1 +\mu_{k})\cdot g\cdot h[/tex]
[tex]h = \frac{v^{2}}{2\cdot (1 + \mu_{k})\cdot g}[/tex] (1)
Where:
[tex]h[/tex] - Maximum height of the wood block, in meters.
[tex]v[/tex] - Initial speed of the block, in meters per second.
[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, no unit.
[tex]g[/tex] - Gravitational acceleration, in meters per square second.
[tex]m[/tex] - Mass, in kilograms.
[tex]s[/tex] - Distance travelled by the wood block along the wooden ramp, in meters.
[tex]\theta[/tex] - Inclination of the wooden ramp, in sexagesimal degrees.
If we know that [tex]v = 10\,\frac{m}{s}[/tex], [tex]\mu_{k} = 0.20[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the height reached by the block above its starting point is:
[tex]h = \frac{\left(10\,\frac{m}{s} \right)^{2}}{2\cdot (1+0.20)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]
[tex]h = 4.249\,m[/tex]
The wood block reaches a height of 4.249 meters above its starting point.
190 students sit in an auditorium listening to a physics lecture. Because they are thinking hard, each is using 125 W of metabolic power, slightly more than they would use at rest. An air conditioner with a COP of 5.0 is being used to keep the room at a constant temperature. What minimum electric power must be used to operate the air conditioner?
Answer:
W = 4.75 KW
Explanation:
First, we will calculate the heat to be removed:
Q = (No. of students)(Metabolic Power of Each Student)
Q = (190)(125 W)
Q = 23750 W = 23.75 KW
Now the formula of COP is:
[tex]COP = \frac{Q}{W}\\\\W = \frac{Q}{COP}\\\\W = \frac{23.75\ KW}{5}\\\\[/tex]
W = 4.75 KW
A bullet with mass 5.35 g is fired horizontally into a 2.174-kg block attached to a horizontal spring. The spring has a constant 6.17 102 N/m and reaches a maximum compression of 6.34 cm.
(a) Find the initial speed of the bullet-block system.
(b) Find the speed of the bullet.
Answer:
a)[tex]V=1.067\: m/s[/tex]
b)[tex]v=434.65\: m/s [/tex]
Explanation:
a)
Using the conservation of energy between the moment when the bullet hit the block and the maximum compression of the spring.
[tex]\frac{1}{2}MV^{2}=\frac{1}{2}k\Delta x^{2}[/tex]
Where:
M is the bullet-block mass (0.00535 kg + 2.174 kg = 2.17935 kg)V is the speed of the systemk is the spring constant (6.17*10² N/m)Δx is the compression of the spring (0.0634 m)Then, let's find the initial speed of the bullet-block system.
[tex]V^{2}=\frac{k\Delta x^{2}}{M}[/tex]
[tex]V=\sqrt{\frac{6.17*10^{2}*0.0634^{2}}{2.17935}}[/tex]
[tex]V=1.067\: m/s[/tex]
b)
Using the conservation of momentum we can find the velocity of the bullet.
[tex]mv=MV[/tex]
[tex]v=\frac{MV}{m}[/tex]
[tex]v=\frac{2.17935*1.067}{0.00535}[/tex]
[tex]v=434.65\: m/s [/tex]
I hope it helps you!
A gymnast falls from a height onto a trampoline. For a moment, both the gymnast’s kinetic energy and gravitational potential energy are zero. How is the gymnast’s mechanical energy stored for that moment? Question 12 options: rest energy chemical energy elastic energy thermal energy
Answer:
elastic energy
Explanation:
When a gymnast falls on a trampoline from a height, after coming in contact with the trampoline, both the gymnast and the trampoline start to move down due to the elastic property of the trampoline.
During this stretching of the trampoline there comes a maximum point up to which the trampoline is stretched. At this point, both the kinetic energy and the gravitational potential energy of the gymnast are zero due to zero speed and zero height, respectively.
The only energy stored in the gymnast's body at this point is the elastic potential energy due to stretching of the trampoline. Hence,the correct option is:
elastic energy
While an object near the earths surface is in free fall, its
A) velocity increases
B) acceleration increases
Answer:
a
Explanation:
The rate of change of an object's location with relation to a reference point is its velocity, which is dependent on time. when an object is dropped from space at rest (t = 0) under the influence of gravity, the velocity of the object changes and increases with time while the acceleration decreases.
А pressure gauge with a measurement range of 0-10 bar has a quoted inaccuracy of £1.0% f.s. (+1% of full-scale reading). (a) What is the maximum measurement error expected for this instrument? (b) What is the likely measurement error expressed as a percentage of the or reading if this pressure gauge is measuring a pressure of 1 bar?
Answer:
I am not able to answer this question please don't mind...Explanation:
please marks me as brainliests...When Peter tosses an egg against a sagging sheet, the egg doesn't break due to
A) reduced impulse.
B) reduced momentum.
C) both of these
D) neither of these
It has to do with impulse or force. Just how the sheet has no volume. There is no sufficient impulse to crack the shell.
What is force?A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.
The sagging sheet gives the impact with the egg additional time, which prevents the egg from breaking when it is hurled against it. This lessens the force the egg would have applied to the wall had it been flung at it.
It has to do with impulse or force. Just how the sheet has no volume. There is no sufficient impulse to crack the shell.
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A uniformly dense solid disk with a mass of 4 kg and a radius of 4 m is free to rotate around an axis that passes through the center of the disk and perpendicular to the plane of the disk. The rotational kinetic energy of the disk is increasing at 21 J/s. If the disk starts from rest through what angular displacement (in rad) will it have rotated after 3.3 s?
Answer:
3.44 rad
Explanation:
The rotational kinetic energy change of the disk is given by ΔK = 1/2I(ω² - ω₀²) where I = rotational inertia of solid sphere = MR²/2 where m = mass of solid disk = 4 kg and R = radius of solid disk = 4 m, ω₀ = initial angular speed of disk = 0 rad/s (since it starts from rest) and ω = final angular speed of disk
Since the kinetic energy is increasing at a rate of 21 J/s, the increase in kinetic energy in 3.3 s is ΔK = 21 J/s × 3.3 s = 69.3 J
So, ΔK = 1/2I(ω² - ω₀²)
Since ω₀ = 0 rad/s
ΔK = 1/2I(ω² - 0)
ΔK = 1/2Iω²
ΔK = 1/2(MR²/2)ω²
ΔK = MR²ω²/4
ω² = (4ΔK/MR²)
ω = √(4ΔK/MR²)
ω = 2√(ΔK/MR²)
Substituting the values of the variables into the equation, we have
ω = 2√(ΔK/MR²)
ω = 2√(69.3 J/( 4 kg × (4 m)²))
ω = 2√(69.3 J/[ 4 kg × 16 m²])
ω = 2√(69.3 J/64 kgm²)
ω = 2√(1.083 J/kgm²)
ω = 2 × 1.041 rad/s
ω = 2.082 rad/s
The angular displacement θ is gotten from
θ = ω₀t + 1/2αt² where ω₀ = initial angular speed = 0 rad/s (since it starts from rest), t = time of rotation = 3.3 s and α = angular acceleration = (ω - ω₀)/t = (2.082 rad/s - 0 rad/s)/3.3 s = 2.082 rad/s ÷ 3.3 s = 0.631 rad/s²
Substituting the values of the variables into the equation, we have
θ = ω₀t + 1/2αt²
θ = 0 rad/s × 3.3 s + 1/2 × 0.631 rad/s² (3.3 s)²
θ = 0 rad + 1/2 × 0.631 rad/s² × 10.89 s²
θ = 1/2 × 6.87159 rad
θ = 3.436 rad
θ ≅ 3.44 rad
A 1,200kg roller coaster car starts rolling up a slope at a speed of 15m/s. What is the highest point it could reach
Answer: 11.36 m
Explanation:
Given
Mass of roller coaster is m=1200 kg
Initial speed of roller coaster is v=15 m/s
Energy at bottom and at the top is same i.e.
[tex]\Rightarrow \dfrac{1}{2}mv^2=mgh\\\\\Rightarrow \dfrac{1}{2}\times 1200\times 15^2=1200\times 9.8\times h\\\\\Rightarrow h=\dfrac{15^2}{2\times 9.8}\\\\\Rightarrow h=11.36\ m[/tex]
Thus, the highest point reach by the roller coaster is 11.36 m
Answer:
11.36m
Explanation:
~~~~NEED HELP ASAP~~~~
Block A slides into block B along a frictionless surface. They are moving in the direction from left o the right.
Block A= 3kg
Block B= 4kg
Block A velocity before collision =30m/s.
Block B velocity before collision = 15 m/s
The velocity of block B after the collision is 20m/s.
a.) What is the velocity of block A after collision?
b.) Is the collision elastic? Show work to explain answer why or why not.
Answer:
Block A velocity is 23.33 m/s and the collission is not elastic.
Explanation:
a) m1v1 + m2v2 = m1v1' + m2v2'
Plug in givens
90+60=3v1'+80
solve for v1'= 23.33m/s
b) Find the initial and final kinetic energy of Block B
Ki= 1/2(4)(15)^2 + 1/2(3)(30)^2 = 1800 J
Kf= 1/2(4)(20)^2 + 1/2(3)*(23.33)^2= 1616.433J
Since Ki does not equal Kf the collision is not elastic
A resident of a lunar colony needs to have her blood pressure checked in one of her legs. Assume that we express the systemic blood pressure as we do on earth and that the density of blood does not change. Suppose also that normal blood pressure on the moon is still 120/80 (which may not actually be true).
Required:
If a lunar colonizer has her blood pressure taken at a point on her ankle that is 1.5 m below her heart, what will be her systemic blood-pressure reading, expressed in the standard way, if she has normal blood pressure? The acceleration due to gravity on the moon is 1.67 m/s^2
Answer:
The pressure is 2505 Pa.
Explanation:
Height, h = 1.5 m
density of blood, d = 1000 kg/cubic meter
Gravity, g = 1.67 m/s^2
let the pressure is P.
The pressure due to the fluid is given by
P = h d g
P = 1.5 x 1000 x 1.67
P = 2505 Pa
Q 26.12: Assume current flows in a cylindrical conductor in such a way that the current density increases linearly with radius, from zero at the center to 1.0 A/m2 at the surface of the conductor. If the conductor has a cross sectional area of 1.0 m2, what can you say about the current in this conductor
Answer:
The current is 0.67 A.
Explanation:
Density, J = 1 A/m^2
Area, A = 1 m^2
Let the radius is r. And outer is R.
Use the formula of current density
[tex]I = \int J dA = \int J 2\pi r dr\\\\I = \int_{0}^{R}\frac{2\pi r^2}{R} dr\\\\I = \frac{2 \pi R^2}{3}.... (1)Now A = \pi R^2\\\\1 =\pi R^2\\\\R^2 = \frac{1}{\pi}\\\\So, \\\\I = \frac{2\pi}{3}\times \frac{1}{\pi}\\\\I = 0.67 A[/tex]
A rope, under a tension of 221 N and fixed at both ends, oscillates in a second-harmonic standing wave pattern. The displacement of the rope is given by y = (0.10 m)(sin πx/2) sin 12πt, where x = 0 at one end of the rope, x is in meters, and t is in seconds.
What are:
a. the length of the rope.
b. the speed of the waves on the rope
c. the mass of the rope
d. If the rope oscillates in a third-harmonic standing wave pattern, what will be the period of oscillation.
Answer:
sup qwertyasdfghjk
Explanation:
Distillation is the separation of multiple Choose... components based on their different Choose... . As the mixture is heated and the first component Choose... , its Choose... form travels through the distillation set-up and Choose... into a different container.
Answer:
Explanation:
Distillation is the separation of multiple LIQUID components based on their different BOILING POINT. As the mixture is heated and the first component SEPARATES, its PURE form travels through the distillation set-up and GOES into a different container
The block in the drawing has dimensions L0×2L0×3L0,where L0 =0.2 m. The block has a thermal conductivity of 150 J/(s·m·C˚). In drawings A, B, and C, heat is conducted through the block in three different directions; in each case the temperature of the warmer surface is 35 ˚C and that of the cooler surface is 16 ˚C Determine the heat that flows in 6 s for each case.
Answer:
1140 J, 6840 J, 10260 J
Explanation:
Lo x 2 Lo x 3 Lo, Lo = 0.2 m, K = 150 J/(s · m · C˚) , T = 35 ˚C, T' = 16 ˚C,
time, t = 6 s
The heat conducted is
[tex]H = \frac{K A (T - T') t}{d}\\\\H = \frac{150\times 3\times 0.2\times 0.2\times (35-16) \times 6}{3\times 0.2}\\\\H = 1140 J[/tex]
The heat conducted is
[tex]H = \frac{K A (T - T') t}{d}\\\\H = \frac{150\times 3\times 0.2\times 2\times0.2\times (35-16) \times 6}{3\times 0.2}\\\\H = 6840 J[/tex]
The heat conducted is
[tex]H = \frac{K A (T - T') t}{d}\\\\H = \frac{150\times 3\times 0.2\times 2\times0.2\times (35-16) \times 6}{2\times 0.2}\\\\H = 10260 J[/tex]
A transverse sine wave with an amplitude of 2.50 mm and a wavelength of 1.80 m travels, from left to right along a long, horizontal stretched string with a speed of 36.0 m s. I Take the origin at the left end of the undisturbed string. At time t = 0 the left end of the string has its maximum upward displacement,
(a) What is the frequency of the wave?
(b) What is the angular frequency of the wave?
(c) What is the wave number of the wave?
(d) What is the function y(x,t) that describes the wave?
(e) What is y(t) for a particle at the left end of the string?
(f) What is y(t) for a particle 1.35 m to the right of the origin?
(g) What is the maximum magnitude of transverse velocity of any particle of the string?
(h) Find the transverse displacement of a particle 1.35 m to the right of the origin at time t = 0.0625 s.
(i) Find the transverse velocity of a particle 1.35 m to the right of the origin at time t = 0.0625 s.
Explanation:
Given that,
Amplitude, A = 2.5 nm
Wavelength,[tex]\lambda=1.8\ m[/tex]
The speed of the wave, v = 36 m/s
At time t = 0 the left end of the string has its maximum upward displacement.
(a) Let f is the frequency. So,
[tex]f=\dfrac{v}{\lambda}\\\\f=\dfrac{36}{1.8}\\\\f=20\ Hz[/tex]
(b) Angular frequency of the wave,
[tex]\omega=2\pi f\\\\=2\pi \times 20\\\\=125.7\ rad/s[/tex]
(c) The wave number of the wave[tex]=\dfrac{1}{\lambda}[/tex]
[tex]=\dfrac{1}{1.8}\\\\=0.56\ m^{-1}[/tex]
What is the total surface charge qint on the interior surface of the conductor (i.e., on the wall of the cavity)
Answer: hello your question is incomplete below is the missing part
A spherical cavity is hollowed out of the interior of a neutral conducting sphere. At the center of the cavity is a point charge, of positive charge q.
answer:
- q
Explanation:
Since the spherical cavity was carved out of a neutral conducting sphere hence the electric field inside this conductor = zero
given that there is a point charge +q at the center of the spherical cavity hence for the electric field inside the conductor to be = zero the total surface charge qint on the wall of the cavity will be -q
Mary and her younger brother Alex decide to ride the carousel at the State Fair. Mary sits on one of the horses in the outer section at a distance of 2.0 m from the center. Alex decides to play it safe and chooses to sit in the inner section at a distance of 1.1 m from the center. The carousel takes 5.8 s to make each complete revolution.
Required:
a. What is Mary's angular speed %u03C9M and tangential speed vM?
b. What is Alex's angular speed %u03C9A and tangential speed vA?
Answer:
you can measure by scale beacause we dont no sorry i cant help u but u can ask me some other Q
Infrared radiation from young stars can pass through the heavy dust clouds surrounding them, allowing astronomers here on Earth to study the earliest stages of star formation, before a star begins to emit visible light. Suppose an infrared telescope is tuned to detect infrared radiation with a frequency of 4.39 THz. Calculate the wavelength of the infrared radiation.
Answer:
[tex]\lambda=6.83\times 10^{-5}\ m[/tex]
Explanation:
Given that,
An infrared telescope is tuned to detect infrared radiation with a frequency of 4.39 THz.
We know that,
1 THz = 10¹² Hz
So,
f = 4.39 × 10¹² Hz
We need to find the wavelength of the infrared radiation.
We know that,
[tex]\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{4.39\times 10^{12}}\\\\=6.83\times 10^{-5}\ m[/tex]
So, the wavelength of the infrared radiation is [tex]6.83\times 10^{-5}\ m[/tex].
as the ball rises the vertical component of it's velocity_____. explain
Answer:
Decreases
Explanation:
because its moving against gravitational attraction and at maximum height its velocity will be and it will decrease until it reaches maximum height and the start to increase again
A room has dimensions of 15 ft by 15 ft by 20 ft contains air with a density of 0.0724 pounds-mass per cubic feet. The weight of air in the room in pounds-force is
Answer:
the weight of the air in pound-force (lb-f) is 325.8 lbf
Explanation:
Given;
dimension of the room, = 15 ft by 15 ft by 20 ft
density of air in the room, ρ = 0.0724 lbm/ft³
The volume of air in the room is calculated as;
Volume = 15 ft x 15 ft x 20 ft = 4,500 ft³
The mass of the air is calculated as;
mass = density x volume
mass = 0.0724 lbm/ft³ x 4,500 ft³
mass = 325.8 lb-m
The weight of the air is calculated as;
Weight = mass x gravity
Weight = 325.8 lb-m x 32.174 ft/s²
Weight = 10482.29 lbm.ft/s²
The weight of the air in pound-force (lb-f) is calculated as;
1 lbf = 32.174 lbm.ft/s²
[tex]Weight =10,482.29\ lbm.ft/s^2\times \frac{1 \ lbf}{32.174 \ lbm.ft/s^2} \\\\Weight = 325.8 \ lbf[/tex]
Therefore, the weight of the air in pound-force (lb-f) is 325.8 lbf
Find a parametric representation for the surface. The plane through the origin that contains the vectors i - j and j - k
Answer:
parametric representation: x = u, y = v - u , z = - v
Explanation:
Given vectors :
i - j , j - k
represent the vector equation of the plane as:
r ( u, v ) = r₀ + ua + vb
where: r₀ = position vector
u and v = real numbers
a and b = nonparallel vectors
expressing the nonparallel vectors as :
a = i -j , b = j - k , r = ( x,y,z ) and r₀ = ( x₀, y₀, z₀ )
hence we can express vector equation of the plane as
r(u,v) = ( x₀ + u, y₀ - u + v, z₀ - v )
Finally the parametric representation of the surface through (0,0,0) i.e. origin = 0
( x, y , z ) = ( x₀ + u, y₀ - u + v, z₀ - v )
x = 0 + u ,
y = 0 - u + v
z = 0 - v
∴ parametric representation: x = u, y = v - u , z = - v
For a research project, a student needs a solenoid that produces an interior magnetic field of 0.0100 T. She decides to use a current of 1.00 A and a wire 0.500 mm in diameter. She winds the solenoid in layers on an insulating form 1.00 cm in diameter and 20.0 cm long.
Determine the number of layers of wire needed. (Round your answer up to the nearest integer.)
Determine the total length of the wire. (Use the integer number of layers and the average layer diameter.)
Answer:
[tex]n=3.8[/tex]
Explanation:
From the question we are told that:
Magnetic Field [tex]B=0.01T[/tex]
Current [tex]I=1.00[/tex]
Wire Diameter [tex]d_w=0.5*10^3m[/tex]
Layers Diameter [tex]d_l=1*10^2m[/tex]
Length [tex]l=0.2m[/tex]
Generally the equation for number of layers is mathematically given by
[tex]n=\frac{Bd_w}{\mu_o I}[/tex]
Where
[tex]Vacuum\ permeability=\mu_0[/tex]
[tex]n= \frac{0.01*0.5*10^3m}{4 \pi *10^{-7}*1 }[/tex]
[tex]n=3.8[/tex]
