Answer:
[tex]36.44\ \text{kg m/s}\hat{i}[/tex]
Explanation:
I = Moment of inertia of sphere = [tex]\dfrac{2}{5}MR^2[/tex]
M = Mass of sphere = 29 kg
R = Radius of sphere = 0.5 m
T = Time taken for one revolution = 0.5 s
[tex]\omega[/tex] = Angular velocity = [tex]\dfrac{2\pi}{T}[/tex]
[tex]L=I\omega\\\Rightarrow L=\dfrac{2}{5}MR^2\dfrac{2\pi}{T}\\\Rightarrow L=\dfrac{4MR^2\pi}{5T}\\\Rightarrow L=\dfrac{4\times 29\times 0.5^2\pi}{5\times 0.5}\\\Rightarrow L=36.44\ \text{kg m/s}[/tex]
The rotational angular momentum of the sphere is [tex]36.44\ \text{kg m/s}\hat{i}[/tex].
HELP URGENT PLEASE!!!!!!!
Answer:
I think c I dont know sorry if I'm wrong
Which of the following best defines
weather?
A. the expanding or contracting of the atmosphere
B. the measurement of the amount of water vapor in the
atmosphere
C. the condition of the atmosphere at a certain time and
place
Help Resources
D. the average air temperature of a specific region
Answer:
I'd say D
Explanation:
because not all weather happens within the atmosphere, and most weather depends on region (lile if your near the equator or not)
Pls help ASAP
Imagine that Maritans launch a rocket toward the Earth at a great speed. While the
rocket is traveling toward us, it will appear
than it actually is.
O more blue
darker
larger
more red
Answer:
The rocket will appear larger than it actually is
What is a transfer of energy called?
A. Displacement
B. Acceleration
C. Work
D. Torque
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
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.
A water balloon weighing 4.5 N rests on a table. The balloon has an area of 2.6 x 10-3
m² in contact with the table. What pressure does the balloon exert on the table?
Answer:
the pressure the balloon exerts on the table is 1,730.77 N/m²
Explanation:
Given;
weight of the water balloon, F = 4.5 N
area of the balloon, A = 2.6 x 10⁻³ m²
The pressure the balloon exerts on the table is calculated as follows;
[tex]P = \frac{F}{A}[/tex]
substitute the given values and solve for pressure, P;
[tex]P = \frac{4.5}{2.6 \times 10^{-3}} \\\\P = 1,730.77 \ N/m^2[/tex]
Therefore, the pressure the balloon exerts on the table is 1,730.77 N/m²
I’ll mark you as brinlist please help.
Answer:
245 divided by 5.14=47.6653696 or 47.66
Explanation:
An object is projected from a height of 100m above the ground at an angle of 300to the
horizontal with a velocity of 100m/s.
Calculate
(4)
(4)
The maximum height reached above the ground
Time of flight
The velocity and the direction of the object 1 sec before it hit the ground
(4)
Answer:
a) y = 127.6 m, b) 11.9s, c) v = 103.6 m / s, θ’= 326.7º
Explanation:
This is a missile throwing exercise
let's start by breaking down the initial velocity
sin 30 = [tex]\frac{v_{oy} }{v_o}[/tex]
cos 30 = v₀ₓ / v₀
v_{oy} = v₀ go sin 30
v₀ₓ = v₀ cos 30
v_{oy} = 100 sin 30 = 50 m / s
v₀ₓ = 100 cos 30 = 86.6 m / s
a) the maximum height is requested.
At this point the vertical velocity is zero (v_y = 0)
v_y² = [tex]v_{oy}^2[/tex] - 2 g y
0 = v_{oy}^2 - 2g y
y = [tex]\frac{v_{oy}^2 }{2g}[/tex]
y = 50² / (2 9.8)
y = 127.6 m
b) Flight time
this is the time it takes to reach the ground, the reference system for this movement is taken on the ground this is a height of y = 0 m and the body is at an initial height of i = 100m
y = y₀ + v₀ t - ½ g t²
0 = 100 + 50 t - ½ 9.8 t²
we solve the quadratic equation
4.9 t² - 50 t - 100 = 0
t = [tex]\frac{50 \pm \sqrt{50^2 + 4 \ 4.9 \ 100} }{2 \ 4.9}[/tex]
t = [tex]\frac{50 \ \pm \ 66.8}{9.8}[/tex]
t₁ = 11.9 s
t₂ = -8.4 s
flight time is 11.9s
c) The time 1 s before hitting the ground is
t1 = 11.9 -1
t1 = 10.9 s
let's find the vertical speed
v_y =[tex]v_{oy}[/tex] - g t
v_y = 50 - 9.8 10.9
v_y = -56.8 m / s
the negative sign indicates that the direction of the velocity is downward.
On the x-axis there is no acceleration therefore the speed is constant.
Let's use the Pythagorean theorem
v = [tex]\sqrt{v_x^2+v_y^2}[/tex]
v = [tex]\sqrt { 86.6^2 + 56.8^2}[/tex]
v = 103.6 m / s
let's use trigonometry
tan θ = [tex]\frac{v_y}{v_x}[/tex]
θ = tan⁻¹ \frac{v_y}{v_x}
θ = tan⁻¹ (-56.8 / 86.60)
θ = -33.3º
the negative sign indicates that it is measured clockwise from the x-axis
for a counterclockwise measurement
θ’= 360 - θ
θ' = 360 - 33.3
θ’= 326.7º
which particle have a mass of 1 u
Answer:
Explanation:
proton
if the density of a napthalene ball is 0.02kg.what is the mass of the napthalene ball if it has a volume of 100m³
7. A particle of mass 3 kg is held in equilibrium by two light unextensible strings. One string is horizontal, as shown in Figure 7.30. The tension in the horizontal string is PN and the tension in the other string is N. Find a) the value of 0 b) the value of P.
The tension in the strings are 31.47 and 19.25 N respectively.
Mass of the block, m = 3 kg
From the figure, consider the vertical components,
T₁ sin45° + T₂ sin30° = mg
(T₁/√2) + (T₂/2) = 3 x 9.8 = 29.4
Also, consider the horizontal components,
T₁ cos45° = T₂ cos30°
T₁/√2 = T₂ x√3/2
T₁ = T₂ x √3/2 x √2
So,
T₁ = 0.612T₂
Applying in the first equation,
(T₁/√2) + (T₂/2) = 29.4
(0.612T₂/1.414) + 0.5T₂ = 29.4
0.434 T₂ + 0.5 T₂ = 29.4
0.934 T₂ = 29.4
Therefore, the tension,
T₂ = 29.4/0.934
T₂ = 31.47 N
So, the tension,
T₁ = 0.612 T₂
T₁ = 0.612 x 31.47
T₁ = 19.25 N
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A student takes the elevator up to the fourth floor to see her favorite physics instructor. She stands on the floor of the elevator, which is horizontal. Both the student and the elevator are solid objects, and they both accelerate upward at 5.80 m/s2. This acceleration only occurs briefly at the beginning of the ride up. Her mass is 87.0 kg. What is the normal force exerted by the floor of the elevator on the student during her brief acceleration? (Assume the j direction is upward.)
The normal force exerted by the floor of the elevator on the student during her brief acceleration is 1357.2 N.
To determine the normal force exerted by the floor of the elevator on the student during the brief acceleration, we need to analyze the forces acting on the student.
The force of gravity acting on the student can be calculated using the equation F_gravity = m * g, where m is the mass of the student and g is the acceleration due to gravity (approximately 9.8 m/s²). Therefore, F_gravity = 87.0 kg * 9.8 m/s² = 852.6 N.
Since the elevator and the student are accelerating upward, there is an additional upward force acting on the student. This force is equal to the mass of the student multiplied by the acceleration of the elevator, which is given as 5.80 m/s². So, the upward force is F_upward = m * a = 87.0 kg * 5.80 m/s² = 504.6 N.
Now, the normal force exerted by the floor of the elevator on the student is equal to the sum of the gravitational force and the upward force. In this case, the normal force is given by F_normal = F_gravity + F_upward = 852.6 N + 504.6 N = 1357.2 N.
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Review please help.
Answer:
1 and 3
Explanation:
because they are going up from 0
A 4.0 kg block is moving at 5.0 m/s along a horizontal frictionless surface toward and ideal spring that is attached to a wall , After the block collides with the spring, the spring is compressed a maximum distance of 0.68m . what is the speed of the block when the spring is compressed to only one-half of the maximum distance?
A 4.0 kg block is moving at 5.0 m/s along a horizontal frictionless surface toward an ideal spring that is attached to a wall, the maximum speed of the block when the spring is compressed to one-half of the maximum distance is 4.33 m/s
From the conservation of energy; the kinetic energy of the mass is equal to the work done on the spring.
i.e.
[tex]\mathbf{\dfrac{1}{2} mv^2 = \dfrac{1}{2}kx^2_{max}}[/tex]
Given that:
the mass of the block = 4.0 kg the speed at which it is moving = 5.0 m/scompression of the spring = 0.68 m∴
From the equation above, multiplying both sides with 2, we have:
[tex]\mathbf{mv^2 =kx^2_{max}}[/tex]
Making (k) the subject of the formula;
[tex]\mathbf{k = \dfrac{mv^2}{x^2_{max}}}[/tex]
[tex]\mathbf{k = \dfrac{4 \times 5^2}{0.68^2}}[/tex]
k = 216.26 N/m
However, when compressed to one-half of the maximum distance; the speed is computed as follows:
x = 0.68/2 = 0.34 m
∴
[tex]\mathbf{\dfrac{1}{2}mv_o^2 - \dfrac{1}{2}mv^2 = \dfrac{1}{2}kx^2}[/tex]
[tex]\mathbf{m(v_o^2 -v^2) =kx^2}[/tex]
[tex]\mathbf{(v_o^2 -v^2) =\dfrac{kx^2}{m}}[/tex]
[tex]\mathbf{(5^2 -v^2) =\dfrac{216.26 \times 0.34^2}{4.0}}[/tex]
25 - v² = 6.25
25 -6.25 = v²
v² = 18.75
[tex]\mathbf{ v= \sqrt{18.75 }}[/tex]
v = 4.33 m/s
Therefore, we can conclude that the speed of the block when the spring is compressed to only one-half of the maximum distance is 4.33 m/s
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_____is the nitrogen base found only in DNA.
Answer:
thymine
Explanation:
The nitrogen base found only in DNA is known as thymine which is also called 5-methyl uracil as thymine is a derivative of uracil.
light of wavelength 485 nm passes through a single slit of width 8.32 *10^-6m. what is the single between the first (m=1) and second (m=2) interference minima?
Answer:
3.35
Explanation:
Got it on Acellus
The light of wavelength 485 nm passes through a single slit. The single between the first (m=1) and second (m=2) interference minima is 3.36°.
What is diffraction?Diffraction is the phenomenon of bending of waves through obstacles.
Given is the wavelength λ= 485 nm, silt width d = 8.32 *10⁻⁶ m, then the angle θ will be
d sinθ =mλ
for m=1, sin θ₁ = λ/d
for m=2, sin θ₂ = 2λ/d
Substitute the values into both expressions to find the angles,
sin θ₁ = 485 x 10⁻⁹ / 8.32 *10⁻⁶
θ₁ = 3.34°
and sin θ₂ = (2 x 485 x 10⁻⁹ )/ 8.32 *10⁻⁶
θ₂ = 6.7°
The angle between m =1 and m=2 will be
θ₂ -θ₁ = 6.7° - 3.34° =3.36°
Thus, angle between the first (m=1) and second (m=2) interference minima is 3.36°.
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Please help me easy question just get mixed up.
Answer:
The mechanical energy of the ball is 112.667 J
Explanation:
Mechanical energy = kinetic energy + potential energy
Where:
kinetic energy = [tex]\frac{1}{2}[/tex]m[tex]v^{2}[/tex]
potential energy = mgh
So that for the given ball, its mechanical energy = [tex]\frac{1}{2}[/tex]m[tex]v^{2}[/tex] + mgh
= m([tex]\frac{1}{2}[/tex][tex]v^{2}[/tex] + gh)
But, m = 400 g = 0.4 kg, v = 23.0 m/s, g = 9.81 m/[tex]s^{2}[/tex] and h = 1.75 m.
Mechanical energy of the ball = 0.4([tex]\frac{1}{2}[/tex]([tex]23^{2}[/tex]) + 1.75*9.81)
= 0.4(264.5 + 17.1675)
= 0.4 x 281.6675
= 112.667
The mechanical energy of the ball is 112.667 J.
Blocks A (mass 5.00 kg) and B (mass 6.50 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block A is moving toward it at 3.00 m/s. The blocks are equipped with ideal spring bumpers. The collision is head-on, so all motion before and after the collision is along a straight line.
(a) Find the maximum energy stored in the spring bumpers and the velocity of each block at that time.
(b) Find the velocity of each block after they have moved apart.
Answer: i believe its B Find the velocity of each block after they have moved apart sorry
Explanation: have a nice day buddy
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.
If you live in Melbourne, Australia, the local magnetic field has a strength of about 4x10-5 T. The magnetic field vector is directed northward, making an angle of 30 deg above the horizontal. An electron in Melbourne is moving parallel to the ground, in the west direction, at a speed of 9x105 m/s. What are the magnitude and direction of the magnetic force on the electron
Answer:
[tex]5.76\times 10^{-18}\ \text{N}[/tex] perpendicular to the velocity and magnetic field
Explanation:
B = Magnetic field = [tex]4\times 10^{-5}\ \text{T}[/tex]
[tex]\theta[/tex] = Angle the magnetic field makes with the horizontal = [tex]30^{\circ}[/tex]
v = Velocity of electron = [tex]9\times 10^5\ \text{m/s}[/tex]
q = Charge of electron = [tex]1.6\times 10^{-19}\ \text{C}[/tex]
Magnetic force is given by
[tex]F=qvB\sin\theta\\\Rightarrow F=1.6\times 10^{-19}\times 9\times 10^5\times 4\times 10^{-5}\sin30^{\circ}\\\Rightarrow F=2.88\times 10^{-18}\ \text{N}[/tex]
The magnitude of the magnetic force is [tex]2.88\times 10^{-18}\ \text{N}[/tex] and the direction is perpendicular to the velocity and magnetic field.
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.
An object undergoing simple harmonic motion takes 0.15 s to travel from one point of zero velocity to the next such point. The distance between those points is 30 cm. (a) Calculate the period of the motion. s (b) Calculate the frequency of the motion. Hz (c) Calculate the amplitude of the motion. cm
Answer:
Explanation:
Point of zero velocity are extreme points situated on either side of equilibrium position .
a )
Time taken to travel between these two points is .15 s
time for half the oscillation = .15 s
Time for full one oscillation = .30 s .
Time period of oscillation = .30 s
b)
frequency = 1 / time period
= 1 / .30s = 3.33 oscillation per second.
c )
Distance between these two point is equal to two times amplitude
2 x amplitude = 30 cm
amplitude = 15 cm
A 1.65-m-long wire having a mass of 0.100 kg is fixed at both ends. The tension in the wire is maintained at 16.0 N. (a) What are the frequencies of the first three allowed modes of vibration
Answer:
Explanation:
mass per unit length ρ = .100 / 1.65 = .0606 . kg /m
length of wire L = 1.65 m
For fundamental frequency , the expression is as follows
n = [tex]\frac{1}{2L} \sqrt{\frac{T}{m} }[/tex]
L = 1.65 , T = 16 n and m = .0606
n = [tex]\frac{1}{2\times 1.65} \sqrt{\frac{16}{.0606} }[/tex]
= 4.9 /s .
This is fundamental frequency .
other mode of vibration ( first three ) will be as follows
4.9 x 2 = 9.8 /s ,
4.9 x 3 = 14.7 /s .
Which of these cubes absorb the most light?
Answera black cube or dark colors cause dark colors suck in heat
A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surface of the water. It takes a time of 2.60 s for the boat to travel from its highest point to its lowest, a total distance of 0.630 m . The fisherman sees that the wave crests are spaced a horizontal distance of 5.70 m apart.
Required:
a. How fast are the waves traveling?
b. What is the amplitude of each wave?
c. If the total vertical distance traveled by the boat were 0.30 m but the other data remained the same, how would the answers to parts (a) and (b) be affected?
Answer:
a) v = 1.1 m/s
b) A = 0.315 m
c) v = 1.1 m/s A= 0.15 m
Explanation:
a)
In any travelling wave, there exists a fixed relationship between the propagation speed, the wavelength and the frequency, as follows:[tex]v = \lambda * f (1)[/tex]
If the wave crests are spaced a horizontal distance of 5.7 m apart, this means that the wavelength of the wave is just the same, i.e., 5.70 m.Regarding the frequency, we know that the frequency is just the inverse of the period, i.e., the time needed to complete one oscillation.If it takes a time of 2.60 s to go from the highest point to the lowest, the time needed to complete an oscillation (the period T) will be just double of this time:⇒ T = 2.60 s * 2 = 5.20 s (2)Since we have now T, we can find the frequency f as follows:[tex]f = \frac{1}{T} = \frac{1}{5.20s} = 0.19 Hz (3)[/tex]
Replacing f and λ in (1) we get:[tex]v = \lambda * f = 5.70 m * 0.19 Hz = 1.10 m/s (4)[/tex]
b)
The amplitude of the wave is just the amount that the water aparts from its equilibrium level, which is just the half of the distance between its highest point and the lowest one, as follows:[tex]A = \frac{0.630m}{2} = 0.315 m (5)[/tex]
c)
Part a) will not be affected by the new amplitude, because we have showed that the speed is independent of the amplitude, so v can be written as follows:v = 1.10 m/s (6)
Part b) will change , due to the amplitude changes. If the total vertical distance traveled by the boat is 0.30 m, by the same token as explained in b), the new amplitude will be just half of this, as follows:[tex]A = \frac{0.30m}{2} = 0.15 m (7)[/tex]
Select the correct answer Which object is an insulator
A. iron
b. cooper
c. plastic
d. salt water
What is Newton's scientific view?
Answer:
Newton's first law of motion concerns any object that has no force applied to it.
Explanation:
three laws of motion and the law of universal gravitation.
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
A student using a stopwatch finds that the time for 10 complete orbits of a ball on the end of a string is 25 seconds. The period of the orbiting ball is
Answer:
T = 2.5 s
Explanation:
Given that,
Number of complete orbits = 10
Time, t = 25 seconds
We need to find the period of the orbiting ball. Let it is T. We know that number of oscillations per unit time is called frequency and the reciprocal of frequency is called period of the ball.
So,
[tex]T=\dfrac{t}{n}\\\\T=\dfrac{25}{10}\\\\T=2.5\ s[/tex]
So, the period of the orbiting ball is equal to 2.5 seconds.