Answer:
The wire would stick to the magnet????????????????????????
Explanation:
Electric fields are MOST associated with ________.
Question 7
Review
Which particles are not affected by the strong force?
A.
hadrons
B.
protons
C.
neutrons
D
electrons
Submit A
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Answer:
Electron
Explanation:
Because electron are not hadrons so electron are not affected by strong force
Particles that can not be affected by strong forces are electrons.
What are electrons?Electrons are the rotating material around the nucleus of an atomic element in orbit.
Atoms have electrostatic energy between their electrons. This force is not broken by a force as strong as nuclear power.
What are strong forces?Strong force is a fundamental interaction of nature that acts between subatomic particles of matter.
There are four basic forces in nature:
Gravity: the gravitational force used between any heavy objects. It has an infinite range.Electrical energy: energy used between electrically charged objects. It can be either attraction or repulsion.Nuclear power: is the magnetic field that responds to the binding of protons and neutrons within the nucleus of an atom. It only works for very short distances.Weak nuclear power: a force that causes nuclear decay. It only works for very short distances.Therefore, particles that are unaffected by strong force are electrons.
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You would like to know whether silicon will float in mercury and you know that can determine this based on their densities. Unfortunately, you have the density of mercury in units of kilogram-meter^3 and the density of silicon in other units: 2.33 gram-centimeter^3. You decide to convert the density of silicon into units of kilogram-meter^3 to perform the comparison.
By which combination of conversion factors will you multiply 2.33 gram-centimeter^3 to perform the unit conversion?
Answer:
Explanation:
To convert gram / centimeter³ to kg / m³
gram / centimeter³
= 10⁻³ kg / centimeter³
= 10⁻³ / (10⁻²)³ kg / m³
= 10⁻³ / 10⁻⁶ kg / m³
= 10⁻³⁺⁶ kg / m³
= 10³ kg / m³
So we shall have to multiply be 10³ with amount in gm / cm³ to convert it into kg/m³
2.33 gram / cm³
= 2.33 x 10³ kg / m³ .
Is mercury (the planet) rocky or gaseous(meaning relating to or having the characteristics of a gas.)
Answer:
Mercury is rocky
Explanation:
Answer:
Rocky
Explanation:
It has no atmosphere so it cannot hold gas.
Which theory states that deviance results not only from what people do, but also from how others respond to those actions?
A.
strain theory
B.
labeling theory
C.
differential association theory
D.
control theory
E.
conflict theory
The theory states that deviance results not only from what people do but also from how others respond to those actions are labeling theory. Hence, the option B is correct.
What is labeling theory?This theory states that deviance and conformity result not so much from what people do but also from how others respond. It is called labeling theory. Eg: Skipping school, and underage drinking.
It also suggests that any deviance results in how society responds to certain behaviors. It defines the behavior of human beings influenced by other members of society.
It also notes that a person is made to act in a negative way by the manner in which society identifies him. If a person is identified as a criminal then he involves in the criminal activities.
Conflict theory refers to the theory linking deviance to social inequality. Anticipating the consequences of a person's behavior is control theory.
Hence, the correct option is B) labeling theory.
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why are brother anoying
Answer:
because they want attention, and big brother loves his younger one
Explanation:
A person drives north 6 blocks, then turns west, and drives 6 blocks. The driver then turns south and drives 6 blocks. How could the driver have made the distance shorter while maintaining the same displacement?
Answer:
Considering that there is no obstructions, he could go west from the start.
Explanation:
Pendulum clock. Your friend is trying to construct a clock for a craft show and asks you for some advice. She has decided to construct the clock with a pendulum. The pendulum will be a very thin, very light wooden bar with a thin, but heavy, brass ring fastened to one end. The length of the rod is 80 cm and the diameter of the ring is 10 cm. She is planning to drill a hole in the bar to place the axis of rotation 15 cm from one end. She wants you to tell her the period of this pendulum.
Answer:
The time period for this pendulum is 1.68 seconds
Explanation:
Solution
Given that:
The length of the pendulum is measured from the axis of rotation to the center of mass of the bob of the pendulum
Now,
In this case, the length becomes:
L= 80 - 15+5
L = 70 cm
The time period = T = 2π √L/g
T = 2* 3.14 *√0.7/9.8
= 1.68 seconds
Note: Kindly find an attached work to the part of the solution of the given question
A 3.6 kg block moving with a velocity of 4.3 m/s makes an elastic collision with a stationary block of mass 2.1 kg.
(a) Use conservation of momentum and the fact that the relative speed of recession equals the relative speed of approach to find the velocity of each block after the collision. 1.1315 m/s (for the 3.6 kg block) 5.43 m/s (for the 2.1 kg block)
(b) Check your answer by calculating the initial and final kinetic energies of each block. 33.282 J (initially for the 3.6 kg block) J (initially for the 2.1 kg block) J (finally for the 3.6 kg block) J (finally for the 2.1 kg block) Are the two total kinetic energies the same?
Answer:
a) Velocity of the block of mass 3.6 kg after collision = 1.13 m/s
Velocity of the block of mass 2.1 kg after collision = 5.43 m/s
b) Initial energy of the 3.6 kg block = 33.282 J
Final energy of the 3.6 kg block = 2.3 J
Initial energy of the 2.1 kg block = 0J
Final energy of the 2.1 kg block = 30.96 J
The two total kinetic energies are the same = 33.30 J
Explanation:
Check the attached files for the complete solution and explanations.
A cube, 1 m on each edge, is held 0.75 m below the surface of a pool of water by a rope. The mass of the cube is 700 kg, and the weight distribution is such that one face remains parallel to the surface of the water. a. Find the tension in the rope. b. Assuming the rope is cut, how far will the cube protrude from the water
Answer:
Explanation:
The mass of cube = 700 kg
volume = 1 m³
density = 700 kg / m²
Its density is less than that of water so it will try to float on the surface .
Tension in rope will be equal to net upward force
upthrust = volume x density of water x g
= 1 x 10³ x 9.8
= 9800 N
weight of cube = mass x g
= 700 x 9.8
= 6860 N .
Net upward force = 9800 - 6860
= 2940 N.
Tension in the rope = 2940 N.
Rope will hold the cube inside and not allow it to go outside water .
b )
If rope is cut , cube being lighter , will float on surface of water .
Part of cube inside water while floating
= 6860 / 9800
= .7
.7 m will remain inside water
part floating outside
= 1 - 0.7
= 0.3 m .
10) Two students want to use a 12-meter long rope to create standing waves. They first measure the speed at which a single wave pulse moves from one end of the rope to another and find that it is 36 m/s. What frequency must they vibrate the rope at to create the second harmonic
Answer:
To create a second harmonic the rope must vibrate at the frequency of 3 Hz
Explanation:
First we find the fundamental frequency of the rope. The fundamental frequency is the frequency of the rope when it vibrates in only 1 loop. Therefore,
f₁ = v/2L
where,
v = speed of wave = 36 m/s
L = Length of rope = 12 m
f₁ = fundamental frequency
Therefore,
f₁ = (36 m/s)/2(12 m)
f₁ = 1.5 Hz
Now the frequency of nth harmonic is given in general, as:
fn = nf₁
where,
fn = frequency of nth harmonic
n = No. of Harmonic = 2
f₁ = fundamental frequency = 1.5 Hz
Therefore,
f₂ = (2)(1.5 Hz)
f₂ = 3 Hz
The main component of all computer memory is
Answer: R.A.M
Explanation:
1. Deri had a large tank of oil (s-0.8) and was requested to determine the viscosity of that fluid. To assist with the process, she was given a 0.25-inch-diameter steel ball (sphere, s=8.0) to conduct the test. From the tests, she found that the terminal velocity of the sphere was 2.5 fpm. What is the viscosity of the oil? Remember, the volume of a sphere is (pi D3 /6). ANS. viscosity is 0.258 lb-s/ft2
Answer:
0.25916 lb-s/ft^2
Explanation:
Given:-
- The specific gravity of oil, SGo = 0.8
- The specific gravity of sphere, SGo = 8
- Terminal velocity of sphere, v = 2.5 fpm
- The diameter of sphere, D = 0.25 in
Find:-
What is the viscosity of the oil?
Solution:-
- Consider a sphere completely submerged into oil and travelling with terminal velocity ( v ).
- Develop a free body diagram for the sphere. There are forces acting on the sphere.
- The downward acting force is due to the weight of the sphere ( W ):
[tex]W = m_s*g[/tex]
Where,
The mass ( m_s ) of the sphere is given as:
[tex]m_s = S.G_s*p_w*V_s[/tex]
Where,
ρ_w : Density of water = 1.940 slugs/ft3
V_s: The volume of object ( sphere )
- The volume of sphere is expressed as a function of radius:
[tex]V_s = \frac{\pi *D^3}{6}[/tex]
Hence,
[tex]W = S.G_s*p_w*\frac{\pi*D^3 }{6}* g\\\\W = 8*1.940*\frac{\pi*(0.25/12)^3 }{6}*32\\\\W = 0.00235 lb[/tex]
- One of the upward acting force is the buoyant force ( Fb ) that is proportional to the volume of fluid displaced by the immersed object.
- The buoyant force ( Fb ) is given by:
[tex]F_b = S.G_o*p_w*V_s*g[/tex]
- Therefore the buoyant force ( Fb ) becomes:
[tex]F_b = 0.8*1.94*\frac{\pi*(0.25/12)^3 }{6} *32\\\\F_b = (4.73451*10^-^6)*(49.664)\\\\F_b = 0.00023 lb[/tex]
- The other upward acting force is the frictional drag ( F_d ) i.e the resistive frictional force acting on the contact points of the sphere and the fluid oil.
- From stokes formulations the drag force acting on a spherical object which is completely immersed in a fluid is given as:
[tex]F_d = 3*\pi*D*u*v[/tex]
Where,
μ: The viscosity of fluid
v : The velocity of object
Therefore,
[tex]F_d = 3*\pi*\frac{0.25}{12} *u*0.041666\\\\F_d = 0.00818*u\\[/tex]
- Apply Newton's second law of motion for the sphere travelling in the fluid:
[tex]F_n_e_t = m_s*a[/tex]
Where,
a: Acceleration of object = 0 ( Terminal velocity condition )
[tex]F_n_e_t = 0[/tex]
- Plug in the three forces acting on the metal sphere:
[tex]F_d + F_b - W = 0\\\\F_d = W - F_b\\\\0.00818*u = 0.00235 - 0.00023\\\\u = \frac{0.00212}{0.00818} = 0.25916 \frac{lb-s}{ft^2}[/tex]
Which elements is malleable and ductile
Answer:
The 38 elements in groups 3 through 12 of the periodic table are called "transition metals."
As with all metals, the transition elements are both ductile and malleable, and conduct electricity and heat.
Their valence electrons are present in more than one shell.
Explanation:
See Attached.
What spectacles are required for reading purposes by a person whose near point is 2.0m
Answer:Convex lens spectacles is required for reading purpose..
Explanation:
I don't say you have to mark my ans as brainliest but if it has really helped you please don't forget to thank me...
An electric dipole consists of a positive and a negative charge of equal magnitude. Consider an electric dipole with each charge having a magnitude of 1 × 10−6 C. The negative charge is located at (3 cm, 0) and the positive charge is located at (−3 cm, 0). Calculate the electric field from each charge at the points A through E, described below. Use symmetry as much as possible! Using the scale 1 cm = 105 N/C, draw the vector to represent the magnitude and direction of the electric field from each charge. (When entering angle values, enter a number greater than or equal to 0° and less than 360° measured counterclockwise from the +x-axis.) • A = (−13 cm, 0) • B = (−3 cm, 10 cm) • C = (0, 10 cm ) • D = (3 cm, 10 cm) • E = (13 cm, 0) For the negative charge:
Answer:
Explanation:
To find the electric field you use the equation for an electrostatic electric field:
[tex]E=k\frac{q_1q_2}{r^2}[/tex]
r: distance in which E is calculated, from each charge
In the of a dipole you have two contributions to E:
[tex]\vec{E}=\vec{E_1}+\vec{E_2}[/tex]
where E1 is the electric field generated by the first charge and E2 by the second one.
A. (-13 cm, 0):
First you calculate the vectors E1 and E2:
[tex]E_1=(8.98*10^9)\frac{(1*10^{-6}C)}{(-0.13-0.03)^2}\hat{j}\\\\E_1=350781.25N/C\\\\E_2=-(8.98*10^9)\frac{(1*10^{-6}C)}{(-0.13+0.03)^2}\hat{j}\\\\E_2=-989000N/C[/tex]
Then, you sum both contributions:
[tex]\vec{E}=-547218.75N/C\hat{j}[/tex]
B. (-3cm, 10cm):
[tex]r_1=\sqrt{(0.06)^2+(0.1)^2}=0.116m\\\\\theta=tan^{-1}(\frac{0.06}{0.1})=30.96\°\\\\r_2=0.1m\\\\E_1=(8.98*10^9Nm^2/C)\frac{(1.6*10^{-6}C)}{(0.116m)^2}[cos(30.96\°)\hat{i}+sin(30.96\°)\hat{j}]\\\\E_1=[-915646\hat{i}-549306.42\hat{j}]N/C\\\\\theta=(90-30.96)+180=239.04\°\\\\[/tex]
the last angle is calculated again because the vector direction is measured from the +x axis.
and for the second vector:
[tex]E_2=(8.98Nm^2/C)\frac{1.6*10^{-6}C}{(0.1m)^2}\hat{j}\\\\E_2=1436800N/C\hat{j}[/tex]
the total E is:
[tex]\vec{E}=[-915646\hat{i}+887493.58\hat{j}]N/C[/tex]
Three packing crates of masses, M1 = 6 kg, M2 = 2 kg
and M3 = 8 kg are connected by a light string of
negligible mass that passes over the pulley as shown.
Masses M1 and M3 lies on a 30o
incline plane which
slides down the plane. The coefficient of kinetic friction
on the incline plane is 0.28.
Determine the acceleration of the system.
Answer:
a = 2.5 m / s²
Explanation:
This is an exercise of Newton's second law, in this case we fix a coordinate system with the x axis parallel to the plane with positive direction
Let's write the second law for bodies in the inclined plane
W₁ₓ + W₃ₓ - fr = (m₁ + m₃) a
N₁ - [tex]W_{1y}[/tex] + N₃- W_{3y} = 0
N₁ + N₃ = W_{1y} + W_{3y}
let's use trigonometry to find the weight components
sin 30 = Wₓ/ W
Wₓ = W sin 30
cos 30 = W_{y} / W
W_{y} = W cos 30
we substitute
N₁+ N₃ = W₁ cos 30 + W₃ cos 30
W₁ₓ + W₃ₓ - μ (m₁ + m₃) g cos30 = (m₁ + m₃) a
a = (m₁g sin 30 + m₃g sin 30 - μ (m₁ + m₃) g cos 30) / (m₁ + m₃)
a = g sin 30 - μ g cos30
let's calculate
a = 9.8 sin 30 - 0.28 9.8 cos 30
a = 4.9 - 2,376
a = 2.5 m / s²
The space probe Deep Space 1 was launched on October 24th, 1998 and it used a type of engine called an ion propulsion drive. An ion propulsion drive generates only a weak force (or thrust), but can do so for long periods of time using only a small amount of fuel. Suppose the probe, which has a mass of 474 kg is travelling at an initial speed of 275 m/s. No forces act on it except the 5.60 x 10⁻² N thrust from the engine. This external force is directed PARALLEL to the displacement. The displacement has a magnitude of 2.42 x 10⁹ m. {PART A} Calculate the INITIAL kinetic energy of the probe [2 marks] {PART B} Find the work done BY THE ENGINE on the space probe [2 marks] {PART C} Calculate the FINAL KINETIC ENERGY of the probe (Hint W=∆E) [2 marks] {PART D} Determine the final speed of the probe, assuming that its mass remains constant [3 marks]
Answer:
Explanation:
mass of probe m = 474 Kg
initial speed u = 275 m /s
force acting on it F = 5.6 x 10⁻² N
displacement s = 2.42 x 10⁹ m
A )
initial kinetic energy = 1/2 m u² , m is mass of probe.
= .5 x 474 x 275²
= 17923125 J
B )
work done by engine
= force x displacement
= 5.6 x 10⁻² x 2.42 x 10⁹
= 13.55 x 10⁷ J
C ) Final kinetic energy
= Initial K E + work done by force on it
= 17923125 +13.55 x 10⁷
= 1.79 x 10⁷ + 13.55 x 10⁷
= 15.34 x 10⁷ J
D ) If v be its velocity
1/2 m v² = 15.34 x 10⁷
1/2 x 474 x v² = 15.34 x 10⁷
v² = 64.72 x 10⁴
v = 8.04 x 10² m /s
= 804 m /s
By which process does the heat from the Sun reach the Earth? (AKS 4b DOK 1) *
A 0.025-kg block on a horizontal frictionless surface is attached to an ideal massless spring whose spring constant is The block is pulled from its equilibrium position at x = 0.00 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the horizontal x-axis. When the displacement is what is the kinetic energy of the block?
Complete Question
A 0.025-kg block on a horizontal frictionless surface is attached to an ideal massless spring whose spring constant is 150 N/m. The block is pulled from its equilibrium position at x = 0.00 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the horizontal x-axis. When the displacement is x = 0.024 m, what is the kinetic energy of the block?
Answer:
The kinetic energy is [tex]KE = 0.4368\ J[/tex]
Explanation:
From the question we are told that
The mass of the block is [tex]m= 0.025\ kg[/tex]
The spring constant is [tex]k = 150 N/m[/tex]
The length of first displacement is [tex]x_1 = 0.80 \ m[/tex]
The length of first displacement is [tex]x_2 = 0.024 \ m[/tex]
At the [tex]x_2[/tex] the kinetic energy is mathematically evaluated as
[tex]KE = \Delta E[/tex]
Where [tex]\Delta E[/tex] is the change in energy stored on the spring which is mathematically represented as
[tex]\Delta E = \frac{1}{2} k (x_1 ^2 - x_2^2)[/tex]
=> [tex]KE = \frac{1}{2} k (x_1 ^2 - x_2^2)[/tex]
Substituting value
[tex]KE = \frac{1}{2} * 150 * (0.08^2 - 0.024^2)[/tex]
[tex]KE = 0.4368\ J[/tex]
How are the elements in the same row similar
Answer:
All elements in a row have the same number of electron shells. Each next element in a period has one more proton and is less metallic than its predecessor. Arranged this way, groups of elements in the same column have similar chemical and physical properties, reflecting the periodic law.
A Texas cockroach of mass 0.157 kg runs counterclockwise around the rim of a lazy Susan (a circular disk mounted on a vertical axle) that has a radius 14.9 cm, rotational inertia 5.92 x 10-3 kg·m2, and frictionless bearings. The cockroach's speed (relative to the ground) is 2.92 m/s, and the lazy Susan turns clockwise with angular velocity ω0 = 3.89 rad/s. The cockroach finds a bread crumb on the rim and, of course, stops. (a) What is the angular speed of the lazy Susan after the cockroach stops? (b) Is mechanical energy conserved as it stops?
Answer:
-7.23 rad/s
Explanation:
Given that
Mass of the cockroach, m = 0.157 kg
Radius of the disk, r = 14.9 cm = 0.149 m
Rotational Inertia, I = 5.92*10^-3 kgm²
Speed of the cockroach, v = 2.92 m/s
Angular velocity of the rim, w = 3.89 rad/s
The initial angular momentum of rim is
Iw = 5.92*10^-3 * 3.89
Iw = 2.3*10^-2 kgm²/s
The initial angular momentum of cockroach about the axle of the disk is
L = -mvr
L = -0.157 * 2.92 * 0.149
L = -0.068 kgm²/s
This means that we can get the initial angular momentum of the system by summing both together
2.3*10^-2 + -0.068
L' = -0.045 kgm²/s
After the cockroach stops, the total inertia of the spinning disk is
I(f) = I + mr²
I(f) = 5.92*10^-3 + 0.157 * 0.149²
I(f) = 5.92*10^-3 + 3.49*10^-3
I(f) = 9.41*10^-3 kgm²
Final angular momentum of the disk is
L'' = I(f).w(f)
L''= 9.41*10^-3w(f)
Using the conservation of total angular momentum, we have
-0.068 = 9.41*10^-3w(f) + 0
w(f) = -0.068 / 9.41*10^-3
w(f) = -7.23 rad/s
Therefore, the speed of the lazy Susan after the cockroach stops is -7.23 and is directed in the opposite direction of the initial lazy Susan angular speed
b)
The mechanical energy of the cockroach is not converted as it stops
1. A block of metal of mass 2kg is resting on
a frictionless floor. It is struck by a jet
releasing water at the rate of 1kg/sec at a
speed of 5ms-1. What will be the initial
acceleration of the block?
Answer:
The acceleration is [tex]a = 2.5 \ m/s^2[/tex]
Explanation:
From the question we are told that
The mass of the metal block is [tex]m_b = 2 \ kg[/tex]
The mass flow rate of the water is [tex]\r m = 1\ kg/s[/tex]
The speed of the water of the water release is [tex]v_w = 5 m/s[/tex]
Generally according to the law of conservation of linear momentum
[tex]p_i = p_f[/tex]
Now [tex]p_i[/tex] is the initial momentum of the system which mathematically represented as
[tex]p_i = m_w * v_w + m_b * v_b[/tex]
Now [tex]m_w[/tex] is the mass of water at the point of contact with the block which can be deduced as [tex]m_w = 1 \ kg[/tex]
Now since at initial the block is at rest
[tex]v_b = 0 \ m/s[/tex]
So
[tex]p_i = 1 * 5[/tex]
[tex]p_i = 5 \ kgm/ s[/tex]
And [tex]p_f[/tex] is the final momentum of the system which mathematically represented as
[tex]p_f = m_w * v__{fw} } + m_b * v__{fb}}[/tex]
So [tex]v__{fw} }[/tex] is the final velocity of water which is zero due to the fact that when the water hits the block it losses its momentum and eventually the velocity becomes zero
So
[tex]5 = 2 * v__{fb }[/tex]
Thus [tex]v__{fb }} = \frac{5}{2}[/tex]
[tex]v__{fb }} = 2.5 \ m/s[/tex]
Thus
[tex]p_f = 2.5 * 2[/tex]
[tex]p_f = 5 \ kgm /s[/tex]
Now the average momentum change is
[tex]p_a = \frac{p_i +p_f}{2}[/tex]
[tex]p_a = \frac{5+5}{2}[/tex]
[tex]p_a =5 kgm/s[/tex]
Now the force acting on the block is
[tex]F = \frac{p_a }{t}[/tex]
and from the question the initial movement of the block took 1s as it is a mass of water moving at a rate of 1kg/s that caused the first movement of the block
So
[tex]F= \frac{5}{1}[/tex]
[tex]F= 5 \ N[/tex]
Now the acceleration is
[tex]a = \frac{F}{m_b}[/tex]
=> [tex]a = \frac{5}{2}[/tex]
[tex]a = 2.5 \ m/s^2[/tex]
List and describe the steps of energy transfers that occur that allow a digital recording to be played through a speaker and ultimately become sound waves.
Answer:
In the production of sound, Electrical, mechanical, and wave energies are involved.
Explanation:
First, all the components of a digital recorder are powered and driven by electrical energy. This relates to all the internal workings which include decoding of the recorded data which is stored magnetically and the transmission of same to the speakers.
As electricity travels through the coil, it causes the coil to vibrate thus converting electrical energy into mechanical energy. The coil is a copper spiral that is inserted into a magnet and is connected to a lightweight paper membrane or a diaphragm.
As electricity travels through the magnet encircled coil, it causes it to vibrate. The vibration, in turn, causes the diaphragm attached it to also vibrate. The vibration triggers a physical wave of energy. This wave energy travels to the listener.
Cheers!
Which of the following statements are characteristics of magnetic fields? Select all that apply.
Magnetic fields point from the north pole to the south pole of a magnet.
The earth's magnetic field has no effect on the electron rays coming from the sun.
An example of the Biot-Savart law is the effect of the earth's maghytic field on the electron rays coming from the sun.
The north pole of a magnet will be attracted to the south pole of the earth.
If a bar magnet is cut in half two magnets with like poles will be created.
Answer:
Magnetic fields point from the north pole to the south pole of a magnet.
An example of the Biot-Savart law is the effect of the earth's maghytic field on the electron rays coming from the sun.
The north pole of a magnet will be attracted to the south pole of the earth.
If a bar magnet is cut in half two magnets with like poles will be created
Explanation:
The magnetic field of Earth is due to the presence of iron in the core of the Earth.
The metal emits the magnetic waves from it and the North and South pole of the planet.
Both the poles emit the magnetic rays which create magnetic sheet around it. The Earth acts like a magnet bar if which is cut into two half, the planet will act like two magnets. Also, Biot Savarts's law states that the magnetic field does not affect the electron rays coming from the Sun.
Thus, the selected options are correct.
Answer:
ACDE
Explanation:
1. You are playing with a jump rope that is tied at both ends. You untie one end, hold it taut and wiggle the end up and down sinusoidally with frequency 2.00Hz and amplitude 0.075m. At time t=0, the end has a maximum positive displacement and is instantaneously at rest. Assume no wave bounces back from the far end to change the pattern. What is the equation for the displacement of the wave? What is the displacement at a point 3.00m from the end .
Answer:
[tex]f(x=3.00m)=0.075mcos(\frac{2\pi(2.00Hz)}{v}(3.00m))[/tex]
Explanation:
To find the equation of the wave you use the general equation for a wave, given by:
[tex]f(x)=Acos(k x-\omega t)[/tex]
A: amplitude of the wave = 0.075m
k: wave number
you select a cosine function because for x=0 and t= 0 you get a maximum displacement.
To find the displacement of the wave for x=0 you can consider that the form of the wave is independent of time t.
Then, you calculate k:
[tex]k=\frac{\omega}{v}=\frac{2\pi f}{v}[/tex]
Thus, you need the value of the speed of the wave (you only have the frequency f), in order to calculate f(x), for x=3.00m:
[tex]f(x=3.00m)=0.075mcos(\frac{2\pi(2.00Hz)}{v}(3.00m))[/tex]
What happens to the brightness of the lightbulb when its resistance is increased? *
Answer:
It will be dimmer than before
Answer:For parallel connection,the brightness would be dimmer, while for series connection it would be brighter
Explanation:
For parallel connection,resistance and brightness are inversely proportional.meaning as resistance increases, brightness decreases.
For series connection,resistance and brightness are directly proportional. Meaning as the resistance increases, brightness also increases.
How is the particle displacement related to the direction of wave movement in a longitude wave?
Answer:
The displacement of particles is perpendicular to the direction of wave motion.
What must x be so that the handle end of the bat remains at rest as the bat begins to move? (Hint: Consider the motion of the center of mass and the rotation about the center of mass. Find x so that these two motions combine to give v=0 for the end of the bat just after the collision. Also, note that integration of equation ∑τ⃗ =dL⃗ dt gives ΔL=∫t1t2(∑τ)dt. )
Complete Question
The complete question is shown on the first uploaded image
Answer:
Explanation:
From the question we are told that
The mass of the bat is [tex]m_b = 0.800 \ kg[/tex]
The bat length is [tex]L_b = 0.900 \ m[/tex]
The distance of the bat's center of mass to the handle end is [tex]z_c = 0.600 \ m[/tex]
The moment of inertia of the bat is [tex]I = 0.0530 \ kg \cdot m^2[/tex]
The objective of the solution is to find x which is the distance from the handle of the bat to the point where the baseball hit the bat
Generally the velocity change at the end of the bat is mathematically represented as
[tex]\Delta v_e = \Delta v_c - \Delta w* z_c[/tex]
Where [tex]\Delta v_c[/tex] is the velocity change at the center of the bat which is mathematically represented as
[tex]\Delta v_c = \frac{Impulse}{m_b }[/tex]
We are told that the impulse is J so
[tex]\Delta v_c = \frac{J}{m_b }[/tex]
And [tex]\Delta w[/tex] is the change in angular velocity which is mathematically represented as
[tex]\Delta w = \frac{J (z -z_c)}{I}[/tex]
Now we have that
[tex]\Delta v_e = \frac{J}{m_b } - [\frac{J (x- z_c)}{I} ] * z_c[/tex]
Before a swing when the bat is at rest the velocity change a the end of the bat handle is zero and the impulse will be 1
So
[tex]0 = \frac{1}{m_b } - [\frac{J (x- z_c)}{I} ] * z_c[/tex]
=> [tex]x = \frac{I}{m_b z_c} + m_b[/tex]
substituting values
[tex]x = \frac{0.530}{0.800 * 0.600} + 0.600[/tex]
[tex]x = 0.710 \ m[/tex]
A physics student standing on the edge of a cliff throws a stone vertically downward with an initial speed of 10 m/s. The instant before the stone hits the ground below, it is traveling at a speed of 30 m/s. if the physics student were to throw the rock horizontally outward from the cliff instead, with the same initial speed of 10 m/s, how fast would the stone be traveling just before it hits the ground
Answer:
The velocity just before hitting the ground is [tex]v_f = 30 m/s[/tex]
Explanation:
From the question we are told that
The initial speed is [tex]u = 10 m/s[/tex]
The final speed is [tex]v = 30 \ m/s[/tex]
From the equations of motion we have that
[tex]v^2 =u^2 + 2as[/tex]
Where s is the distance travelled which is the height of the cliff
So making it the subject of the the formula we have that
[tex]s = \frac{v^2 - u^2 }{2a}[/tex]
Where a is the acceleration due to gravity with a value [tex]a = 9.8m/s^2[/tex]
So
[tex]s = \frac{30^2 - 10^2 }{2 * 9.8 }[/tex]
[tex]s = 40.8 \ m[/tex]
Now we are told that was through horizontally with a speed of
[tex]v_x =10 m/s[/tex]
Which implies that this would be its velocity horizontally through out the motion
Now it final velocity vertically can be mathematically evaluated as
[tex]v_y = \sqrt{2as}[/tex]
Substituting values
[tex]v_y = \sqrt{(2 * 9.8 * 40.8)}[/tex]
[tex]v_y = 28.3 \ m/s[/tex]
The resultant final velocity is mathematically evaluated as
[tex]v_f = \sqrt{v_x^2 + v_y^2}[/tex]
Substituting values
[tex]v_f = \sqrt{10^2 + 28.3^2}[/tex]
[tex]v_f = 30 m/s[/tex]
