Answer:
the pearls have an electrical charge induced by contact with the ions of the solution and these charges are attracted by the electrode by a force electric
Explanation:
The pearls are suspended in a solution, when connecting the power source, it is subjected to an electric shock, the pearls have an electrical charge induced by contact with the ions of the solution and these charges are attracted by the electrode by a force electric
F = q E
In a region of space there is a uniform magnetic field pointing in the positive z direction. In what direction should a negative charge move to experience a force in the positive x direction
Answer:
the speed in -y
Explanation:
For this exercise we must use the right hand rule. The motion of a positive charge is given by.
Thumb points in the direction of speed
fingers extended in the direction of the magnetic field, + z axis
the palm in the direction of the force, as the charge is negative in the opposite direction of the force, axis + x
therefore the thumb is in the direction - y
the speed in -y
Suppose an astronomer observes a binary star system where the stars are separated by 2.0 AU , and they have an orbital period of 7.0 years . Using Newton's version of Kepler's Third Law, find the combined mass of the stars.
Answer:
4.408 [tex]\mathsf{M_{sun}}[/tex]
Explanation:
According to Kelper's Third Law, the equation of the combined mass (m₁+m₂) can be expressed as:
[tex](m_1 + m_2) = \dfrac{\text{(distance between stars)}^3}{\text{(orbital period)}^2}[/tex]
[tex]\text{combined mass}(m_1+m_2)} =\dfrac{(6.0)^3}{(7)^2} \ M_{sun}[/tex]
[tex]\text{combined mass}(m_1+m_2)} =\dfrac{216}{49} \ M_{sun}[/tex]
combined mass (m₁+m₂) = 4.408 [tex]\mathsf{M_{sun}}[/tex]
different between pressure and force
Force is mass into acceleration
and pressure is force applied per unit area.
HELP ME PLEASEEEEEEEEEEEEEE
Answer: The correct statements are:
The atoms are very attracted to one another.The atoms are held tightly together.Explanation:
Solid state: In this state, the molecules are closely packed and cannot move freely from one place to another that means no space between them and the intermolecular force of attraction between the molecules are strong.
In solid substance, the particles are very close to each other due to this the intermolecular forces of attraction are strongest.
The key point about solid are:
The atoms are very attracted to one another.The atoms are not moving freely.It will not spread out evenly to fill any container.The atoms are held tightly together.The forces of attraction are strong to bring molecules together.The atoms are close and in fixed positions.How much work can a motor with a power output of 0.70 hp do in 2 s?
Answer:
the work done by the motor is 1,044 J.
Explanation:
Given;
the output power of the motor, P = 0.7 hp
duration of the work, t = 2 s
The relationship between horse-power and watt is given as;
1 hp = 745.7 W
0.7 hp = ?
0.7 hp = 522 W = 522 J/s
The work done by the motor is calculated as;
W = Power x time
W = 522 J/s x 2 s
W = 1,044 J
Therefore, the work done by the motor is 1,044 J.
Astronauts use a centrifuge to simulate the acceleration of a rocket launch. The centrifuge takes 40.0 ss to speed up from rest to its top speed of 1 rotation every 1.30 ss . The astronaut is strapped into a seat 5.90 mm from the axis. What is the astronaut's tangential acceleration during the first 40.0 s?
How many g's of acceleration does the astronaut experience when the device is rotating at top speed? Each 9.80 m/s^2 of acceleration is 1 g.
Answer:
speed = 0.9 mm/s
Explanation:
time, t = 40 s
initial angular speed, wo = 0 rad/s
final frequency, f = 1/1.03 rps = 0.97 rps
final angular speed, w = 2 x 3.14 x 0.97 = 6.1 rad/s
time, t = 40 s
distance, r = 5.9 mm
The angular acceleration is given y the first equation of motion.
[tex]w =wo + \alpha t\\6.1 = 0 +\alpha \times 40\\\alpha = 0.1525 rad/s^{2}[/tex]
The linear velocity is
[tex]v =5.9\times 10^{-3}\times 0.1525 = 9\times 10^{-4} m/s[/tex]
speed, v = 0.9 mm/s
A train with mass 3.3 x 107 kg starts from rest and accelerates to a speed of 42
m/s. What is the initial kinetic energy of the train?
Answer:
kinetic energy of the train = 2,910.6 x 10⁷ joule
Explanation:
Given:
Mass of train = 3.3 x 10⁷ kg
Speed of train = 42 m/s
Find:
kinetic energy of the train
Computation:
kinetic energy = (1/2)(m)(v²)
kinetic energy of the train = (1/2)(3.3 x 10⁷)(42²)
kinetic energy of the train = (1/2)(3.3 x 10⁷)(1,764)
kinetic energy of the train = (3.3 x 10⁷)(882)
kinetic energy of the train = 2,910.6 x 10⁷ joule
Answer: The initial kinetic energy of the train is [tex]2910.6 \times 10^{7} J[/tex].
Explanation:
Given: Mass = [tex]3.3 \times 10^{7} kg[/tex]
Speed = 42 m/s
Kinetic energy is the energy acquired by an object due to its motion.
Formula to calculate kinetic energy is as follows.
[tex]K.E = \frac{1}{2}mv^{2}[/tex]
where,
m = mass of object
v = speed of object
Substitute the values into above formula as follows.
[tex]K.E = \frac{1}{2}mv^{2}\\= \frac{1}{2} \times 3.3 \times 10^{7} kg \times (42 m/s)^{2}\\= 2910.6 \times 10^{7} kg m^{2}/s^{2} (1 J = 1 kg m^{2}/s^{2})\\= 2910.6 \times 10^{7} J[/tex]
Thus, we can conclude that the initial kinetic energy of the train is [tex]2910.6 \times 10^{7} J[/tex].
Walking at a brisk pace, you cover 10 m in 5.0 s . How many seconds will you need to cover 50 m ?
An atom has 20 protons and 22 neutrons and 18 electrons. The charge of this atom is: ________
Answer:
the number of electrons should equal to the the number of protons in a neutral atom
if there is a inequality between the numbers it means the atom has a + or - charge
The charge of this atom=+(20-18)=+2If a second ball were dropped from rest from height ymax, how long would it take to reach the ground
Answer:
[tex](b)\ t_1 - t_0[/tex]
[tex](d)\ t_2 - t_1[/tex]
[tex](e)\ \frac{t_2 - t_0}{2}[/tex]
Explanation:
Given
See attachment for complete question
Required
How long to reach the ground from the maximum height
First, calculate the time of flight (T)
[tex]T =t_2 - t_0[/tex]
The time taken (t) from maximum height to the ground is:
[tex]t = \frac{1}{2}T[/tex]
So, we have:
[tex]t = \frac{t_2 - t_0}{2}[/tex]
Another representation is:
At ymax, the time is: t1
On the ground, the time is t2
The difference between these times is the time taken.
So;
[tex]t = t_2 - t_1[/tex]
Since air resistance is to be ignored, then
[tex]t_2 - t_1 = t_1 - t_0[/tex] --- i.e. time to reach the maximum height from the ground equals time to reach the ground from the maximum height
Dr. John Paul Stapp was a U.S. Air Force officer who studied the effects of extreme deceleration on the human body. On December 10, 1954, Stapp rode a rocket sled, accelerating from rest to a top speed of 282 m/s (1015 km/h) in 5.00 s, and was brought jarringly back to rest in only 1.40 s. Calculate his:
a. acceleration in his direction of motion
b. acceleration opposite to his direction of motion.
Answer: [tex]56.4\ m/s^2, 201.42\ m/s^2[/tex]
Explanation:
Given
Rocket attain a velocity of [tex]v=282\ m/s[/tex] in a time period of [tex]t=5\ s[/tex]
It was brought jarringly back to rest in only [tex]t'=1.4\ s[/tex]
Acceleration is the change in velocity of the object over a period of time
(a) Acceleration in his direction of motion
[tex]\Rightarrow a=\dfrac{v-0}{t}\\\\\Rightarrow a=\dfrac{282}{5}\\\\\Rightarrow a=56.4\ m/s^2[/tex]
(b) acceleration opposite to his direction of motion i.e. deceleration is
[tex]\Rightarrow a_d=\dfrac{0-v}{t'}\\\\\Rightarrow a_d=\dfrac{-282}{1.4}\\\\\Rightarrow a_d=-201.42 \ m/s^2\\\Rightarrow a_d=201.42\ \text{decelration}[/tex]
If two charged objects each have 2.5 C of charge on them and are located 100 m apart, how strong is the electrostatic force between them?
Answer:
5.619×10⁶ N
Explanation:
Applying,
F = kqq'/r²................... Equation 1
Where F = electrostatic force between the charges, k = coulomb's constant, q = first charge, q' = second charge, r = distance btween the charges
From the questiion,
Given: q = 2.5 C, q' = 2.5 C, r = 100 m
Constant: 8.99×10⁹ Nm²/C²
Substitute these values into equation 1
F = (2.5×2.5×8.99×10⁹)/100²
F = 56.19×10⁵
F = 5.619×10⁶ N
6. In an integrated circuit, each wafer is cut into sections, which
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A. have multiple circuits and are placed in individual cases.
B. carry a single circuit and are placed in individual cases.
C. carry a single circuit and are placed all together in one case.
D. have multiple circuits and are placed all together in one case.
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Answer:
B. carry a single circuit and are placed in individual cases.
Explanation:
An electric circuit can be defined as an interconnection of electrical components which creates a path for the flow of electric charge (electrons) due to a driving voltage.
Generally, an electric circuit consists of electrical components such as resistors, capacitors, battery, transistors, switches, inductors, etc.
Similarly, an integrated circuit (IC) also referred to as microchip can be defined as a semiconductor-based electronic component that comprises of many other tiny electronic components such as capacitors, resistors, transistors, and inductors.
Integrated circuits (ICs) are often used in virtually all modern electronic devices to carry out specific tasks or functions such as amplification, timer, oscillation, computer memory, microprocessor, etc.
A wafer can be defined as a thin slice of crystalline semiconductor such as silicon and germanium used typically for the construction of an integrated circuit.
In an integrated circuit, each wafer is cut into sections, which generally comprises of a single circuit that are placed in individual cases.
Additionally, a semiconductor can be defined as a crystalline solid substance that has its conductivity lying between that of a metal and an insulator, due to the effects of temperature or an addition of an impurity.
Answer: B got it right on the test just now
Explanation:
You drive past a potential parking space in center city. Your new car is travelling at 85% the speed of light. If your car is 6.0 m long (which you measured the day you bought it) and you observe the space to be 3.0 m, should you try to park? Why is your friend on the sidewalk (who hasn't studied relativity) so sure that you can park? How does the situation appear to him?
Answer:
We should not try to park the car because its rest length is greater than the space available.
The car seems to be approximately equal to the friend (L = 3.16 m). Due to this reason he is sure to park.
Explanation:
We should not try to park the car because its rest length is greater than the space available.
The friend is sure about parking because the car appears short in length to him. For this, we will solve Einstein's length contraction formula from theory of relativity:
[tex]L = L_o\sqrt{1-\frac{v^2}{c^2}}[/tex]
where,
L = Relative length observed by friend = ?
L₀ = rest length = 6 m
v = relative speed = 85% of speed of light = 0.85c
Therefore,
[tex]L = (6\ m)\sqrt{1-\frac{(0.85c)^2}{c^2}}[/tex]
L = 3.16 m
Hence, the car seems to be approximately equal to the friend. Due to this reason he is sure to park.
How many x-ray photons per second are created by an x-ray tube that produces a flux of x rays having a power of 1.00 W
Complete question:
How many x-ray photons per second are created by an x-ray tube that produces a flux of x rays having a power of 1.00 W. Assume the average energy per photon in 78.0 keV.
Answer:
The number of x-ray photons per second created by the x-ray tube is 8.01 x 10¹³ photons/sec
Explanation:
Given;
power of the flux produced, P = 1 W = 1 J/s
energy per photon, E = 78 keV
Convert the energy per photon to J
E = 78 x 10³ x 1.6 x 10⁻¹⁹ = 1.248 x 10⁻¹⁴ J / photon
let the number of photons = n
n(1.248 x 10⁻¹⁴ J / photon) = 1 J/s
[tex]n = \frac{1 \ J/s}{1.248 \times 10^{-14}\ J/photon } = 8.01 \times 10^{13} \ photons/s[/tex]
Therefore, the number of x-ray photons per second created by the x-ray tube is 8.01 x 10¹³ photons/sec
What is the speed acquired by a freely falling object 4 seconds after being dropped from a rest position? Use units of meter per second (m/s) and assume acceleration from gravity is 10 m/s2.
speed = 40 m/s
Explanation:
Since the object is dropped, V0y = 0.
Vy = V0y - gt
= -(10 m/s^2)(4 s)
= -40 m/s
This means that its velocity is 40 m/s downwards. Its speed is simply 40 m/s.
The speed acquired by a freely falling object 4 seconds after being dropped from a rest position would be 40 meters/seconds.
What are the three equations of motion?There are three equations of motion given by Newton
The first equation is given as follows
v = u + at
the second equation is given as follows
S = ut + 1/2×a×t²
the third equation is given as follows
v² - u² = 2×a×s
Keep in mind that these calculations only apply to uniform acceleration.
As given in the problem, we have to find the speed acquired by a freely falling object 4 seconds after being dropped from a rest position,
By using the first equation of motion,
v = u + at
initial velocity(u) = 0 m/s
acceleration(a) = 10 m/s²
v = 0 + 10×4
v = 40 meters/seconds
Thus, the speed acquired by a freely falling object 4 seconds after being dropped from a rest position would be 40 meters/seconds.
Learn more about equations of motion from here,
brainly.com/question/5955789
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What are impact and non-impact printers?
Impact printers involve mechanical components for conducting printing. It is a type of printer that works by direct contact of an ink ribbon with paper.
In Non-Impact printers, no mechanical moving component is used.
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as a mercury atom absorbs a photon of energy as electron in the atom changes from energy level B to energy level E. calculate the frequency of the absorb photon.
Answer:
2.00x 10 14th Hz
Explanation:
Answer:
2.99 x 10^14 Hz
Explanation:
E photon= hf (you have to solve for f)
f= E photon/h
f= 1.98 x 10^-19 J / 6.63 x 10^-34 J x s
f=2.99 x 10^14 Hz
an object moves clockwise around a circle centered at the origin with radius m beginning at the point (0,). a. find a position function r that describes the motion of the object moves with a constant speed, completing 1 lap every s. b. find a position function r that describes the motion if it occurs with speed .
Answer:
Answer to An object moves clockwise around a circle centered at the origin with radius 6 m beginning at ... 6 M Beginning At The Point (0,6) B. Find A Position Function R That Describes The Motion If It Occurs With Speed E T A. R(t)= S The Motion Of The Object Moves With A Constant Speed, Completing 1 Lap Every 12 S.
Explanation:
You throw a glob of putty straight up toward the ceiling, which is 3.50 mm above the point where the putty leaves your hand. The initial speed of the putty as it leaves your hand is 9.50 m/sm/s. Part A What is the speed of the putty just before it strikes the ceiling
Answer: [tex]4.65\ m/s[/tex]
Explanation:
Given
Distance putty has to travel is 3.5 m
The initial speed of putty is 9.50 m/s
Using equation of motion to determine the velocity of putty just before it hits ceiling
[tex]v^2-u^2=2as[/tex]
[tex]\Rightarrow v^2-(9.5)^2=2(-9.8)(3.5)\\\\\Rightarrow v^2=9.5^2-68.6\\\Rightarrow v=\sqrt{90.25-68.6}\\\Rightarrow v=4.65\ m/s[/tex]
So, the velocity of putty just before hitting is [tex]4.65\ m/s[/tex]
what is the light synthesis ?
Answer:
Photosynthesis, the process by which green plants and certain other organisms transform light energy into chemical energy. ... During photosynthesis in green plants, light energy is captured and used to convert water, carbon dioxide, and minerals into oxygen and energy-rich organic compounds.
Explanation:
thank me later
Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.390 mm wide. The diffraction pattern is observed on a screen 3.10 m away. Define the width of a bright fringe as the distance between the minima on either side.
Answer:
Y = 5.03 x 10⁻³ m = 5.03 mm
Explanation:
Using Young's Double-slit formula:
[tex]Y = \frac{\lambda L}{d}[/tex]
where,
Y = Fringe Spacing = Width of bright fringe = ?
λ = wavelength = 633 nm = 6.33 x 10⁻⁷ m
L = Screen distance = 3.1 m
d = slit width = 0.39 mm = 3.9 x 10⁻⁴ m
Therefore,
[tex]Y = \frac{(6.33\ x\ 10^{-7}\ m)(3.1\ m)}{3.9\ x\ 10^{-4}\ m}[/tex]
Y = 5.03 x 10⁻³ m = 5.03 mm
A body of mass 5 kg is moved by a horizontal force of 0.5 N on a smooth frictionless table for 20 seconds. Calculate the change in kinetic energy.
A. 5 J
B. 20 J
C. 10 J
D. 30 J
Answer: 10 J
Explanation:
Trust ;)
A car travels at a constant speed around a circular track whose radiu is 2.6 km. The goes once arond the track in 360s . What is the magnitude
Answer:
Centripetal acceleration = 0.79 m/s²
Explanation:
Given the following data;
Radius, r = 2.6 km
Time = 360 seconds
Conversion:
2.6 km to meters = 2.6 * 1000 = 2600 meters
To find the magnitude of centripetal acceleration;
First of all, we would determine the circular speed of the car using the formula;
[tex] Circular \; speed (V) = \frac {2 \pi r}{t}[/tex]
Where;
r represents the radius and t is the time.Substituting into the formula, we have;
[tex] Circular \; speed (V) = \frac {2*3.142*2600}{360} [/tex]
[tex] Circular \; speed (V) = \frac {16338.4}{360} [/tex]
Circular speed, V = 45.38 m/s
Next, we find the centripetal acceleration;
Mathematically, centripetal acceleration is given by the formula;
[tex] Centripetal \; acceleration = \frac {V^{2}}{r}[/tex]
Where;
V is the circular speed (velocity) of an object.r is the radius of circular path.Substituting into the formula, we have;
[tex] Centripetal \; acceleration = \frac {45.38^{2}}{2.6}[/tex]
[tex] Centripetal \; acceleration = \frac {2059.34}{2600}[/tex]
Centripetal acceleration = 0.79 m/s²
The electric field of a negative infinite line of charge: Group of answer choices Points perpendicularly away from the line of charge and decreases in strength at larger distances from the line charge Points parallel to the line of charge and decreases in strength at larger distances from the line charge Points parallel to the line of charge and increases in strength at larger distances from the line charge Points perpendicularly away from the line of charge and increases in strength at larger distances from the line charge Points perpendicularly toward the line of charge and increases in strength at larger distances from the line charge Points perpendicularly toward the line of charge and decreases in strength at larger distances from the line charge
Answer:
Points perpendicularly toward the line of charge and decreases in strength at larger distances from the line charge
Explanation:
The electric field for a uniform line of charge is given by E = λ/2πε₀r where λ = charge density and r = distance from line of charge.
If λ is negative, E is negative so it points in the negative direction towards the line of charge.
Also, since for negative charges, electric field lines end up in them, the electric field for an infinitely long negative line of charge points towards the charge perpendicular to it.
Also as r increases, E decreases since E ∝ 1/r
So, the electric field decreases at larger distances from the line of charge.
So, the electric field of a negative infinite line of charge Points perpendicularly toward the line of charge and decreases in strength at larger distances from the line charge.
Use your understanding of heat loss to ESTIMATE the cost of the lost energy through one standard window during an average summer day in Maryland. Use $0.17 per kWh as your energy cost.
Answer:
The cost of energy is $ 0.34.
Explanation:
The energy is the capacity to do work.
The energy is a scalar quantity and its SI unit is Joule.
The commercial unit of energy is kWh.
Cost of 1 kWh energy = $ 0.17
energy loss by standard window is 2 kWh .
So, the cost of lost of energy is
Cost = $ 0.17 x 2 = $ 0.34
A cart weighing 40 pounds is placed on a ramp incline 15 degrees to the horizon. The cart is held in place by a rope inclined 60 degrees to the horizontal. find the force that the rope must exert on the cart to keep it from rolling down the ramp.
Answer: [tex]14.64\ N[/tex]
Explanation:
Given
Inclination of ramp is [tex]\theta=15^{\circ}[/tex]
Rope is inclined [tex]\phi=60^{\circ}[/tex] to the horizontal
Weight of cart [tex]W=40\ lb[/tex]
from the diagram, rope is at angle of [tex]45^{\circ}[/tex] w.r.t ramp
Sine component of weight pulls down the cart Cosine component of force applied through rope held it at the position
[tex]\Rightarrow 40\sin 15^{\circ}=F\cos 45^{\circ}\\\\\Rightarrow F=40\cdot \dfrac{\sin 15^{\circ}}{\cos 45^{\circ}}\\\\\Rightarrow F=40\times 0.366\\\Rightarrow F=14.64\ N[/tex]
Which of these is NOT an effect of humor?
strengthened immune system
reduced stress levels
reduced feelings of anxiety
feelings of jealousy and envy
One of the earliest vertebrate animal groups that evolved in the early Paleozoic Era
are
An athlete training for an event does 6.53 104 J of work during a workout and gives off 5.97 105 J of heat. Consider the athlete to be like a heat engine. (a) Determine the magnitude of the change in internal energy of the athlete. J (b) What is the efficiency of the athlete
Answer:
(a) The magnitude of the change in internal energy is 6.623 x 10⁵ J
(b) the efficiency of the athlete is 10.94 %
Explanation:
Given;
work done by the athlete (system), W = 6.53 x 10⁴ J
the heat given off by the athlete (system), Q = 5.97 x 10⁵ J
The simple diagram below will be used to illustrate the direction of the energy flow assuming a heat engine.
Q← ⊕ →W
The work, W, points away from the system since the system does the work
The heat, Q, points away from the system since heat is given off
Apply first law of thermodynamic;
ΔU = Q + W
where;
q is the heat flowing into or out of the system
(+q if the heat is flowing into the system
(-q if the heat is leaving the system
w is the work done by or on the system
(+w if the work is done on the system by the surrounding
(-w if the work is done by the system to the surrounding
Thus, from the above explanation, the change in internal energy of the system is calculated as;
ΔU = -Q - W
ΔU = - 5.97 x 10⁵ J - 6.53 x 10⁴ J
ΔU = -6.623 x 10⁵ J
The magnitude of the change in internal energy = 6.623 x 10⁵ J
(b) the efficiency of the athlete;
[tex]Efficiency = \frac{W}{Q} \times 100\%\\\\Efficiency = \frac{6.53 \times 10^4}{5.97 \times 10^5} \times 100\%\\\\Efficiency = 10.94 \ \%[/tex]