Answer:
Current through them, voltage across them.
Explanation:
Ohm's law states that at constant temperature, the current flowing in an electrical circuit is directly proportional to the voltage applied across the two points and inversely proportional to the resistance in the electrical circuit.
Mathematically, Ohm's law is given by the formula;
[tex] V = IR[/tex]
Where;
V represents voltage measured in voltage.
I represents current measured in amperes.
R represents resistance measured in ohms.
A resistor can be defined as an electronic component that opposes the free flow of current from one point to another in an electrical circuit.
In a series circuit of two resistors, the resistors have the same current through them; in a parallel circuit of two resistors, the resistors have the same voltage across them.
In a liquid with a density of 1500 kg/m3, longitudinal waves with a frequency of 410 Hz are found to have a wavelength of 7.80 m. Calculate the bulk modulus of the liquid.
Answer:
The bulk modulus of the liquid is 1.534 x 10¹⁰ N/m²
Explanation:
Given;
density of the liquid, ρ = 1500 kg/m³
frequency of the wave, F = 410 Hz
wavelength of the sound, λ = 7.80 m
The speed of the wave is calculated as;
v = Fλ
v = 410 x 7.8
v = 3,198 m/s
The bulk modulus of the liquid is calculated as;
[tex]V = \sqrt{\frac{B}{\rho} } \\\\V^2 = \frac{B}{\rho}\\\\B = V^2 \rho\\\\B = (3,198 \ m/s)^2 \times 1500 \ kg/m^3\\\\B = 1.534 \ \times 10^{10} \ N/m^2[/tex]
Therefore, the bulk modulus of the liquid is 1.534 x 10¹⁰ N/m²
As the distance between the sun and earth decreases, the gravity force between them
a
Increases
b
decreases
c
stays the same
Answer:
a increases
Explanation:
as distance between two objects increases the gravitational force decreases so when distance decreases the gravitational force increases
In January 2017, when Clemson won the football championship, Coach Dabo Swinney decided to buy pizza for all students at Clemson to celebrate. So, From his office(Death Valley), he drove 100 m North to reach HWY 93, 200 m East to reach the 133 Junction(Downtown) and 500 m North to reach Papa John's to place his order. What was the total displacement of Coach Swinney
Answer:
Explanation:
We shall represent each move of coach in vector form , considering unit vector i towards east and unit vector j towards north .
he drove 100 m North
First displacement D₁ = 100 j
200 m East to reach the 133
second displacement D₂ = 200 i
500 m North to reach Papa John's
third displacement
D₃ = 500 j
Total displacement ( resultant displacement )
= D₁ + D₂ + D₃
= 100 j + 200 i + 500 j
= 200 i + 600 j
magnitude of resultant displacement
= √ ( 200² + 600² )
= √ 40000 + 360000
= 632.45 m
keli learned that an air mass is a very large body of air with similar temperature humidity and pressure and the air mass are constantly in motion she knows that you're messing depending on the temperature and moisture content tent of region where they form she looked up more information about what makes them move what are the major causes for moving & Masten North America choose two that apply.
Answer choices
A. changing humidity
B. low temperature
C. jet storm
D. prevailing westerlies
Air masses from the tropics and the equator are warm as they form over lower latitudes. The major causes for moving air masses North America exists jet storm.
What is meant by air mass?An air mass is a volume of air that in meteorology is identified by its temperature and humidity. Many hundreds or thousands of square miles are covered by air masses, which adjust to the properties of the land underneath them. Latitude and their continental or maritime source regions are used to categories them.
Warmer air masses are referred to as tropical, whilst colder air masses are referred to as polar or arctic. Superior and maritime air masses are moist, whereas continental and superior air masses are dry. Air masses with various densities are divided by weather fronts. Once an air mass has left its original location, nearby plants and bodies of water can quickly change the way it behaves. Classification systems address both the properties and modification of an air mass.
Air masses from the tropics and the equator are warm as they form over lower latitudes. They move poleward along the southern edge of the subtropical ridge and are drier and hotter than those that originate over seas. Trade air masses are another name for tropical maritime air masses. The Caribbean Sea, southern Gulf of Mexico, and tropical Atlantic Oceans, east of Florida via the Bahamas, are the origins of maritime tropical air masses that have an impact on the United States.
Monsoon air masses are moist and unstable. Rarely do dry superior air masses touch the ground. A trade wind inversion, which is a warmer and drier layer over the more moderately moist air mass below, is typically created over maritime tropical air masses when they are located above them.
Therefore, the correct answer is option C. jet storm.
To learn more about Air mass refer to:
https://brainly.com/question/19626802
#SPJ2
The range is the horizontal distance from the cannon when the pumpkin hits the ground. This distance is given by the product of the horizontal velocity (which is constant) and the amount of time the pumpkin is in the air (which is determined by the vertical component of the initial velocity, as you just discovered). Set the initial speed to 14 m/s, and fire the pumpkin several times while varying the angle between the cannon and the horizontal.
Required:
For which angle is the range a maximum (with the initial speed held constant)?
Answer:
Explanation:
For range o a projectile , the formula is as follows
R = u² sin2Ф / g where u is initial velocity of throw , Ф is angle of throw and g is acceleration due to gravity .
Here u = 14 m /s
R = 14² sin2Ф / 9.8
R = 20 sin2Ф
Now R will have maximum value when sin2Ф has maximum value .
Maximum value of sin2Ф = 1
sin2Ф = 1 = sin 90°
Ф = 45°
So when throw is aimed at 45° , range will be maximum .
I need help please will mark brainliest
Answer:
c
Explanation:
a childs weight is 331 N. what is the childs mass in kg?
Answer:
the child's mass is 33.1 kg
As the distance between the sun and earth decreases, the force of gravity
a
Increases
b
decreases
c
stays the same
Answer:
B Decrease
Explanation:
ik im right cuz i looked up the answer
What would cause surface ocean water to have a higher salt content?
A.
Surface ocean water will have a higher salt content from the melting of sea ice
B.
Surface ocean water will have a higher salt content from low rates of evaporation and high rates of precipitation.
C.
Surface ocean water will have a higher salt content from water flowing out of a river into the ocean
D.
Surface ocean water will have a higher salt content from high rates of evaporation and low rates of precipitation
Answer:
d I think? not sure I don't know much abt the ocean
What is the efficiency of a machine that has an output work of 1675 J and an input work
of 1895 J?
Answer:
1.13%
explanation:
work output/work input =100%
g In an historical movie, two knights on horseback start from rest 84.1 m apart and ride directly toward each other to do battle. Sir George's acceleration has a magnitude of 0.316 m/s2, while Sir Alfred's has a magnitude of 0.289 m/s2. Relative to Sir George's starting point, where do the knights collide?
Answer:
The knights will collide at 43.854 meters relative to Sir George's starting point.
Explanation:
Let suppose that initial positions of Sir George and Sir Alfred are 0 and 84.1 meters, respectively. If both knights accelerate uniformly, then we have the following kinematic formulas:
Sir George
[tex]x_{G} = x_{G,o}+v_{o,G}\cdot t + \frac{1}{2}\cdot a_{G}\cdot t^{2}[/tex] (1)
Sir Alfred
[tex]x_{A} = x_{A,o}+v_{o,A}\cdot t + \frac{1}{2}\cdot a_{A}\cdot t^{2}[/tex] (2)
Where:
[tex]x_{G,o}[/tex], [tex]x_{A,o }[/tex] - Initial position of Sir George and Sir Alfred, measured in meters.
[tex]x_{G}[/tex], [tex]x_{A}[/tex] - Final position of Sir George and Sir Alfred, measured in meters.
[tex]v_{o,G}[/tex], [tex]v_{o,A}[/tex] - Initial velocity of Sir George and Sir Alfred, measured in meters per second.
[tex]t[/tex] - Time, measured in seconds.
[tex]a_{G}[/tex], [tex]a_{A}[/tex] - Acceleration of Sir George and Sir Alfred, measured in meters per square second.
Both knights collide when [tex]x_{G} = x_{A}[/tex], then we simplify this system of equations below:
[tex]x_{G,o} + v_{o,G}\cdot t + \frac{1}{2}\cdot a_{G}\cdot t^{2} = x_{A,o}+v_{o,A}\cdot t + \frac{1}{2}\cdot a_{A}\cdot t^{2}[/tex]
[tex](x_{A,o}-x_{G,o}) +(v_{o,A}-v_{o,G})\cdot t +\frac{1}{2}\cdot (a_{A}-a_{G})\cdot t^{2} = 0[/tex] (3)
If we know that [tex]x_{A,o} = 84.1\,m[/tex], [tex]x_{G,o} = 0\,m[/tex], [tex]v_{o,A} = 0\,\frac{m}{s}[/tex], [tex]v_{o,G} = 0\,\frac{m}{s}[/tex], [tex]a_{A} = -0.289\,\frac{m}{s^{2}}[/tex] and [tex]a_{G} = 0.316\,\frac{m}{s^{2}}[/tex], then we have the following formula:
[tex]84.1 -0.303\cdot t^{2} = 0[/tex] (4)
The time associated with collision is:
[tex]t \approx 16.660\,s[/tex]
And the point of collision is:
[tex]x_{G} = 0\,m + \left(0\,\frac{m}{s} \right)\cdot (16.660\,s)+ \frac{1}{2}\cdot \left(0.316\,\frac{m}{s^{2}} \right) \cdot (16.660\,s)^{2}[/tex]
[tex]x_{G} = 43.854\,m[/tex]
The knights will collide at 43.854 meters relative to Sir George's starting point.
A 45945990 prism is immersed in water. A ray of light is incident normally on one of its shorter faces. What is the minimum index of refraction that the prism must have if this ray is to be totally reflected within the glass at the long face of the prism
Answer:
Explanation:
Angle of incidence at the longer face = 45⁰ ( see the figure in the attached file )
For total reflection from longer face
i = C where C is critical angle .
And relation between critical angle and refractive index is as follows .
μ = 1 / sinC
= 1 / sin45
= √2
= 1.414 .
why would the bulb not light?
Answer:
It's not connected to the negative end of the battery
Explanation:
To turn on it would need to connect to the positive (+) and negative (-) ends of the battery
A space vehicle is coasting at a constant velocity of 22.3 m/s in the y direction relative to a space station. The pilot of the vehicle fires a RCS (reaction control system) thruster, which causes it to accelerate at 0.203 m/s2 in the x direction. After 56.7 s, the pilot shuts off the RCS thruster. After the RCS thruster is turned off, find (a) the magnitude and (b) the direction of the vehicle's velocity relative to the space station. Express the direction as an angle (in degrees) measured from the y direction.
Answer:
Explanation:
Initial velocity in y direction Vy = 22.3 m /s
initial acceleration in x direction ax = .203 m /s ²
time of acceleration t = 56.7 s
final velocity in x direction
v = u + a t
Vx = 0 + .203 x 56.7 = 11.51 m /s
Final velocity in y direction will remain same as initial velocity in y direction = 22.3 m /s because there is no acceleration in y direction .
Magnitude of final velocity
= √ ( Vx² + Vy²)
= √ (22.3² + 11.51² )
= √ ( 497.29 + 132.48)
= 25.1 m /s
Direction of final velocity from y direction be Ф
TanФ = Vx / Vy = 11.51 / 22.3 = .516
Ф = 27.3° .
Compare and contrast climate and weather
A 5.1 g bullet is fired into a 2.3 kg ballistic pendulum. The bullet emerges from the block with a speed of 221 m/s, and the block rises to a maximum height of 7 cm . Find the initial speed of the bullet. The acceleration due to gravity is 9.8 m/s 2 . Answer in units of m/s.
Answer:
748.62 m /s.
Explanation:
mass of bullet m = .0051 kg .
mass of block M = 2.3 kg
block rises to a height of .07 m so velocity of block after collision
V = √ 2 gh
=√ (2 x 9.8 x .07 )
= 1.17 m /s
velocity of bullet after collision v = 221 m /s
Now we shall apply law of conservation of momentum to find out the velocity of bullet before collision .
Let it be Vx . then
5.1 x 10⁻³ x Vx + 0 = 5.1 x 10⁻³ x 221 + 2.3 x 1.17
= 1.127 + 2.691 = 3.818
Vx = 748.62 m /s
What is the car's acceleration from 0 to 1 second?
A. 8 mph/s
B. 20 mph/s
C. 60 mph/s
D. 10 mph/s
The electric field between two parallel plates is uniform, with magnitude 646 N/C. A proton is held stationary at the positive plate, and an electron is held stationary at the negative plate. The plate separation is 4.26 cm. At the same moment, both particles are released.
Required:
Determine the distance from the positive plate at which the two pass each other.
Answer:
The distance from the positive plate at which the two pass each other is 0.0023 cm.
Explanation:
We need to find the acceleration of each particle first. Let's use the electric force equation.
[tex]F=Eq[/tex]
[tex]ma=Eq[/tex]
For the proton
[tex]m_{p}a_{p}=Eq_{p}[/tex]
[tex]a_{p}=\frac{Eq_{p}}{m_{p}}[/tex]
[tex]a_{p}=\frac{646*1.6*10^{-19}}{1.67*10^{−27}}[/tex]
[tex]a_{p}=6.19*10^{10}\: m/s^{2}[/tex]
For the electron
[tex]m_{e}a_{e}=Eq_{e}[/tex]
[tex]a_{e}=\frac{Eq_{e}}{m_{e}}[/tex]
[tex]a_{e}=\frac{646*1.6*10^{-19}}{9.1*10^{−31}}[/tex]
[tex]a_{e}=1.14*10^{14}\: m/s^{2}[/tex]
Now we know that the plate separation is 4.26 cm or 0.0426 m. The travel distance of the proton plus the travel distance of the electron is 0.0426 m.
[tex]x_{p}+x_{e}=0.0426[/tex]
Both of them have an initial speed equal to zero. So we have:
[tex]\frac{1}{2}a_{p}t^{2}+\frac{1}{2}a_{e}t^{2}=0.0426[/tex]
[tex]t^{2}(a_{p}+a_{e})=2*0.0426[/tex]
[tex]t^{2}=\frac{2*0.0426}{a_{p}+a_{e}}[/tex]
[tex]t=\sqrt{\frac{2*0.0426}{6.19*10^{10}+1.14*10^{14}}}[/tex]
[tex]t=2.73*10^{-8}\: s[/tex]
With this time we can find the distance from the positive plate (x(p)).
[tex]x_{p}=\frac{1}{2}a_{p}t^{2}[/tex]
[tex]x_{p}=\frac{1}{2}6.19*10^{10}*(2.73*10^{-8})^{2}[/tex]
[tex]x_{p}=0.0023\: cm[/tex]
Therefore, the distance from the positive plate at which the two pass each other is 0.0023 cm.
I hope it helps you!
Explain how you could use iron filings and a piece of paper to help reveal the effect of a magnetic field.
Answer:
you could put the iron filings on the peace of paper and hover a magnet over top of the paper and the iron filings would stand up, or even stick to the magnet
Explanation:
What Is a Sound Wave? Learning Goal: To understand the nature of a sound wave, including its properties: frequency wavelength, loudness, pitch, and timbre. Sound is a phenomenon that we experience constantly in our everyday life. Therefore, it is important to understand the physical nature of a sound wave and its properties to correct common misconceptions about sound propagation Most generally, a sound wave is a longitudinal wave that propagates in a medium (ie, air) The particles in the medium oscillate back and forth along the direction of motion of the wave. This displacement of the particles generates a sequence of compressions and rarefactions of the medium Thus, a sound wave can also be described in terms of pressure variations that travel through the medium. The pressure fluctuates at the same frequency with which the particles positions oscillate When the human ear perceives sound. It recognizes a series of pressure fluctuations rather than displacements of individual air particles. Part 1 Figure 1 of 2 > Fi MA length Part A Based on the information presented in the introduction of this problem, what is a sound wave? Propagation of sound particles that are offerent from the particles that comprise the medium Propagation of energy that does not require a medium Propagation of pressure fluctuations in a medium Propagation of energy that passes through empty spaces between the partides that com Submit Request Anst Part B Complete previous parts) Part hall to the other? Does air play a role in the propagation View Available Hints) SUITE Part D The graphs shown in (Figure 1) represent pressure variation versus time recorded by Enter the letters of all the correct answers in alphabetical order.
Answer:
A) Propagation of pressure fluctuations in a medium
B) air is the medium in which the wave is transported,
Explanation:
Part A.
A sound wave is a longitudinal oscillation of the molecules that forms in a material medium, they can be solid, liquid or gases, therefore the wave propagates in the same direction as the oscillation of the particles.
The most correct answer is:
* Propagation of pressure fluctuations in a medium
Part b
air is the medium in which the wave is transported, otherwise it cannot propagate
A toy rocket is launched with an initial velocity of 10.0 m/s in the horizontal direction from the roof of a 38.0-m-tall building. The rocket's engine produces a horizontal acceleration of (1.60 m/s^3)t, in the same direction as the initial velocity, but in the vertical direction the acceleration is g, downward. Air resistance can be neglected.
Required:
What horizontal distance does the rocket travel before reaching the ground?
Answer:
Explanation:
Consider downward displacement first .
downward displacement h = 38 m .
downward acceleration = g
downward initial velocity u = 0
h = ut + 1/2 g t ²
38 = 0 + 9.8 t ²
t = 1.97 s
During this period , there will be horizontal displacement with initial velocity u = 10 m /s and acceleration a = 1.6 m /s²
s = ut + 1/2 a t²
= 10 x 1.97 + .5 x 1.6 x 1.97²
= 19.7 + 3.10
= 22.8 m
As the distance between the sun and earth decreases, the speed of the planet
a
increases
b
decreases
c
stays the same
Answer:
Explanation:
Increases. The force of gravity is distance dependent. Therefore, a smaller 'r' value will result in a larger force. Net force is proportional to the acceleration, so the planet will increase its speed.
What happens to the force attraction of the distance two objects is increased?
Answer:
Explanation:
The attraction weakens. Two objects that are farther apart are not drawn together as strongly as if they were close together.
What is the average speed of a cheetah that sprints 100 m in 4s? How about if it sprints 50 m in 2 s
Speed = (distance covered) / (time took)
A characteristic of a nebula is that it-
Answer:
Center of solar system
Explanation:
Answer: b
Explanation:
What is the main cause of ocean currents? Question 2 options:
The prevailing winds
The Coriolis effect
Waves
The sun and the moon
An ideal spring is lying horizontally on a frictionless surface. One end of the spring is attached to a wall. The other end is attached to a moveable block that has a mass of 5 kg. The block is pulled so that the spring stretches from its equilibrium position by 0.65 m. Then the block is released (from rest), and as a result the system oscillates with a frequency of 0.40 Hz (that's 0.40 rev/sec) Find:
a) the acceleratiuon of the block when the spring is stretched by 0.28 m.
b) the maximum force magnitude exerted by the spring on the block.
c) the oscillation frequency of a 2.5 kg blcok under the same circumstances (i.e. with the same spring and initial displacement).
Answer:
a) a = - 1.76 m / s², b) F = 20.5 N, c) w = 3.55 rad / s
Explanation:
a) a simple harmonic motion is described by the expression
x = A cos (wt + Ф)
in this case they give us the frequency
w = 2π f
w = 2π 0.40
w = 2.51 rad / s
as the maximum elongation is 0.65 m this corresponds to the amplitude of the movement
A = 0.65 m
to find the phase constant (Ф) we use the initial condition that for t = 0 v = 0 and x = A, we substitute
A = A cos (0+ Ф)
cos Ф = 1
Ф = 0
the resulting equation is
x = 0.65 cos (2.51 t)
Let's find the time it takes to get to x = 0.28 m
0.28 = 0.65 cos 2.51 t
2.51 t = cos-1 (0.28 / 0.65)
remember angles are in radians
t = 1.1254 / 2.51
t = 0.448 s
the acceleration is
[tex]a = \frac{dv}{dt} = \frac{d^2x}{dt^2}[/tex]
a = -A w² cos wt
we subtitle
a = - 0.65 2.51² cos (2.51 0.448)
a = - 1.76 m / s²
b) the maximum acceleration occurs when the cosine is ±1
a = A w²
a = 0.65 2.51²
a = 4.10 m / s²
Let's use Newton's second law
F = m a
F = 5 4.1
F = 20.5 N
c) The angular velocity is given by
w² = k / m
let's find the spring constant
k = m w²
k = 5 2.51²
k = 31.5 N / m
therefore if the block is exchanged for another with mass m'= 2.5 kg
w = √(31.5 / 2.5)
w = 3.55 rad / s
A baseball is hit when it is 2.5 ft above the ground. It leaves the bat with an initial velocity of 145 ft/sec at a launch angle of 23°. At the instant the ball is hit, an instantaneous gust of wind blows against the ball, adding a component of -14i (ft/sec) to the ball’s initial velocity. A 15-ft-high fence lies 300 ft from home plate in the direction of the flight.
a. Find a vector equation for the path of the baseball.
b. How high does the baseball go, and when does it reach maxi-mum height?
c. Find the range and flight time of the baseball, assuming that the ball is not caught.
d. When is the baseball 20 ft high? How far (ground distance) is the baseball from home plate at that height?
e. Has the batter hit a home run? Explain.
Answer:
Explanation:
Take base of the ground as origin .
component of initial velocity along i and j direction is 145 con23 and 145 sin23 . Along j , gravity acts but along i , no force acts .
The path of ball in vector form
s = (145 cos23- 14 )t i + ( 2.5 + 145sin23 t - 1/2 g t² ) j
t is time period .
b )
vertical component of initial velocity = 145 sin 23 =
for vertical displacement
v² = u² - 2gH
For maximum height , v = 0
0 = (145 sin 23 )² - 2 g H , H is maximum height attained .
H = 3209.56 / 2 x 9.8
= 163.75 m
Total height attained = 163.75 + 2.5 = 166.25 m
if time be t for reaching maximum height
v = u -gt
0 = 145 sin 23 - gt
t = 145 sin23 / g
= 5.78 s
c )
For time of flight , vertical displacement = 2.5 m
2.5 = - 145 sin 23 t + 1/2 g t²
2.5 = -56.65 t + 4.9 t²
4.9 t² - 56.65 t - 2.5 = 0
t = 11.60s
horizontal displacement during this period = 145 cos23 x 11.60 = 1548.28 m
Range = 1548.28 m.
If two people, mass of 70 kg and 85 kg respectively, approach each other with speeds of 4 m/s and 7 m/s, what is the total momentum of the two person system? Give the momentum of each and then the total momentum.
Answer:
a. Momentum A = 280 Kgm/s.
b. Momentum B = 595 Kgm/s.
c. Total momentum = 875 Kgm/s.
Explanation:
Let the two people be A and B respectively.
Given the following data;
Mass A = 70kg
Mass B = 85kg
Velocity A = 4m/s
Velocity B = 7m/s
Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.
Mathematically, momentum is given by the formula;
[tex] Momentum = mass * velocity [/tex]
a. To find the momentum of A;
[tex] Momentum \; A = 70 * 4 [/tex]
Momentum A = 280 Kgm/s.
b. To find the momentum of B;
[tex] Momentum \; B = 85 * 7 [/tex]
Momentum B = 595 Kgm/s.
c. To find the total momentum of the two persons;
[tex] Total \; momentum = Momentum \; A + Momentum \; B [/tex]
Substituting into the equation, we have;
[tex] Total momentum = 280 + 595 [/tex]
Total momentum = 875 Kgm/s.
Two masses are being pulled up a 30.0-degree incline by a force F parallel to the incline. The acceleration up the incline is 1.00 m/s2 and the velocity is down the incline. The force is applied to a 200-kg mass and a string connects the 200-kg mass to a 150-kg mass. The coefficient of kinetic friction is 0.200. The force F is
Answer:
Explanation:
The acceleration up the incline is 1.00 m/s²
Net force acting on two masses = total mass x acceleration
= 350 x 1 = 350 N
weight acting down the plane = m g sinФ
= 350 x 9.8 x sin30 = 1715 N
Friction force acting down the plane = mg cosФ x μ where μ is coefficient of friction
= 350 x 9.8 x cos30 x .2 = 594N
Net force acting on total mass
= F - 1715 - 594 = 350 , where F is required force
F = 2659 N .