Answer:
Answer:
Speed of the wave in the string will be 3.2 m/sec
Explanation:
We have given frequency in the string fixed at both ends is 80 Hz
Distance between adjacent antipodes is 20 cm
We know that distance between two adjacent anti nodes is equal to half of the wavelength
So \frac{\lambda }{2}=20cm
2
λ
=20cm
\lambda =40cmλ=40cm
We have to find the speed of the wave in the string
Speed is equal to v=\lambda f=0.04\times 80=3.2m/secv=λf=0.04×80=3.2m/sec
So speed of the wave in the string will be 3.2 m/sec
What is the current in a 120V circuit if the resistance is 20Ω?
We have: [tex]I=\frac{U}{R}=\frac{120}{20}=6A[/tex]
ok done. Thank to me :>
Answer:
The Current is 6 Ampere.
Step-by-step explanation:
Given :
[tex]\small\red\bull[/tex] Voltage = 120V[tex]\small\red\bull[/tex] Resistance = 20ΩTo Find :
[tex]\small\red\bull[/tex] CurrentUsing Formula :
[tex] \star{\small{\underline{\boxed{\sf{ I = \dfrac{V}{R}}}}}}[/tex]
[tex]\small\blue\star[/tex] I = Current[tex]\small\blue\star[/tex] V = Voltage[tex]\small\blue\star[/tex] R = ResistanceSolution :
Substituting all the given values in the formula to find Current :
[tex]{\dashrightarrow{\pmb{\sf{ I = \dfrac{V}{R}}}}}[/tex]
[tex]{\dashrightarrow{\sf{ Current = \dfrac{Voltage}{Resistance}}}}[/tex]
[tex]{\dashrightarrow{\sf{ Current = \dfrac{120}{20}}}}[/tex]
[tex]{\dashrightarrow{\sf{ Current = \cancel{\dfrac{120}{20}}}}}[/tex]
[tex]{\dashrightarrow{\sf{ Current = 6A}}}[/tex]
[tex]{\star{\underline{\boxed{\sf{\red{ Current = 6A}}}}}}[/tex]
Hence, the Current is 6 Ampere.
[tex]\rule{300}{1.5}[/tex]
How many volts would it take to push 1 amp through a resistance of 1 ohm?
Answer:
[tex]1\; \rm V[/tex].
Explanation:
Ohm's Law give the following relationship between the potential [tex]V[/tex] across a conductor, the electrical resistance [tex]R[/tex] of that conductor, and the current [tex]I[/tex] through that conductor:
[tex]V = I \, R[/tex].
It is given that the current through this conductor is [tex]I = 1\; {\rm A}[/tex], and that the electrical resistance of this conductor is [tex]R = 1\; \rm \Omega[/tex].
Note, that the unit of electrical resistance Ohm is a composite unit; [tex]1\; {\rm \Omega} = 1\; {\rm V \cdot A^{-1}}[/tex] (volts per ampere.)
Substitute the value of [tex]I[/tex] and [tex]R[/tex] into the equation [tex]V = I \, R[/tex] from Ohm's Law to find the potential across this conductor:
[tex]\begin{aligned}V &= I \, R \\ &= 1\; {\rm A} \times 1\; {\rm \Omega} \\ &= 1\; {\rm A} \times 1\; {\rm V \cdot A^{-1}} \\ &= 1\; {\rm V}\end{aligned}[/tex].
what class of motion did aristotle attribute to the moon?
Over a period of 2.50 seconds, a speedboat accelerates uniformly from 18.5 m/s to 46.5 m/s. What is the acceleration of the speedboat relative to the shore?
According to the question, the acceleration of the speedboat relative to the shore is found to be 11.2 [tex]m/sec^2.[/tex]
What is Acceleration?Acceleration may be characterized as the rate of change of the velocity of an object with respect to time. It is a vector quantity, as it defined both magnitude and direction. This allows you to measure how fast velocity changes in meters per second squared (m/s^2).
According to the question,
The initial velocity of the boat = 18.5 m/s.
The final velocity of the boat = 46.5 m/s.
Difference in velocity = Final velocity - Initial velocity = 46.5-18.5 = 28 m/s.
The formula for calculating the acceleration of the boat is as follows:
a = Difference in velocity/time taken.= 28/2.50 = 11.2 [tex]m/sec^2[/tex].
Therefore, the acceleration of the speedboat relative to the shore is found to be 11.2 [tex]m/sec^2.[/tex]
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what is the static friction force and kinetic friction for the image
According to the plot, static friction force has a maximum magnitude of around 3.0 N, and kinetic friction has a magnitude of about 1.5 N.
The plot appears to be telling you the force required to get the yellow block moving along the table. If one applies less than 3.0 N of force, the block remains motionless. But as soon as it starts to slide, one need only apply 1.5 N of force to keep it moving (presumably at a constant speed).
why is it less air pressure up in the sky and why does it affect our ears
Answer:
When you in a airplane you are in high altitude and thats where air density drops. Thats where ear pops come from.
Explanation:
the motion of a particle along a straight line is represented by the position versus time graph above. at which of the labeled points on the graph is the magnitude of the acceleration of the particle greatest?
Point A has the largest magnitude of acceleration as compared to other points on the position verses time graph.
On the graph, A is the point where magnitude of the acceleration of the particle is greatest as compared to other positions on the graph because the height of point A is the largest as compared to other points of the graph.
The graph shows at which point acceleration of an object is higher and lower so we can conclude that point A has the largest magnitude of acceleration as compared to other points on the position verses time graph.
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name an instrument that is used to measure atmospheric pressure
Answer:
Barometer also called barometric pressure.
Answer:
A barometer
Explanation:
There are two types of barometers;
Mercury barometerAneroid barometer[tex]{}[/tex]
White light is an _____ mixture of all the colors.
Answer:
White light is a combination of all colors in the color spectrum.
Explanation:
It has all the colors of the rainbow. Combining primary colors of light like red, blue, and green creates secondary colors: yellow, cyan, and magenta
This is not meant to be a long, mathematically-intensive problem. If you find yourself going down that road, you might want to rethink your approach.
Answer:
rethink your approach to what?
Explanation:
7. A drag car takes off at the green light heading in a straight line. The car goes
from a complete stop to 45m/s in 9 seconds. What is its acceleration?
Answer:
405
Explanation:
If the car stops at 45m/s in 9 seconds then it took the car 405 meters to stop
The International Space Station (ISS) orbits Earth at an altitude of 4.08 × 105 m above the surface of the planet. At what velocity must the ISS be moving in order to stay in its orbit?(1 point)
A) 7.91 × 10^3 m/s
B) 3.12 × 10^4 m/s
C) 7.66 × 10^3 m/s
D) 8.17 × 10^3 m/s
Satellite A is orbiting Earth at an altitude of 500 km and Satellite B is orbiting 800 km above the surface. How does the velocity of Satellite A compare to the velocity of Satellite B?(1 point)
A) It depends on the masses of the satellites.
B) The velocity of Satellite A is greater than the velocity of Satellite B.
C) The velocity of Satellite B is equal to the velocity of Satellite A.
D) The velocity of Satellite A is less than the velocity of Satellite B.
What is the direction of the net force acting on a satellite as it orbits Earth at a constant speed?(1 point)
A) in the direction the satellite is moving
B) toward the center of Earth
C) away from the center of Earth
D) opposite the direction the satellite is moving
What happens to the gravitational force and orbital velocity of a satellite as the satellite transfers to an orbit that is closer to Earth?(1 point)
A) The gravitational force decreases and the velocity increases.
B) The gravitational force increases and the velocity increases.
C) The gravitational force decreases and the velocity decreases.
D) The gravitational force increases and the velocity decreases.
Newton's second law and universal gravitation law allow to find the results for questions about the motion of satellites in orbit are:
1) The correct answer is D: v = 8.17 10³ m/s
2) The correct answer is B
The velocity of Satellite A is greater than the velocity of Satellite B.
3) The correct answer is B
Toward the center of Earth
4) The correct answer is B
The gravitational force increases and the velocity increases.
Newton's second law establishes a relationship between force, mass, and the acceleration of bodies.
F = m a
In the case of the satellite the force is given by the law of universal gravitation.
[tex]F = - G \frac{Mm}{r^2 }[/tex]
Where G is the constant of universal gravitation. M and m the mass of each object and r the distance between them.
In this case the satellite is in a circular orbit, therefore the acceleration is centripetal.
[tex]a = \frac{v^2}{r}[/tex]
We substitute.
[tex]G \frac{Mm}{r^2} = m \frac{v^2}{r}[/tex]
[tex]\frac{GM}{r} = v^2[/tex]
Let's analyze the answers to find the correct one.
1) They indicate the height of the space station r = 4.08 10⁵ m and ask the speed.
[tex]v= \sqrt{ \frac{6.67 \ 10 ^{-11} 5.9 \ 10^{24}}{4.08 \ 10^6 } }[/tex]
v = 9.82 10³ m / s
The correct answer is D.
2) Satellite A has an orbit of hₐ = 500 km and satellite b an orbit of
h_b = 800 km
The distance from the center of the earth to each satellite is:
rₐ = R + hₐ
r_b = R + h_b
rₐ = 6.37 106 + 500 10³ = 6.87 10⁶ m
r_b = 6.37 10⁶ + 800 10³ = 7.17 10⁶ m
Let's find the ratio of the speeds
[tex]\frac{v_a}{v_b} = \sqrt{ \frac{r_b}{r_a} } \\\frac{v_a}{v_b} = \sqrt{ \frac{7.17}{6.87} }[/tex]
[tex]\frac{v_a}{v_b}[/tex] = 1,022
we see that the speed of satellite a is slightly greater than the speed of satellite b.
Let's analyze the claims.
A) False. The speed does not depend on the mass of the satellites.
B) True. The velocity of a is slightly greater than the velocity of b.
C) False. The speed of a is greater.
D) False. The speed of a is greater.
3) As the orbit is circular, the force must be radial, that is, it points towards the center of the earth.
Let's analyze the claims.
A) False. The speed modulus does not change, therefore there is no acceleration in the direction of the satellite.
B) True. Aim for the center of the Earth, change the direction of the velocity.
C) False. Aim for the scepter of the earth.
D) false. The modulus of velocity is constant and the direction changes towards the center of the earth, therefore the force must go towards the center of the earth.
4) The force in the law of universal gravitation increased as the distance decreased.
When a satellite approaches the earth its speed must increase since the speed is proportional in inverse of the square root of the distance.
Let's examine the claims.
A) False. The attraction force increases.
B) True. You agree with the explanation.
C) False. The gravitational force increases.
D) False. Speed increases.
In conclusion, using Newton's second law and the universal law of gravitation we can find the results for the questions about the movement of satellites in orbit are:
1) The correct answer is D: v = 8.17 10³ m/s
2) The correct answer is B
The velocity of Satellite A is greater than the velocity of Satellite B.
3) The correct answer is B
Toward the center of Earth
4) The correct answer is B
The gravitational force increases and the velocity increases.
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Answer:
1. 7.66 × 10^3 m/s
2. The velocity of Satellite A is greater than the velocity of Satellite B.
3. toward the center of Earth
4. The gravitational force increases and the velocity increases.
Explanation:
Calculate the elastic potential energy stored in a spring if it has a force constant of 150 N/m. the spring is extended to a length of 0.30m
Answer:
6.75J
Explanation:
U=1/2KΔx²
U=0.5* 150*0.30^2
How did Thomson's model get its name?
Answer:
thats actually a really good question. sadly i don't know the answer to it. but im sure that there are others who can!
which term is defined by the interaction between two charged particles?
Answer: Electromagnetism
The term which is defined by the interaction between two charged particles is known as electromagnetism.
What is an electromagnet?The study of charge and the fields and forces it generates is known as electromagnetism. Electromagnetism has two components: electricity and magnetism.
The special theory of relativity, developed by Albert Einstein in 1905, proved beyond a shadow of a doubt that both are components of the same reality. But in reality, magnetic and electric forces behave very differently and are modeled by various equations. Electric charges can produce forces whether they are stationary or moving. On the other hand, magnetic forces are only created by moving charges and only affect charges that are moving.
Thus, the term is known as electromagnetism.
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What is the frequency of a wave with a wavelength of 15 m and a wavespeed of
300 m/s?
Answer: f=20 (i think)
Explanation:
all I did was divide 300 and 15.
300/15= 20
Please help my physics teacher cannot teach pls hurry pls help uwu
Hi there!
To find how much the spring stretches with the mass attached, we must find the spring's equilibrium point.
The equilibrium point satisfies the following:
[tex]\large\boxed{\Sigma F = 0}[/tex]
The two forces acting in this situation are the weight of the mass and the spring force.
The spring force (restoring force) is in the opposite direction of the weight, so:
[tex]\Sigma F = 0 = Mg- k\Delta x[/tex]
Mg = weight
k = Spring Constant (N/m)
Δx = distance to equilibrium point (m)
Now, we can set the two equal to each other:
[tex]Mg = k\Delta x\\\\544 = 1146\Delta x\\\\\boxed{\Delta x= .475 m}[/tex]
How far does an object travel if it is moving at a speed of 4 m/s for 3.2 seconds?
Answer:
12.8 meters
Explanation:
m/s means meters per second. So, to find how far it would travel in 3.2 seconds, you would multiply 4 meters every second by 3.2 seconds.
4×3.2=12.8 meters
On average, how many stars would we have to search before we would expect to hear a signal? assume there are 500 billion stars in the galaxy.
We would have to search at least 5,000,000,000 (5 billion) stars before we would expect to hear a signal.
To find out the number of stars that we will need to search to find a signal, we need to use the following formula:
total of stars/civilizations500,000,000,000 (500 billion) stars / 100 civilization = 5,000,000,000 (5 billion)
This shows it is expected to find a civilization every 5 billion stars, and therefore it is necessary to search at least 5 billion stars before hearing a signal from any civilization.
Note: This question is incomplete; here is the complete question.
On average, how many stars would we have to search before we would expect to hear a signal? Assume there are 500 billion stars in the galaxy.
Assuming 100 civilizations existed.
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7. If the speed of light in medium-1 and medium-2 are 2.5x 10 m/s and 2x 10 m/s respectively then the refractive index of medium-1 with respect to medium-2 is (a) 3/2.5 (b) 2/2.5 (c) 2.5/3
Answer:
Answer to the question is 4 / 5
To know the answer with proof see the above attachmentHelpppppppp pleaseeeee!!
The mass of an atom is concentrated in its
a. nucleus
b. electron
c. neutron
d. angstrom
e. isotope
What is the mass of a box if it accelerated 5 m/s from a force of 12 N?
Answer:
60
Explanation:
You multiply the acceleration and the force and you get your answer
Mechanical energy is similar to thermal energy because they both?
Answer:
Mechanical energy is the ordered movement of the molecules as a single unit. Thermal energy is the random movement of the molecules. Mechanical energy can be 100% converted to thermal energy, but thermal energy cannot be fully converted to mechanical energy.
explain how the spray painting process makes use of static electricity
Answer:
The spray paint gun is charged positively, which causes every paint particle to become positively charged. Since the paint particles all have the same charge, they repel each other and produce a fine mist of paint which is sprayed towards the wall/car.
Answer:
xerography can i get brainliest
Explanation:
Jeff Gordon leads his race and must drive into a curve at top speed to win it all. The radius of the curve is 1000 m and the coefficient of static friction between his tires and the dry pavement is 0.50.
Answer:
Apparently the curve is not banked.
The maximum frictional force is -
Ff = m g * .5
Fc = m v^2 / R where Fc is the centripetal force that must be present in
order for the car not to slip.
.5 g = v^2 / R equating forces
v^2 = .5 R g = .5 ^ 1000 * 9.80
v = (500 * 9.80)^1/2 = 70 m/s
(about 230 ft/sec = 230/88 * 60 = 157 mph
A 10.0-kg crate slides along a raised horizontal frictionless surface at a constant speed of 4.0 m/s. The crate then slides down a frictionless incline and across a second, roughened horizontal surface as shown in the figure. What is the kinetic energy of the crate as it reaches the lower surface?
The kinetic energy of the crate as it reaches the lower surface is 80 J
Kinetic energy is the energy possed by an object in motion. Mathematically, the kinetic energy can be expressed as follow:
KE = ½mv²
With the above formula, the kinetic energy of the crate can be obtained as follow:
Mass (m) = 10 KgVelocity (v) = 4 m/sKinetic energy (KE) =?KE = ½mv²
KE = ½ × 10 × 4²
KE = 5 × 16
KE = 80 J
Therefore, the kinetic energy of the crate is 80 J
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1. What do machines do? Change the amount of force one exerts. Decrease the amount of work that is done. Increase the amount of work that is done. Eliminate friction.
Answer: increase the amount of work that is done?
Explanation:
I’m not entirely sure to be honest but I’ll just use my logic here. Machines are made to increase efficiency and production. Machines are made to save time and energy as a pose to human work. So several machines can mass produce a lot of things and keep business booming. I’m not very good at explaining things but I hope this makes sense eventually.
an object moves 15.0 m north and then 11.0 m south. find both the distance it has traveled and the magnitude of its displacement.
What is an object's mass if it accelerates at 5 m/s2 when a force of 0.5 N is applied?
Answer:
0.1kg
Explanation:
mass = force / acceleration
0.5 / 5
0.1kg
PLEASE HELP WILL GIVE BRAINLIEST
Answer:
D, A
Explanation:
#1
Height is same for all cars
So they all will arrive at same speed but A will come first
Option D