Answer:
Speed and frequency are directly proportional to each other.
Explanation:
The relationship between speed, frequency and wavelength is given by :
[tex]v=f\lambda[/tex]
Where
v is speed
f is frequency
[tex]\lambda[/tex] is wavelength
It can be seen that the relation between the speed and the frequency is direct. It means as the speed increases, the frequency will increase and vice versa.
If a dog is running at 4.38 m/s for 37.34 seconds, how far will it go?
Answer:
163.5m
this is how far the doggo would go
(In theory)
If I'm wrong I apologize
Charge flows from low potential to high potential.
O True
or
O False
Answer:
False
Explanation:
Think of the electric potential in terms of potential energy. If you imagine a place with high elevation (A) and another one at sea level (B), a ball will roll from high potential to low potential (A-->B).
Everything in our universe wants to reach a lower state of energy if no external force is acted upon it. Every object tends to slow down (friction), a radioactive element dissipates energy (an unstable element releases energy to get to a stable state), water in the clouds comes down to the ground (rain experiencing difference in potential energy).
Electric potential is exactly the same, you just can't see it! It flows from higher voltage (which is a synonym for electric potential) to lower voltage.
Can we always see the same amount of the illuminated side of the Moon from Earth? Explain. Thanks so much for everyone's help! :)
Answer:
Nope.
Explanation:
No. The Moon rotates on its own axis at the same rate that it orbits around Earth. That means we always see the same side of the Moon from our position on Earth. The side we don't see gets just as much light, so a more accurate name for that part of the Moon is the "far side."
No, we can't see the similar quantity of the illuminated side of the Moon from the planet (Earth).
The gravitation of the moon somewhat distorts the form of our entire globe as well as provides us tidal waves. Earth also taps upon that moon, generating a stony and high-threshold "bulge" that faces us.It finished functioning like something of brakes and slowed right down the moon's rotation to the common material, and therefore high moonlight tide confronts us constantly.
Thus the above answer is correct.
Learn more:
https://brainly.com/question/11606610
Serena is mixing a material into a beaker filled with a liquid. She notices that the material seems to disappear into the liquid. What physical property of the material is Serena most likely observing? solubility conductivity magnetism density
Answer:
Solubility
Explanation:
Material becomes soluble in solvent and disappears.
The physical property of the material is Serena most likely observing is solubility.
Explanation:
Solubility is defined as the property of a substance to get dissolve in another substance. A small amount of substance is termed as solute and a substance in a larger amount is called a solvent.
Intermolecular interaction of solute particles with solvent particles results in the dissolving of a solute in the solvent.
Conductivity is the property of the substance to conduct heat, electric current, or sound through them.
magnetism is a physical phenomenon that arises due to the motion of electric current through a conductor resulting in the production of a magnetic field.
Density is a measure of the mass of a substance present in the unit volume of the substance.
So, when Serena observed the disappearance of a material into a beaker filled with liquid she is observing the property of solubility of that material.
Learn more about solubility here:
brainly.com/question/13620168?referrer=searchResults
brainly.com/question/3110135?referrer=searchResults
If a crane uses 500 Joules to raise bricks 20 meters how much Newton’s of force were used
Answer:
25N
Explanation:
Given parameters:
Potential energy expended = 500J
Height = 20m
Unknown:
The force used = ?
Solution:
To solve this problem, we use the expression below:
Potential energy = Force applied x height
500 = Force applied x 20
Force applied = 25N
A shearing of 50N is applied to an aluminum rod with a length of 10m a cross sectional area of 1.0×10-5 and a shear modulus of 2.5×1010 as result the rod is sheared through a distance
Answer:
0.002 m or 2 mm
Explanation:
Given that:
Force, F = 50N
Area = 1 * 10^-5
Length, L = 10m
Shear modulus, = 2.5 * 10^10
Using the relation ;
D = (50 ÷ 1*10^-5) ÷ (2.5 * 10^10 ÷ 10)
D = 5000000 ÷ 2.5 * 10^9
D = 5 * 10^6 ÷ 2.5 * 10^9
D = (5/2.5) * 10^(6-9)
D = 2 * 10^-3
D = 0.002 m
1m = 1000 mm
0.002m = (1000 * 0.002) = 2 mm
Xavier is administering medication to his patient. He administers four cups 4 points
of liquid with two ounces in each and one cup of liquid with three ounces.
How many ounces of liquid did the patient get? *
9 ounces
11 ounces
14 ounces
24 ounces
Answer:
11 ounces
Explanation:
The patient gets 11 ounces of liquid.
From the description of this problem:
Xavier administers four cups of liquid each containing two ounces
The number of ounces in the four cups = 2 x 4 = 8ounces
Also,
One cup of the liquid contains 3 ounces.
Total number of ounces administered = 8 + 3 = 11 ounces.
here is it u can see it
Answer:
The first one is true.
Explanation:
Here it is
Josie has a bag of ice that weighs 5 pounds. she left it in a sealed container and it melted. how much does the resulting water weigh? how do you know this?
true or false
A clue that a chemical change had occurred is that heat is given off.
Answer:
true
Explanation:
What is the energy transfer that takes place in a coal burning power plant?
HELP NEEDED
On a particle level, when does the least amount of thermal conduction occur
in a liquid?
A. When particles stop moving
B. When the average speed of particles is the same throughout the
liquid
C. When slow particles sink to the bottom
D. When fast particles rise to the top
Answer:
B. When the average speed of particles is the same throughout the liquid
2. Suppose your car has a maximum braking acceleration of -5 m/s2. Calculate the stopping distance for an initial speed of 25 m/s.
Answer:
s=62.5m
Explanation:
Use the equation v²=u²+2as, where v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance.
0²=25²+2(-5)s
10s=625
s=62.5m
What happens to sediment as a result of erosion and deposition?
(Basically what happens when Erosion or Deposition occur I think. Pls tell me if I was right along with your answer.)
Answer:
In the explanation
Explanation:
After erosion and eruptions which is a primary occuration for sediments, this will create heat and tough pressure leading the sediment to be extremely stable and Hard.
Henrietta manages to swing at a softball that was pitched very low - it almost hit the home
plate! Starting from very near the ground, the ball is launched upward at an angle with
amazing speed. 30 m/s upwards and 20 m/s across. 100m away from home plate is the outfield fence that's 24m tall. Will
Hentrietta get a home run?
a) How long will the ball take to get to the top?
b) How far from home plate will the ball be, once it's at the top?
c) Did the ball reach the fence yet, when it gets to the top?
d) How high up is the softball when it's at the top?
e) how long in seconds will it take for the ball to get 100m away from home plate?
F) How high above the ground is the ball when it’s 100m away from home plate?
G) finally, when the ball is 100m away from home plate, how high above the ground is it?
Answer:
a) The time it takes the ball to reach maximum height is approximately 3.06 seconds
b) The distance of the ball from the home plate once it is at the top is approximately 38.78 m
c) No
d) The height of the ball when it is at the top is approximately 45.92 m
e) The time it takes the ball to travel 100 m horizontally is 5 seconds
F) The height of the ball when it is 100 m away from the home plate is 27.5 m
G) The height of the ball above the ground when it 100 m away from the home plate, is 27.5 m
The height of the ball above the wall at 100 m from home plate is 3.5 m
Explanation:
The given parameters for the motion of the ball are;
The upward (vertical) velocity of the ball, [tex]u_y[/tex] = 30 m/s
The across (horizontal) velocity of the ball, uₓ = 20 m/s
The distance from the home plate to the outfield fence = 100 m
The height of the outfield fence = 24 m
a) The time, '[tex]t_{mh}[/tex]', it takes the ball to reach maximum height is by the following kinematic equation of motion;
[tex]v_y[/tex] = [tex]u_y[/tex] - g·t
Where;
[tex]u_y[/tex] = 30 m/s
g = The acceleration due to gravity ≈ 9.8 m/s²
At maximum height, we have;
[tex]v_y[/tex] = 0 m/s, t = [tex]t_{mh}[/tex]
∴ 0 m/s = [tex]u_y[/tex] - g·[tex]t_{mh}[/tex]
[tex]u_y[/tex] = g·[tex]t_{mh}[/tex]
[tex]t_{mh}[/tex] = [tex]u_y[/tex]/g
∴ [tex]t_{mh}[/tex] = [tex]u_y[/tex]/g = (30 m/s)/(9.8 m/s²) = 150/49 s ≈ 3.06 seconds
The time it takes the ball to reach maximum height, [tex]t_{mh}[/tex] ≈ 3.06 seconds
b) The horizontal distance travelled by the ball, when it is at maximum height, [tex]x_{mh}[/tex], is given as follows;
[tex]x_{mh}[/tex] = uₓ × [tex]t_{mh}[/tex] = 20 m/s × 150/49 s
[tex]x_{mh}[/tex] = 20 m/s × 150/49 s = 3000/49 m
[tex]x_{mh}[/tex] = 3000/49 m
The distance of the ball from the home plate, 'd', once it is at the top (maximum height), is given as follows;
d = 100 m - [tex]x_{mh}[/tex]
∴ d = 100 m - 3000/49 m = 1900/49 m ≈ 38.78 m
The distance of the ball from the home plate once it is at the top = d ≈ 38.78 m
c) The ball is yet to reach the fence when it gets to the top
The ball has approximately 38.78 meters to reach the fence
d) The height of the ball, [tex]h_{max}[/tex], when it is at the top is given by the following kinematic equation
[tex]h_{max}[/tex] = [tex]u_y[/tex]²/(2·g)
Where;
[tex]u_y[/tex] = 30 m/s
g = The acceleration due to gravity ≈ 9.8 m/s²
∴ [tex]h_{max}[/tex] = (30 m/s)²/(2×9.8 m/s²) = 2,250/49 m ≈ 45.92 m
The height of the ball when it is at the top, [tex]h_{max}[/tex] ≈ 45.92 m
e) The time, t₁₀₀, it will take the ball to get 100 m from home plate is found from the following relationship;
Velocity, u = Distance, x/(time, t)
∴ Time = Distance/(Velocity)
t = x/v
The given distance is the horizontal distance, therefore, we use the horizontal component of the velocity as follows;
v = Horizontal velocity of the softball = uₓ = 20 m/s
x = The given horizontal distance travelled by the ball = 100 m
t₁₀₀ = The time it takes the ball to travel 100 m horizontally
∴ t₁₀₀ = x/(uₓ) = 100 m/(20 m/s) = 5 s
The time it takes the ball to travel 100 m horizontally, t₁₀₀ = 5 seconds
F) The height of the ball above ground when it is 100 m away from the home plate is given as follows;
When the ball is 100 m from the home plate, the time, t₁₀₀ = 5 seconds
The height of the ball at 100m, [tex]h_{t_{(100 \, m)}}[/tex], is given according to the kinematic equation of motion for vertical height reached by an object as follows;
h = [tex]u_y[/tex]·t - 1/2·g·t²
∴ [tex]h_{t_{(100 \, m)}}[/tex] = [tex]u_y[/tex]·t₁₀₀ - 1/2·g·t₁₀₀²
Plugging in the known values gives;
[tex]h_{t_{(100 \, m)}}[/tex] = 30 m/s × 5 s - 1/2 × 9.8 m/s × (5 s)² = 27.5 m
The height of the ball when it is 100 m away from the home plate, [tex]h_{t_{(100 \, m)}}[/tex] = 27.5 m.
G) The height of the ball above the ground when it 100 m away from the home plate, is [tex]h_{t_{(100 \, m)}}[/tex] = 27.5 m
The height of the ball above the wall at 100 m from the home plate = The height of the ball when it is 100 m away from the home plate - The height of the wall at 100 m away from the home plate
∴ The height of the ball above the wall at 100 m from home plate = 27.5 m - 24 m = 3.5 m.
can a physical quantity have different dimensions in different systems of units?
Answer:
yes.
Explanation:
becausehjjkvfdssaajkkbcsssddhkkgddd
What are the characteristics of stars on
the main sequence of the Hertzsprung-
Russell diagram?
Answer:
The main sequence stretching from the upper left (hot, luminous stars) to the bottom right (cool, faint stars) dominates the HR diagram. It is here that stars spend about 90% of their lives burning hydrogen into helium in their cores. Main sequence stars have a Morgan-Keenan luminosity class labelled V.
Explanation:
On Distance Displacement Position vs. Time graphs. Velocity vs. Time graphs, and Acceleration vs. Time graphs, which vanable is always on the x-axis? A:Distance/Displacement/Position C:Acceleration B:Time D:Velocity
Answer:
your question dosent make sense?
Explanation:
Answer:
The time is always in the x axis.
Explanation:
A 1600kg cannon fires a 5kg cannonball horizontally. The exit velocity of the cannonball is 80m/s and the barrel length is 2m. What is the average acceleration of the cannonball before it leaves the barrel?
Answer:
a = 1600 m / s²
Explanation:
For this exercise we use the kinematics relations,
v² = v₀² + 2 a x
where v₀ is the initial velocity of the bullet, which as part of rest is zero, for the distance (x) we can assume that the gases accelerate along the entire trajectory of the cannon x = 2m
a = [tex]\frac{v^2}{2x}[/tex]
let's calculate
a = [tex]\frac{80^2}{2 \ 2}[/tex]
a = 1600 m / s²
HELP PLZZZ!!!
An archer pulls back the string of the bow (spring constant of 250 N/m) a distance of 1 meter. How much elastic energy is stored?
Answer:
im soo sad brainly wouldnt let me in the account that asked this question now i have to start everything over
Explanation:
A 98 N ball is suspended from a cable so that it hangs 3.5 m above the earth. Find the mass of the ball and the
gravitational PE of the ball.
Answer:
Yes
Explanation:
Sue and Shannon are having an argument about diet myths. In order to settle the argument, select the "myth" that is really a TRUE statement.
Answer:
To lose weight, you must use more weight than you consume. (D.)
Explanation:
I took the test and got a 100
A potential difference of 12Voltage is applied across a resistor of 120resistance the current in the circuit is
Answer:
the current flowing in the circuit is 0.1 A.
Explanation:
Given;
potential difference, V = 12 V
resistance of the circuit, R = 120 ohms
The current flowing in the circuit is calculated as;
V = IR
I = V/R
where;
I is the current flowing in the circuit
I = 12 / 120
I = 0.1 A
Therefore, the current flowing in the circuit is 0.1 A.
A 900 kg car has a momentum of 11,700 kg-m/s towards the west. What is the magnitude and direction of its velocity?
Answer:
mv=11700
m=900
v=11700/900=13m/s in west
A moving sidewalk has a velocity of 1.3 m/s north. If a man walks south on the sidewalk at a speed of 0.9 m/s, how long does it take him to travel 20 m relative to a stationary observer? O A. 21.9 s B. 15.4 s O C. 33.3 O D. 50.0 S
help please!!!
Answer:
D: 50 s
Explanation:
We are told that the sidewalk has a velocity of 1.3 m/s north.
Also, we are told that the man walks south on the sidewalk at a speed of 0.9 m/s.
Thus, it means they are moving in opposite directions. The net velocity of the man will be;
v_net = 1.3 - 0.9
v_net = 0.4 m/s
We want to find how long does it take him to travel 20 m relative to a stationary observer,thus;
We we know that;
Time = distance/speed
Time = 20/0.4
Time = 50 s
Which is true about using energy? There is an unlimited amount of energy in the universe. When energy is used, it disappears forever. When energy is used, it can transform to new types but can never disappear. A falling object increases its total energy.
Answer:
3. When energy is used, it can transform to new types but can never disappear.
Explanation: it can transform into heat, light, ect and will never disapear unlessed turned of.
What happens to an object when there are no forces acting on it or all the forces acting on it are balanced?
Answer:
It stays and remains at rest
Explanation:
When no forces acts on an object or all the forces on it are balanced, the body stays at rest or remains stationary.
This is based on the Newton's law of inertia which states that "an object will remain in a state of uniform motion or at rest unless an external force acts on it".
If the forces acting on a body is balanced, the object will perpetually remain and keep at rest
why the student should not heat the oil directly with a flame.
it is possible that, when oil reaches a high enough temperature, it can ignite and burn very fiercely, being difficult to extinguish.
How far does Andre have to push a box to do 430 J of work, if he applies a force of 15 Newtons
Answer:
28.7m
Explanation:
Given parameters;
Work done = 430J
Force applied = 15N
Unknown:
Distance moved = ?
Solution:
Work done is the force applied to move a body through a certain distance.
Work done = Force x distance
So;
430 = 15 x distance
Distance = 28.7m
When you pick something up do you do work to it
Answer:
Simple answer: Yes
Explanation:
Even if you touch an item with a stick you re still doing work to it, most of the time something sitting on a table not being disturbed is having work done to it. Everything has the force of gravity working on it to essentially keep the items from floating away so workis being done to it.
Work done can be something so small (e.g) a pencil sitting on a table) or as big as an earthquake or kicking a ball through a window and smashing the glass.