Storing music digitally requires less storage room than analog records or tapes. Digital music is easier to copy and the copies are the same as the original. The quality of the signal does not degrade over long periods of time.
Sample Response: Storing music digitally requires less storage room than analog records or tapes. Digital music is easier to copy and the copies are the same as the original. The quality of the signal does not degrade over long periods of time.
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
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.
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:
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.
Which of the following is NOT accelerating?
A ball being juggled.
A women walking at a constant speed of 2.5 m/s along a straight path.
A rocket taking off.
A car slowing down toward a stop sign.
Answer:
3 of them are not accelerating . only one of them is and that one is the rocket taking off because its speed increases
Answer:
Second Option
Explanation:
The answer is the second option or "a women walking at a constant speed of 2.5 m/s along a straight path." Remember that constant means always staying the same so if a women is walking at a constant speed of 2.5 miles per hour on a straight path, she will remain at her current speed at 2.5 mils per hour which isn't acceleration.
Hope this helps.
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.
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:
Milk in a hot cup can be cooled by pouring its back and fourth between the bigger cup.How does this action help to cool milk?
Do the atoms in salt follow a pattern
Answer:
Explanation:
Yes, the atoms in a salt do follow a pattern knows as crystalline pattern or crystalline structure. It means it has an ordered arrangement of atoms or ions. As salt is an ionic compound of Sodium and Chlorine, it is formed through ionic bond. This pattern keeps on repeating for each molecule of NaCl
Answer:
yes they do crystalline pattern as well as othe ionic compounds.
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
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
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
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?
To date, Saturn is known to have 13 rings made from which of the following?
A. Dust
B. Dark particles
C. Moons
D. Stars
Answer:
it is believed to be shattered moons, asteroids, and comets, so C
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
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.
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:
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 weightlifter slowly lifts a large amount of weight with a force of 500 N and an acceleration
m/s2. What is the amount of weight being lifted in grams?
Explanation:
Given parameters:
Force = 500N
Acceleration =
Unknown:
Weight can be used to this problem = ?
Solution:
Weight is the vertical force on a body;
Weight = mass x acceleration
So from this we can find the mass and then convert to the second wealth.
One role of a producer in a food chain is to provide? A) water for plants B) sunlight for plants C) nutrients for animals D) shelter for animals
Answer:
Nutrients for animals
Explanation:
Producers are organisms that produce their food, they do not need to depend on other animals for their food. They possess photosynthetic properties that enable them to produce their food.
These set of organisms are located at the bottom of the food chain, they can convert components like water, sunlight and air into carbohydrate through the unique process of photosynthesis.
Without these producers, the food chain would not exist as they form the base and foundation of the food chain, some lower animals depend on them as their source of food.
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
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
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
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
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.
can a physical quantity have different dimensions in different systems of units?
Answer:
yes.
Explanation:
becausehjjkvfdssaajkkbcsssddhkkgddd
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
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.
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
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.