Answer:
C requires the most work and A requires the least work
Answer:
Which scenario requires the most work?
✔ C
Which scenario requires the least work?
✔ A
Which describes how the spring constant affects the potential energy of an object for a given displacement from an equilibrium position?
A the higher the spring constant, the greater the gravitational potential energy.
B The lower the spring constant, the greater the gravitational potential energy.
C The higher the spring constant, the greater the elastic potential energy.
D The lower the spring constant, the greater the elastic potential energy
Answer:
Based on the answer choices provided, the correct answer is:
C, The higher the spring constant, the greater the elastic potential energy.
Explanation:
The spring constant is best defined as the measure of a spring's resistance to force (in regards to factors such as stretching and compression).
Thus, the higher the spring constant, the stiffer the string - this equates to greater elastic potential energy (think of stretching a rubber band or using it as a slingshot).
Hope this helps!
Answer:
C.) The higher the spring constant, the greater the elastic potential energy.
Explanation:
The lower the angle of the slope, ________ the acceleration along the ramp, therefore, the speed at the bottom of a slope will be_______, and, consequently, the control will be better.
The acceleration along the ramp, therefore, the speed at the bottom of a slope will be:__________
a. lower
b. gcostheta
c. higher
d. gsintheta
Answer:
Lower
Lower
gsintheta (gsinθ)
Explanation:
The sum of forces resolved parallel to the inclined plane is given by;
F - mgsinθ = 0
ma - mgsinθ = 0
ma = mgsinθ
a = gsinθ
Acceleration is proportional to angle of inclination, thus the lower the angle of the slope, lower the acceleration along the ramp.
therefore, the speed at the bottom of a slope will be lower, (velocity is directly proportional to acceleration) and, consequently, the control will be better.
The acceleration along the ramp, is gsintheta (gsinθ)
A mass of 0.4 kg is dropped from rest in a medium offering a resistance of 0.2|v|, where is measured in meters per second. Acceleration due to gravity is 9.8 m/s^2.
Required:
a. If the mass is dropped from a height of 25 m, find its velocity when it hits the ground.
b. If the mass is to attain a velocity of no more than 8 m/s, find the maximum height fromwhich it can be dropped.
c. Suppose that the resistive force isk|v|, wherevis measured in m/s, andkis constant.If the mass is dropped from a height of 25 m and must hit the ground with a velocity of no more than 8 m/s, determine the coefficient of resistance that is required.
Answer:
A.) 22.12 m/s
B.) 3.27 m
C.) 0.57
Explanation:
Given that a mass of 0.4 kg is dropped from rest in a medium offering a resistance of 0.2|v|, where is measured in meters per second. Acceleration due to gravity is 9.8 m/s^2.
a. If the mass is dropped from a height of 25 m, its initial velocity U will be equal to zero. when it hits the ground, the final velocity V will be:
V^2 = U^2 + 2gH
Substitute g and the height into the formula
V^2 = 2 × 9.8 × 25
V^2 = 490
V = 22.12 m/s
b. If the mass is to attain a velocity of no more than 8 m/s, the maximum height from which it can be dropped will be calculated by using the same formula.
V^2 = U^2 + 2gH
Where V = 8 m/s and U = 0
8^2 = 0 + 2 × 9.8 × H
64 = 19.6H
H = 64/19.6
H = 3.27 m
c. Suppose that the resistive force is k|v|, where v is measured in m/s, and k is constant. If the mass is dropped from a height of 25 m and must hit the ground with a velocity of no more than 8 m/s, determine the coefficient of resistance that is required.
The energy at 25m heigh will be:
E = mgh
E = 0.4 × 9.8 × 25
E = 98 J
1/2mv^2 = 98
1/2 × 0.4 × ( 8 - v )^2 = 98
0.2( 8 - v )^2 = 98
( 8-v)^2 = 98/0.2
( 8 - v )^2 = 490
8 - v = 22.12
v = 22.14 - 8
v = 14.14
Coefficient of resistance required will be 8/ 14.14 = 0.57
the volume of a cube is measured in grams, true or false
Answer: false
Explanation: They are measured by cubic ______. ______ is feet, meters, inches, any way to measure
Particles in a warmer block will have
the particles in the colder block.
kinetic energy and move
than
more / slower
less / slower
less / faster
more / faster
Answer:
more/ faster
Explanation:
I got it right on my test
Particles in a warmer block will have more kinetic energy and move faster than the particles in a colder block option (D) more/ faster is correct.
What is kinetic energy?The energy which an item has as a function of motion is determined by the initial energy. It is described as the effort required to propel a material body from rest to the given velocity.
As we know,
As the temperature rises, the particles' kinetic energy rises and they begin to move more quickly. Particles have higher kinetic energy in gases than in solids.
Thus, particles in a warmer block will have more kinetic energy and move faster than the particles in a colder block option (D) more/ faster is correct.
The question is incomplete.
The complete question is:
Particles in a warmer block will have ________ kinetic energy and move ______than the particles in the colder block.
more / slowerless / slowerless / fastermore / fasterLearn more about kinetic energy here:
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Which of the vectors below is the sum of vectors à and b, shown here?
Answer:
B
Explanation:
Answer:
B
Explanation:
Just took the quiz
a car with a mass of 2000kg is traveling at a velocity of 50 m/s. the engine supplies a force of 1000N forward.the force of friction is 1000 N backwards. What is the velocity after 5 seconds
Answer:
no
Explanation:
ok
Mina finds out that a company is pumping water from an aquifer near her home. What can Mina conclude about the water being pumped? The water being pumped is under a lot of pressure and comes from a confined aquifer. The water being pumped is under a lot of pressure and comes from an unconfined aquifer. The water being pumped is at atmospheric pressure and comes from a confined aquifer. The water being pumped is at atmospheric pressure and comes from an unconfined aquifer.
Answer:
The water being pumped is at atmospheric pressure and comes from an unconfined aquifer.
Answer:
The water being pumped is at atmospheric pressure and comes from an unconfined aquifer.
Explanation:
It is the final seconds of an ice hockey game between the Flyers and the Bruins. The Bruins are down by 1 point. With 20 s left in the game, the Bruins pull the goalie and have him play as a forward in an attempt to tie the game.
The Flyers successfully defend their goal for 9 s. With only 1.25 s remaining on the game clock, a Flyer shoots the puck on the ice past the skates and sticks of the other players and toward the Bruins' goal. The puck is 37 m from the goal when it leaves the stick with an initial horizontal velocity of 30 m/s. The shot is perfectly directed toward the empty goal, but the ice slows the puck down at a constant rate of 0.50 m/s2 as it slides toward the goal. None of the Bruins can stop the puck before it reaches the goal.
a. Where is the puck when the game clock reaches zero and the horn sounds to end the game?
b. Do the Flyers win the game by 1 or 2 points?
Answer:
a
[tex]P =0.1094 \ m[/tex] past the goal post
Yes the Flyers won the match
Explanation:
Generally the distance covered by the pluck during the last 1.25 seconds is mathematically represented as
[tex]D = ut + \frac{1}{2} * a * t^2[/tex]
=> [tex]D =30 * 1.25 + \frac{1}{2} * (-0.5) * (1.25)^2[/tex]
=> [tex]D =37.1094 \ m [/tex]
Generally the position of the puck when the game clock reaches zero is mathematically represented as
[tex]P = 37.1094 -37[/tex]
[tex]P =0.1094 \ m[/tex] past the goal post
Given that the Bruins where one point down and Flyers scored another goal it means that the flyer are now two point up hence they won the match
Explain why the driver's reaction time affects the thinking distance.
Answer:
The thinking distance depends on the reaction time of the driver which could can affected by alcohol, distractions and tiredness. A faster speed increases both thinking distance, increasing the total stopping distance.
I hope this is helpful. Would appreciate if you add me as brainliest.A faster pace extends both the thinking distance and the total stopping distance.
Given that,
To justify, the driver's reaction time affects the thinking distance.
The reaction time of an organism is a measure of how quickly it responds to a signal. The time interval between the presentation of the stimulus and the appearance of the appropriate voluntary response in the subject is defined as the RT.
The thinking distance is determined by the driver's reaction time, which can be influenced by drinking, distractions, and fatigue. A higher speed increases both thinking distance and total stopping distance.
Thus, A faster pace extends both the thinking distance and the total stopping distance.
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3. What would happen if Earth did not rotate?
Answer:
If Earth were to stop spinning on its axis, gradually the oceans would migrate towards the poles from the equator. The atmosphere would still be in motion with the Earth's original 1100 mile per hour rotation speed at the equator. And this means, rocks, topsoil, trees, buildings, and so on, would be swept away into the atmosphere.
Hope that helped <3
Analysis of an accident scene indicates a car was traveling at a velocity of 70.0 mph (31.3 m/s) along the positive x-axis at the instant the brakes were applied and then traveled a distance of 115 m before hitting an obstacle at a velocity of 4.65 m/s. Determine the car's acceleration a (assumed constant) and the time t between applying the brakes and hitting the obstacle. (Assume the car is moving in the positive direction. Indicate the direction with the sign of your answer.)
Answer:
The acceleration of the car is -4.166 m/s²
The time between applying the brakes and hitting the obstacle is 6.4 s
Explanation:
Given;
initial velocity of the car when the brake was applied, u = 31.3 m/s
final velocity of the car after hitting an obstacle, v = 4.65 m/s
distance traveled before stopping, s = 115 m
The acceleration of the car is given by;
v² = u² + 2as
2as = v² - u²
a = (v² - u²) / 2s
a = (4.65² - 31.3²) / (2 x 115)
a = -4.166 m/s²
The time between applying the brakes and hitting the obstacle;
v = u + at
at = v - u
t = (v - u) / a
t = (4.65 - 31.3) / -4.166
t = 6.4 s
What is the mass of a 2 kg object on the Earth and on the moon?
Answer:
Same
Explanation:
Mass is the quantity of matter in a certain object.
WHEREVER you take a 2kg object, the mass will remain 2kg. All that changes is the Weight ..Weight the force which the centre of a Planet uses to pull everything towards itself.
On earth, it is 9.81 whereas on the Moon it is 1.6
Here, we are required to determine the mass of a 2kg object on the Earth and on the moon.
The mass of the 2kg object is the same on earth and on the moon.The mass of an object is simply defined as the quantity of matter present in the object and this quantity remains constant.
The mass of an object is simply defined as the quantity of matter present in the object and this quantity remains constant.In essence, if the mass of an object is 2kg, this mass remains constant on planet earth and the moon.
PS: The weight of an object which is a measure of the acceleration due to gravity which differs with location and position is the constant which may be different about the object on the earth and the moon respectively.
The acceleration due to gravity on earth and the moon is 9.8m/s² and 1.6m/s² respectively.
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Which method of measurement is most likely to be precise but lacks
accuracy?
A. Measuring the distance across a pond by counting how many
strokes you make as you swim across
B. Measuring time by counting the number of breaths you take as
you do an activity
O C. Measuring your weight with a scale that measures to the nearest
tenth of a kg but adds 2 kg to each measurement
D. Measuring the length of a bus using a tape measure that
measures to the nearest millimeter
Answer:
C
Explanation:
Measuring your weight with a scale that measures to the nearest
tenth of a kg but adds 2 kg to each measurement
Does voltage remain the same throughout a series circuit
It depends on the circuit. Sometimes it becomes a bit weaker, sometimes it stays the same.
Answer:
In a series circuit, the current that flows through each of the components is the same, and the voltage across the circuit is the sum of the individual voltage drops across each component.
At the bifurcation of the femoral artery the pressure is 100 mm Hg. The venous pressure after either the left or right femoral artery is 10 mm Hg. The resistance in the right femoral artery is 1.67 times the resistance through the left femoral artery. The total flow rate just upstream of the bifurcation is 800 mL/min. How does the flow divide into the left and right femoral arteries
Answer:
The values are [tex]I_L = 500.37 \ mL / minutes [/tex]
and [tex]I_r = 299.63 \ mL / minutes [/tex]
Explanation:
From the question we are told that
The pressure at the bifurcation of the femoral artery is [tex]P = 100\ mm\ Hg[/tex]
The pressure at the left or right femoral artery is [tex]P_r = P_l = 10 \ mm \ Hg[/tex]
The resistance in the left femoral artery is [tex]R_L[/tex]
The resistance in the right femoral artery is [tex]R_r = 1.67 R_L[/tex]
The total flow rate is [tex]I = 800 mL/minute[/tex]
The diagram illustrating this question is shown on the first uploaded image
Generally this flow of blood through the artery can be compared to the a circuit as shown on the first uploaded image
Generally the rate at which blood flows through the left femoral artery is mathematically represented as
[tex]I_L = I * \frac{R_r }{R_L + R_r}[/tex]
=> [tex]I_L = 800 * \frac{1.67 R_L }{R_L + 1.67 R_L}[/tex]
=> [tex]I_L = 800 * \frac{1.67 R_L }{2.67 R_L}[/tex]
=> [tex]I_L = 500.37 \ mL / minutes [/tex]
Generally the rate at which blood flows through the right femoral artery is mathematically represented as
[tex]I_r = I * \frac{R_L }{R_L + R_r}[/tex]
=> [tex]I_r = 800 * \frac{R_L }{R_L + 1.67 R_L}[/tex]
=> [tex]I_r = 800 * \frac{ R_L }{2.67 R_L}[/tex]
=> [tex]I_r = 299.63 \ mL / minutes [/tex]
How to call the police?
Answer:
dail 911
tell problem and location
plz mark brainliest
A gymnast on the uneven parallel bars is at rest, tipped at a 45∘ angle from the vertical. The distance from her hands to her feet is 1.8 m. If we model her body as having a uniform cross section and assume that her center of gravity is midway between her hands and her feet, what is her initial angular acceleration?
The initial angular acceleration is 5.78 rad/s^2.
Calculation of the angular acceleration;Since
A gymnast on the uneven parallel bars is at rest, tipped at a 45∘ angle from the vertical. The distance from her hands to her feet is 1.8 m.
So,
[tex]Torque \tau = r \times F\\\\ \tau = l\alpha\\\\so, r \times F = l\alpha\\\\mg\ cos\ 45 \times r = (1/3)mR^2 \times \alpha\\\\9.8\times cos\ 45 \times 0.9 = (1/3)\times (1.8)^2 \times \alpha\\\\\alpha = 5.78\ rad/s^2[/tex]
hence, The initial angular acceleration is 5.78 rad/s^2.
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Which would be harder to move: a kilogram of feathers or a kilogram of bricks? Explain your answer.
Answer:
The kilogram (kg) is the SI unit for mass. Not weight.
So when you say « heavier » and mention kg, I would assume you’re talking about mass.
The mass of 1 kg of feathers and 1 kg ok bricks are identical.
And if you want an answer to your question comparing the weight on Earth of those 2 items, in vacuum of one kilogram of feathers would have exactly the same weight as one kilogram of bricks.
Explanation:
Hope this helps! <3
Q1: An object with a charge of 1.2 C is located 4.5 m away from a second object that has a charge of 0.36 C. Find the electrical force that the objects exert on each other. Use 8.988100 N·m2/C2 for the value of Coulomb's constant. Give your answer to two decimal places.
Answer:
a) F= 0,19 [N] according to problem statement
b) F = 0,19*10⁹ [N] using the right value of K
Explanation:
The force between two electric charges is according to Coulomb´s law is:
F = K * q₁*q₂ / d² where q₁ and q₂ are the charges on body one and body 2 respectively, d is the distance between the two bodies and K is a constant K = 8,988100*10⁹ N.m²/C². The problem establishes to use K = 8,988100 N.m²/C².
NOTE: To value of is : K = 8,988100*10⁹ N.m²/C². I am going to solve the problem using K = 8,988100 N.m²/C² if that information was an error, all we need to get the right answer is multiply the result by 10⁹
Then:
F = 8,988100 * 1,2* 0,36 / (4,5)² [ N*m²/C² ] * [ C*C*/m²]
F = 3,882859/ 20,25 [N]
F= 0,19 [N]
The force is of repulsion since the two charges are positive and in the direction of the straight line which passes through the centers of the bodies
A ball tossed vertically upward rises, reaches its highest point, and then falls back to its
starting point. While the ball is rising, what is happening to its speed.
O it is increasing by 10 m/s^2
O it is increasing, but not at a constant rate
it is decreasing by 10 m/s^2
it is decreasing, but not at a constant rate
O it moves with constant speed
Answer:
Third Option: It is decreasing by 10m/s^2
Explanation:
When a ball is thrown in the air, it's fastest velocity occurs just as it leaves its source (e.g. the hand that throws it), and just before it hits the starting point. This is because this is when it has the most kinetic energy which translated into speed. As it rises, it starts to lose kinetic energy as it transforms into gravitational potential energy as the ball is fighting against the force of gravity as it moves further away from the Earth's surface. This means it begins to slow down until its speed is 0, its highest point, before falling and gaining kinetic energy again. Because of this, we can eliminate options 1, 2 and 5 as we know the ball's velocity decreases as it rises.
As the ball rises it's velocity decreases at a constant rate, as the amount of kinetic energy it has is proportional to its height. Therefore, the best option is option 3, as it states it is decreasing and gives a constant deceleration rate.
Hope this helped!
A 10.0-cm-diameter and a 20.0-cm-diameter charged ring are arranged concentrically (so they share the same axis). Assume both are centered at the origin. The 10 cm ring is charged to 30.0nC, while the 20 cm ring is charged to 60.0nC. Using the library electric field formula for Ering, compute the total electric field strength numerically at:
a, The origin
b. 40.0 cm along the axis of the 10 cm ring.
Answer:
a) Eₓ = 0 , b) c) Ex_total = 3.96 10⁻⁶ N
Explanation:
For this problem we will look for the expression for the electric field of a ring of charge, at a point on its central axis.
From the symmetry of the ring, the resulting field is in the direction of the axis, we suppose that it corresponds to the x and y axis and the perpendicular direction but since the two sides of the annulus are offset.
dE = k dq / r²
The component of this field in the direction to x is
d Eₓ = dE cos θ
the distance of ring to the point
r² = x² + a²
where a is the radius of each ring
the cosine is the adjacent leg between the hypotenuse
Cos θ = x /r
we substitute
d Ex = k / (x² + a²) dq x /√ (x² + a²)
we integrate
Ex = ∫ k / (x² + a²)^{3/2} x dq
Ex = k x/ (x² + a²)^{1.5} Q
the total field is the sum of the fields created by each ring, since they are on the same line (x-axis), you can perform an algebraic sum
Eₓ_total = Eₓ₁ + Eₓ₂
a) at the origin x = 0
in field is zero, since the numerator e makes zero
Eₓ = 0
b) for point x = 40.0cm = 0.400 m
we substitute the values in our equation
Ex_total = k x [Q (x² + 0.1²)^{1.5} + 2Q / (x² + 0.2²2) ^3/2]
Ex_total = k Q x [1 / (x² + 0.01)^1.5 + 2 / (x² + 0.04)^1.52]
let's calculate
Ex_total = 9 10⁹ 30 10⁻⁹ 0.40 [1 / (0.4 2 + 0.01) 3/2 + 2 / (0.4 2 + 0.04) 3/2] 10⁻⁹
Ex_total = 108 [14.2668 + 22.36] 10⁻⁹
Ex_total = 3.96 10⁻⁶ N
What does your ecological foot print tell you?
Answer:
How big your foot is. (Big Brain)
Explanation:
Use Percentages Given that the molecular mass of magnesium
hydroxide (MOH.) is 58.32 amu and the atomic mass of an atom of
oxygen is 15.999 amu, what percentage of this compound is oxygen?
Answer:
27%
Explanation:
15.999 divided by 58.32 = .27433128
Move the decimal place over 2 places.
27%
Make use of percentages. Given that the molecular mass of magnesium hydroxide (MOH.) is 58.32 amu and the atomic mass of an oxygen atom is 15.999 amu, oxygen accounts for 27% of this compound.
Molecular versions of many substancesChemical substances known as molecular compounds assume the shape of distinct molecules. Examples include common compounds like carbon dioxide (CO2) and water (H2O) (Figure 3.1.1 ).
These substances differ significantly from ionic substances like sodium chloride (NaCl). When metal atoms give up one or more of their electrons to non-metal atoms, ionic compounds are created. The cations and anions that result
Therefore,
58.32 / 15.999 (27.3%) =.27433128
Add two places to the decimal place.
27%
The molar mass, which is stated in g/mol, is defined as the mass of a specific material divided by the amount of a substance. In light of this, the molar mass is average. Amu, short for atomic mass units, is the unit used to quantify molecular mass.
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A man throws a football straight into the air Which type of energy is the
football gaining as it rises
Answer:
It is gaining potential energy
Answer: it could also be gravitational potential energy
This is because as it rises in the air it will gain more energy to fall back down
Which change will decrease the electric force between two positively charged objects?
1.moving them closer together
2.moving them farther apart
3.adding neutrons
4.removing electrons
Answer:
2
Explanation:
Answer:
The answer is B
Explanation:
Tension is actually a force that tends to
Answer:
Pull towards itself.
Explanation:
That's what I know.
Tension is a force that tends to pull or stretch an object. A flexible material, like a rope, cable, or string, is used to transmit the force.
A force known as tension has a tendency to pull or stretch an object. When a flexible medium, like a rope, cable, or string, is under strain and supporting an imposed load or weight, it transmits force via the flexible medium.
A rope or cable that is under tension receives equal and opposing forces pulling it in different directions at either end. For instance, the tension in a rope will be the same along its whole length if both ends are pulled with equal effort.
When you use a pulley system to hoist a weight, suspend an object from a rope, or support the weight of traffic on a bridge, tension can be present.
Therefore, Tension is a force that tends to pull or stretch an object.
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You observe two cars traveling in the same direction on a long, straight section of Highway 5. The red car is moving at a constant vR equal to 34 m/s and the blue car is moving at constant vB equal to 28 m/s. At the moment you fist see them, the blue car is 29.0 m ahead of the red car.
Required:
a. How long after you first see the cars does the red car catch up to the blue car?
b. How far did the red car travel between when you fist saw it and when it caught up to the blue car?
c. Suppose the red car started to accelerate at a rate of a equal to 5/3 m/s^2 just at the moment you saw the cars.
d. How long after that would the red car catch up to the blue car?
Answer:
a) 3.66 s
b) 124.4 m
c) 3.12s
Explanation:
Given that
Speed of the Red Car, v₁ = 34 m/s
Speed of the Blue Car, v₂ = 28 m/s
Distance between the two cars, d = 22 m
The difference between the speed of the cars is: 34 - 28 = 6 m/s
From this, we can deduce that the red car will be catching up to the blue car at a speed of 6 m/s.
1)
If we divide the distance by the difference in speed. This becomes
d/v = 22/6 = 3.66 s. Which means, It takes 3.66 seconds for the red car to meet up with the blue car.
2
From the previous part, we were able to confirm that it took 3.66 seconds for the red car to meet up the blue car. Also, the speed with which it were traveling was travelling at, was constant, so we only need to multiply it by the time from (1) to get the distance.
v = d * t = 34 * 3.66 = 124.4
Therefore the red car travels at 124.4 m before catching up to the blue car.
3
If the red car starts to accelerate the moment we see it, the time it will take, to get to the blue car will be less than what we had gotten. We can find this using one of the equations of motion.
S = ut + ½gt², where
S = 22
u = 6
t = ?
g = 2/3
22 = 6t + 1/3t²
By using the quadratic formula, we find out the two answers listed below
t1 = 3.12 s
t2 = - 21.12 s
We all know that negative time is not possible, so the answer is t1. At 3.12 seconds
If the point B was initially at the same level as the center of the pulley C, what is the direction of the velocity of point B after the pulley at A has undergone 2 revolutions? Give this in degrees from the horizontal (positive in the counterclockwise direction, negative in the clockwise direction)
Answer:
θ = 720º
Explanation:
Let's use the rotational kinematics relations for this exercise, remember that all angles must be given in radians.
Let's reduce the magnitudes to the SI system
θ = 2 rev (2π rad / 1 rev) = 4π rad
we assume that the pulley has a radius r
linear and angular coordinates are related
y = θ r
let's calculate
y = 4π r
to convert the distance in angles to radians
θ = y / r
θ = 4π
let's change this angle to degrees
π rad = 180º
θ = 4π rad (180º /π rad)
θ = 720º
A man runs 800m due north in 110 seconds followed by 400m due suoth in 90 seconds calculate his average speed and his average velocity
Answer:
Average speed is 6m/s
Average velocity is 5.859m/s
Explanation:
Average speed, s = d/t
d is distant
t is time
speed north is 800m and 110s
speed due south is 400m and 90s
Average speed = (800+400) / (110+90)
= 1200/200
=6m/s
Average velocity V = (v + u)/2
v = final velocity
u = initial velocity
V = (7.273 + 4.444)/2 = 11.717/2
V = 5.859m/s